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Author Message
Ed
PostPosted: Tue Jul 08, 2008 1:38 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Old SSv3.2.1 scores: 
Killer rating table      
Rounded Score from SSv3.2.1 
! = 0.10+ change from previous Score
pg# on this thread - PART B
(E) = Easy  (H) = Hard
======================================================================
|A ##       Rate  Score|A ##       Rate  Score|A ##       Rate  Score|
|----------------------+----------------------+----------------------|
|A.70      1.00    0.90|A.71v2    1.50    1.35|A.72v2     1.50  !4.15|
|A.70v2    1.75    1.55|A.71v1.5 H1.25    1.30|A.73       1.00   0.90|
|A.70v3    1.25    1.15|CS        1.25    1.40|                      |
|A.71     H1.50    1.50|A.72      1.00    1.15|                      |
|====================================================================|
                              page #6
Old scores SSv3.3.0: 
Rounded Score from SSv3.3.0 
! = 0.10 change from previous Score
pg# on this thread - PART B
(E) = Easy  (H) = Hard
======================================================================
|A ##       Rate  Score|A ##       Rate  Score|A ##       Rate  Score|
|----------------------+----------------------+----------------------|
|A.70      1.00   !1.00|A.71v2   E1.50   !1.45|A.72v2     1.50  !5.40|
|A.70v2    1.75   !1.80|A.71v1.5 H1.25    1.25|A.73       1.00  !1.00|
|A.70v3    1.25    1.15|CS        1.25    1.35|                      |
|A.71     H1.50   !1.65|A.72      1.00    1.20|                      |
|====================================================================|
                              page #6
Killer rating table
SudokuSolver Target range v3.6.3
Rating.....Score
0.50   =   0.85
0.75   =   0.90-0.95
1.00   =   1.00-1.20
1.25   =   1.25-1.45
1.50   =   1.50-1.70   (E) = Easy  (H) = Hard
===========================================================================================
|A ##     by     Rate    Score|A ##     by     Rate    Score|A ##     by     Rate    Score|
|-----------------------------+-----------------------------+-----------------------------|
|A.70    Ruud    1.00     1.00|A.71v2  Ruud    1.50     1.40|A.72v2  Para    1.50     3.80|
|A.70v2  Ruud    1.75     1.65|A71v1.5 mhp    H1.25     1.40|A.73    Ruud    1.00     1.00|
|A.70v3  Ed      1.25     1.15|CS      mhp     1.25     1.25|                             |
|A.71    Ruud   H1.50     1.60|A.72    Ruud    1.00     0.95|                             |
|=========================================================================================|
                                         page #6


Assassin 70 by Ruud (Sept 07)
Puzzle pic:
Attachment:
a70.JPG
a70.JPG [ 101.31 KiB | Viewed 11682 times ]
Code: Select, Copy & Paste into solver:
3x3::k:4864:2817:2818:2818:4868:3333:3333:4359:3592:4864:2817:2818:4364:4868:2318:3333:4359:3592:4864:3859:3859:4364:4868:2318:5400:5400:3592:4864:3868:3859:4364:7455:2318:5400:3874:3592:3868:3868:3868:7455:7455:7455:3874:3874:3874:2349:4654:2863:2864:1585:2354:3379:2100:3893:2349:4654:2863:2864:1585:2354:3379:2100:3893:3903:4654:4417:4417:1585:2884:2884:2100:5447:3903:3903:4417:4171:4171:4171:2884:5447:5447:
Solution: 
+-------+-------+-------+
| 8 2 3 | 1 6 7 | 5 9 4 |
| 6 9 7 | 2 5 4 | 1 8 3 |
| 4 5 1 | 9 8 3 | 6 7 2 |
+-------+-------+-------+
| 1 4 9 | 6 7 2 | 8 3 5 |
| 3 6 2 | 8 9 5 | 7 4 1 |
| 7 8 5 | 4 3 1 | 9 2 6 |
+-------+-------+-------+
| 2 3 6 | 7 1 8 | 4 5 9 |
| 9 7 4 | 5 2 6 | 3 1 8 |
| 5 1 8 | 3 4 9 | 2 6 7 |
+-------+-------+-------+
Quote:
Afmob: This one was a bit strange since there were many solving paths you could go, but I tried to keep it as short as possible. Overall, a rather easy assassin
mhparker: Rating? 0.75? It was probably significantly easier than Ruud intended it to be
sudokuEd: Definitely only a 0.75 rating - maybe even 0.5 - except I used 1 little contradiction move. Maybe that speeded things up
Andrew: I did miss something important. .... However it still comes out and would only be rated between 1.0 and 1.25 that way. An excellent solving path by Afmob!
azpaull: I didn't think it was doable without that key, but since it was difficult for (most of) us to find that key, I think we need to give it at least a 1.25..Edit: Make that a 1.5, I just reviewed the rating scale
Walkthrough by Afmob using killer pair blocks:
This one was a bit strange since there were many solving paths you could go, but I tried to keep it as short as possible. Overall, a rather easy assassin. I'm looking forward to V2.

1. C1234
a) Outies of C123 = 6(2) = {15/24}
b) 29(4) = {5789} locked for N5
c) Innies of C1234 = 11(2) = [56/74/83/92] -> R9C4 = (2346)
d) 11(2) @ N5: R7C4 = (5789)
e) 9(2) @ N5: R7C6 <> 1,2,4

2. C6789
a) 17(2) = {89} locked for C8 + N3
b) 8(3) = 1{25/34} -> 1 locked for C8
c) Innies C789 = 11(4) = {1235} -> locked for C6
d) 13(2) = {49/67}
e) Outies of C789 = 13(2) = {49/58/67}, R1C6 <> 4
f) Innies C6789 = 14(2) = {59/68}, R9C6 <> 8
g) Killer pair (58) of Innies C6789 blocks {58} of Outies of C789

3. C7
a) 21(3) must have 8 or 9 -> only possible @ R4C7
-> R4C7 = (89), 21(3) <> 4 ( {489} not possible )
b) R3C7 = (67) blocks {67} of 13(2)
c) 13(2) = {49} -> locked
d) R4C7 = 8
e) 21(3) = {678} -> 6,7 locked for N3 + R3
f) 15(2) @ C9: R7C9 <> 7
g) 15(4) <> 9 because 15(4) must have 6 xor 7 because of R5C7 = (67)
-> R45C8+R5C9 <> 6,7

4. C9
a) Innies+Outies: 5 = R9C8 - R5C9 -> R9C8 = (67), R5C9 = (12)
b) 4 locked in R123C9 @ 14(4)
c) 14(4) = 34{16/25} <> 7 -> other combinations blocked by R5C9 = (12)
d) 21(3): R89C9 <> 6 since R89C9 would be {68} -> blocked by killer pair (68) of 15(2)

5. N69
a) Innies+Outies: 1 = R8C6 - R4C9 -> R4C9 = (356)

6. C5
a) 6(3) = {123} locked

7. C6
a) Innies = 27(4) = 69{48/57} -> 6 locked
b) Killer pair (45) in 9(3) blocks [45] of 9(2)
c) 9(2) = [18/27]

8. C4
a) 6 locked in R46C4

9. C6 !
a) 16(3): R9C5 <> 6 since R9C6 has no 2,3,7,8 ( 6{28/37} )
b) 6 locked in R89C6
c) 6 locked in 19(3) @ C5 -> 19(3) <> 7
d) 13(3) = 1{39/57} -> no 2

10. N3
a) 1 locked in 13(3) for C7 + N3
b) 14(4) = {2345} -> locked for C9
c) R5C9 = 1

11. C9
a) 6 locked in 15(2) -> 15(2) = {69} locked
b) 21(3) = {678} locked for N9
c) R7C9 = 9 -> R6C9 = 6 -> R5C7 = 7 -> R3C7 = 6 -> R3C8 = 7 -> R9C8 = 6
d) R7C7 = 4 -> R6C7 = 9
e) Hidden Singles: R8C6 = 6 @ C6, R4C4 = 6 @ C4

12. N9
a) 11(3) = {236} -> 2,3 locked for C7 + N9
b) 8(3) = {125} locked in C8 -> R6C8 = 2 -> R6C6 = 1 -> R6C5 = 3 -> R6C4 = 4
c) R4C6 = 2, R7C4 = 7, R7C6 = 8

13. C456
a) Hidden Singles: R4C5 = 7 @ C5, R1C6 = 7 @ C6
b) 9(3) = {234} -> 3,4 locked for N2
c) 19(3) = {568} -> locked for C5 + N2

14. Rest is clean-up and singles
Outline by garyw including "consider" steps:
Again just my outline of solving path.


1. I/O N8 r8c4+r8c6=r6c456+3 > r8c46 min 9
2. r1c4+r8c4=6 <>3/3
3. r5c4+r9c4=11 no.1
4. r5c6+r9c5=14
5. I/O c9r9c8=r5c9+5 > r9c8=6/7 (8/9 blocked by r12c8=8/9) r5c9=1/2
6. r1c6+r4c79=20 ; r4c9 max 6 >r1c6 min =5
7. r67c6 9/2 cage<> 3/6 blocked by 9/3 cage r234c6
8. r459c5=20; r9c5<> 9

Consider position of 1 in c4

<> r23(blocked by 29/4 cage) <> r567 and <>r9 (see 3 above).
But 1 in r4c4 forces 1 into r78c5 and thus into r23c6 and 3 into r679c4/r678c5.This x-wing on 3 forces 3 into r23c 6 also.This places a 5 in r5c6 which is impossible.

Thus r18c4=1/5.
But,from 1. above a 1 in r8c4 > 8 in r8c6 > r6c456 =1/2/3.But a 3 cannot be placed in any of these cells (see 7. above)

Therefore r8c4=5 and r1c4=1

Hidden single in N5. A 6 must go in r4c4

In N8 r9c4= 2/3/4 (see 3. above)
But if a 2 cannot place a 9 in N8.
If r9c4=3 r9c56=49 (6/7 blocked by r9c8)
If r9c4=4 r9c56=39 but blocked by 6/3 cage
Therefore r9c456=349.

Just a mop up now.


A nice puzzle.

Regards

Gary

P.S. Did killersudokuonline number 90 in about 1 hr. Definitely easier than any assassins.
Analysis of garyw outline by mhparker using closed loops:
Hi all,

Not much discussion going on here! :( Where are all the regulars (Cathy?, Para?, Andrew?, ...)? So think I'll liven things up by adding my two cents worth:
Afmob wrote:
This one was a bit strange since there were many solving paths you could go... Overall, a rather easy assassin.
Rating? 0.75? It was probably significantly easier than Ruud intended it to be for the following reason:

There's an early key move, which Afmob used, namely the C789 innies = 11(4) = {1235}, which effectively cracked the puzzle. Those who saw this move early on will probably have found the puzzle considerably easier than those who didn't. Interestingly, however, neither Sumocue nor JSudoku (in the default state, with all solvers enabled) pick up this move. Presumably, quadruple innies and outies are relatively low down in the solving technique pecking order, so they found the less productive but "simpler" moves first, causing them to make heavier going of the puzzle than necessary. Therefore, for this reason, it's possible that this move escaped Ruud's attention.
gary w wrote:
Again just my outline of solving path.
Thanks, Gary. BTW, you chose the harder route in comparison to Afmob. However, you had some good ideas there! Here are a few comments you might find interesting:

Your move involving the " x-wing on 3" is something I would have expressed in a simpler form, as follows:

9. R23C6 cannot contain both of {13}, due to 5 being unavailable in R2C6
9a. The only other place for {13} in N2 is R123C4
9b. -> either R1C4 = 1, or...
9c. ...17(3) cage contains a 3 within R23C4 (or both)
9d. Either way, R4C4 cannot contain a 1

Also, I couldn't follow your contradiction: "But a 3 cannot be placed in any of these cells". I still had at least one 3 in R6C456 at this point. However, this contradiction could be avoided completely by eliminating the 8 from R8C6 in a simpler (separate) step. For example:

10. 1 in C6 locked in 9(3) cage at R234C6 or 9(2) cage at R67C6
10a. either 1 in 9(3) cage, then 9(3) = {135} ({126} blocked because 3 in C6 locked in this cage) -> R23C6 must contain a 5...
10b. ...or 1 is in 9(2) cage = [18] -> R7C6 = 8
10c. -> R237C6 must contain one of {58}
10d. -> {58} combo blocked for C789 outies at R18C6
10e. -> no 5,8 in R18C6

I also liked your logic in rejecting the [286] permutation for the 16(3) cage at R9C456. As mentioned several times before, I like to express such contradictions as closed loops, VIZ:

12. 17(3) cage at R9C456 = {2..} -> R9C4 = 2 -> R67C4 <> [29] -> R9C6 = 9 -> 17(3) cage at R9C456 = {9..}
12a. i.e., if 17(3) cage at R9C456 contains a 2, it must also contain a 9
12b. -> [286] combination/permutation blocked.
Afmob wrote:
I'm looking forward to V2.
Me, too! Hopefully, Ruud will post it soon.
Walkthrough by sudokuEd including killer pairs & "when" contradiction:
Para wrote:
What about posting a walk-through for Assassin 70
More penance :wink: . Definitely only a 0.75 rating - maybe even 0.5 - except I used 1 little contradiction move. Maybe that speeded things up.
gary w wrote:
Actually found a70v3 easier than version 1
Wow - we must solve in very different ways! I haven't looked at any of your outlines yet, or Afmod's walk-through. But am very curious since I'm still stuck on both V2 & V3. Later.

Cheers
Ed
(ps Please let me know of any corrections for the WT [edit: thanks Andrew])

Walk-through for Assassin 70

Prelims
i. 11(2)n1: no 1
ii. 11(3)n1 & n8: no 9
iii. 19(3)n2: no 1
iv. 17(2)n3 = {89}
v. 14(4)n3: no 9
vi. 9(3)n2: no 789
viii. 21(3)n3:no 123
ix. 29(4)n5 = {5789}
x. 9(2)n4 & n5: no 9
xi. 11(2)n4 & n5: no 1
xii. 6(3)n5 = {123}
xiii. 13(2)n6: no 123
xiv. 8(3)n6 = {125/134}
xv. 15(2)n6 = {69/78}
xvi. 21(3)n9 = {489/579/678}

1. 17(2)n3 = {89}: both locked for n3 & c8

2. "45"c9: r9c8 - 5 = r5c9
2a. r9c8 = {67}
2b. r5c9 = {12}

3. 21(3)n9 must have 6/7 = {579/678}(no 4)
3a. = 7{..}: 7 locked for n9
3b. no 8 r6c9
3c. no 6 r6c7

4. 14(4)n3 = {1346/2345}(no 7,8) (combo's with {12..} blocked by r5c9)

5. 8 in c9 only in n9: 8 locked for n9
5a. no 5 r6c7

6. "45" c789: 4 innies r1289c7 = h11(4) = {1235}: all locked for c7
6aa. forgot some clean-up: -> no 8 r6c7

6a. "45"c789:2 outies r18c6 = h13(2)
6b. = {49/58/67}(no 1,2,3)
6c. max r8c6 = 8 -> min r1c6 = 5

7. "45"c6: r59c6 = h14(2) = {59}/[86] = [5/8,6/9..]
7a. r5c6 = {589}
7b. r9c6 = {569}

8. h13(2)r18c6 = [94]/{67} = [6/9..]({58} clashes with r59c6)

9. Killer pair [69] in h14(2) & h13(2): both locked for c6

10. 9(3)n2 = {135/234} = 3[4/5..]
10a. 3 locked for c6
10b. -> 9(2) = {18/27}(no 45: blocked by 9(3))
[edit thanks Andrew]

11.29(4)n5 = {5789} all locked for n5
11a. no 12 in r7c6
11b. no 2346 in r7c4

12. 6(3)n5 = {123}: all locked for c5
12a. 19(3)n2 = {469/478/568}
12b. r45c5 must have 2 of {5789} -> 9 locked in 19(3) or r45c5
12c. no 9 r9c5

13. "45"c123: r18c4 = h6(2) = {15/24}

14. "45" c4: r59c4 = h11(2)
14a. = [56/74/83/92]
14b. r9c4 = {2346}

15. "45"c89: r3c8 = r5c7 = {467}

16. 21(3)n3 must have 2 of {467}
16a. = {678} only
16b. r5c7 = {67} (step 15)

17. r3c78 = {67}: both locked for r3 & n3

18. r4c7 = 8

19. r39c8 & r35c7 = {67}: both locked for c8 & c7

20. 8 in n5 only in r5: 8 locked for r5

21. "45" n69: r8c6 - 1 = r4c9
21a. r8c6 = {467} -> r4c9 = {356}

22. when 14(4)n3 = {1346} -> 6 must be in r4c9 -> 7 in r5c7 -> r67c9 = [96]; but this means 2 6's in c9
22a. 14(4) = {2345}(no 1,6): all locked for c9
22b. no 6 in r4c9 -> no 7 in r8c6 (step 21)
22c. -> no 6 in r1c6 (h13(2)r18c6

23. r5c9 = 1, r9c8 = 6 (i/oc9)

24. r3c78 = [67], r5c7 = 7

25. r89c9 = {78}(last combo): both locked for c9

26. r67c9 = [69], r67c7 = [94]

27. h14(2)r59c6 = {59}(last combo): both locked for c6

28. h13(2)r18c6 = [76]

29. r12c7 = {15}(last combo): both locked for c7 & n3

30. "45" n3: r4c9 = 5

31. "45" n6 1 innie r6c8 = 2


All simple from here
Walkthrough by Andrew to show it can still come out without Afmob's key step:
1. R12C2 = {29/38/47/56}, no 1

2. R12C8 = {89}, locked for C8 and N3

3. R67C1 = {18/27/36/45}, no 9

4. R67C3 = {29/38/47/56}, no 1

5. R67C4 = {29/38/47/56}, no 1

6. R67C6 = {18/27/36/45}, no 9

7. R67C7 = {49/58/67}, no 1,2,3

8. R67C9 = {69/78}

9. 11(3) cage at R1C3 = {128/137/146/236/245}, no 9

10. R123C5 = {289/379/469/478/568}, no 1

11. R234C6 = {126/135/234}, no 7,8,9

12. 21(3) cage at R3C7 = {579/678} (cannot be {489} because 8,9 only in R4C7) = 7{59/68}, no 4
12a. CPE no 7 in R12C7

13. R678C5 = {123}, locked for C5

14. R678C8 = 1{25/34}, 1 locked for C8

15. 11(3) cage at R8C6 = {128/137/146/236/245}, no 9

16. 21(3) cage in N9 = {489/579/678}, no 1,2,3

17. R1234C9 = {1238/1247/1256/1346/2345}, no 9

18. 29(4) cage in N5 = {5789}, locked for N5, clean-up: R7C4 = {5789}, no 1,2,4 in R7C6

19. 21(3) cage at R3C7 (step 12) = 7{59/68}
19a. 8,9 only in R4C7 -> R4C7 = {89}
19b. 7 locked in R3C78, locked for R3 and N3
19c. Max R12C7 = 10 (cannot be {56} which clashes with R3C78) -> min R1C6 = 3

20. 21(3) cage in N9 = {579/678} (cannot be {489} because R89C9 = {89} clashes with R67C9) = 7{59/68}, no 4, 7 locked for N9, clean-up: no 6 in R6C7, no 8 in R6C9

21. Killer pair 8,9 in R67C9 and R89C9, locked for C9

22. 45 rule on C5 3 innies R459C5 = 20 = {479/569/578}
22a. 4 of {479} and 6 of {569} must be in R9C5 -> no 9 in R9C5
[Alternatively min R45C5 = 12 -> max R9C5 = 8]

23. 45 rule on C123 2 outies R18C4 = 6 = {15/24}

24. 45 rule on C789 2 outies R18C6 = 13 = {58/67}/[94], no 1,2,3, no 4 in R1C6
[At this stage I missed 45 rule on C789 4 innies R1289C7 = 11 = {1235}, locked for C7, which breaks the puzzle wide open.
When I mentioned it to Ed he commented “I missed this early on too - put into the posted walk-through earlier than I originally found it. I often miss those sorts of innies. Get blinded by those 2 pointy outies”.]

25. 45 rule on C1234 2 innies R59C4 = 11 = [56/74/83/92], R9C4 = {2346}

26. 45 rule on C6789 2 innies R59C6 = 14 = [59/86/95], no 7 in R5C6, R9C5 = {569}

27. R9C456 = {259/268/349/358/367/457}
27a. 6 of {268/367} must be in R9C6 -> no 6 in R9C45, clean-up: no 5 in R5C4 (step 25)

28. 6 in N8 locked in R789C6, locked for C6, clean-up: no 3 in R7C6, no 7 in R8C6 (step 24)

29. R234C6 (step 11) = {135/234} = 3{15/24}, 3 locked for C6, clean-up: no 6 in R7C6

30. 6 in C6 locked in R89C6
30a. CPE no 6 in R9C7

31. R459C5 (step 22) = {479/578} = 7{49/58}, 7 locked for C5

32. 6 in C5 locked in R123C5, locked for N2
32a. R123C5 (step 10) = 6{49/58}

33. 45 rule on R89 3 innies R8C258 = 10 = {127/136/145/235}, no 8,9
33a. 7 of {127}and 6 of {136} must be in R8C2
33b. 1 of {145} must be in R8C5
33c. Taking steps 33a and 33b together -> no 1 in R8C2

34. 45 rule on C1 1 innie R5C1 – 2 = 1 outie R9C2, no 1,2 in R5C1, no 8,9 in R9C2

35. 45 rule on C12 1 innie R3C2 – 3 = 1 outie R5C3, no 1,2,3 in R3C2, no 4,7,8,9 in R5C3

36. 45 rule on C9 1 outie R9C8 – 5 = 1 innie R5C9 -> R5C9 = {12}, R9C8 = {67}
[At this stage I missed R1234C9 (step 17) = {1346/2345} (cannot be {1247/1256} which clash with R5C9). In my solving path {1238} had already been eliminated by the killer pair in step 21 while {1247} is eliminated by the killer pair in step 38 and {1256} by step 39 so missing this contradiction had little effect on my solving path.]

37. 21(3) cage in N9 (step 20) = 7{59/68}
37a. {678} must have 6 in R9C8 (R89C9 = {68} clashes with R67C9) -> no 6 in R89C9
37b. R89C9 = {59/78}

38. Killer pair 7,9 in R67C9 and R89C9, locked for C9

39. 3,4 in C9 locked in R1234C9 = 34{16/25}

40. 45 rule on C89 1 innie R3C8 = 1 outie R5C7 -> R5C7 = {567}

41. 11(3) cage at R8C6, min R8C6 = 4 -> max R89C7 = 7, no 8

42. Hidden killer pair 8,9 for C7 in R4C7 and R67C7 -> R67C7 = {49/58}, no 6,7
[Alternatively 7 in R3 locked in R3C78, R3C8 = R5C3 -> 7 in C7 locked in R35C7.]

43. 21(3) cage at R3C7 (step 12) = 7{59/68}
43a. 5 of {579} must be in R3C7 (R3C8 cannot be 5 because R34C7 = {79}, R5C7 = 5 (step 40) clash with R67C7) -> no 5 in R3C8, clean-up: no 5 in R5C7 (step 40)

44. Killer pair 6,7 in R39C8, locked for C8

45. R234C4 = {179/269/368/467} (cannot be {278} which clashes with R59C4, cannot be {458} which clashes with R18C4, cannot be {359} => R59C4 = [74] so R234C4 and R59C4 clash with R18C4), no 5
45a. 1 of {179} must be in R4C4 -> no 1 in R23C4
45b. 6 of {269/368/467} must be in R4C4
45c. Taking steps 45a and 45b together -> R4C4 = {16}
45d. 7 of {467} must be in R2C4 -> no 4 in R2C4

46. 45 rule on R1234 3 innies R4C258 = 14, min R4C58 = 7 -> max R4C2 = 7

47. 45 rule on N3 3 outies R1C6 + R4C79 = 20, max R1C6 + R4C7 = 18 -> no 1 in R4C9

48. 17(3) cage at R8C3 max R8C4 = 5 -> min R89C3 = 12, no 1,2

49. 45 rule on N7 2 outies R6C2 + R8C4 – 5 = 2 innies R7C13, max R6C2 + R8C4 = 14 -> max R7C13 = 9, no 9 in R7C3, clean-up: no 2 in R6C3

50. 15(4) cage at R4C8 = {1257/1347/2346} (cannot be {1356} because R45C8 = {35} clashes with R678C8)

51. 45 rule on N2 2 innies R1C46 = 2 outies R4C46, max R4C46 = 10 -> max R1C46 = 10 -> no 5 in R1C4, clean-up: no 1 in R8C4 (step 23)
51a. Min R1C46 = 6 but cannot form R4C46 = 6 -> min R4C46 = 7
51b. R4C46 cannot be 7 = [61] because R23C6 = {35}, R1C46 = 7 = [25] -> R23C6 clashes with R1C6 -> no 1 in R4C6
51c. R4C46 = 8,9,10 = [62/63/64] -> R4C4 = 6, clean-up: no 5 in R7C4
51d. For R1C46 = R4C46 = 9, R4C46 = [63] -> R23C6 = {15/24} -> R1C46 cannot be [45] -> no 4 in R1C4, clean-up: no 2 in R8C4 (step 23)
51e. R1C46 = 8,9,10 = [17/18/19/27/28], no 5 in R1C6, clean-up: no 8 in R8C6 (step 24)
51f. 1 in N5 locked in R6C56, locked for R6, clean-up: no 8 in R7C1

52. R4C4 = 6 -> R23C4 = {29/38/47} (step 45)

53. Killer triple 2,3,4 in R23C4, R6C4 and R9C4, locked for C4 -> R1C4 = 1, R8C4 = 5, clean-up: no 8 in R1C6 (step 24), no 4 in R6C6
53a. R12C3 = {28/37/46}, no 5
[R8C4 has probably been a hidden single for C4 since step 45. That would have bypassed most or all of step 51.]

54. R6C6 = 1 (hidden single in C6), R7C6 = 8, clean-up: no 3 in R6C3, no 3 in R6C4, no 5 in R6C7, no 7 in R6C9

55. Naked pair {69} in R67C9, locked for C9

56. R1234C9 (step 17) = {2345} (only remaining combination), locked for C9 -> R5C9 = 1
[Alternatively R5C9 = 1 has been a hidden single in N6 since step 51f.]

57. 15(4) cage at R4C8 (step 50) = {1257/1347} -> R5C7 = 7, R45C8 = {25/34}, R3C8 = 7 (step 40), R9C8 = 6, R67C9 = [69], R67C4 = [47], clean-up: no 2 in R6C1, no 3,5 in R7C1, no 5 in R7C3

58. R9C6 = 9 (naked single), R1C6 = 7, R5C6 = 5, clean-up: no 4 in R2C2, no 3 in R2C3 (step 53a), no 2 in R4C8 (step 57)

59. Naked pair {89} in R5C45, locked for R5 and N5 -> R4C5 = 7

60. 5 in N2 locked in R123C5 = {568} (step 32a), locked for C5 and N2 -> R5C5 = 9, R5C4 = 8, R9C5 = 4, R9C4 = 3, R8C6 = 6
60a. R8C4 = 5 -> R89C4 = 12 = [48] (only remaining permutation), R89C9 = [87], clean-up: no 5 in R6C1, no 7 in R6C3, no 3 in R7C3

61. R8C1 = 9 (hidden single in R8) -> R8C2 = 7 (hidden single in R8), R7C2 = 3 (hidden single in N7), R6C2 = 8, R6C7 = 9, R7C7 = 4, R4C7 = 8, R3C7 = 6, R67C3 = [56], clean-up: no 4 in R1C2, no 3 in R6C1, no 1 in R7C1

62. R67C1 = [72], R7C5 = 1, R68C5 = [32], R7C8 = 5, R6C8 = 2, R8C8 = 1, R78C7 = [32]

63. Naked pair {23} in R15C3, locked for C3 -> R2C3 = 7, R1C3 = 3, R5C3 = 2

64. R4C3 = 9 (hidden single in R4), R3C3 = 1, R3C2 = 5, clean-up: no 6 in R12C2

65. Naked pair {29} in R2C24, locked for R2

and the rest is naked singles


Last edited by Ed on Sun Dec 28, 2008 9:10 am, edited 2 times in total.

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Ed
PostPosted: Tue Jul 08, 2008 1:45 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 70v2 by Ruud (Sept 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:5888:1793:4866:4866:4356:3845:3845:1287:4872:5888:1793:4866:5132:4356:3598:3845:1287:4872:5888:2579:2579:5132:4356:3598:4888:4888:4872:5888:6684:2579:5132:4383:3598:4888:3618:4872:6684:6684:6684:4383:4383:4383:3618:3618:3618:2349:5166:815:1328:3633:3634:2355:5172:2869:2349:5166:815:1328:3633:3634:2355:5172:2869:2367:5166:4929:4929:3633:2884:2884:5172:3911:2367:2367:4929:5451:5451:5451:2884:3911:3911:
Solution: 
+-------+-------+-------+
| 9 6 4 | 8 7 2 | 5 1 3 |
| 5 1 7 | 6 9 3 | 8 4 2 |
| 8 2 3 | 5 1 4 | 9 7 6 |
+-------+-------+-------+
| 1 4 5 | 9 2 7 | 3 6 8 |
| 7 9 6 | 3 4 8 | 1 2 5 |
| 3 8 2 | 1 6 5 | 7 9 4 |
+-------+-------+-------+
| 6 5 1 | 4 3 9 | 2 8 7 |
| 4 7 8 | 2 5 1 | 6 3 9 |
| 2 3 9 | 7 8 6 | 4 5 1 |
+-------+-------+-------+
Quote:
sudokuEd: I'm still stuck on both V2 & V3
gary w: Obviously this one turned out to require afearsome amount of combination work
mhparker: I discovered that this puzzle could be unlocked without an awful lot of combination crunching
Andrew in 2011: I'll rate my walkthrough for A70 V2 at 1.75. Mike's alternative ending was the simplest; a neat way to get the final breakthrough! My solving path was quite a bit different from that in the "tag"

A 2022 forum Revisit to this puzzle here
Tag solution by sudokuEd, Afmob:
sudokuEd:
OK - this is as far as I could get on the V2. [Thanks to Andrew for some corrections and clarifications]

Don't like tt for tag, so hope no-one minds staying big.

I wouldn't normally go this many steps - but it's already typed, so.... BTW - some of the combo clashes I could never have found without Sudoku Solver's help in showing all the mutual exclusions.

Assassin 70V2
i. 3(2)n4 = {12}
ii. 5(2)n3 & n5 = {14/23}
iii. 14(2)n8 = {59/68}
iv. 7(2)n1: no 789
v. 9(2)n4 & n6: no 9
vi. 11(2)n6: no 1
vii.9(3) n7: no 789
viii. 21(3)n8: no 123
ix. 10(3)n1: no 89
x. 20(3)n2, n4 & n6: no 12
xi. 11(3)n8: no 9
xii. 19(3)n1 & n3 &n7:no 1
xiii.14(4)n6: no 9
xiv. 26(4)n4:no 1

1. 3(2)n4 = {12}: both locked for c3

2. min r34c3 in 10(3) = {35} = 8 -> max r3c2 = 2 ({34} blocked by 3 required in r3c2)
2a. 10(3)n1 can only have 1 of 1/2 = {136/145/235}(no 7)
2b. r3c2 = {12}
2c. r34c3 = {36/45/35}

3. "45" n147: r18c4 = h10(2)
3a. = {28/37/46}(no 5,9)

4. "45" n369: r18c6 = h3(2) = {12}: both locked for r6
4a. 15(3)r1c6 must have 1/2 = {159/168/249/258/267}(no 3) = [1/2] but not both
4b. ->no 1,2 in r12c7

5. 14(2)n5 = {59/68} = [5/6..]

6. 14(3)n2 = {347} ({356} clashes with [5/6..] in 14(2)n5)
6a. 3,4,7 all locked for c6
6b. 17(3)n2 can have at most 1 of 3,4,7 because of r23c6 -> {467} blocked

7. "45" c6: r59c6 = h14(2) = {59/68}

8. "45" c4: r59c4 = h10(2)
8a. no 5
8b. min r9c4 = 4 -> max r5c4 = 6

9. 5 in c4 only in 20(3) = 5{69/78}(no 3,4)

10. "45" r89: r8c258 = h15(3)
10a. min r8c28 = {34} = 7 -> max r8c5 = 8

11. "45" c1: r9c2 + 4 = r5c1
11a. max r9c2 = 5
11a. min r5c1 = 5

12. 9(3)n7 = {135/234}(no 6) = [1/2..] ({126} blocked by r7c3)
12a. = 3{..}: 3 locked for n7
12aa. add clean-up -> no 6 r6c1 (from Afmob's next post: thanks)
12b. no 1 in r9c2 since this would force 5 into both r5c1 (step 11) and r89c1 ({135} combo in 9(3))
12c. no 5 r5c1 (step 11)
12d. Killer pair [1/2] in 9(3) and r7c3: both locked for n7
12e. r6c1, no 78

13. 1 in c2 only in n1: 1 locked for n1

14. "45" c9: r5c9 = r9c8
14a. no 9 in r5c9 -> no 9 in r9c8
14b. no 1 in r123c7 -> no 1 in r5c9 or r9c8 (nishio?)

15. "45" c12: r5c3 - 4 = r3c2.
15a. r3c2 = {12} -> r5c3 = {56}
15b. -> {145} combo blocked from 10(3)n1 since it would clash with r5c3
15c. 10(3) = {136/235}(no 4)
15d. r345c3 = {356} naked triple: all locked for c3

16. 19(3)n1 & n7: must have 2 of {4789}
16a. = {289/379/469/478}
16b. h10(2)r18c4 must have 1 of {4789}
16c. -> when 20(3)n2 = {569} -> 5(2) n5 must have the remaining {4789} for c34 = {14}
16d. -> when 20(3)n2 = {578} -> all {4789} taken for c34 -> 5(2)n5 = {23}. These might be important later.

17. "45" c89: r5c7 + 6 = r3c8
17a. max. r5c7 = 3
17b. min r3c8 = 7
[I think it's here I should have seen CPE on 9's in c7 -> no 9 in r3c8. I get rid of it sometime anyway - but you can save a bit of head room by doing it now. I should have bookmarked each step on Sudoku Solver to check this. Learned now!]

18. "45" c89: r345c7 = h13(3)
18a. = {39/48}[1]/{47/56}[2]/{28/46}[3] ({57}[1] blocked by 7 in r3c8 step 17; {38}[2] blocked by 8 in r3c8; {19}[3] blocked by 9 in r3c8))
18b. = {139/148/238/247/256/346}

19. "45" n3: 3 outies = 13
19a. r1c6 = {12} -> r4c79 = 12/11
19b. no 1 r4c9

20. "45" n36: r1c6 + 18 = r6c789
20a. r6c789 = 19/20
20. -> no 1 r6c7; no 8 r7c7

21. 1 in n6 only in 14(4)
21a. {2345} combo blocked

22. "45" r1234: r4c258 = h12(3)

23. 21(3)n8 = {489/579/678} & r9c3 = {4789}
23a. -> 7 locked in these 4 cells for r9
23b. Common Peer Elimination (CPE): r8c4 'sees' all 7's in r9 -> no 7 r8c4 [SS found this one for me - annoyed that I missed it and annoyed that I forgot to turn off "Placement feedback".]
23c. no 3 r1c4
23d. {379} blocked from 19(3)n1

24. Since 21(3)n8 = [4/7,5/8,6/9,7/8,..]
24a. -> 19(3)n7, r8c3 + r8c4 + r9c3 = [469/478/874] all clash with 21(3)
24b. r8c3 + r8c4 + r9c3 = [829/928/739/937/964/748/784]
24c. no 4 r8c3
24d. 9 locked in r9c3456 for r9
24e. 15(3)n9; 7/9 in {249/267} must be in r8c9 & for {258},[2]{58} would clash with 21(3)n8)
24f. -> no 2 in r8c9

25. no 1 in r6c5 because of 1's in c4. Here's how.
25a. 1 in r7c4 -> 1 in 3(2)n4 in r6c3 -> no 1 r6c5
25b. 1 in r56c4 -> no 1 r6c5

26. "45"c789: r1289c7 = h23(4)
26a. r12c7 + r89c7 = {59-18/59-36/68-45/49-28/49-37/58-37/58-46/67-28} ({59-27/68-27} blocked by combo's in 11(3)n8)
26b. h23(4) = {1589/2489/2678/3479/3569/3578/4568} = [2/3/8..]

27. h13(3)r345c7 = {139/148/247/256/346} ({238} blocked by h23(4) step 26b)
27a. from 18a. = h13(3) {39/48}[1]/{47/56}[2]/{46}[3]
27b. no 2 r34c7
27c. no {289} combo in 19(3)n3
27d. no 2 in r5c9 or r9c8 (step 14b)
27e. h13(3) = [3/5/7/8..]

28. 19(4)n3 must have 1/2 for n3
28a. {3457} blocked

29. from step 26b. h23(4) = {1589/2489/2678/3479/3569/4568} ({3578} blocked by h13(3) step 27e)
29a. = r12c7 + r89c7 = {59-18/59-36/68-45/49-28/49-37/58-46/ }

30. when h13(3)r345c7 = {346} = [3]{46} the only combo in h23(4)r1289c7 = {1589} = {59-18}. But this means 2 9s in n3 (i/oc89)
30a. h13(3) = {139/148/247/256} = [2/4/9..]
30b. = h13(3) = {39/48}[1]/{47/56}[2]
30c. r5c7 = {12}
30d. no 9 r3c8
30e. 19(3)n3 = {379/478/568} = [6/7..]

31. 15(3)n2 = {159/168/249/258}(no 7) ({267} blocked by 19(3)n3 step 30e)

32.from step 29 h23(4) = {1589/3479/3569/4568}(no 2) ({2489} blocked by h13(3) step 30a; {2678} blocked by no 7 r12c7 step 29a)
32a. = r12c7 + r89c7 = {59-18/59-36/68-45/49-37/58-46}
32b. no 2 r89c7
32c. h23(4) = [4/5..] -> {45} blocked from 9(2)n6

33. r45c8 must have [1/2] for c8
33a. Killer pair [12] with r5c7: both locked for n6
33b. no 7 r7c7
33c. no 9 r7c9

34. 9 in c8 in 20(3)n6
34a. = 9{38/47/56}

35. 15(3)n9 = {159/168/249/258/267/357/456} ({348} clashes with r89c7 step 32a)

36. "45" n3: 3 outies = 13 -> r4c79 = 11/12
36a. min r4c79 = 3 -> r4c79 = 3/4/5..]

37. h12(3)r4c258 = {129/138/147/156/237/246} ({345} blocked by r4c79 step 36a)

38. "45" n2: r4c46 - 6 = r1c46

39.19(4)n3 {1279} must have 1 & 2 in n3 which clashes with 5(2)
39a. {1378} must have 3 in r4c9 so that it won't clash with 5(2): but {78} clashes with r3c8
39b. {2458} must have 4 in r4c9 so that it won't clash with 5(2): but {258}+{14} in 5(2) clashes with [4/5/8/] needed in r12c7 (step 32a)
39c. = {1369/1459/1468/1567/2359/2368/2467} = [5/6;4/5/6..]

40. 11(2) = {29/38/47} ({56} blocked by 19(4))

41. OK - should have seen this ages ago! Since r5c9 = r9c8: if r9c8 = 5/6 -> r5c9 = 5/6 -> r5c8 <>5/6 as this would clash with r5c3
41a. r5c8: no 5,6
41b. no 7 in r5c9 since no 7 in r9c8

42. 15(3)n9: {456} is the only combo with 4 in r89c9: but {456} in r5c9 + r89c9 (remembering that r5c9 = r9c8) clashes with 19(4)n3
42a. no {456} combo
42b. 15(3) = {159/168/249/258/267/357}
42c. no 4 r89c9

43. Better try the same thing for c1
43a. remembering r9c2 + 4 = r5c1
43b. r589c1 = [6]{34}/[7]{15/24}/[8]{23}/[9]{13}

44. 23(4)n1 = {1589/1679/2489/2579/2678/3569/4568} ({3479/3578} blocked by r589c1)

45. "45" n1: r4c1 + 15 = 4 innies in n1
45a. r12c3 = 11, 12, 13, 15, 17 (cage sum - r1c4)
45b. r3c23 = 4, 5 or 7 (cage sum - r4c3)
45c. cannot sum to 23 -> no 8 r4c1

46. 26(4)n4 must have 5/6 = {3689/4589/4679/5678}(no 2)

47. From step 36. "45" n3: 3 outies = 13 -> r4c79 = 11/12
& 36a. min r4c79 = 3 -> r4c79 = 3/4/5..]
47a. -> when h12(3)r4c258 + r4c46 = {147}[53/63] leaves no combinations with 3/4/5 in r4c79 that sum to 11/12 ({38/47/56/39/48/57})
47b. -> h12(3)r4c258 = {129/138/156/237/246}
47c. 9 in {129} must be in r4c2 -> no 9 r4c5

48. 19(3)n1 = {289/469/478}
48a. -> r12c3 = {89/78/49/48/47} = [4/8..]
48b. -> combo's with 4 & 8 in 23(4)n1 must have 4 in r4c1 -> r12c3 = [7/9..] not 2 of -> the 23(4) must have 1 of [7/9] for n1
48c. -> {4568} combo blocked from 23(4)n1. Whew.

49. r89c3 = {89/79/78/49/47} = [7/9..]
49a. -> for {479} combo in 20(3), must have 7/9 in r6c2
49b. no other combo with 4 -> no 4 in r6c2

50. "45" c2: r3459c2 = h18(4) = {1269/1278/1368/1467/2349/2358/2457} ({1359/1458} clashes with 7(2)n1);{2367} clashes with 6 in r5c3 when r3c2 = 6(i/o c12)

Marks here. Updated to include cleanup;no 6 in r6c1. Thanks Afmob.
Code:
.-------------------------------.-------------------------------.-------------------------------.
| 23456789  123456    4789      | 24678     123456789 12        | 45689     1234      123456789 |
| 23456789  123456    4789      | 56789     123456789 347       | 45689     1234      123456789 |
| 23456789  12        356       | 56789     123456789 347       | 3456789   78        123456789 |
:-------------------------------+-------------------------------+-------------------------------:
| 12345679  3456789   356       | 56789     12345678  347       | 3456789   12345678  3456789   |
| 6789      3456789   56        | 12346     123456789 5689      | 12        123478    34568     |
| 12345     356789    12        | 1234      23456789  5689      | 3678      3456789   34789     |
:-------------------------------+-------------------------------+-------------------------------:
| 45678     456789    12        | 1234      123456789 5689      | 1236      3456789   23478     |
| 12345     456789    789       | 23468     12345678  12        | 1345678   3456789   1356789   |
| 12345     2345      4789      | 46789     456789    5689      | 134568    34568     123568    |
'-------------------------------.-------------------------------.-------------------------------'

Afmob:
51. C123: 26(4) <> {5678} because
- If 26(4) = {5678} -> R4C3 = 3 -> 9(2) = [18/27/45] -> R6C2 = (49)
Now we show that both candidates for R6C2 are impossible:

a) R6C2 = 4 -> R78C2 = {79} @ 20(3) -> R89C3 = {48} @ 19(3) -> impossible since R8C4 <> 7
b) R6C2 = 9 -> R78C2 = {47/56} @ 20(3) -> Killer pair (45) in 9(3)
+ 20(3) @ N9 locked -> 9(2) = [18/27] + R89C3 <> 4
i. If R78C2 = {47} -> R7C1 = 8 -> R8C3 = 9 = R9C3
ii. If R78C2 = {56} -> 7(2) = {34} -> C3 <> 4

Therefore 26(4) <> {5678}.

52. 26(4) = 9{368/458/467} -> 9 locked for N4

PS: If my walkthrough for V3 is correct then V2 is more difficult than V3 because of step 51 which is quite complex.
Assassin V2 should be cracked now if you apply Killer pairs but that's up to someone else :wink:.

sudokuEd:
Afmob wrote:
26(4) <> {5678}
Hmm. OK. I've tried very hard to avoid these (looking both ways) sort of contradiction moves. But if there is no nicer way to proceed, how about expressing it this way [edit: note - it is still a "looking both ways" type of move.] [but] when I break it down like this I can 'see' it in my head better and makes it more satisfying. Would rather find another way though.

i. r6c2 = {356789}(no 4 see step 49)
ii. 9 in r6c2 -> r78c2 = {47} only. {56} in r78c2 is blocked because it forces 4 in n7 into 9(3) (c12) and 4 into 7(2)(c2) in n1: but, this leaves no 4 for c3.
iii. [9]{47} in 21(3) -> r89c3 = {89} -> r8c4 = 2 (cage sum)
iv. "45" n7: 4 outies = 15
v. 9 in r6c2 -> 2 in r8c4 -> r6c13 = 4 = [31]
vi. In summary, 9 in r6c2 -> 3 in r6c1
vii. when 26(4)n4 = {5678} -> r4c3 = 3 -> r6c2 = 9 (last candidate)
viii. but from V. when r6c2 = 9 -> r6c1 = 3: -> 2 3's in n4
ix. -> {5678} blocked from 26(4).

Cheers
Ed
Alt endings by goooders & Para:
goooders:
for what its worth i did v2 by realising the 20 cage in column 2 couldnt be 956 as it led to a contradiction this meant the only place for a 6 in the bottom left nonet was in the 9(2) cage after that its fairly straightforward
however for me that was a bit too much like trial and error unless anyone disagrees

Para:
Hi

Surely there is a nicer way. At least i think so.
Let's turn gooders solution into something a bit less T&E.(how to make that elimination in a logical way). Also it looks a bit better than Afmob's breaking move, which is going a bit towards T&E as well.
I did say no walk-throughs(which i didn't keep while solving this V2), but let's just call this a TAG instead :wink:

52. 6's in N7: Either R67C1 = [36] or 20(3) at R6C2 = {569} -> R6C2: no 3.
52a. 20(3) = {479/569/578}

53. 45 on C2: 4 innies: R3459C2 = 18. (A quick note, this hidden cage translates into the 26(4) cage at R4C2 through I/O difference of C1 and C12)
53a. R3459C2 = [1]{78}[2]/[1]{69}[2]/[1]{68}[3]/[2]{38}[5]/[2]{58}[3]/[2]{39}[4]/[2]{49}[3](other combinations blocked by other cages in C2)
53b. Translating to 26(4) cage: 26(4) at R4C2(R5C1,R45C2,R5C3) = [6]{78}[5]/[7]{68}[5]/[9]{38}[6]/[7]{58}[6]/[8]{39}[6]/[7]{49}[6] = {3689/4679/5678}= {3|7..},{4|8..},{3|4|5..}: 6 locked for N4; R3459C2: [1]{69}[2] blocked.

54. 45 on N1: 3 outies: R1C4 + R4C13 = 14 = [2][93/75]/[6]{35}/[7][25/43]/[8][15]: [4][73] blocked by 26(4) at R4C2: R1C4: no 4
54a. Clean up: R8C4: no 6

55. 19(3) at R8C3 = {89}[2]/{79}[3]/{78}[4]/{47}[8]: R89C3 = {47/78/79/89} = {7|9..}
55a. 20(3) at R6C2: [4]{79} blocked by R89C3: R6C2: no 4
Actually Ed already did this, i just took it from approximately Ed's position. Too lazy to read his walk-through.

56. 45 on N7: 4 outies: R6C123 + R8C4 = 15
56a. 26(4) at R4C2 = {3|7..},{4|8..},{3/4/5..}; R4C3 = {35}: R6C123 can't have any combination with both {37},{48},{34},{35} or {45}
56b. R6C123 + R8C4 = [182/281][4]/[192/291/381/471][3]/[391/382/472/571][2]: [371][4]/[372][3]/[481][2] blocked by 26(4), [452][4] blocked by 26(4) + R4C3: R6C2: no 5; R8C3: no 8
56c. Clean up: R1C4: no 2

57. 19(3) at R8C3 = {89}[2]/{79}[3]/{78}[4]: R89C3: no 4
57a. 4 in C3 locked for N1
57b. Clean up: R12C2 = {16/25} = {5|6..} : no 3

58. 20(3) at R6C2 = {479/578}:{569} blocked by R12C2: no 6
58a. R7C1 = 6(hidden); R6C1 = 3; R4C3 = 5

And now it is just basics(singles and last cage combos) till the end.
Contradiction path by garyw:
Obviously this one turned out to require afearsome amount of combination work and I gave up on it... :roll: however,on route,I did find a rather nice contradiction path that you might (or might not!!) want to see.

1.Usual prelims on inies/outies.Also the 14/3 cage N2/5 =3/4/7 ..no other possibilities given other cages in row and innie/outie totals.r6c123+r8c4=15.
2.In c4 9 must be in the 20/3 cage or at r9
3.IF at r4c9
a) 20/3 cage c4 is 5/7/8
b) r4c18 =46 or 64
c) 5/2 cage c4=2/3 -> kp 2 in r67c34 -> 11/2 cage r67c9 <> 2/9
d) Now consider the two 19/3 cages N1/7.From 2. above 4 is either at r8c4 or r1c4. If at r1c4 ->4 at r9c3 (can't be anywhere else in r3)Either way a 4 "looks" at the 21/3 cage N8.
4.So,the 21/3 cage N8 =9 (postulate 3)75 (57 blocked by 14/3 cage c6)
5.With r9c6=5 -> r5c6=9 (these 2 cells=14 from prelim work).
6.With r8c4=a 4/6 r6c2<>9 (otherwise r6c123+r8c4>15..thus 9 is in 26/4 cage N4.Thus,from 5. above r4c2=9 -> for N7 9 at r8c3.
7.Now the 9s placed in N56 and N78 force a 9 into,respectively,r6c8 and r7c8 ( cannot go into 11/2 cage..3c above).Contradiction.
8.Therefore postulate 3 is wrong the 9 in c4 must be in the 20/3 cage which is then 5/6/9.

After this still not easy but does solve fairly readily.

Wouldn't normally post a hypothetical like this but I was struck by its rather lovely symmetry re the 4/6 @ r18c4,the KP on the 2 eliminating a 9 from the 11/2 cage c9 and then the 4 9s -> a contradiction.

The other thing I noted was the two totally different soving pathways in v1 and v3 just determined by the broken cage at r4c258.Killers are fascinating!!

Regards

Gary
Alt ending by mhparker:
gary w wrote:
Obviously this one turned out to require a fearsome amount of combination work
Or does it? I discovered that this puzzle could be unlocked without an awful lot of combination crunching by looking at the N4 innies. Therefore, here are some alternative steps, starting from Ed's marks pic. Note that the first step in the sequence happens to be the same as Para's step 52.

51. 6 in n7 locked in r7c1+r78c2
51a. -> either 6 in r78c2, or...
51b. ...6 in r7c1 -> 3 in r6c1
51c. Either way: no 3 in r6c2

52. Innies n4: r4c46+r6c123 = 19(5) = {12349/12358/12367/12457}
(Note: {13456} blocked by r5c3)
52a. if {12349}, 9 must go in r6c2
52b. -> no 9 in r4c1
52c. if {12358}, 8 must go in r6c2
52d. if {12457}, 5 must go in r4c3
52e. -> no 5 in r6c2

53. 20(3)n47 = {479/569/578} = {(4/5)..}
53a. {45} of 20(3)n47 now only available within r78c2
53b. -> r78c2 and 9(3)n7 (step 12) form killer pair on {45} within n7
53c. -> no 4,5 elsewhere in n7
53d. cleanup: no 4,5 in r6c1

54. 4 in c3 locked in n1 -> not elsewhere in n1
54a. 4 locked within 19(3)n12 within r12c3
54b. -> no 2,4 in r1c4 (no 2 because otherwise 19(3) cage sum unreachable)
54c. cleanup: no 3 in r12c2; no 6,8 in r8c4 (step 3)
54d. 7(2)n1 = {16/25} = {(5/6)..}

55. 7(2)n1 blocks {569} combo for 20(3)n47
55a. -> 20(3)n47 = {479/578} (no 6)
55b. 7 locked for c2

56. Hidden single in n7 at r7c1 = 6
56a. -> r6c1 = 3

and so on...
2011 Walkthrough by Andrew:
Thanks Ruud for a challenging V2.

Good work Ed for making so much progress before making it a "tag". Then Mike's alternative ending was the simplest; a neat way to get the final breakthrough!

My solving path was quite a bit different from that in the "tag", probably because I used a fairly early contradiction move; the "tag" used more combination analysis.

Here is my walkthrough for A70 V2.

Prelims

a) R12C2 = {16/25/34}, no 7,8,9
b) R12C8 = {14/23}
c) R67C1 = {18/27/36/45}, no 9
d) R67C3 = {12}
d) R67C4 = {14/23}
e) R67C6 = {59/68}
f) R67C7 = {18/27/36/45}, no 9
g) R67C9 = {29/38/47/56}, no 1
h) 19(3) cage at R1C3 = {289/379/469/478/568}, no 1
i) 20(3) cage at R2C4 = {389/479/569/578}, no 1,2
j) 10(3) cage at R3C2 = {127/136/145/235}, no 8,9
k) 19(3) cage at R3C7 = {289/379/469/478/568}, no 1
l) 20(3) cage at R6C2 = {389/479/569/578}, no 1,2
m) 20(3) cage at R6C8 = {389/479/569/578}, no 1,2
n) 9(3) cage in N7 = {126/135/234}, no 7,8,9
o) 19(3) cage at R8C3 = {289/379/469/478/568}, no 1
p) 11(3) cage at R8C6 = {128/137/146/236/245}, no 9
q) 21(3) cage in N8 = {489/579/678}, no 1,2,3
r) 26(4) cage in N4 = {2789/3689/4589/4679/5678}, no 1
s) 14(4) cage in N6 = {1238/1247/1256/1346/2345}, no 9

1. Naked pair {12} in R67C3, locked for C3
1a. 10(3) cage at R3C2 = {136/145/235} (cannot be {127} because 1,2 only in R3C2), no 7
1b. 1,2 only in R3C2 -> R3C2 = {12}
1c. 9(3) cage in N7 = {135/234} (cannot be {126} which clashes with R7C3), no 6, 3 locked for N7, clean-up: no 6 in R6C1
1d. Killer pair 1,2 in R7C3 and 9(3) cage, locked for N7, clean-up: no 7,8 in R6C1

2. 45 rule on C1 1 innie R5C1 = 1 outie R9C2 + 4, no 2,3,4 in R5C1

3. 45 rule on C12 1 outie R5C3 = 1 innie R3C2 + 4 -> R5C3 = {56}
3a. Killer pair 5,6 in 10(3) cage at R3C2 and R5C3, locked for C3
3b. 26(4) cage in N4 = {3689/4589/4679/5678} (cannot be {2789} because R5C3 only contains 5,6), no 2

4. 45 rule on C123 2 outies R18C4 = 10 = {28/37/46}, no 5,9

5. 45 rule on C1234 2 innies R59C4 = 10 = [19/28/37/46/64], no 5, no 7,8,9 in R5C4
5a. 1 in C4 only in R59C4 = [19] or R67C4 = {14} -> no 4 in R59C4 (locking-out cages) -> R59C4 = [19/28/37], no 4,6
5b. 5 in C4 only in 20(3) cage at R2C4 = {569/578}, no 3,4

6. 45 rule on C789 2 outies R18C6 = 3 = {12}, locked for C6

7. 45 rule on C6789 2 innies R59C6 = 14 = {59/68}
7a. Naked quad {5689} in R5679C6, locked for C6

8. 45 rule on C89 1 innie R3C8 = 1 outie R5C7 + 6 -> R3C8 = {789}, R5C7 = {123}
[After looking at the “tag”, I realised that I had also missed 9 in C7 only in R1234C7, CPE no 9 in R3C8, which would have simplified this step and probably later ones.]

9. 45 rule on C9 1 innie R5C9 = 1 outie R9C8, no 9 in R9C8
[Ed found a neat elimination. After my next step, 1 in N3 only in R12C8 and R123C9 -> no 1 in R5C9 + R9C8.]

10. 15(3) cage at R1C6 = {159/168/249/258/267} (cannot be {348/357/456} because R1C6 only contains 1,2), no 3
10a. R1C6 = {12} -> no 1,2 in R12C7

11. 45 rule on C1 3 innies R589C1 = 13 must contain one of 6,7,8,9 -> R5C1 = {6789}, clean-up: no 1 in R9C2 (step 2)
11a. 1 in C2 only in R123C2, locked for N1

12. 45 rule on C123 4 innies R1289C3 = 28 = {4789} (only remaining combination), locked for C3

13. 45 rule on C89 3 outies R345C7 = 13 = {139/148/238/247/256/346} (cannot be {157} because 19(3) cage at R3C7 cannot be {57}7)
13a. 45 rule on C789 4 innies R1289C7 = 23 = {1589/2678/3479/3569/4568} (cannot be {1679/2579} which aren’t compatible with combinations for 15(3) cage at R1C6 and 11(3) cage at R8C6 – in the case of {2579} because 11(3) cage cannot be {27}2, cannot be {2489/3578} which clash with R345C7)
13b. R345C7 = {139/148/247/256/346} (cannot be {238} which clashes with R1289C4)
13c. 2 of {247/256} must be in R5C7 -> no 2 in R34C7

14. 45 rule on N3 3(1+2) outies R1C6 + R4C79 = 13
14a. Max R1C6 + R4C7 = 11 -> min R4C9 = 2

15. 45 rule on N9 4(3+1) outies R6C789 + R8C6 = 21
15a. Max R8C6 = 2 -> min R6C789 = 19, no 1, clean-up: no 8 in R7C7

16. 45 rule on R1234 3 innies R4C258 = 12
16a. Min R4C28 = 4 -> max R4C5 = 8

17. 6 in N7 only in R67C1 = [36] or in 20(3) cage at R6C2 = {569} -> no 3 in R6C2 (locking-out cages)

18. 1 in N6 only in 14(4) cage = {1238/1247/1256/1346}
18a. 8 of {1238} must be in R5C89 (R5C789 cannot be {123} which clashes with R5C4) -> no 8 in R4C8

[I can’t see any more routine steps at the moment so I’ll try something a bit harder but still fairly short.]
19. 20(3) cage at R6C2 = {479/569/578} cannot be {479}, here’s how
20(3) cage = {479}, locked for C2, R7C1 = 6 (hidden single in N7), R6C1 = 3, R9C2 = 3 (hidden single in N7), R5C1 = 7 (step 2), R3C3 = 3 (hidden single in N1) => no remaining combination for 26(4) cage in N4 because {5678} clashes with R4C3
19a. -> 20(3) cage at R6C2 = {569/578}, no 4, 5 locked for C2, clean-up: no 2 in R12C2, no 9 in R5C1 (step 2)

20. R12C2 = {16/34}
20a. Consider the combinations for R12C2
R12C2 = {16}, locked for C2 and N1
or R12C2 = {34}, R3C2 = 1 (hidden single in C2), R9C2 = 2 (hidden single in C2), R5C1 = 6 (step 2)
20b. -> 6 in R12C2 or R5C1, CPE no 6 in R123C1 + R456C2

21. Again consider the same combinations
21a. R12C2 = {16}, locked for C2 -> R3C2 = 2, R34C3 = {35} (step 1a), R5C3 = 6
or R12C2 = {34}, R3C2 = 1 (hidden single in C2), R9C2 = 2 (hidden single in C2), R5C1 = 6 (step 2)
21b. -> 6 must be in R5C13, locked for R5 and N4, clean-up: no 8 in R9C6 (step 7), no 6 in R9C8 (step 9)

22. 21(3) cage in N8 = {489/579/678}
22a. 6 of {678} must be in R9C6 -> no 6 in R9C5

23. 9 in C1 only in 23(4) cage at R1C1 = {1589/2489/2579/3479}
23a. R67C1 = [18/27/36] (cannot be {45} which clashes with 23(4) cage at R3C1), no 4,5

24. 45 rule on N7 4(3+1) outies R6C123 + R8C4 = 15
24a. Min R6C123 = {125} = 8 -> max R8C4 = 7, clean-up: no 2 in R1C4 (step 4)

25. Hidden killer pair 1,2 in 17(3) cage and R1C6 for N2, R1C6 = {12} -> 17(3) cage in N2 must contain one of 1,2 = {179/269/278}, no 3,4,5
25a. 5 in N2 only in R23C4, locked for 20(3) cage at R2C4, no 5 in R4C4

26. R4C258 = 12 (step 16) = {129/138/147/237/246} (cannot be {156} because no 1,5,6 in R4C2, cannot be {345} which clashes with R4C3), no 5

27. 45 rule on C5 3 innies R459C5 = 14
27a. 17(4) cage in N5 = {1259/1268/1349/1358/2348/2357/2456} (cannot be {1367} because R5C6 only contains 5,8,9, cannot be {1457} because R459C5 = 14 and no 3 in R9C5)

28. R459C5 = 14 (step 27) = {149/158/248/347} (cannot be {167/239/257} which clash with 17(3) cage in N2, cannot be {356} = [635] because no combinations of 17(4) cage in N5 contains both of 3,6), no 6

29. R9C3 = {4789}, 21(3) cage in N8 = {489/579/678} -> 7 must be in R9C3 + 21(3) cage, locked for R9, clean-up: no 7 in R5C9 (step 9)
[I missed another CPE here, 7 in R9 only in R9C345, CPE no 7 in R8C4.]

30. 45 rule on N1 3(2+1) outies R1C4 + R4C13 = 14
30a. Min R1C4 + R4C3 = [33] = 6 -> max R4C1 = 8

31. 9 in C1 only in R123C1, locked for N1
31a. 9 in C3 only in R89C3, locked for N7

32. 19(3) cage at R8C3 must contain 9 = {289/379/469}
32a. 2,3,6 only in R8C4 -> R8C4 = {236}, clean-up: no 3,6 in R1C4 (step 4)

33. 19(3) cage at R1C3 = {478}, CPE no 4,7,8 in R1C12, clean-up: no 3 in R2C2
33a. 3 in N2 only in R23C6, locked for C6

34. R1C4 + R4C13 = 14 (step 30)
34a. R1C4 = {478} -> R4C13 = 6,7,10 = [15/25/43/73] -> R4C1 = {1247}

35. R4C258 (step 26) = {129/138/237/246} (cannot be {147} which clashes with R4C6)
35a. 4 of {246} must be in R4C2 -> no 4 in R4C58

36. 23(4) cage at R1C1 (step 23) = {1589/2489/2579/3479}
36a. 9(3) cage in N7 (step 1c) = {135/234}
Consider candidates for R9C2
R9C2 = 2 => R89C1 = {34}, locked for C1 => no {3479} in 23(4) cage
or R9C2 = 3 => R5C1 = 7 (step 2) => no {3479} in 23(4) cage
or R9C2 = 4 => R89C1 = {23}, locked for C1 => no {3479} in 23(4) cage
-> no {3479} in 23(4) cage
36b. 23(4) cage at R1C1 = {1589/2489/2579}, no 3

37. 23(4) cage at R1C1 (step 36b) = {1589/2489/2579}
37a. 4 of {2489} must be in R4C1 (cannot be 9{48}2 which clashes with R12C3, ALS block) -> no 4 in R23C1
37b. R12C2 = {16}/[34]
Consider combinations for R12C2
R12C2 = {16}, locked for C2 => R3C2 = 2 => no {2489} in 23(4) cage (which must be {289}4)
or R12C2 = [34] => R12C3 = {78}, locked for C2 => no {2489} in 23(4) cage
-> no {2489} in 23(4) cage
37c. -> 23(4) cage at R1C1 = {1589/2579}, no 4, 5 locked for C1 and N1
37d. 5 in C3 only in R45C3, locked for N4

38. 4 in C1 only in R89C1, locked for N7
38a. 9(3) cage in N7 (step 1c) = {234} (only remaining combination), locked for N7 -> R7C3 = 1, R6C3 = 2, clean-up: no 4 in R6C4, no 8 in R6C7, no 7 in R7C1, no 3 in R7C4, no 7 in R7C7, no 9 in R7C9

39. 4 in C3 only in R12C3, locked for N1 and 19(3) cage at R1C3, no 4 in R1C4, clean-up: no 3 in R1C2, no 6 in R8C4 (step 4)

40. Naked pair {16} in R12C2, locked for C2 and N1 -> R3C2 = 2, R3C3 = 3, R4C3 = 5, R5C3 = 6, R9C2 = 3, R5C1 = 7 (step 2), R4C1 = 1, R6C1 = 3, R7C1 = 6, R6C4 = 1, R7C4 = 4, clean-up: no 8 in R6C6, no 5 in R6C7, no 5,7 in R6C9, no 8 in R7C9

41. 20(3) cage at R6C2 (step 19) = {578} (only remaining combination) -> R6C2 = 8, R78C2 = {57}, locked for N7, clean-up: no 3 in R7C9

42. 19(3) cage at R8C3 (step 32) = {289} (only remaining combination) -> R8C4 = 2, R1C4 = 8 (step 4), R89C1 = [42], R5C4 = 3, R9C4 = 7 (step 5), R8C6 = 1, R1C6 = 2, clean-up: no 3 in R2C8, no 2 in R5C9 (step 9)

43. 17(3) cage in N2 (step 25) = {179} (only remaining combination), locked for C5 and N2, 14(3) cage at R2C6 = [347]

44. R4C258 (step 35) = {246} (only remaining combination) = [426], R4C4 = 9, R5C2 = 9, clean-up: no 5 in R7C6, no 3 in R7C7, no 5 in R7C9

45. R459C5 (step 28) = {248} (only remaining combination) = [248], R89C3 = [89], R7C6 = 9, R6C6 = 5

46. 19(3) cage at R3C7 = {379} (only remaining combination, cannot be {568} because 5,6 only in R3C7) -> R4C7 = 3, R3C78 = {79}, locked for R3 and N3

47. R1C6 = 2 -> R12C7 = 13 = [58], R7C7 = 2, R6C7 = 7, R7C9 = 7, R6C9 = 4

and the rest is naked singles.


Rating Comment. I'll rate my walkthrough for A70 V2 at 1.75. I used a fairly short contradiction move and several short forcing chains; each of them may have been in the 1.5 range but since I used several I've rated my walkthrough at 1.75. Also, on looking at my contradiction move in step 19 again, I see that it uses singles in three different nonets so maybe it should be in the 1.75 range.
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Ed
PostPosted: Tue Jul 08, 2008 1:49 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 70v3 by sudokuEd (Sept 07)
Puzzle pic courtesy of Borge:
Image

NOTE: remote 14(3) cage at r4c258
Code: Select, Copy & Paste into solver:
3x3::k:4864:2817:2818:2818:4868:3333:3333:4359:3592:4864:2817:2818:4364:4868:2318:3333:4359:3592:4864:3859:3859:4364:4868:2318:5400:5400:3592:4864:3612:3859:4364:3612:2318:5400:3612:3592:2852:2852:2852:5671:5671:5671:3114:3114:3114:2349:4654:2863:2864:1585:2354:3379:2100:3893:2349:4654:2863:2864:1585:2354:3379:2100:3893:3903:4654:4417:4417:1585:2884:2884:2100:5447:3903:3903:4417:4171:4171:4171:2884:5447:5447:
Solution: 
+-------+-------+-------+
| 8 2 3 | 1 6 7 | 5 9 4 |
| 6 9 7 | 2 5 4 | 1 8 3 |
| 4 5 1 | 9 8 3 | 6 7 2 |
+-------+-------+-------+
| 1 4 9 | 6 7 2 | 8 3 5 |
| 3 6 2 | 8 9 5 | 7 4 1 |
| 7 8 5 | 4 3 1 | 9 2 6 |
+-------+-------+-------+
| 2 3 6 | 7 1 8 | 4 5 9 |
| 9 7 4 | 5 2 6 | 3 1 8 |
| 5 1 8 | 3 4 9 | 2 6 7 |
+-------+-------+-------+
Quote:
garyw: Didn't find it too bad..no worse than version 1 ..I noted .. the two totally different soving pathways in v1 and v3 just determined by the broken cage at r4c258. Killers are fascinating!!
Afmob: it was nearly as difficult as V2 though the moves were less complicated. I guess rating would be 1.25
sudokuEd: Really good solution Afmob. 9c is a ripper! Good call on the rating - once you find one of the keys it falls very easily
Andrew in 2011: Actually I found it as hard to solve as the V2.. I'll rate my walkthrough for A70 V3 at 1.5. I ... managed to avoid any contradiction moves
Ed in 2011: Very happy with this alternate start. Feels simpler than both the previous walkthroughs.
Walkthrough by Afmob:
After taking a short break, I solved A70 V3 today. I say it was nearly as difficult as V2 though the moves were less complicated. I guess rating would be 1.25.

Walkthrough for Assassin 70 V3:

1. C789
a) 17(2) = {89} locked for C8 + N3
b) 21(3) @ N3 must have 8 or 9, only possible @ R4C7 -> R4C7 = (89)
c) 21(3) @ N3 = 7{59/68} <> 4 because of b), 7 locked for R3 + N3
d) 8(3) = 1{25/34} -> 1 locked for C8
e) Innies+Outies C9: 5 = R9C8 - R5C9 -> R9C8 = (67), R5C9 = (12)
f) 21(3) @ N9 <> 4 because {489} is impossible since R9C8 = (67)
g) 3,4 locked in 14(4) = 34{16/25} <> 7,8
h) 21(3) @ N9: R89C9 <> 6 since R89C9 would be {68} which is blocked by killer pair (68) of 15(2)

2. C789
a) 21(3) = 7{59/68} -> 7 locked for N9
b) 15(2): R6C9 <> 8; 13(2): R6C7 <> 6
c) 8 locked in R456C7 for C7
d) 13(2): R6C7 <> 5
e) Innies+Outies C89: 18 = R345C7 - R4C8 -> R4C8 <> 7
-> R4C8 = (23456) -> R345C7 = 20/21/22/23/24
f) Innies+Outies C89: R5C7 <> 1,2,3 since R345C7 would be smaller than 20

3. R5
a) 22(3) = 9{58/67} -> 9 locked for R5 + N5
b) 11(2) @ N5: R7C4 <> 2
c) 11(3) <> 8 because {128} blocked by R5C9 = (12)
d) 12(3) <> {345} because of Killer pair (34) of 11(3)
e) Killer pair (12) of 11(3) + R5C9 = (12) -> R5C8 <> 2
f) 12(3) <> 5 -> {156} blocked by Killer pair (56) of 22(3)

4. R89
a) Innies = 10(3) <> 8,9; R8C2 <> 1 since R8C5 + R8C8 <= 8 (3+5)

5. C56
a) Innies C6 = 27(4) <> 1,2
b) 9(2) <> 3,6 since {36} is blocked by Killer pair (36) of 9(3)
c) 6(3) = {123} locked for C5

6. C12
a) Innies+Outies C1: -2 = R9C2 - R5C1 -> R5C1 <> 1,2; R9C2 <> 6,7,8,9
b) 18(3) <> 1 because R8C2 has no 1,8,9

7. N3
a) Innies+Outies: -1 = R1C6 + R4C9 - R3C78; R3C78 = 12/13 (see step 1c)
-> R1C7 + R4C9 = 11/12
-> R1C6 <> 3,4 since R1C6 + R4C9 would be smaller than 11

8. C789 !
a) Innies = 14(4+1): 1,2,3 locked in R1289C7 = 123{4/5/6} = 10/11/12
-> R4C8 = (234)

9. C789
a) 14(4): R4C9 <> 1,2 since 14(4) has 1 xor 2, so if R4C9 = (12)
R12C7 @ 13(3) must have 1 and 2 which is impossible (13(3) = 1 + 2 + ?)
-> 13(3) must have 1 xor 2
b) 11(3): R8C6 <> 8 because R89C7 would be {12} so that R12C7 <> 1,2
c) ! 13(2) <> 6,7 because if 13(2) = [76] -> 15(2) = {69}
-> no possible combination for 21(3) @ N9 (no 6,9)
d) Killer pair (89) in 13(2) + R4C7 = (89) locked for C7

10. N5
a) 8 locked in 22(3) -> 22(3) = {589} locked for R5 + N5
b) 11(2): R6C4 <> 3,6
c) 9(2): R6C6 <> 1,4

11. N6 !
a) 8(3): R6C8 <> 5 because R78C8 would be {12} -> blocked by
Killer pair (12) of 11(3) @ N9
b) Hidden Single: R4C9 = 5

12. C789
a) 21(3) @ N9 = {678} locked for N9
b) R9C8 = 6, R7C9 = 9 -> R6C9 = 6
c) 14(4) = {2345} -> 2,3,4 locked for C9 + N3
d) R5C9 = 1
e) 13(3) = {157} -> R1C6 = 7; 1,5 locked for C7 + N3
f) R3C8 = 7, R3C7 = 6, R4C7 = 8
g) Hidden Single: R6C7 = 9 @ N6 -> R7C7 = 4, R5C7 = 7, R5C8 = 4
h) 11(3) = {236} -> R8C6 = 6; 2,3 locked for N9
i) 8(3) = {125} -> R6C8 = 2 -> R4C8 = 3

13. C456
a) 9(3) = 3{15/24} -> Killer pair (45) blocks {45} of 9(2)
b) R6C6 = 1, R7C6 = 8, R6C5 = 3
c) 11(2) = [47] -> R6C4 = 4, R7C4 = 7
d) R4C6 = 2 -> R4C4 = 6 -> R4C5 = 7 -> R4C2 = 4
e) 6(3) = {123} -> 1,2 locked for N8
f) 16(3) = {349} locked for R9 + N8
g) R8C4 = 5, R9C7 = 2, R8C7 = 3, R8C8 = 1, R7C8 = 5, R8C5 = 2, R7C5 = 1
h) 17(3) @ N8 = {458} -> R8C3 = 4, R9C3 = 8

14. Rest is clean-up and singles

I still hope to see more walkthroughs for V2.
2011 Walkthrough by Andrew:
Thanks Ed for this variant of Ruud's A70 and A70 V2. Maybe you should have called it V1.5, rather than V3? Actually I found it as hard to solve as the V2; I think I found some steps harder to spot but I used fewer "high tariff" steps. Also it was hard because of the mistake I originally made before step 17.

Same puzzle as Assassin 70 except that disjoint 14(3) cage in R4C258 has been split out from the R45 cages and the cage totals have been changed to give a different solution. Many relationships for Assassin 70 still apply. However I’ve started again rather than just using as many previous steps as possible.

Here is my walkthrough for A70 V3.

Prelims

a) R12C2 = {29/38/47/56}, no 1
b) R12C8 = {89}
c) R67C1 = {18/27/36/45}, no 9
d) R67C3 = {29/38/47/56}, no 1
e) R67C4 = {29/38/47/56}, no 1
f) R67C6 = {18/27/36/45}, no 9
g) R67C7 = {49/58/67}, no 1,2,3
h) R67C9 = {69/78}
i) 11(3) cage at R1C3 = {128/137/146/236/245}, no 9
j) 19(3) cage in N2 = {289/379/469/478/568}, no 1
k) 9(3) cage at R2C6 = {126/135/234}, no 7,8,9
l) 21(3) cage at R3C7 = {489/579/678}
m) 11(3) cage in N4 = {128/137/146/236/245}, no 9
n) 22(3) cage in N5 = {589/679}
m) 6(3) cage at R6C5 = {123}
n) 8(3) cage at R6C8 = {125/134}
o) 11(3) cage at R8C6 = {128/137/146/236/245}, no 9
p) 21(3) cage in N9 = {489/579/678}, no 1,2,3
q) 14(4) cage at R1C9 = {1238/1247/1256/1346/2345}, no 9

Steps resulting from Prelims
1a. Naked pair {89} in R12C8, locked for C8 and N3
1b. Naked triple {123} in 6(3) cage at R6C5, locked for C5
1c. 22(3) cage in N5 = {589/679}, locked for R5 and N5, clean-up: no 2 in R7C4
1d. 8(3) cage at R6C8 = {125/134}, 1 locked for C8

2. 21(3) cage at R3C7 = {579/678} (cannot be {489} because 8,9 only in R4C7), no 4
2a. 8,9 only in R4C7 -> R4C7 = {89}
2b. 21(3) cage at R3C7 = {579/678}, 7 locked for R3 and N3

3. 45 rule on C1 1 innie R5C1 = 1 outie R9C2 + 2, no 1,2 in R5C1, no 7,8,9 in R9C2

4. 45 rule on C9 1 outie R9C8 = 1 innie R5C9 + 5, R5C9 = {12}, R9C8 = {67}
4a. 21(3) cage in N9 = {579/678} (cannot be {489} because R9C8 only contains 6,7), no 4, 7 locked for N9, clean-up: no 6 in R6C7, no 8 in R6C9
4b. 6 of {678} must be in R9C8 (R89C9 cannot be {68} which clashes with R67C9)
-> no 6 in R89C9
4c. Killer pair 8,9 in R67C9 and 21(3) cage, locked for C9
4d. 8 in C9 only in R789C9, locked for N9, clean-up: no 5 in R6C7

5. 3,4 in C9 only in 14(4) cage at R1C9 = {1346/2345}, no 7

6. 11(3) cage in N4 = {137/146/236/245} (cannot be {128} which clashes with R5C9), no 8, clean-up: no 6 in R9C2 (step 3)
6a. Killer pair 1,2 in 11(3) cage and R5C9, locked for R5

7. 45 rule on N3 3(2+1) outies R1C6 + R4C79 = 20
7a. Max R4C7 = 9 -> min R1C6 + R4C9 = 11, no 1,2,3,4 in R1C6, no 1 in R4C9

8. 45 rule on C789 5(4+1) innies R1289C7 + R4C8 = 14
8a. Min R1289C7 = 10 -> max R4C8 = 4

9. 14(3) disjoint cage R4C258
9a. Max R4C58 = [84] = 12 -> min R4C2 = 2
9b. Min R4C58 = [24] = 6 -> max R4C2 = 8

10. 45 rule on C789 2 outies R18C6 = 1 innie R4C8 + 10
10a. Min R4C8 = 2 -> min R18C6 = 12, no 1,2 in R8C6

11. 45 rule on C6789 3(2+1) innies R4C8 + R59C6 = 17
11a. Max R4C8 = 4 -> min R59C6 = 13, no 1,2,3 in R9C6
11b. Min R9C56 = 9 -> max R9C4 = 7

12. 45 rule on C6 4 innies R1589C6 = 27 = {3789/4689/5679}
12a. R67C6 = {18/27/45} (cannot be {36} which clashes with R1589C6), no 3,6

13. 45 rule on R89 3 innies R8C258 = 10 = {127/136/145/235}, no 8,9
13a. 6,7 of {127/136} must be in R8C2, 1 of {145} must be in R8C5 -> no 1 in R8C2

14. 45 rule on C89 2 innies R34C8 = 1 outie R5C7 + 3
14a. Min R34C8 = 7 -> min R5C7 = 4

15. 45 rule on C89 4(3+1) innies R345C8 + R5C9 = 15 = {257}1/{347}1/{346}2 (cannot be {356}1 which clashes with 8(3) cage at R6C8, cannot be {247/256}2 because both 2s would be in N6)
15a. 6 of {346}2 must be in R3C8 -> no 6 in R5C8

16. 12(3) cage in N6 = {138/147/156/237/246} (cannot be {345} because R5C9 only contains 1,2)
16a. 6 of {156} must be in R5C7 -> no 5 in R5C7

[At this stage I originally thought that I could use
45 rule on C89 3 outies R345C7 = 1 innie R4C8 + 18
R4C8 = {234} -> R345C7 = 20,21,22 = {479/569/579/678/589/679} (cannot be {578} = [587] because 21(3) cage at R3C7 cannot contain both of 5,8)
Initially I mistakenly thought that {678} could only have one permutation but later realised that there are two permutations so this step doesn’t help at this stage. However with hindsight, see comment before step 33, I could have added a contradiction move to make this the key breakthrough.]

[Then I found]
17. 21(3) cage in N9 (step 4a) = {579/678}
17a. Consider combinations for 21(3) cage
21(3) cage = {579} => R89C9 = {59}, locked for C9 => R67C9 = [78] => no 7 in R6C7
or 21(3) cage = {678}, locked for N9 => R67C9 = [69], no 6 in R7C7
17b. -> R67C7 = [49/85/94] (cannot be [76]), no 6,7, R67C9 = [69/78], no 9 in R6C9, no 6 in R7C9

18. 9 in C9 only in R789C9, locked for N9, clean-up: no 4 in R6C7
18a. Naked pair {89} in R46C7, locked for C7

19. 8 in R5 only in 22(3) cage = {589} (only remaining combination), locked for R5 and N5, clean-up: no 3,6 in R7C4, no 1,4 in R7C6, no 3 in R9C2 (step 3)

20. 12(3) cage in N6 (step 16) = {147/237/246}
20a. Killer pair 6,7 in 12(3) cage and R6C9, locked for N6

21. R46C7 = {89} = 17
21a. 45 rule on N6 4 remaining innies R46C89 = 16 = {1357/1456/2356} (cannot be {1267/2347} which clash with 12(3) cage)
21b. 4 of {1456} must be in R4C8 -> no 4 in R4C9 + R6C8
21c. 4 in C9 only in R123C9, locked for N3

22. 14(3) disjoint cage R4C258 = {248/257/347/356}
22a. 5,8 of {248/257/356} must be in R4C2 -> no 2,6 in R4C2

23. R1C6 + R4C79 = 20 (step 7)
23a. Max R4C79 = [95] = 14 -> min R1C6 = 6

24. 13(3) cage at R1C6 = {139/157/238/256}
24a. 6 of {256} must be in R1C6 -> no 6 in R12C7

25. Hidden killer pair 1,2 in 13(3) cage at R1C6 and R123C9 for N3, 13(3) cage contains one of 1,2 in R12C7 -> R123C9 must contain one of 1,2
25a. Killer pair 1,2 in R123C9 and R5C9, locked for C9

26. 14(3) disjoint cage R4C258 = {248/257/347} (cannot be {356} which clashes with R4C9), no 6

27. 45 rule on C5 3 innies R459C5 = 20 = {479/578} (cannot be {569} because R4C5 only contains 4,7), no 6, 7 locked for C5
27a. 9 of {479} must be in R5C5 -> no 9 in R9C5

28. 6 in C5 only in 19(3) cage, locked for N2

29. R18C6 = R4C8 + 10 (step 10)
29a. Max R4C8 = 4 -> max R18C6 = 14, min R1C6 = 7 -> max R8C6 = 6 (R18C6 cannot be [77])

30. 16(3) cage in N8 = {178/259/268/349/358/457} (cannot be {169} because no 1,6,9 in R9C5, cannot be {367} which clashes with R9C8)
30a. 2 of {268} must be in R9C4 -> no 6 in R9C4

[Now to use part of that step I tried originally.]
31. 45 rule on C89 3 outies R345C7 = 1 innie R4C8 + 18
R4C8 = {234} -> R345C7 = 20,21,22 = {479/678/679} (cannot be {569/579/589} which clash with R67C7, cannot be {578} = [587] because 21(3) cage at R3C7 cannot contain both of 5,8), no 5 in R3C7

32. 45 rule on N7 4(3+1) outies R6C123 + R8C4 = 25
32a. Max R6C123 = 22 (cannot be {689/789} which clash with R6C7) -> min R8C4 = 3

[Looking again at R345C7 (step 31) = {479/678/679} I can now see that
{479} can only be [794] => R3C8 = 5
{678} can only be [687] (cannot be [786] => R3C8 = 6 and cannot place 6 in N9)
{679} can only be [796] because 21(3) cage at R3C7 cannot contain both of 6,9 => R3C8 = 5
which taken together eliminate 6 from R3C8.
However these days I prefer to use a forcing chain, rather than a contradiction move, where possible.]
33. R345C7 (step 31) = {479/678/679}
33a. Consider placement for 6 in N9
6 in R89C7 => R345C7 = {479} = [794] => R3C8 = 5
or 6 in R9C8 => R3C8 = {57}
33b. -> R3C8 = {57}

[At last I’m basically back where I thought I was after my incorrect analysis, so the puzzle should now be cracked.]
34. R9C8 = 6 (hidden single in C8)
34a. 21(3) cage in N9 (step 4a) = {678} (only remaining combination) -> R89C9 = {78}, locked for C9 and N9 -> R67C9 = [69], clean-up: no 2 in R6C3, no 2 in R6C4, no 3 in R7C1, no 5 in R7C3, no 5 in R7C4

35. 14(4) cage at R1C9 (step 5) = {2345} (only remaining combination), locked for C9 -> R5C9 = 1, 2 locked for N3
35a. 12(3) cage in N6 (step 20) = {147} (only remaining combination), locked for R5 and N6, clean-up: no 2,5 in R9C2 (step 3)
35b. Naked triple {236} in 11(3) cage, locked for N4, clean-up: no 6,7 in R7C1, no 8 in R7C3

36. 8(3) cage at R6C8 = {125/134}, 1 locked for N9
36a. 1 in N3 only in 13(3) cage at R1C6 (step 24) = {139/157}, no 8

37. R3C7 = 6 (hidden single in C7)
37a. 21(3) cage at R3C7 (step 2) = {678} (only remaining combination) -> R3C8 = 7, R4C7 = 8, R5C78 = [74], R6C7 = 9, R7C7 = 4, clean-up: no 5 in R6C1, no 7 in R6C4

38. 8(3) cage at R6C8 = {125} (only remaining combination), locked for C8 -> R4C8 = 3, R4C9 = 5, R6C8 = 2, clean-up: no 7 in R7C6
38a. Naked pair {15} in R78C8, locked for N9
38b. Naked pair {23} in R89C7, locked for C7, R8C6 = 6 (cage sum)
38c. R12C7 = {15}, R1C6 = 7 (step 36a), clean-up: no 4 in R2C2, no 2 in R7C6

39. Naked pair {47} in R4C25, locked for R4
39a. Naked pair {19} in R4C13, locked for R4 and N4 -> R4C46 = [62], clean-up: no 8 in R7C1
39b. Naked triple {134} in R6C456, locked for R6 and N5 -> R4C5 = 7, R4C2 = 4, R9C2 = 1, R5C1 = 3 (step 3), clean-up: no 7 in R2C2, no 8 in R6C1, no 5 in R7C1, no 7 in R7C3
39c. R67C1 = [72]

40. 15(3) cage in N7 = {159} (only remaining combination), R89C1 = {59}, locked for C1 and N7 -> R4C1 = 1, R4C3 = 9
40a. Naked triple {468} in R123C1, locked for N1, clean-up: no 3,5 in R12C2
40b. Naked pair {29} in R12C2, locked for C2 and N1 -> R5C23 = [62]
40c. R4C3 = 9 -> R3C23 = 6 = [51], R12C3 = [37], R1C4 = 1 (cage sum), R6C23 = [85], R7C3 = 6
40d. R89C3 = {48}, R8C4 = 5 (cage sum), R7C6 = 8, R6C6 = 1, R9C5 = 4, R9C6 = 9

41. 19(3) cage in N2 = {568} (only remaining combination) -> R3C5 = 8

and the rest is naked singles.


Rating Comment. I'll rate my walkthrough for A70 V3 at 1.5. I used a couple of short forcing chains but managed to avoid any contradiction moves.
2011 Alt start by Ed:
Thanks very much Andrew for getting me interested in this old puzzle. Very happy with this alternate start. Feels simpler than both the previous walkthroughs.

Prelims: see Andrew's list

NOTE: this is an optimized walkthrough so some obvious eliminations are not included. However, I try and do clean-up as I go.
1. 17(2)n3 = {89} both locked for c8 & n3

2. "45" on c9: 1 outie r9c8 - 5 = 1 innie r5c9
2a. r9c8 = (67); r5c9 = (12)

3. 22(3)n5 = {589/679} = [5/6..]
3a. must have 9: locked for r5 & n5
3b. no 2 in r7c4

4. 12(3)n6 must have 1/2 for r5c9
4a. but {156} blocked by 22(3)n3 (step 3.)
4b. = {138/147/237/246}(no 5)
4c. only has one of 1/2 -> no 1,2 in r5c78

5. 21(3)n9 must have 6/7 for r9c8
5a. = {579/678}(no 4) = [8/9 in c9..]
5b. must have 7 -> 7 locked for n9
5c. no 8 in r6c9
5d. no 6 in r6c7

6. Killer pair 8,9 in 21(3)n9 & 15(2)r6c9: 8 locked for c9

7. 8 in n6 only in c7: 8 locked for c7
7a. no 5 in r6c7

8. 21(3)r3c7: can only have one of 8/9 in r4c7 = {579/678}(no 4)
8a. must have 8/9 -> r4c7 = (89)
8b. must have 7 -> 7 locked for r3 & n3

9. "45" on c789: 5 innies r1289c7 + r4c8 = 14
9a. min. r1289c7 = 10 -> max. r4c8 = 4

10. Hidden killer pair in c7: 1 & 2 only in r1289c7
10a. r12c7 cannot have both 1&2 or the 13(3)r1c6 won't reach the cage sum
10b. -> r89c7 must have at least one of 1/2 (no eliminations yet)

11. 8(3)r6c8 = {125/134}:
11a. but 5 blocked from r6c8 since {12} in r78c8 would clash with r89c7 (step 10b)

12. r4c9 = 5 (hidden single n6)

13. 21(3)n9 = {678} only: 6&8 locked for n9

Cracked


Last edited by Ed on Sat Nov 07, 2009 8:17 pm, edited 2 times in total.

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Ed
PostPosted: Tue Jul 08, 2008 1:56 am 
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Location: Sydney, Australia
Assassin 71 by Ruud (Oct 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:5376:5376:5890:5890:5124:5124:3590:3590:3590:4873:5376:5376:5890:5890:5124:5124:4368:3590:4873:4873:3092:2581:4886:2583:3352:4368:4368:4891:4873:3092:2581:4886:2583:3352:6946:4368:4891:4891:2086:2086:4886:2089:2089:6946:6946:4141:4891:2351:2864:2865:3378:1331:4660:6946:4141:4141:2351:2864:2865:3378:1331:4660:4660:6463:4141:3649:3649:5699:5699:5445:5445:4660:6463:6463:6463:3649:3649:5699:5699:5445:5445:
Solution: 
+-------+-------+-------+
| 8 9 2 | 7 5 4 | 6 3 1 |
| 7 3 1 | 8 6 2 | 9 5 4 |
| 6 5 4 | 1 9 3 | 8 2 7 |
+-------+-------+-------+
| 4 1 8 | 9 2 7 | 5 6 3 |
| 2 6 5 | 3 8 1 | 7 4 9 |
| 9 7 3 | 6 4 5 | 2 1 8 |
+-------+-------+-------+
| 1 4 6 | 5 7 8 | 3 9 2 |
| 5 2 7 | 4 3 9 | 1 8 6 |
| 3 8 9 | 2 1 6 | 4 7 5 |
+-------+-------+-------+
Quote:
CathyW: This is definitely a tough one. 45 steps (about 3 hours) and counting. :shock: ..Finally! Took about 4 hours in total. At least 1.5 rating. I can't say I particularly want to go through it again
gary w: this was a tough one..sure was..took me 5 hrs plus :!:
Afmob: What a monster. I didn't find a nice way to solve to puzzle so be prepared for some contradiction moves. I think it was more difficult than A69 V1.5 so I guess rating is about 1.75
Andrew: A71 must be one of the hardest ever V1s!... I'll go along with Afmob's rating of 1.75. I think this is the first time that a V1 has been harder than the V2
sudokuEd: I'd rate the puzzle a 1.5 since it took me less than 10 hours to get a nice solution
mhparker, in A74 thread: there's been a tendency for people to overrate puzzles recently. The A71 was a good example of this...Sure, it was very tough. But does it really deserve a 1.75 rating?
sudokuEd in A74 thread: I agree (about overrating puzzles) - especially about A71
2022 forum Revisit here
Walkthrough by CathyW:
No, I haven't solved it yet! But the fact that no-one else has either shows either it's really difficult or you haven't had time yet. I'll have another look at it this evening. :)

Edit: This is definitely a tough one. 45 steps (about 3 hours) and counting. :shock:
I have c7 fixed and a few other placements and thought it would start falling into place but there's still work to do!

Edit 2: Finally! Took about 4 hours in total. At least 1.5 rating. I can't say I particularly want to go through it again so if there's any corrections required, I'll just take your word for it.

Prelims

a) 14(4) @ r1c7 and r8c3: no 9
b) 12(3) r34c3 = {39/48/57}
c) 10(2) @ r3c4 and r3c6: no 5
d) 19(3) @ r3c5: no 1
e) 13(2) @ r3c7 and r6c6 = {49/58/67}
f) 8(2) @ r5c3 and r5c6 = {17/26/35}
g) 27(4) @ r4c8 = {3789/4689/5679} -> 9 not elsewhere in N6 -> r3c7 <> 4
h) 9(2) r67c3: no 9
i) 11(2) r67c4 and r67c5: no 1
j) 5(2) r67c7 = {14/23}

1. Innies r12: r2c18 = 12 = {39/48/57}

2. Innies r89: r8c29 = 8 = {17/26/35}

3. Outies c12: r29c3 = 10 = {19/28/37/46} -> r158c3 = 14

4. Outies c89: r18c7 = 7 = {16/25} ({34} blocked by 5(2))
a) KP with 5(2) -> 1,2 not elsewhere in c7 -> r5c6 <> 6,7
b) r259c7 = 20 = {389/479/578} ({569} blocked by 13(2) -> r259c7 <> 6 -> r5c6 <> 2

5. Innies r6789: r6c29 = 15 = {69/78}

6. Outies c1234: r29c5 = 7 = {16/25/34}

7. Outies c6789: r18c5 = 8 = {17/26/35}

8. O-I N1: r4c23 – r1c3 = 7 -> r4c2 <> 7

9. O-I N3: r2c7 – r4c79 = 1
a) r4c79 max 8 -> r4c7 = (4567), r4c9 = (123) -> r3c7 <> 5
b) r2c7 min 6 -> r2c7 = (789)

10. O-I N7: r6c13 – r8c3 = 5 -> r6c1 <> 5

11. O-I N9: r9c7 – r6c78 = 1
a) r6c78 is max 8 -> r6c8 <> 8
b) 8 locked to 27(4) in N6 = {3789/4689} (eliminate 5)

12. O-I r1234: r4c18 – r5c5 = 2 -> r4c1 <> 9

13. Innies N69: r4c9 + r459c6 = 19
a) 5(2) r67c7 must have one of 3,4
b) split 19(4) must have one of 3,4 within r459c6 -> split 19(4) <> {1567}

14. 27(4) N6 must have one of 3,4
a) From step 9, O-I N3 <> [843]

15. Split 19(4) = 1[639/738/459], 2[539/638/458], 3{57}4 -> r9c7 <> 5,7

16. Split 20(3) r259c7 = [839/938/974/758]

17. 27(4) N6 must have one of 3,6
a) Split 19(4) can’t have both 3,6 within N6 -> 19(4) <> [1639/2638]
b) r4c7 <> 6, r3c7 <> 7

18. Killer combo N6: 27(4) must have one of 3,4; split 19(4) must have one of 3,4 within N6
a) r6c78 <> 3,4 -> r7c7 <> 1,2
b) 21(4) N9 and 18(4) @ r6c8 can’t have both 3,4
c) 16(4) @r6c1 can’t have both 3,4 within r7c12

19. Innies N36: r256c7 + r6c8 = 19
a) Can’t have both 3,6 in N6 because of 27(4) -> {1369/2368} blocked.
b) Options for split 19(4) = {1279/1378/1567/2359/2377}

20. Innies N14: r6c1 + r156c3 = 19 -> can’t have both 1,2 within r6c13

21. Innies N47: r4c2 + r458c3 = 21

22. Innies r1234: r34c5 + r4c18 = 21

23. Conflicting combo: Split 19(4) r256c7, r6c8 <> [7516] since would block both options for split 7(2) r18c7
a) r5c7 <> 5 -> r5c6 <> 3
b) r6c8 <> 6

24. 27(4) N6 = {4689}(only available combo since r5c7 = (37)) -> r4c7 <> 4 -> r3c7 <> 9
a) Clean up: r6c2 <> 8

25. 4 locked r79c7 n/e N9

26. Split 20(3) r259c7 = [839/938/974] -> r2c7 <> 7
a) 7 locked r45c7 -> r6c8 <> 7
b) Split 19(4) r4c9 + r569c7 must have 7 = [1738/3574] -> r9c7 <> 9

27. HS r2c7 = 9
a) split 12(2): r2c18 <> 3
b) split 10(2) r29c3: r9c3 <> 1
c) r4c79 = [53/71] -> 2 locked r6c78 n/e r6 -> clean up: r7c45 <> 9, r7c3 <> 7

28: r4c79 = [53/71], r5c67 = [17/53]
a) r4c6 <> 1 -> r3c6 <> 9
b) r4c5 <> 5
c) r4c4 <> 1 -> r3c4 <> 9

29. 9 locked r789c6 -> r46c6 <> 9 -> r7c6 <> 4; r3c6 <> 1

30. 1 locked r5c46 -> r5c123 <> 1 -> r5c4 <> 7

31. Conflicting combo N6:
a) r4c9 = 1 -> r4c7 = 7, r5c7 = 3 -> Conflict for 5(2)
b) Therefore: r4c9 = 3 -> r4c7 = 5, r5c7 = 7, r5c6 = 1, r3c7 = 8, r9c7 = 4, r7c7 = 3, r6c7 = 2, r6c8 = 1
c) Clean up: r2c1 <> 4, r3c46 <> 7, r4c46 <> 2, r6c45 <> 8, r6c3 <> 6, r7c3 <> 8, r3c3 <> 7,9, r4c3 <> 4,

32. HS r1c8 = 3

33. 14(4) N3 = {1346} only possible combo, -> r2c8 = (57), r3c89 = (257)
a) 2 locked r3c89 n/e r3
b) Clean up: r4c46 <> 8
c) 4 locked r45c8 -> r5c9 <> 4
d) Split 12(2) r2c18 = {57} n/e r2

34. 10(2) r34c6 = [37]/{46} -> KP 13(2) r67c6 <> {67}

35. 1 locked r7c123 n/e N7

36. Split 8(2) r8c29 = {26} only possible combo -> n/e r8
a) r8c7 = 1 -> r1c7 = 6
b) 25(4) N7 <> {2689} -> r9c123 <> 2
c) 2 locked r9c456 -> r7c45 <> 2 -> r6c45 <> 9

37. 1 locked r4c12 -> r3c1 <> 1

38. Grouped Turbot (5): r2c1 = r2c8 – r89c8 = r9c9
-> r9c1 <> 5

39. 9 locked to 19(3) r345c5 = {289/379/469}

40. 5 locked r1c456 -> r1c123 <> 5

41. Split 8(2) r18c5 = [17/53]

42. 14(4) @ r8c3 = {1238/1247/1346/2345} Must have one of 3,7
a) Since r8c5 = (37), r8c6 <> 3,7

43. Grouped Turbot (9): r3c12 = r3c5 – r45c5 = r4c4
-> r4c2 <> 9

44. Split 10(2) r29c3 = [19/28/37/46]

45. Split 7(2) r29c5 = [16/25/43/61]

46. Pointing cells: 6 locked r2c2, r3c12 -> r4c2 <> 6

47. 14(4) @ r8c3 must have 1 within r9c45 -> {2345} blocked -> 14(4) <> 5
a) clean up: r2c5 <> 2

48. 2 locked r45c5 n/e N5 – r5c3 <> 6
a) 19(3) r345c5 = 9{28}
b) 11(2) r67c5 <> [38]
c) 13(2) r67c6 <> [85]

49. HS r4c4 = 9 -> r3c4 = 1, r1c5 = 5, r8c5 = 3

50. 11(2) r67c5 = {47} -> r2c5 = 6, r9c5 = 1

51. r12c6 of 20(4) = {24} n/e N2 -> r3c6 = 3, r4c6 = 7, …

All singles from here.
Walkthrough by Afmob:
What a monster. I didn't find a nice way to solve to puzzle so be prepared for some contradiction moves. I think it was more difficult than A69 V1.5 so I guess rating is about 1.75.

Assassin 71 Walkthrough:

1. R6789
a) Innies R89 = 8(2) -> no 4,8,9
b) Innies R6789 = 15(2) = {69/78}

2. R12
a) Innies R12 = 12(2) -> no 1,2,6

3. C1234
a) Outies C12 = 10(2) -> no 5
b) Outies C1234 = 7(2) -> no 7,8,9

4. C6789
a) Outies C89 = 7(2) -> no 7,8,9
b) Innies C789 = 20(3) -> no 1,2
c) Innies C789 = 20(3) -> R29C7 <> 3 because R5C7 <= 7
d) 8(2) @ N6: R5C6 <> 6,7
e) Outies C6789 = 8(2) -> no 4,8,9
f) 27(4) = 9{378/468/567} -> 9 locked for N6
g) 13(2) @ N3: R3C7 <> 4

5. N3
a) Innies+Outies: -14 = R4C9 - R23C7 -> R23C7 = 15/16/17 -> R23C7 <> 4,5
-> R4C9 = (123)
b) 13(2): R4C7 <> 8

6. N69
a) Innies+Outies N9: -1 = R6C78 - R9C7; R9C7 = (456789)
-> R6C78 = 3/4/5/6/7/8 -> R6C8 <> 8
b) 8,9 locked in 27(4) = 89{37/46} -> no 5

7. C789
a) Outies C89 = 7(2): {34} blocked by Killer pair (34) of 5(2)
b) Innies C789 = 20(3) <> 6 since {569} blocked by Killer pair (56) of Outies of C89
c) 8(2) @ N6 = [17/35/53]

8. C5
a) Innies = 15(4) = 7(2) + 8(2) (see 3b,4e) = 1{257/347/356}
because 7(2), 8(2) <> 8,9 and 19(3), 11(2) <> 1
-> Outies C6789 = 8(2) = {17/35}
b) 19(3) <> 3 because {379} blocked by Killer pair (37) of Innies

9. C123
a) Outies 10(2) <> {37} because if 10(2) = {37} -> 12(2) = {48}
-> no combination for 9(2)

10. C7+N6
a) 13(2): R4C7 <> 6 because if 13(2) = {67} -> Innies 20(3) = {389}
-> R5C7 = 3 -> R4C7 <> 6 (Killer pair (36) of 27(4))
b) 13(2): R3C7 <> 7
c) Consider position of 6 in N6: Either 27(4) = {4689} or R6C8 = 6
-> R6C8 <> 4
d) R9C7 <> 5 -> 20(3) = [875] -> 13(2) = [94] -> R45C7 = [47]
-> blocks both combos of 27(4)

11. C7+N69 !
a) Innies N69 = 19(3+1) = R9C7 + R45C7 + R4C9; R9C7 <> 9 because:
If R9C7 = 9 -> R45C7 + R4C9 = 10(3) = {127/235/145}
- i. <> {127} because R45C7 has no 1,2
- ii. <> {235} -> R4C7 = 5, R5C7 = 3 -> Innies of C7 would be {389}
- but is impossible because of 13(2) = [85]
- iii. <> {145} R5C7 = 5 -> 13(2) = [94] -> not possible because R9C7 = 9
-> R9C7 <> 9
b) 9 locked in R23C7 for N3
c) Innies R12: R2C1 <> 3
d) Innies+Outies N9: -1 = R6C78 - R9C7; R9C7 = (478)
-> R6C78 = 3/6/7 -> R6C8 <> 7, R6C7 <> 3 (3 + ? = 3/6/7)
e) 5(2): R7C7 <> 2
f) Innies N69 = 19(3+1): R5C7 <> 5 because
If R5C7 = 5 -> Innies C7 = 20(3) = {578},
Innies N69 = 14(2+1) = R4C79 + R9C7 = {24+8/34+7}:
- i. 34+7 blocked by Killer pair (34) of 27(4)
- ii. 24+7 -> R4C79 = [42] -> R6C1 = 1 -> 5(2) = [14] -> 2 4's in C7
-> R5C7 <> 5
g) 8(2) = [17/53]

12. C7+N6
a) 9 locked in Innies = 20(3) = 9[38/74] -> R2C7 = 9, R9C7 <> 7
b) 13(2) = [67/85], XOuties C12: R9C3 <> 1
c) 7 locked in R45C7 for N6
d) 27(4) = {4689} locked
e) 4 locked in R79C7 for N9
f) Innies R12 = 12(2) = {48/57}

13. N36
a) Innies+Outies N3: -14 = R4C9 - R23C7, R2C7 = 9, R3C7 = (68)
-> R4C9 = (13)
b) 2 locked in R6C78 for R6
c) 11(2) @ C5: R7C5 <> 9, 11(2) @ C4: R7C4 <> 9, 9(2) @ C3: R7C3 <> 7

14. C6
a) 9 locked in R789C6
b) 10(2) <> 1, 13(2): R7C6 <> 4

15. N369 !
a) Innies N36 = 10(3): R6C8 <> 3 because R5C7 would be 7
b) ! Innies+Outies N9: -1 = R6C78 - R9C7; R9C7 = (48)
-> R6C78 = 3/7 = {12/25}
But if R6C7 = 1 then R6C78 must be [12] -> R7C7 = 4 = R9C7
-> R6C7 <> 1
c) R6C7 = 2, R7C7 = 3, R5C7 = 7, R5C6 = 1, R4C7 = 5, R3C7 = 8, R9C7 = 4
d) R6C8 = 1, R4C9 = 3
e) Hidden Single: R1C8 = 3 @ C8
f) 14(4) = 3{146} locked in N3 because R1C7 = (16)
g) 4 locked in R12C9 for C9

16. R3
a) 2 locked in R3C89
b) 12(2) = [39/48/57], 10(2) @ C4 = [19/37/46/64], 10(2) @ C6 = [37/46/64]

17. N9
a) 18(4) = 12{69/78} -> 2 locked

18. R7
a) 1 locked in R7C123 for N7
b) 9(2): R6C3 <> 6, R6C4 <> 8; 11(2) @ C4: R6C4 <> 8; 11(2) @ C5: R6C5 <> 8

19. C6789
a) 13(2) <> 6,7 which is blocked by Killer pair (67) of 10(2)
b) Outies C6789 = 8(2) = [17/53/71]

20. R12
a) Innies 12(2) = {57} locked for R2
b) 20(4): R1C5 <> 7 since R1C6 <> 1,3
c) 20(4) = 9{128/137/146/245} -> has 1 xor 5 -> R1C6 <> 5 because R5C1 = (15)

21. C1234
a) Outies C12 = 10(2) = [19/28/46/82]
b) Outies C1234 = 7(2) = [16/25/43/61]

22. R89
a) Innies = 8(2) = {26} locked for R8
b) R8C7 = 1, R1C7 = 6
c) 25(2) <> 2 since {2689} is blocked by R8C2 = (26)
d) Outies C12 = 10(2) = [19/28/46]
e) 2 locked in R9C46 for N8
f) 11(2) @ C4: R6C4 <> 9, 11(2) @ C5: R6C5 <> 9

23. C5
a) 9 locked in 19(3) -> no 5,7

24. R12
a) 5,7 locked in R1C456
b) 7 locked in 19(4) -> R4C2 <> 7
c) 21(4) must have 3 or 6 -> only possible @ R2C2 -> R2C2 = (36)
d) 8 locked in R1C123 for R1
e) 21(4) = 9{138/246} -> 9 locked for R1 and N1
f) Hidden Single: R4C4 = 9 @ C4
g) R3C4 = 1, R1C5 = 5

25. C12345
a) Hidden Single: R9C5 = 1 @ C5, R9C3 = 9 @ C3, R3C5 = 9 @ N2
b) Outies C1234 = 7(2) = [61] -> R2C5 = 6
c) R2C2 = 3
d) Naked quad (2478) locked in N2
e) R3C6 = 3 -> R4C6 = 7, R4C3 = 8, R3C3 = 4
f) Hidden Single: R1C4 = 7 @ N2
g) 19(4) = {1567} -> R4C2 = 1
h) 23(4) = {2678} -> R1C3 = 2, R2C4 = 8

26. Rest is clean-up and singles.

Don't want to think about taking on the V2 right now since V1 was so relentless.
Walkthrough by Andrew:
Ruud wrote:
The border came first. In my original design, the interior had bigger cages, but to make this Assassin solvable, I needed to compensate the border complexity.
Solvable. Yes! But only after a very, very long time!!
CathyW wrote:
Finally! Took about 4 hours in total.
gary w wrote:
Cathy said this was a tough one..sure was..took me 5 hrs plus :!:
Congratulations to both of you and to Afmob for solving it so quickly!

I don't know how long I took, over several days, but it must have been at least twice that long. I must have sat staring at the position after step 35 for several hours wondering if I was ever going to make any more progress. ](*,) Then I eventually found a contradiction move that provided the first breakthrough. Still a lot of hard work after that. I'll stick with my earlier rating of 1.75.

I haven't yet gone through the posted walkthroughs, I need a break from this puzzle before I do that, but I'll assume that their solving paths will be different so here is my walkthrough

Walkthrough for A71

1. R34C3 = {39/48/57}, no 1,2,6

2. R34C4 = {19/28/37/46}, no 5

3. R34C6 = {19/28/37/46}, no 5

4. R34C7 = {49/58/67}, no 1,2,3

5. R5C34 = {17/26/35}, no 4,8,9

6. R5C67 = {17/26/35}, no 4,8,9

7. R67C3 = {18/27/36/45}, no 9

8. R67C4 = {29/38/47/56}, no 1

9. R67C5 = {29/38/47/56}, no 1

10. R67C6 = {49/58/67}, no 1,2,3

11. R67C7 = {14/23}

12. R345C5 = {289/379/469/478/568}, no 1

13. 14(4) cage in N3 = {1238/1247/1256/1346/2345}, no 9

14. 27(4) cage in N6 = {3789/4689/5679}, 9 locked for N6, clean-up: no 4 in R3C7

15. 14(4) cage at R8C3 = {1238/1247/1256/1346/2345}, no 9

16. 45 rule on R12 2 innies R2C18 = 12 = {39/48/57}, no 1,2,6

17. 45 rule on R6789 2 innies R6C29 = 15 = {69/78}

18. 45 rule on R89 2 innies R8C29 = 8 = {17/26/35}, no 4,8,9

19. 45 rule on C12 2 outies R29C3 = 10 = {19/28/37/46}, no 5

20. 45 rule on C1234 2 outies R29C5 = 7 = {16/25/34}, no 7,8,9

21. 45 rule on C89 2 outies R18C7 = 7 = {16/25} (cannot be {34} which clashes with R67C7)

22. Killer pair 1,2 in R18C7 and R67C7, locked for C7, clean-up: no 6,7 in R5C6

23. 45 rule on C789 3 innies R259C7 = 20 = {389/479/578} (cannot be {569} which clashes with R34C7), no 6, clean-up: no 2 in R5C6
23a. 3 of {389} must be in R5C7 -> no 3 in R29C7

24. 45 rule on C6789 2 outies R18C5 = 8 = {17/26/35}, no 4,8,9

25. 45 rule on N3 2 innies R23C7 – 14 = 1 outie R4C9 -> min R23C7 = 15, R23C7 = {6789}, max R23C7 = 17 -> max R4C9 = 3, clean-up: no 8 in R4C7

26. 45 rule on N9 2 innies R79C7 – 6 = 1 outie R6C8, max R79C7 = 13 -> max R6C8 = 7

27. 45 rule on N7 2 outies R6C13 – 5 = 1 innie R8C3
27a. IOU no 5 in R6C1

28. 8,9 in C5 locked in R34567C5 = 30 = 89{247/256/346}
28a. 3 of {34689} must be in R67C5 -> no 3 in R345C5
28b. R18C5 (step 24) = {17/35} (cannot be {26} which clashes with R34567C5), no 2,6

29. 45 rule on C123 3 innies R158C3 = 14 = {149/158/167/239/248/257/356} (cannot be {347} which clashes with R34C5)

30. 45 rule on C6 5 innies R12589C6 = 22 = {12379/12469/12568/13459/13567/23458/23467} (cannot be {12478/13468} which clash with R67C6)

31. 45 rule on C4 5 innies R12589C4 = 24 = {12489/12579/13479/13569/13578/14568/23478/23568/24567} (cannot be {12678/23469} which clash with R34C4)

32. 8,9 in N6 locked in 27(4) cage (step 14) = {3789/4689}, no 5

33. 45 rule on N1 2 outies R4C23 – 7 = 1 innie R1C3
33a. IOU no 7 in R4C2

34. 45 rule on R5 5 innies R5C12589 = 29 = {14789/24689/34589} [1/2/3, 5/6/7]

35. 45 rule on C3 5 innies R12589C3 = 24 = {12489/12579/12678/13569/14568/23469/24567} (cannot be {13479/13578/23478/23568} which clash with R34C4)

36. No 4 in R4C7, here’s how
36a. 27(4) cage in N6 (step 32) = {3789/4689}
36b. If 27(4) = {3789} and R4C9 = 1 => R23C7 = 15 (step 25) = {78}/[96] -> no 9 in R3C7 -> no 4 in R4C7
36c. If 27(4) = {3789} and R4C9 = 2 => R23C7 = {79} => R4C7 = {46}
36ca. Then if R4C7 = 4, R5C7 = 5, R6C8 = 6, R67C7 = [14] clashes with R4C7
36cb. Then if R4C7 = 6 -> no 4 in R4C7
36d. If 27(4) = {4689} -> no 4 in R4C7
36e. -> no 4 in R4C7, clean-up: no 9 in R3C7

37. 9 in C7 locked in R29C7
37a. R259C7 (step 23) = {389/479}, no 5, clean-up: no 3 in R6C5
37b. 7 of {479} must be in R5C7 -> no 7 in R29C7

38. 27(4) cage in N6 (step 32) = {4689} (only remaining combination, cannot be {3789} which clashes with R5C7), locked for N6, clean-up: no 7 in R3C7, no 8 in R6C2 (step 17), no 1 in R7C7
38a. 7 in C7 locked in R45C7, locked for N6
38b. R23C7 cannot total 16 -> no 2 in R4C9 (step 25)
38c. 2 in N6 locked in R6C78, locked for R6, clean-up: no 7 in R7C3, no 9 in R7C45
38d. 4 in C7 locked in R79C7, locked for N9

39. R23C7 – 14 = R4C9 (step 25)
39a. R4C9 = 1 => R23C7 = [96]
39b. R4C9 = 3 => R23C7 = [98]
39c. -> R2C7 = 9, clean-up: no 3 in R2C18 (step 16), no 1 in R9C3 (step 19)

40. 20(4) cage at R1C5 = {1289/1379/1469/2369/2459}
40a. 5 of {2459} must be in R1C5 -> no 5 in R12C6

41. 17(4) cage at R2C8 = {1358/1367/1457/2357} (cannot be {1268} which clashes with R3C7, cannot be {2348} which clashes with 14(4) cage, cannot be {2456} because R4C9 only contains 1,3)
41a. 14(4) cage in N3 (step 13) = {1238/1247/1346/2345} (cannot be {1256} which clashes with 17(4) cage)

42. R79C7 – 6 = R6C8 (step 26)
42a. R79C7 cannot total 8 with 4 locked in R79C7 -> no 2 in R6C8

43. R6C7 = 2 (hidden single in R7), R7C7 = 3, R5C7 = 7, R4C7 = 5, R3C7 = 8, R9C7 = 4 (hidden single in C7), R6C8 = 1 (step 26), R4C9 = 3, R5C6 = 1, clean-up: no 4 in R2C1 (step 16), no 6 in R2C3 (step 19), no 3 in R2C5 (step 20), no 7,9 in R3C3, no 7,9 in R3C46, no 4 in R4C3, no 2 in R4C46, no 9 in R4C6, no 6 in R6C3, no 8 in R6C45, no 8 in R7C3, no 5,7 in R8C2 (step 18)
43a. 1 in N4 locked in R4C12
43b. CPE no 1 in R3C1

44. R1C8 = 3 (hidden single in N3), clean-up: no 5 in R8C5 (step 24)
44a. 14(4) cage in N3 (step 41a) = {1346} (only remaining combination, cannot be {2345} because R1C7 only contains 1,6), locked for N3, clean-up: no 8 in R2C1 (step 16)
44b. 4 in N3 locked in R12C9, locked for C9
44c. 2 in N3 locked in R3C89, locked for R3, clean-up: no 8 in R4C46

45. Naked pair {57} in R2C18, locked for R2, clean-up: no 3 in R9C3 (step 19), no 2 in R9C5 (step 20)

46. R67C6 (step 10) = {49/58} (cannot be {67} which clashes with R34C6)

47. 22(4) cage at R8C5 = {2479/3469/3478/4567} (cannot be {1489} which clashes with R67C6), no 1, clean-up: no 7 in R1C5 (step 24)

48. 22(4) cage at R8C5 cannot be {3478}, here’s how
48a. If R8C5 = 3 => R89C6 = {78} => R34C6 = {46} and R67C6 = {49} clash
48b. If R8C5 = 7 => R89C6 = {38} => R34C6 = {46} and R67C6 = {49} clash
48c. -> 22(4) cage at R8C5 = {2479/3469/4567}, no 8
48d. R89C6 = {29/69/56}, no 3,7

49. 1 in N8 locked in 14(4) cage at R8C3 within N8 -> no 1 in R8C3

50. 18(4) cage at R6C8 = {1269/1278}, no 5, 2 locked for N9, clean-up: no 3 in R8C2 (step 18)
50a. 21(4) cage in N9 = {1569/1578}

51. 1 in R7 locked in R7C123, locked for N7, clean-up: no 7 in R8C9 (step 18)

52. Naked pair {26} in R8C29, locked for R8 -> R8C7 = 1, R1C7 = 6, clean-up: no 5,9 in R9C6 (step 48d)

53. 25(4) cage in N7 = {3589/3679/4579/4678} (cannot be {2689} which clashes with R8C2), no 2

54. 2 in R9 locked in R9C46, locked for N8, clean-up: no 9 in R6C45

55. 9 in N8 locked in R78C6, locked for C6, clean-up: no 4 in R7C6

56. 9 in C5 locked in R345C5
56a. R345C5 (step 12) = {289/469}, no 5,7

57. 7 in N2 locked in R1C46, locked for R1

58. 7 in N1 locked in 19(4) cage at R2C1 = {1279/1378/1567/2467/3457}
58a. 2 of {1279} must be in R4C2 and the 1 must be in R3C2 -> no 9 in R34C2

59. R158C3 (step 29) = {158/167/239/248/257/356} (cannot be {149} because no 1,4,9 in R5C3
59a. 3 of {239} must be in R8C3 and 6 of {356} must be in R5C3 -> no 3 in R5C3, clean-up: no 5 in R5C4
59b. 8 of {158}, 7 of {257} and 3 of {356} must be in R8C3 -> no 5 in R8C3

60. 14(4) cage at R8C3 (step 15) = {1238/1247/1346} (cannot be {1256} because 1,2,6 only in R9C45), no 5, clean-up: no 2 in R2C5 (step 20)
60a. 2 of {1238/1247} must be in R9C4 -> no 7,8 in R9C4
60b. R8C34 = {34/38/47} -> no 3 in R9C45, clean-up: no 4 in R2C5 (step 20)

61. Killer pair 3,7 in R8C34 and R8C5, locked for R8

62. Naked triple {126} in R9C456, locked for R9 and N8, clean-up: no 5 in R6C45, no 9 in R8C8 + R9C89 (step 50a)

63. Naked triple {578} locked in R8C8 + R9C89, locked for N9
63a. 7 in N9 locked in R9C89, locked for R9, clean-up: no 4 in R8C1 (step 53)
63b. 9 in N9 locked in R7C89, locked for R7, clean-up: no 4 in R6C6
63c. 25(4) cage in N7 (step 53) = {3589}, locked for N7, clean-up: no 4 in R6C3

64. Naked pair {16} in R29C5, locked for C5 -> R1C5 = 5, clean-up: no 4 in R345C5 (step 56a)
64a. R3C5 = 9 (naked single)
64b. Naked pair {28} in R45C5, locked for C5 and N5 -> R67C6 = [58], clean-up: no 3 in R6C5, no 4 in R7C3
64c. R7C5 = 3 (hidden single in C5)

65. 20(4) cage at R1C5 (step 40) = {2459} (only remaining combination)
65a. Naked pair {24} in R12C6, locked for C6 and N2 -> R9C6 = 6, R8C6 = 9, R9C45 = [21], R2C5 = 6, clean-up: no 4,6 in R4C4, no 6 in R5C3
65b. R34C6 = [37] -> R34C4 = [19], R34C3 = [48], R45C5 = [28], R67C5 = [47], R12C4 = [78], R1C3 = 2, R12C6 = [42], R12C9 = [14], R8C34 = [74], R56C3 = [53], R2C23 = [31], R7C3 = 6, R567C4 = [365], R9C3 = 9, R8C2 = 2, R8C9 = 6, R56C9 = [98], R7C89 = [92]

66. R3C8 = 2 (hidden single in R3), R5C1 = 2 (hidden single in R5), R9C1 = 3 (hidden single in R9)

67. R2C1 + R3C12 = {567} -> R4C2 = 1 (cage sum), R7C12 = [14], R6C1 = 9 (cage sum)

and the rest is naked singles
Walkthrough by sudokuEd:
About A71 Andrew wrote:
Whether it's actually the hardest ever I don't know. There were several very difficult ones about a year ago; A24, A26, A31, A33 and A34.
Here's a few very tentative scores comparing these puzzles and the hardest 4 v1s and a couple of others of interest. These are from yesterdays version of the ever evolving scoring method that Richard and me are developing using Sudoku Solver . I'm getting more confident - Richard is still ....not. :wink:

If anyone tries these earlier ones, would love to hear how you rate these scores.

Code:
A18 1.59 (may be unreliable)

A24 1.32

A26 0.94

A31 1.00

A33 1.10

A34 0.96

A50 1.64

A60 1.65

A60 RP Lite 2.56

A71 1.90

A71 V2 1.61
Been through this walk-through 4 times to get it correct and simple enough :? . Still found some little mistakes. This puzzle drove me nuts - hate all the clean-up and fiddly stuff required. Can't bear to look at any other WTs for now. I'd rate the puzzle a 1.5 since it took me less than 10 hours to get a nice solution.

Step 23 is a really great piece of light relief - thanks to Andrew for finding it for me.

Assassin 71
Prelims
i. 14(4)n3: no 9
ii. 12(2)n1: no 126
iii. 10(2)r3c4 & r3c6: no 5
iv. 19(3)n2: no 1
v. 13(2)n3 & n5: no 123
vi. 27(4)n6: no 12
vii. 8(2)r5c3 & r5c6: no 489
viii. 9(2)n4: no 9
ix. 11(2)r6c4 & r6c5: no 1
x. 5(2)n6 = {14/23}
xi. 14(4)n7: no 9

1. "45" r89: r8c29 = h8(2): no 489

2. "45" r12: r2c18 = h12(2): no 126

3. "45" c12: r29c3 = h10(2): no 5

4. "45" c89: r18c7 = h7(2): no 789
4a. = {16/25}: {34} clashes with [3/4] required in 5(2)n6
4b. Killer Pair [12] in h7(2) & 5(2): both locked for c7
4c. no 67 in r5c6

5. "45" c789: r259c7 = h20(3) = {389/479/569/578}
5a. 3 in {389} must be in r5c7 -> no 3 in r29c7
5b. {569} clashes with h7(2)r18c7
5c. = {389/479/578}(no 6)
5d. no 2 r5c6

6. "45" c1234: r29c5 = h7(2): no 789

7. "45" c5: r18c5 = h8(2): no 489
7a. {26} clashes with 19(3) + 11(2) in c5 (thanks to Sudoku Solver for combining cages for me)
7b. = {17/35} = [3/7..]

8. 19(3)n2 = {289/469/478/568}(no 3) ({379} clashes with h8(2)r18c5

9. "45" r6789: r6c29 = h15(2) = {69/78}

10. 9 must be in 27(4)n6: 9 locked for n6
10a. no 4 r3c7

11. "45"n7: r6c13 - r8c3 = 5
11a. Since 1 innie r8c3 is in the same c as 1 outie r6c3 -> the 2nd outie r6c1 must be unequal to the difference betweeen the innie and outies (IOU) -> no 5 r6c1

12. "45" n1: 2 outies - r1c3 = 7
12a. IOU -> no 7 in r4c2

13. "45" n3: 2 outies + 1 = r2c7
13a. min r4c79 = [41] = 5 -> min r2c7 = 6
13b. r2c7 = {789}
13c. max. r2c7 = 9 -> max r4c79 = 8 -> max r4c9 = 3 (can't have {44} = 8)
13d. max r4c79 = 8 -> max r4c7 = 7 -> no 8 r4c7 -> no 5 r3c7

14. "45" n3: r4c9 + 14 = r23c7 = 15/16/17 = {78}/[96]/{79}/{89}

15. h20(3)r259c7 = {389/479/578} = [839/938/974/758/875/857]

16. "45" n9: r6c78 + 1 = r9c7
16a. max r9c7 = 9 -> max r6c78 = 8
16b. no 8 r6c8

17. 8 in n6 only in 27(4) = 89{37/46}(no 5)
17a. if = 89{37} -> r5c7 = 5 -> r5c6 = 3 -> 3 in 27(4) only in r4c8
17b. no 3 in r5c89
17c. no 7 in r4c8 (must be 3 in {37..})

18. "45" c123: r158c3 = h14(3)
18a. {347} clashes with 12(2) n1

19. h10(2)r29c3 = {19/28/46}(no 37) ({37} clashes with 12(2) + 9(2) in c3)

20. "45" n69: 4 innies = 19: r4c9 + r459c7 =
20a. [1][738/459/657] ([1][639/675] clash with 27(4) = [3/6;6/7..]
20b. [2][539/458] ([2][638/674] clash with 27(4))
20c. [3][574/754] ([457/475] clash with 27(4)
20d. r9c7: no 5

21. 2 of the cells from step 20, r59c7, must overlap with 2 of the cells from h20(3)r259c7
21a. h20(3)r259c7=[839/938/974/758/857] (step 15) ([875] blocked by no 5 in r9c7
21b. -> from step 20, r459c7: [459/754] do not have a double cell overlap
21c. -> r459c7 = [738/657/539/458/574]

22. because of 13(2)n3: combo's in r459c7 that have 6 in r4c7 cannot have 7 etc
22a. -> r459c7 [657] is blocked
22b. ->r3459c7 = [6738/8539/9458/8574] = 8{..}
22c. no 7 r39c7
22d. no 6 r4c7
22e. 8 locked in r39c7 for c7

23. OK. Here's a fun one from Andrew. No 9 in r9c7. Here's how.
23a R1C5 and R2C7 both odd so R12C6 even (cage sum is even)
23b. R34C6 even, R5C6 odd, R67C6 odd -> R89C6 odd (since "45" is odd)
23c. since R8C5 odd and r89c6 odd (and an even cage sum) -> R9C7 must be even
23d. -> no 9 r9c7
23e. of course, combo placing in h20(3)r259c7 would have done the same thing. But much more fun!

24. 9 in c7 only in n3: 9 locked for n3

25. h20(3)r259c7 = {389/479/578} = [938/974/758]
25a. from 20a. innies n69 =
25b. [1][738]
25c. [2][458]
25d. [3][574]

26. innies n6 = 18(5)
26a. (from step 25) r4c9 + r45c7 + r6c78 =
26b. [1][73] = 11 -> r6c78 = 7 = [25]
26c. [2][45]= 11 -> r6c78 = 7 = [16]
26d. [3][57]= 15 -> r6c78 = 3 = {12}
26e. r6c7 = {12} -> r7c7 = {34}
26f. r6c8 = {1256}

27. from step 26b,c,d: the 5(2) cage restricts r45c7 <>3 with 2 in r6c7; <> 4 with 1 in r6c7
27a. -> r456c7 = [732/451] blocked
27b. -> r456c7 = [57]{12} only
27c. no 1 r5c34
27d. no 8 r6c2 (h15(2)r6c29)

28. r4c9 = 3 (step 26d)
28a. no 7 r3c46
28b. no 7, 9 r3c3

29. r3c7 = 8
29a. no 2 r4c46
29b. no 4 r4c3
29c. no 49 r2c1 (h12(2)r2c18)

30. r5c6 = 1
30a. no 9 r34c6
30b. no 9 r3c4

31. r2c7 = 9
31a. no 3 r2c1 (h12(2)r2c18)
31b. no 1 r9c3 (h10(2)r29c3)

32. r9c7 = 4 (h20(3)r259c7)
32a. no 3 r2c5 (h7(2)r29c5)
32b. no 6 r2c3 (h10(2)r29c3)

33. r67c7 = [23]
33a. no 8 r6c45
33b. no 6 r6c3
33c. no 9 r7c45
33d. no 7 r7c3
33e. no 5 r8c2 (h8(2)r8c29)

34. r6c8 = 1 (hsingle n6)
34a. no 8 r7c3

35. r1c8 = 3 (Hsingle n3)
35a. no 5 r8c5 (h8(2)r18c5)

36. 1 in r7 only in n7: 1 locked for n7

37. 17(4)n3 = 3{257} (3{167} blocked by r1c7)
37a. 2,5,7 all locked for n3 and 2 locked for r3
37b. no 8 r4c46
37c. no 8 r2c1 (h12(2)r2c18)

38. 4 in n3 only in c9: 4 locked for c9

39. h12(2) r2c18 = {57}: both locked for r2
39a. no 2 r9c5 (h7(2)r29c5)

40.deleted

41. 10(2)r3c6 = [37]/{46}(no 2) = [6/7..]

42. 13(2)n5 = {49/58} ({67} blocked by 10(2) step 41)
42a. = [8/9..]

43. 22(4)n8 must have 4 = 4{279/369/378/567}(no 1) ([1]{89} blocked by 13(2) step 42a)
43a. no 7 r1c5 (h8(2)r18c5)
43b. {378} combo must have 7 in r8c5 since {78} in r89c6 would clash with 10(2) + 13(2)c6 (thanks SS)
43c. -> no 7 r89c6

44. 21(4)n9 must have 1 for n9
44a. = 1{569/578}(no 2)
44b. 5 locked for n9

45. h8(2) r8c29 = {26}: both locked for r8
45a. r18c7 = [61]

46. 25(4)n7 = {3589/3679/4579/4678}(no 2)({2689} blocked by r8c2

47. 2 in r9 only in n8: 2 locked for n8
47a. no 9 r6c45

48. h10(2) r29c3, r9c3 = {689}
48b. r2c3 = {124}

49. 20(4)n2 = 9{128/137/146/245}
49a. = 1/5, not both
49b. no 5 r1c6

50. h7(2)r29c5 = {16}/[25/43]

51. 9 in n8 only in c6: 9 locked for c6
51a. no 4 r7c6

52. 14(4)n7 must have 1 in r9c45
52a. = 1{238/247/346}(no 5) ({1256} blocked by r8c34)
52b. no 2 r2c5 (h7(2)r29c5)
52c. h7(2)r29c5 = {16}/[43] = [4/6..]

53. 19(3)n2 must have 9 for c5 = {289} ({469} blocked by h7(2)
53a. all locked for c5 & 8 locked for n5
53b. no 3 r6c5
53c. no 5 r7c6
53d. no 2 r3c4

54. 18(4)r6c8 must have 1 and 2 = 12{69/78}
54a. r7c89 = [8/9..] -> Killer pair [89] with r7c6: both locked for r7
54b. no 3 r6c4

55. r5c4 = 3 (hsingle n5)
55a. no 7 r4c4

56. r5c3 = 5
56a. no 4 r67c3
56b. no 7 r4c3

57. should have seen this before: 2 in n5 only in r45c5 -> no 2 in r3c5
57a. r3c5 = 9
57b. r4c4 = 9

rest is straightforward. Finally
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Ed
PostPosted: Tue Jul 08, 2008 1:58 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 71v2 Full Border by Ruud (Oct 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:4608:4608:4354:4354:6404:6404:4358:4358:4358:4617:4608:4608:4354:4354:6404:6404:5648:4358:4617:4617:2580:3349:2326:2583:3864:5648:5648:5915:4617:2580:3349:2326:2583:3864:5154:5648:5915:5915:3366:3366:2326:809:809:5154:5154:4653:5915:2863:3376:1585:1586:3891:5940:5154:4653:4653:2863:3376:1585:1586:3891:5940:5940:4927:4653:5953:5953:5955:5955:3909:3909:5940:4927:4927:4927:5953:5953:5955:5955:3909:3909:
Solution: 
+-------+-------+-------+
| 1 7 5 | 2 6 8 | 3 9 4 |
| 8 6 4 | 3 7 9 | 2 5 1 |
| 3 2 9 | 5 1 4 | 6 7 8 |
+-------+-------+-------+
| 7 5 1 | 8 3 6 | 9 4 2 |
| 9 4 6 | 7 5 2 | 1 8 3 |
| 2 3 8 | 9 4 1 | 7 6 5 |
+-------+-------+-------+
| 6 9 3 | 4 2 5 | 8 1 7 |
| 5 1 7 | 6 8 3 | 4 2 9 |
| 4 8 2 | 1 9 7 | 5 3 6 |
+-------+-------+-------+
Quote:
Ruud: This puzzle could be a nice addition to the unsolvables list.
Andrew: It still seems easier than A71
mhparker: Overall, I would rate the A71V2 at around 1.5
gary w: Took me less than 1.5 hours...So there Ruud..I reckon this one was much easier than V1 !!!
Afmob: Another difficult V2 though it was a bit easier than V1
Andrew (in 2012): Mike started A71 V2 as a “tag” solution. I added 5 moves and then dropped out to work on A71 without getting hints from V2. Mike then finished V2 well before I had finished V1. I’m now having another try at the V2.
Rating: Easy 1.5.
Tag Solution by mhparker & Andrew:
mhparker:
Hi folks,
Ruud wrote:
The initial candidates in the border look very promising.
Don't they just?! :wink:

Maybe we should do this as a team tag solution?

To get the show in motion, here are my first 20 steps. I'd welcome it if anyone else could take over and add a few more. :D


Assassin 71 V2 Tag Walkthrough

Preliminaries:

10(2)n14 and 10(2)n25 = {19/28/37/46}: no 5
13(2)n25, 13(2)n45 and 13(2)n58 = {49/58/67}: no 1..3
9(3)n25 = {126/135/234}: no 7,8,9
15(2)n36 and 15(2)n69 = {69/78}: no 1..5
3(2)n56 = {12}, locked for r5
11(2)n47 = {29/38/47/56}: no 1
6(2)c5 and 6(2)c6 = {15/24}: no 3,6,7..9

1. 15(2)n36 and 15(2)n69 form Naked Quad on {6789} in c7 -> no 6..9 elsewhere in c7

2. r5c6 and r67c6 form killer pair on {12} in c6 -> no 1,2 elsewhere in c6
2a. cleanup: no 8,9 in r34c6
2b. 10(2)n25 = {37/46}

3. Outies c12: r29c3 = 6(2) = {15/24}: no 3,6,7..9

4. Outies c89: r18c7 = 7(2) = {25/34} (no 1)

5. Innies r12: r2c18 = 13(2) = {49/58/67}: no 1..3

6. Innies r89: r8c29 = 10(2) = {19/28/37/46}: no 5

7. Innies c123: r158c3 = 18(3) = {189/279/369/378/459/468/567} (no eliminations yet)

8. Innies c789: r259c7 = 8(3) = {1(25/34)} (no eliminations yet)

9. Outies c1234: r29c5 = 16(2) = {79}, locked for c5

10. Outies c6789: r18c5 = 14(2) = {68}, locked for c5
10a. 9(3)n25 = {135/234}

11. Innies r6789: r6c29 = 8(2) = {17/26/35}: no 4,8,9

12. r4c5+r56c56 forms Naked Quint on {1..5} in n5 -> no 1..5 elsewhere in n5
12a. cleanup: no 8,9 in r37c4 and r5c3; no 6,7 in r3c6
12b. 3 in n5 locked in 9(3) within r45c5 -> no 3 in r3c5

13. {89} in n5 locked in c4 -> not elsewhere in c4

14. r1289c4 must contain all 3 of {123}
14a. 23(4)n78 cannot contain 2 of {123}, otherwise cage sum unreachable
14b. -> r12c4 must contain 2 of {123}, r89c4 contains 1 of {123} + 1 of {4..7}
14c. -> both of r12c4 = {123} (no 4..9)
14d. 17(4) must contain 2 of {123} (in r12c4)
14e. -> {1457} and {2456} both blocked
14f. must contain 1 of {79} (on r2c5)
14g. -> {1268}, {1358}, and {2348} all blocked
14h. -> poss. combos are: {1259/1349/1367/2357}
14i. -> r1c3 = {456}

15. Innies c123 (step 7): max. r15c3 = 13 -> min. r8c3 = 5
15a. -> no 1..4 in r8c3

16. 8 in n2 locked in 25(4)n23 = {8..} = {2689/3589/4678} (no 1) = {(7/9)..}
(Note: {1789} combo blocked by r2c5)
16a. if {4678}, 4 must go in r2c7
16b. -> no 4 in r12c6
16c. 25(4) and r2c5 form killer pair on {79} within n2
16d. -> no 7 in r3c4
16e. cleanup: no 6 in r4c4

17. 4 in n2 locked in r3 -> not elsewhere in r3
17a. cleanup: no 6 in r4c3

18. I/O diff. n1: r13c3 = r4c2 + 9
18a. min. r4c2 = 1 -> min. r13c3 = 10
18b. -> no 1..3 in r3c3
18c. cleanup: no 7..9 in r4c3
18d. max. r13c3 = 15 -> max. r4c2 = 6
18e. -> no 7..9 in r4c2

19. I/O diff. n3: r23c7 = r4c9 + 6
19a. max. r23c7 = 14 -> max. r4c9 = 8
19b. -> no 9 in r4c9

20. I/O diff. n9: r79c7 = r6c8 + 7
20a. max. r79c7 = 14 -> max. r6c8 = 7
20b. -> no 8,9 in r6c8


Marks pic after step 20:

Code:
.-----------------------.-----------------------.-----------------------.-----------------------------------.
| 123456789   123456789 | 456         123       | 68          356789    | 2345        123456789   123456789 |
:-----------.           '-----------.           '-----------.           '-----------.-----------.           |
| 456789    | 123456789   1245      | 123         79        | 356789      2345      | 456789    | 123456789 |
|           '-----------.-----------+-----------.-----------+-----------.-----------:           '-----------:
| 12356789    12356789  | 6789      | 456       | 1245      | 34        | 6789      | 12356789    12356789  |
:-----------.           |           |           |           |           |           :-----------.           |
| 123456789 | 123456    | 1234      | 789       | 12345     | 67        | 6789      | 123456789 | 12345678  |
|           '-----------+-----------'-----------:           :-----------'-----------:           '-----------:
| 3456789     3456789   | 4567        6789      | 345       | 12          12        | 3456789     3456789   |
:-----------.           :-----------.-----------+-----------+-----------.-----------+-----------.           |
| 123456789 | 123567    | 23456789  | 6789      | 1245      | 1245      | 6789      | 1234567   | 123567    |
|           '-----------:           |           |           |           |           |           '-----------:
| 123456789   123456789 | 23456789  | 4567      | 1245      | 1245      | 6789      | 123456789   123456789 |
:-----------.           :-----------'-----------+-----------'-----------+-----------'-----------.           |
| 123456789 | 12346789  | 56789       1234567   | 68          3456789   | 2345        123456789 | 12346789  |
|           '-----------'-----------.           '-----------.           '-----------.           '-----------:
| 123456789   123456789   1245      | 1234567     79        | 3456789     12345     | 123456789   123456789 |
'-----------------------------------'-----------------------'-----------------------'-----------------------'
[/quote]

Andrew:
Don't know if I'm going to be able to take much part in this. I still haven't finished A71.

From what Mike has posted so far, the early moves are easier, meaning they provide more progress, that the early ones for A71.

Anyway here are a couple more steps; I hope nobody objects that I've used normal text rather than Mike's slightly smaller text.

21. I/O diff. n1: r4c23 = r1c3 + 1
21a. IOU no 1 in r4c2

22. I/O diff. N7: r6c13 = r8c3 + 3
22a. IOU no 3 in r6c1

The IOUs on the other two corner nonets have already been done by Mike's I/O steps.

And a few more moves

23. Innies c6: r12589c6 = 29 = {15689/23789/24689} (cannot be {14789/25679/34589} which clash with r67c6, cannot be {34679/35678} which clash with r34c6)

No point in doing a similar one for c4; that would just repeat what is already known from step 14 and the two 13(2) cages.

Also, to complete step 15
15b. r158c3 = {459/468/567}

24. r158c3 = {468/567} (cannot be {459} which clashes with r29c3)
24a. -> no 9 in r8c3
24b. 6 locked in r158c3, not elsewhere in c3
24c. cleanup: no 4 in r4c3, no 5 in r67c3

25. Killer pair 4,5 in r158c3 and r29c3, not elsewhere in c3
25a. cleanup: no 7 in r67c3

It still seems easier than A71. It's already got further than I've reached with A71 where I haven't yet managed anything in c3.

I'm going to have to drop out of the "tag" now and stop reading this thread. Otherwise I risk not being able to solve A71 myself; not sure at the moment if I can anyway.

I hope that someone else will pick up the "tag" and work with Mike on it!

mhparker:
Thanks, Andrew. :D

Carrying on...

26. 11(2)n47 = {29/38} = {(2/8)..}
26a. -> {28} combo blocked for 10(2)n14
26b. 10(3)n14 = [73/91]

[Note: last 2 steps could have been done more directly as follows:
25/26. Hidden killer pair on {39} in c3 (i.e., {39} in c3 locked within 10(2) and 11(2) cages)
25/26a. neither cage can contain both of {39}, so each must contain exactly 1 of {39}
25/26b. -> 10(2) = {(3/9)..}, 10(2) = {(3/9)..}
25/26c. -> 10(2) = [73/91], 11(2) = {29/38}]

27. From step 14h, 17(4)n12 = {1259/1349/1367/2357} = {(3/9)..}
27a. -> {39} only available in 17(4)n12 within n2
27b. -> 25(4)n23 cannot contain both of {39} within n2
27c. from step 16, poss. combos for 25(4)n23 are {2689/3589/4678}
27d. -> (from step 27b) if 25(4)n23 = {3589}, 3 must go in r2c7
27e. -> no 3 in r12c6, no 5 in r2c7

28. Innies c789 (step 8) = r259c7 = [215/314/413]
28a. -> no 2 in r5c7; no 1,2 in r9c7

29. Naked single (NS) at r5c7 = 1
29a. -> r5c6 = 2
29b. Cleanup: no 4 in r67c6, no 4 in r37c5

30. Naked pair (NP) on {15} at r67c6 -> no 5 elsewhere in c6 (1 already gone)

31. 5 in n2 locked in r3 -> not elsewhere in r3

32. 5 no longer available to 25(4)n23 (step 27c)
32a. -> 25(4)n23 = {2689/4678} (no 3)
32b. 6 locked within r1c56+r2c6 for n2
32c. cleanup: no 7 in r4c4

33. Innies c789 (step 28) = r259c7 = [215/413]
33a. -> no 4 in r9c7

34. {12} unavailable to 23(4)n89, 8 locked
34a. -> 23(4)n89 = {3578/4568} (no 9)
34b. 5 locked
34c. -> r9c7 = 5
34d. -> r2c7 = 2 (step 33)

35. Hidden single (HS) in n8 at r9c5 = 9
35a. -> r2c5 = 7

Looks like things are going to be pretty straightforward from here on.
36. 17(4)n12 = [5237/6137/6317]
36a. -> no 4 in r1c3
36b. 3 locked in r12c4 -> not elsewhere in c4 and n2

37. 10(2)n25 = [46]
37a. cleanup: no 7 in r5c3 and r7c4, no 9 in r3c7

38. r34c4 = [58]
38a. cleanup: no 7 in r3c7, no 5 in r5c3

39. HS in n2 at r1c5 = 6
39a. -> r8c5 = 8

40. NS at r1c3 = 5
40a. -> r12c4 = [23] (step 36), r58c3 = [67] (step 24)
40b. -> r5c4 = 7

41. r34c3 = [91]; r29c3 = [42]; r67c4 = [94]; r89c6 = [37]; r18c7 = [34]
41a. cleanup: no 6 in r7c7

42. NS at r3c5 = 1

43. r67c5 = [42]

44. r67c6 = [15] (hidden singles n58)

45. HS in r8/n7 at r8c1 = 5

46. Split 14(3) at r1c12+r2c2 = {167} (only remaining combo)
46a. -> r2c2 = 6
46b. r1c12 = {17}, locked for r1 and n1

47. r2c1 = 8
47a. -> r12c6 = [89]
47b. -> r2c89 = [51]

48. r3c12 = {23} = 5 total
48a. -> r4c2 = 5 (cage split, but could also be obtained by I/O diff. n1 (step 18))
48b. -> r45c5 = [35]

49. HS in r9/c6/n6 at r6c9 = 5

50. 9 of n9 locked in 18(4)n47 = {1269} (no 3,7,8) (last remaining combo)
50a. -> r67c1 = [26]
50b. r78c2 = {19}, locked for c2 and n7

51. r13c12 = [1732]; r6c23 = [38]; r7c3 = 3; r9c12 = [48]; r5c12 = [94]; r4c1 = 7
51a. cleanup: no 7 in r7c7

All singles and one cage sum from here.
Solving Outline by garyw using UR:
Full wt looks very impressive.I did it like this;

1.Prelims include
a)r259c7=8 -> 1{25/34}
b)r29c5 ={79}
c)r18c5={68}

2. r2c5 <>1 -> r1c56r2c6={789} blocked by r2c5

3. But,can see "by inspection" that if r9c7=1 -> (r5c7=2,r2c7=5) then

a)r9c5=7 (because the 9 must be in the 22/3 block at r8c56r9c6)
b)r2c5=9
c)cannot complete the 25/4 cage N2/3.

4. So r5c7=1 r5c6=2 r67c6={15} r67c5=42 (unique rectangle...that again !!)

Mop up now.

Took me less than 1.5 hours.

To me this seems more like spotting a conflicting combo rather than T&E..certainly no more difficult to see than some "conflicting combos" in earlier wts.

So there Ruud..I reckon this one was much easier than V1 !!! Laughing
Walkthrough by Afmob:
Another difficult V2 though it was a bit easier than V1. There were quite some interesting moves you could apply.

Assassin 71 V2 Walkthrough:

1. R12
a) Innies R12 = 13(2) -> no 1,2,3

2. R6789
a) Innies R89 = 10(2) -> no 5
b) Innies R6789 = 8(2) -> no 4,8,9

3. C1234
a) Outies C12 = 6(2) = {15/24}
b) Outies C1234 = 16(2) = {79} locked for C5

4. C6789
a) Innies C789 = 8(3) = 1{25/34} -> 1 locked for C7
b) Outies C89 = 7(2) = {25/34}
c) Outies C6789 = 14(2) = {68} locked for C5
d) Killer pair (12) of 6(2) + R5C6 = (12) -> 1,2 locked for C6
e) 10(2) = {37/46}

5. R5
a) 3(2) = {12} locked

6. N5
a) Hidden Quad (6789) in R456C4 + R4C6 -> no other candidates
b) All 13(2): R37C4 <> 8,9; R5C3 <> 8,9
c) 10(2): R3C6 <> 6,7

7. C4
a) 8,9 locked in R456C4
b) 17(4) must have 7 xor 9 since R2C5 = (79) -> 7,9 not elsewhere
-> R1C3 <> 8 because R1C3 + R12C4 = 8/10

8. N1
a) Innies+Outies: 1 = R4C23 - R1C3; R1C3 = (123456)
-> R4C23 = 3/4/5/6/7
-> R4C23 <> 7,8,9 and R1C3 <> 1
b) 10(2): R3C3 <> 1,2,3

9. N3
a) 25(4) <> 1 because {1789} not possible since R2C5 = (79)

10. C3
a) Innies = 18(3): R8C3 <> 1,2,3,4 since R15C3 <= 13

11. N5+C5
a) 3 locked in R45C5 for C5
b) 8 locked in 25(4) = 8{269/359/467}
-> R12C6 <> 4 since R2C7 <> 6,7,8
c) 8 locked in 23(4) @ C6 = 8{159/249/267/357/456}

12. C567 !
a) ! 6(2) @ C6 = {15} locked because of following conflict:
If 6(2) = {24} -> 3(2) = [12], 10(2) = [37] -> R259C7 = [521]
-> 25(4) = {3589} -> 3 must be in R12C6 -> 2 3's in C6
b) R5C6 = 2, R5C7 = 1
c) 4 locked in R456C5 for C5
d) 23(4): R9C7 <> 3,4 since both combos with 3,4 need a 5 which is only
possible @ R9C7 and R89C6 has no 2
e) Innies C789 = {125}

14. N2
a) 25(4) = 9{268/358} -> no 7; 9 locked for C6+N2
b) R2C5 = 7, R9C5 = 9
c) 13(2): R4C4 <> 6

15. R12
a) Innies = 13(2) = {49/58}

16. C1234 !
a) ! Innies C4 = 19(5) = 123{49/58/67}: R1289C4 = 123{4/5/6/7} (8,9 only @ R5C4)
-> R12C4 <> 4,5,6 since R89C4 would be {12/13/23} -> no possible combination for 23(4)
-> R1C3 = (56) (only possible there in 17(4))

17. R3
a) 4,5 locked in R3C456
b) 10(2): R4C3 <> 6

18. C3
a) 11(2) <> 5,6 because {56} is blocked by R1C3 = (56)

19. N2+C4
a) 17(4) = 37{16/25} -> 3 locked for C4+N2
b) R3C6 = 4, R4C6 = 6
c) Naked triple (789) in R456C4 locked for C4
d) 13(2) @ N2 = [58/67]
e) 25(4) = {2689} -> R1C5 = 6, R2C7 = 2, R9C7 = 5
-> 8,9 locked for C6
f) R8C5 = 8, R1C3 = 5, R3C4 = 5, R4C4 = 8
g) 17(4) = {2357} -> R2C4 = 3, R1C4 = 2, R3C5 = 1

20. C5
a) 6(2) = {42} -> R6C5 = 4, R7C5 = 2
b) 23(4)@R8C3 = {1679} -> R8C3 = 7
c) R8C6 = 3, R9C6 = 7, R8C7 = 4, R1C7 = 3
d) 9(3) = {135} -> 3,5 locked for N5
e) R6C6 = 1, R7C6 = 5

21. R8
a) Hidden Single: R8C1 = 5
b) 19(4) = {2458} -> 2,4,8 locked for R9+N7
c) 15(4) = {2346} -> R8C8 = 2; R9C89 = {36} locked for R9+N9
d) R9C4 = 1, R8C4 = 6, R7C4 = 4, R6C4 = 9, R5C4 = 7, R5C3 = 6

22. C3
a) 10(2) = [82/91]
b) Killer pair (29) of 10(2) blocks [29] of 11(2)
c) R7C3 = 3, R6C3 = 8, R3C3 = 9, R4C3 = 1, R2C3 = 4, R9C3 = 2
d) R2C1 = 8, R9C1 = 4, R9C2 = 8, R2C6 = 9, R1C6 = 8, R2C8 = 5
e) 18(4) = {1467} -> R2C2 = 6; 1,7 locked for R1+N1

23. Rest is clean-up and singles.
Walkthrough by Andrew (in 2012):
Mike started A71 V2 as a “tag” solution. I added 5 moves and then dropped out to work on A71 without getting hints from V2. Mike then finished V2 well before I had finished V1. I’m now having another try at the V2.

Prelims

a) R34C3 = {19/28/37/46}, no 5
b) R34C4 = {49/58/67}, no 1,2,3
c) R34C6 = {19/28/37/46}, no 5
d) R34C7 = {69/78}
e) R5C34 = {49/58/67}, no 1,2,3
f) R5C67 = {12}
g) R67C3 = {29/38/47/56}, no 1
h) R67C4 = {49/58/67}, no 1,2,3
i) R67C5 = {15/24}
j) R67C6 = {15/24}
k) R67C7 = {69/78}
l) 9(3) cage at R3C5 = {126/135/234}, no 7,8,9

Steps resulting from Prelims
1a. Naked pair {12} in R5C67, locked for R5
1b. Naked quad {6789} in R3467C7, locked for C7

2. Killer pair 1,2 in R5C6 and R67C6, locked for C6, clean-up: no 8,9 in R34C6
2a. 8,9 in N5 only in R456C4, locked for C4, clean-up: no 4,5 in R46C4

3. 9(3) cage at R3C5 = {135/234} (cannot be {126} which clashes with R67C5), no 6
3a. Naked quint {12345} in R34567C5, locked for C5
3b. Naked quint {12345} in R456C5 + R56C6, locked for N5, 3 also locked for C5, clean-up: no 6,7 in R3C6, no 8,9 in R5C3

4. 45 rule on C1234 2 outies R29C5 = 16 = {79}, locked for C5
4a. 17(4) cage at R1C3 = {1259/1349/1367/1457/2357} (cannot be {1268/1358/2348/2456} because R2C5 only contains 7,9), no 8
4b. R2C5 = {79} -> no 7,9 in R1C34 + R2C4

5. 45 rule on R12 2 innies R2C18 = 13 = {49/58/67}, no 1,2,3

6. 45 rule on R89 2 innies R8C29 = 10 = {19/28/37/46}, no 5

7. 45 rule on C12 2 outies R29C3 = 6 = {15/24}

8. 45 rule on C89 2 outies R18C7 = 7 = {25/34}, no 1

9. 45 rule on R6789 2 innies R6C29 = 8 = {17/26/35}, no 4,8,9

10. 8 in N2 only in 25(4) cage at R1C5 = {2689/3589/4678} (cannot be {1789} which clashes with R2C5), no 1
10a. 4 of {4678} must be in R2C7 -> no 4 in R12C6
10b. Killer pair 7,9 in 25(4) cage and R2C5, locked for N2, clean-up: no 6 in R4C4

11. 45 rule on C123 3 innies R158C3 = 18 = {279/369/378/468/567} (cannot be {189} because 8,9 only in R8C3, cannot be {459} which clashes with R29C3), no 1
11a. 8,9 of {279/369/378/468} must be in R8C3 -> no 2,3,4 in R8C3

12. 45 rule on N1 2 outies R4C23 = 1 innie R1C3 + 1, IOU no 1 in R4C2
12a. 45 rule on N1 2 innies R13C3 = 1 outie R4C2 + 9
12b. Max R13C3 = 15 -> max R4C2 = 6
12c. Min R13C3 = 11, no 1,2,3,4 in R3C3, clean-up: no 6,7,8,9 in R4C3

13. 45 rule on N3 2 innies R23C7 = 1 outie R4C9 + 6
13a. Min R23C7 = 8 -> min R4C9 = 2
13b. Max R23C7 = 14 -> max R4C9 = 8

14. 45 rule on N7 2 outies R6C13 = 1 innie R8C3 + 3, IOU no 3 in R6C1

15. 45 rule on N9 2 innies R79C7 = 1 outie R6C8 + 7
15a. Max R79C7 = 14 -> max R6C8 = 7

16. 23(4) cage at R8C3 = {1589/1679/2489/2579/3479/3569/3578} (cannot be {2678/4568} which clash with R8C5)
16a. Hidden killer triple 1,2,3 in R12C4 and 23(4) cage for C4, 23(4) cage contains one of 1,2,3 -> R12C4 must contain two of 1,2,3 -> R12C4 = {123}

17. 17(4) cage at R1C3 (step 4a) = {1259/1349/1367/2357} (cannot be {1457} because 4,5 only in R1C3)
17a. 4,5,6 only in R1C3 -> R1C3 = {456}

18. R158C3 (step 11) = {468/567}, no 9, 6 locked for C3, clean-up: no 4 in R4C3, no 5 in R67C3
18a. Killer pair 4,5 in R158C3 and R29C3, locked for C3, clean-up: no 7 in R67C3
18b. R34C3 = [73/91] (cannot be [82] which clashes with R67C3), no 2,8

19. 45 rule on N7 2 innies R78C3 = 1 outie R6C1 + 8
19a. R78C3 cannot total 12 -> no 4 in R6C1

20. R4C23 = R1C3 + 1 (step 12)
20a. R1C3 = {456} -> R4C23 = 5,6,7 = [23/41/51/43/61], no 3 in R4C2

[I couldn’t see any more routine steps, then I spotted a short forcing chain and this puzzle is cracked …]
21. Consider placements for R2C5 = {79}
R2C5 = 7 => 17(4) cage at R1C3 (step 17) = {1367/2357}, 3 locked for N2 => R3C6 = 4, R4C6 = 6
or R2C6 = 9 => 7 in N2 only 25(4) cage at R1C5, 7 locked for C6 => R4C6 = 6, R3C6 = 4
-> R34C6 = [46], clean-up: no 9 in R3C7, no 9 in R4C4, no 7 in R5C3, no 2 in R67C6, no 7 in R7C4

22. Naked pair {15} in R67C6, locked for C6 -> R5C67 = [21], clean-up: no 7 in R6C2 (step 9), no 4 in R7C5
22a. Naked triple {789} in R456C4, locked for C4
22b. 5 in N2 only in R3C45, locked for R3

23. R158C3 (step 18) = {468/567}
23a. 7,8 only in R8C3 -> R8C3 = {78}

24. 8 in N8 only in 23(4) cage at R8C5 = {2678/3578} (cannot be {2489/4568} because 2,4,5 only in R9C7), no 4,9, 7 locked for C6 and N8 -> R9C5 = 9 (hidden single in N8), R2C5 = 7, clean-up: no 6 in R2C18 (step 5)
24a. 2,5 of 23(4) cage only in R9C7 -> R9C7 = {25}

25. R2C5 = 7 -> 17(4) cage at R1C3 (step 17) = {1367/2357}, no 4, 3 locked for C4 and N2

26. Naked pair {89} in R12C6, locked for C6 and N2 -> R1C5 = 6, R8C5 = 8, R1C3 = 5, R8C3 = 7, R5C3 = 6 (hidden single in C3), R5C4 = 7, R4C4 = 8, R3C4 = 5, R6C4 = 9, R7C4 = 4, R89C6 = [37], R9C7 = 5 (cage sum), R3C3 = 9, R4C3 = 1
26a. R1C5 = 6, R12C6 = {89} = 17 -> R2C7 = 2 (cage sum), R8C7 = 4, R1C7 = 3, R2C3 = 4, R2C1 = 8, R12C6 = [89], R2C8 = 5, R9C3 = 2
26b. Clean-up: no 7 in R3C7, no 6 in R7C7, no 2,7 in R6C9 (step 9), no 2,6 in R8C29 (step 6)

27. R1C3 = 5, R2C5 = 7 -> R12C4 = 5 = [23], R3C5 = 1, clean-up: no 5 in R67C5
27a. R67C5 = [42]
27b. R67C6 = [15] (hidden singles in N5 and N8)

28. Naked pair {19} in R8C29, locked for R8 -> R89C4 = [61], R8C18 = [52]

29. Naked pair {17} in R1C12, locked for R1 and N1 -> R2C29 = [61], R8C29 = [19], clean-up: no 6 in R6C7

30. Naked pair {78} in R67C7, locked for C7 -> R34C7 = [69]

31. R8C78 = [42] = 6 -> R9C89 = 9 = {36}, locked for R9 and N9 -> R9C12 = [48], R67C3 = [83], R7C12 = [69], R6C1 = 2 (cage sum), R67C7 = [78]

32. Naked pair {36} in R69C8, locked for C8 -> R4C8 = 4

and the rest is naked singles.


I'll rate my walkthrough for A71 V2 at Easy 1.5; I used a short forcing chain.
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Ed
PostPosted: Tue Jul 08, 2008 2:01 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 71v1.5 by mhparker (Oct 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:5376:5376:3586:3586:5124:5124:6406:6406:6406:4105:5376:5376:3586:3586:5124:5124:3344:6406:4105:4105:2836:2837:4886:1815:2328:3344:3344:4635:4105:2836:2837:4886:1815:2328:6178:3344:4635:4635:3622:3622:4886:3369:3369:6178:6178:5933:4635:2607:2096:2097:1842:2867:5428:6178:5933:5933:2607:2096:2097:1842:2867:5428:5428:5439:5933:5697:5697:4419:4419:5701:5701:5428:5439:5439:5439:5697:5697:4419:4419:5701:5701:
Solution: 
+-------+-------+-------+
| 3 2 5 | 6 4 8 | 7 1 9 |
| 6 9 7 | 2 1 3 | 5 4 8 |
| 1 4 8 | 7 9 5 | 3 2 6 |
+-------+-------+-------+
| 8 5 3 | 4 7 2 | 6 9 1 |
| 2 1 6 | 8 3 9 | 4 5 7 |
| 4 7 9 | 5 6 1 | 2 8 3 |
+-------+-------+-------+
| 5 8 1 | 3 2 6 | 9 7 4 |
| 9 6 4 | 1 5 7 | 8 3 2 |
| 7 3 2 | 9 8 4 | 1 6 5 |
+-------+-------+-------+
Quote:
garyw: I for one thoroughly enjoyed renewing my acquaintance with this cage...Slightly easier than v1 but...about 3 hours,in fact
Andrew: A lot easier than V1 ..I'll rate it at least a 1.25. Maybe it should be rated higher because it was a fairly narrow solving path and a lot of work after the breakthrough
Afmob: was the easiest of the bunch. It didn't need any complicate or long contradiction moves so given rating of 1.25 should be right. By the way, it was the most fun to solve
mhparker, in Rating Thread:..1.25 rating is possibly too low.
Walkthrough by Andrew with one hypothetical:
Thanks for the V1.5 Mike. Nice puzzle!

A lot easier than V1 but don't know how it compares with V2 because I didn't try to solve that one myself and I haven't yet looked at the later moves for that one or at Afmob's walkthrough.

I'll rate it at least a 1.25. Maybe it should be rated higher because it was a fairly narrow solving path and a lot of work after the breakthrough.

I had to use one hypothetical (step 37) and have commented on another one that I saw but didn't need to use. I'll be interested to see if anyone else posts a walkthrough without hypotheticals or "advanced" moves.

Here is my walkthrough for A71V1.5

1. R34C3 = {29/38/47/56}, no 1

2. R45C4 = {29/38/47/56}, no 1

3. R45C6 = {16/25/34}, no 7,8,9

4. R45C7 = {18/27/36/45}, no 9

5. R5C34 = {59/68}

6. R5C67 = {49/67} (cannot be {58} which clashes with R5C34)

7. R67C3 = {19/28/37/46}, no 5

8. R67C4 = {17/26/35}, no 4,8,9

9. R67C5 = {17/26/35}, no 4,8,9

10. R67C6 = {16/25/34}, no 7,8,9

11. R67C7 = {29/38/47/56}, no 1

12. R345C5 = {289/379/469/478/568}, no 1

13. 14(4) cage at R1C3 = {1238/1247/1256/1346/2345}, no 9

14. 13(4) cage at R2C8 = {1237/1246/1345}, no 8,9
14a. CPE no 1 in R12C9

15. Killer pair 6,9 in R5C34 and R5C67, locked for R5

16. 45 rule on R12 2 innies R2C18 = 10 = {37/46}/[81/92], no 5

17. 45 rule on R1234 2 innies R4C18 – 14 = 1 outie R5C5, max R4C18 = 17 -> max R5C5 = 3, min R5C5 = 2 -> min R4C18 = 16 -> R4C18 = {79/89}, 9 locked for R4, clean-up: no 2 in R3C34

18. R345C5 (step 12) with R5C5 = {23} = {289/379} -> R3C5 = 9, R4C5 = {78}, clean-up: no 2 in R4C34

19. Naked triple {789} in R4C158, locked for R4, clean-up: no 3,4 in R3C34, no 1,2 in R3C7

20. 45 rule on R6789 2 innies R6C29 = 10 = {19/28/37/46}, no 5

21. 45 rule on R89 2 innies R8C29 = 8 = {17/26/35}, no 4,8,9

22. 45 rule on C12 2 outies R29C3 = 9 = {18/27/36/45}, no 9

23. 45 rule on C1234 2 outies R29C5 = 9 = {18/27/36/45}

24. 45 rule on C6789 2 outies R18C5 = 9 = {18/27/36/45}

25. 45 rule on C89 2 outies R18C7 = 15 = {69/78}

26. 45 rule on C789 3 innies R259C7 = 10 = {127/136/145} (cannot be {235} because no 2,3,5 in R5C7) = 1{27/36/45}, no 8,9, 1 locked for C7, clean-up: no 8 in R3C7, no 4 in R5C6
26a. R5C7 = {467} -> no 4,6,7 in R29C7

27. Hidden killer pair 8,9 in R18C7 and R67C7 -> R67C7 = {29/38}, no 4,5,6,7

28. 45 rule on N1 2 outies R4C23 – 3 = 1 innie R1C3
28a. IOU no 3 in R4C2

29. 45 rule on N1 2 innies R13C3 – 8 = 1 outie R4C2

30. 45 rule on N3 2 outies R4C79 – 2 = 1 innie R2C7
30a. IOU no 2 in R4C9
30b. R2C7 = {1235} -> R4C79 = 3,4,5,7 = [21/31/23/41/25/34/43/61], no 5 in R4C7, no 6 in R4C9, clean-up: no 4 in R3C7
30c. 4 in C7 locked in R45C7, locked for N6, clean-up: no 6 in R6C2 (step 20)

31. 45 rule on N3 2 innies R23C7 – 7 = 1 outie R4C9

32. 45 rule on N7 2 outies R6C13 – 9 = 1 innie R8C3
32a. IOU no 9 in R6C1

33. 45 rule on N7 2 innies R78C3 – 1 = 1 outie R6C1, min R78C3 = 3 -> min R6C1 = 2

34. 45 rule on N9 2 outies R6C78 – 9 = 1 innie R9C7
34a. IOU no 9 in R6C8

35. 45 rule on N9 2 innies R79C7 – 2 = 1 outie R6C8

36. 18(4) cage at R4C1, min R4C1 = 7 -> max R5C12 + R6C2 = 11, no 9 in R6C2, clean-up: no 1 in R6C9 (step 20)

[There’s an interesting hypothetical here but it isn’t needed.
1 in C5 must be in R67C6 or in one of the 9(2) split-cages
If one of the 9(2) split-cages = {18} => R45C5 = [73] => R67C6 = {26}
-> R67C6 = {17/26}, no 3,5]

37. No 4 in R6C2, here’s how
37a. 18(4) cage in N4 = {1278/1359/1458/2349/2358/2457}
37b. If {1458} with R6C2 = 4 => R5C12 = {58} clashes with R34C5
37c. If {2349} with R6C2 = 4 => R5C12 = {23} clashes with R5C5
37d. If {2457} with R6C2 = 4 => R4C1 = 7, R5C12 = {25} => R5C5 = 3, R4C5 = 7 clashes with R4C1
37e. -> no 4 in R6C2, no 6 in R6C9 (step 20)

38. 24(4) cage in N6 = {2589/3579} = 59{28/37}, no 1, 5,9 locked for N6, clean-up: no 2 in R7C7

39. 5 in N6 locked in R5C89, locked for R5, clean-up: no 9 in R5C34
39a. Naked pair {68} in R5C34, locked for R5, clean-up: no 7 in R5C67 -> R5C67 = [94], clean-up: no 5 in R3C7
39b. 9 in N8 locked in R89C4, locked for 22(4) cage -> no 9 in R8C3

40. R259C7 (step 26) = {145} (only remaining combination) -> R29C7 = {15}
[Alternatively R29C7 = {15} (hidden pair in C7).]

41. 1 in R5 locked in R5C12, locked for N4, clean-up: no 9 in R6C9 (step 20), no 9 in R7C3

42. R4C8 = 9 (hidden single in N6), R6C3 = 9 (hidden single in R6), R7C3 = 1, clean-up: no 8 in R29C3 (step 22), no 7 in R6C45, no 6 in R6C6, no 7 in R8C9 (step 21)

43. 18(4) cage in N4 (step 38a) = {1278} (only remaining combination), locked for N4 -> R5C34 = [68], R45C5 = [73], R4C1 = 8, clean-up: no 3 in R29C3 (step 22), no 5 in R34C3, no 4 in R3C6, no 1 in R6C5, no 5 in R67C5, no 2,7 in R6C9 (step 20), no 5 in R7C4, no 4 in R7C6

44. Naked pair {26} in R67C5, locked for C5

45. R6C13 – 9 = R8C3 (step 32), R6C3 = 9 -> R6C1 = R8C3 -> R8C3 = {345}

46. 45 rule on C123 2 remaining innies R18C3 = 9 = {45} (only remaining combination), clean-up: no 3 in R6C1 (step 45)
46a. Naked pair {45} in R18C3, locked for C3
46b. R3C3 = 8 (hidden single in C3), R4C3 = 3, R4C9 = 1, clean-up: no 6 in R3C67

47. R6C78 – 9 (step 34) = R9C7, R9C7 = {15} -> R6C78 = 10,14 = [28/37/82/86], no 3 in R6C8

48. 45 rule on N6 3 remaining innies R4C7 + R6C78 = 16 = {268/367}
48a. R46C7 cannot be [68] which would make R3C7 and R7C7 clash -> no 2 in R6C8

49. 13(4) cage at R2C8 (step 14) = {1246/1345} (cannot be {1237} which clashes with R3C7), no 7

50. 14(4) cage at R1C3, min R1C3 = 4 -> max R1C4 + R2C45 = 10, no 8, clean-up: no 1 in R9C5 (step 23)

51. 8 in N2 locked in 20(4) cage at R1C5 = {1478/1568/3458} (cannot be {2378/2468} because R2C7 only contains 1,5), no 2

52. 22(4) cage at R8C3 must contain 9 and at least two of 4,5,8 = {1489} (only remaining combination) -> R8C3 = 4, R9C5 = 8, R2C5 = 1 (step 23), R29C7 = [51], R18C5 = [45], R89C4 = [19], R1C3 = 5, clean-up: no 6 in R4C6, no 2 in R6C6, no 7 in R7C4

53. Naked triple {236} in R7C456, locked for R7 and N8 -> R89C6 = [74], clean-up: no 8 in R1C7 (step 25), no 3 in R37C6, no 8 in R6C7

54. R6C6 = 1 (hidden single in R6), R7C6 = 6, R67C5 = [62], R67C4 = [53], clean-up: no 6 in R34C4

55. R34C4 = [74], R34C7 = [36], R6C1 = 4, R4C2 = 5, R34C6 = [52]

56. R6C9 = 3 (hidden single in R6), R6C2 = 7 (step 20), R6C8 = 8, R67C7 = [29], R18C7 = [78], R7C2 = 8

57. Naked triple {246} in R2C89 + R3C9, locked for N3 -> R1C8 = 1

58. 23(4) cage at R6C1, R6C1 = 4, R7C2 = 8 -> R7C1 + R8C2 = 11 = [56], R8C89 = [32], R8C1 = 9
58a. Naked triple {237} in R9C123, locked for R9

59. 16(4) cage at R2C1 = {1456/2356} (cannot be {1357} because 3,7 only in R2C1), no 7
59a. 4 of {1456} must be in R3C2 -> no 1 in R3C2
59b. 2 of {2356} must be in R3C2 -> no 2 in R3C1

60. R9C1 = 7 (hidden single in C1), R9C23 = [32], R2C3 = 7

61. R5C2 = 1 (hidden single in C2), R5C1 = 2

62. R3C1 = 1 (hidden single in C1)
62a. 16(4) cage at R2C1, R3C1 = 1, R4C2 = 5 -> R2C1 + R3C2 = 10 = [64]

and the rest is naked singles
Walkthrough by Afmob:
A71 V1.5 was the easiest of the bunch. It didn't need any complicate or long contradiction moves so given rating of 1.25 should be right.

By the way, it was the most fun to solve, since the other versions had a point after there was no going further and you could see the next step only after long thinking :? .

Assassin 71 V1.5 Walkthrough

1. R1234
a) Innies R12 = 10(2) = [37/46/64/73/82/91]
b) Innies+Outies: -14 = R5C5 - R4C18 -> R5C5 = (23), R4C18 = 16/17 = {79/89}
-> 9 locked for R4
c) 19(3) = 9{28/37} since R5C5 = (23) -> R3C5 = 9, R4C5 = (78)
d) Naked triple (789) in R4C158 locked for R4
e) Both 11(2) = [56/65/74/83]
f) 9(2): R3C7 <> 1,2

2. R6789
a) Innies R89 = 8(2) -> no 4,8,9
b) Innies R6789 = 10(2) -> no 5

3. R5
a) Killer pair (58) of 14(2) blocks {58} of 13(2)
b) 9 locked in R5C46
c) 14(2): R5C4 <> 5
d) 13(2): R5C6 <> 4
e) Killer pair (69) in 14(2) + 13(2) locked

4. C1234
a) Outies C12 = 9(2) -> no 9
b) Outies C1234 = 9(2)

5. C6789
a) Outies C89 = 15(2) = {69/78}
b) Innies C789 = 10(3) = 1{27/36/45}; {235} impossible because R5C7 = (467)
c) Innies C789 = 10(3): R29C7 <> 4,6,7 since R5C7 is at least 4 and R5C7 <> 5
d) 1 locked in Innies C789 for C7
e) 9(2): R3C7 <> 8
f) Outies C6789 = 9(2)

6. C5
a) Innies = 18(4) = 9(2) + 9(2) (step 4b, 5f); 4 locked in 18(4)
-> since 4 is locked, one of the 9(2) must be {45} -> 5 locked
-> 18(4) = 45{18/27/36}
b) 8(2) = {17/26}
c) Killer pair (27) in 19(3) + 8(2) locked -> 18(4) = 45{18/36}

7. N7
a) Innies+Outies: 9 = R6C13 - R8C3; R6C13 <= 17 -> R8C3 <> 9
b) 9 locked in 10(2) for C3 -> 10(2) = {19} locked for C3
c) Innies+Outies: 9 = R6C13 - R8C3; R8C3 >= 2 -> R6C13 >= 11
-> R6C13 <> 1
d) R6C3 = 9, R7C3 = 1
e) Hidden Single: R4C8 = 9 @ R4
f) Outies C12 <> 8

8. R6789
a) Innies R6789 = 10(2): R6C29 = {28/37/46} -> no 1
b) 8(2) @ C4: R6C4 <> 7
c) 8(2) @ C5: R6C5 <> 7
d) 7(2): R6C6 <> 6
e) 11(2): R7C7 <> 2

9. N5
a) Hidden triple (789) in R4C5+R5C4+R5C6 -> no other candidates
b) 14(2) = [59/68]
c) 13(2) = [76/94]

10. C123
a) 11(2) <> {56} because R5C3 = (56)
b) Innies = 15(3) <> 3 because:
- <> {348} since R5C3 = (56)
- <> {357} is blocked by Killer pair (37) of 11(2)

11. C789
a) Innies = 10(3) = 1{36/45} -> no 2 since R5C7 = (46)
b) 2 locked in R46C7 for N6
c) Consider positions of 2 in N6:
- Either 9(2) = [72] -> 11(2) <> {47}
- Or 11(2) = [29] -> 11(2) <> {47}
-> 11(2) <> {47}

12. N1
a) Innies+Outies: 3 = R4C23 - R1C3; R4C23 = 4/5/6/7/8/9/10; R1C3 = (245678)
-> R4C23 = 5/6/7/8/9/10
-> R1C3 <> 8

13. N69 !
a) ! Innies = 12(1+3) = R9C7 + R45C7+R4C9
= 1+{146/236/245} 3+{126/234} 5+{124} -> other combos impossible since R5C7 = (46)
-> R4C9 <> 4,6 since R4C79+R5C7 would be at least 12
-> R4C7 <> 5 since Innies would be 1+{245} and R4C9 <> 2,4
-> R4C7 <> 3 since Innies would be 1+{236} and R4C9 <> 2,6
b) 9(2) = [36/54/72]
c) 4 locked in R45C7 for N6
d) 24(4) = 9{168/357}: R6C9 <> 8 because R5C89 <> 6
e) Innies R6789 = 10(2) = [37/46/73]

14. R456
a) Killer pair (58) of 14(2) + 24(4) in R5C89 locked
b) 18(4) = {1278} locked for N4 -> R4C1 = 8, R6C2 = 7
c) R4C5 = 7, R5C5 = 3, R5C6 = 9, R5C7 = 4, R5C4 = 8, R5C3 = 6
d) 24(4) = {3579} locked for N6 -> R6C9 = 3
e) R4C9 = 1

15. C789
a) 9(2) = [36/72]
b) Innies 10(3) = {145} locked
c) 11(2) = [29/83]

16. R67
a) 8(2) @ C4: R7C4 <> 5
b) 8(2) @ C5 = {26} locked for C5
c) 7(2): R7C6 <> 4

17. C1234
a) Innies C123 = 15(3) = 6{27/45} -> R8C3 <> 8
b) Hidden Single: R3C3 = 8 @ C3
c) R4C3 = 3

18. R6
a) Naked triple (268) in R6C278 locked
b) 8(2) @ C4 = [17/53]

19. N8
a) 17(4) = 15{38/47} -> other combos blocked by R7C4 = (37) and R7C5 = (26)
and it must have 1 or 5 since R9C7 = (15)
-> 1,5 locked -> R9C45 <> 1,5
b) 9 locked in 22(4) = 9{148/238/247/346} -> it must have 4 or 8 since R9C5 = (48)
-> no 5

20. N7
a) Innies+Outies: R6C1 = R8C3 = 4

21. After clean-up and singles:
-> 16(4) @ N1 <> {1357} since R3C12 <> 3,7
-> R2C1 <> 7
-> Hidden Single: R2C3 = 7 @ N1
-> 21(4) @ N7 = {2379} -> R9C3 = 2, R9C1 = 7, R9C2 = 3
-> 21(4) @ N1 = {2379}

22. Rest is singles.

I got a bit lazy in the end because there were lots of singles but it still wasn't finished after that
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Ed
PostPosted: Tue Jul 08, 2008 2:03 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Concentric Squares Killer by mhparker (Oct 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:4608:4608:4608:1795:1795:3077:3077:4359:4359:4873:4608:2571:2571:4109:4109:2575:2575:4359:4873:3091:2571:4373:4373:4375:4375:2575:1562:4873:3091:2333:4373:5407:5407:4375:2338:1562:4132:1317:2333:4391:5407:5407:2858:2338:2348:4132:1317:2863:4391:4391:2858:2858:4660:2348:4132:5431:2863:2863:3386:3386:4660:4660:5950:2623:5431:5431:3138:3138:2628:2628:5950:5950:2623:2623:3658:3658:3658:3917:3917:3917:5950:
Solution: 
+-------+-------+-------+
| 5 2 3 | 6 1 4 | 8 9 7 |
| 4 8 6 | 3 7 9 | 5 2 1 |
| 7 9 1 | 5 8 2 | 6 3 4 |
+-------+-------+-------+
| 8 3 7 | 4 6 1 | 9 5 2 |
| 6 1 2 | 8 9 5 | 7 4 3 |
| 9 4 5 | 7 2 3 | 1 8 6 |
+-------+-------+-------+
| 1 6 4 | 2 5 8 | 3 7 9 |
| 2 7 8 | 9 3 6 | 4 1 5 |
| 3 5 9 | 1 4 7 | 2 6 8 |
+-------+-------+-------+
Quote:
mhparker, lead-in: (Est. rating: 1.25)
gary w: this turned out to be a rather lovely puzzle
Afmob: It required lots of litte steps but nothing to complicated
Andrew: I'll agree with your rating of 1.25
Outline by garyw:
Hi Mike,

I take it back..this turned out to be a rather lovely puzzle.I've outlined my solution.



Prelims

1.r18c3=11 -> r3c1<>9/8/1 cannot have a 2/3 at r8c3).In c3 9 must be at r89.r2c56={79}
2.r7c29=15-> 13/2 cage N8 <> {67}
3.Outies on r123-> r279c4+r9c5=10 r2c4 max 4 r23c3 min 6
4.In N8 one of 1/2 must go in r7c4 and r9c45.Thus from 4a below r7c4=1/2......4a Outies N9 -> r6c8+r89c6=21


5.From 2. 13/2 cage N8 = {58} or (49}-> 4/8 so 12/2 cage N8 <>{48}
6.So 12/2 cage N8 = {57} or {39} -> 3/7 so 10/2 cage N89 <> {37}
7.Consider placement of 3 in N8.In all cases,whether it is at r8c45 or r9c456 it must enforce r9c3=9.Also now 21/3 cage N7= {678} and r7c13=5
8.Now in N8 6 is at r89c6 ..only places
9Thus from 4a r6c8<>9 so in N6 HS 9 at r4c7
10.N4 9 must be at r56c1
11.N3 9 at r1c89
12.HS 9 at r3c2 -> r4c2=3 -> r56c2={14} 9/2 cage N4= {27} r7c1<>4 and thus from 7. r7c3<>1So r7c3 <> 1 or 2 so r6c3<>8
so HS c3 r8c3=8
13 c3 1 at r23and N1 8 at r12c2 so 18/4 cage N1 ={2358} thus other number in r23c3=6 ..cannot be 4 because of constraints on 10/3 cage
14.R2C4=3 R7C4=2 (R2C4+ R7C134 =10 outies on 1/4)
15. r7c1=1 r7c3=4

Mop up now.
The effect of the placements of the 9s was very attractive.
And,of course,the last placements were the "centre" square.


Now I can enjoy the rest of my weekend!!Seriously though Mike..thanks for a very nice puzzle.

Regards

Gary
Walkthrough by Afmob:
Cool cage pattern. It required lots of litte steps but nothing to complicated.

CS Walkthrough:

1. C1234
a) Outies C12 = 11(2): R1C3 <> 1,8,9
b) Outies C1234 = 18(5) = 12{348/357/456} -> 1,2 locked for C5

2. R12
a) 16(2) = {79} locked for C2+N2
b) Outies R1 = 9(2) = {18/36/45}
c) 12(2) @ R1: R1C7 <> 3,5

3. C9
a) Outies C9 = 10(2) -> no 5
b) Killer pair (45) of 6(2) blocks {45} of 9(2) @ C9

4. R6789
a) Innies+Outies R9: -6 = R8C1 - R9C9 -> R8C1 = (123); R9C9 = (789)
b) 12(2): R8C4 <> 3
c) Outies R89 = 15(2) = {69/78}
d) Killer pair (67) of Outies R89 blocks {67} of 13(2)
e) Killer pair (89) locked in 13(2) + Outies R89 for R7
f) Innies+Outies R789: 12 = R6C38 - R7C1
-> R7C1 = (12345)
-> R6C38 = 13/14/15/16/17 -> no 1,2,3; R6C8 <> 4 since R6C3 <= 8
g) 9 locked in R89C3 for N7
h) 21(3): R8C3 <> 4,5 because R78C2 <> 9

5. N5
a) Innies N5 = 7(2) -> no 7,8,9
b) 17(3) @ R3C4 = 8{36/45} -> 8 locked for C3+N2
c) Innies N5 = 7(2): R6C6 <> 5,6

6. N8
a) Killer pair (48) of 13(2) blocks {48} of 12(2)
b) Killer pair (59) locked in 13(2) + 12(2)
c) 11(3): R7C34 <> 7 because R6C3 is at least 4
d) 10(2): R8C7 <> 1

7. N9
a) Outies = 21(2+1) <> 1,2
b) 10(2): R8C7 <> 8,9

8. N3
a) 12(2) = [39/48/57]

9. C1234
a) Outies C12 = 11(2) = [29/38/47/56]
b) 12(2) @ C2: R4C2 <> 4
c) Innies+Outies C123: -4 = R27C4 - R9C3
-> R9C3 = (789) since R27C4 is at least 3
-> R27C4 = 3/4/5 -> no 5,6
d) 21(3) = {489/579/678} -> has 2 of (789) + R9C3 = (789) -> locked for N7

10. R789
a) Killer pair (37) of 12(2) blocks {37} of 10(2) -> 10(2) = [46/64/82]
b) 21(3): R7C2 <> 7 since R8C23 would be {59/68}
-> blocked by Killer pairs (59,68) of 12(2) and 10(2)
c) 21(3) <> 5 because R7C2 = (68) -> 21(3) = 8{49/67}, 8 locked for N7
d) 7 locked in R7C789 for N9
e) 14(3): R9C45 <> 7,8 since R9C3 = (79)
f) Outies R89 = 15(2) = [69/87]
g) Innies+Outies R9: -6 = R8C1 - R9C9; R9C9 = (89) -> R8C1 = (23)
h) 10(3) <> 4 because R8C1 = (23)
i) 10(3) = 3{16/25} -> 3 locked for N7
j) 1 locked in 23(4) for N9 -> 23(4) = 19{58/67} -> 9 locked for N9

11. N8 !
a) ! 15(3) <> {456} since it must have 7 xor 8 because 15(3) + R9C39 are
the only positions in R9 where 7,8,9 are possible
-> 15(3) = {258/267/348/357}
b) 14(3) <> 2 because {239} is blocked by Killer pair (23) of 15(3)
-> 14(3) = {149/167/347}
c) Hidden Single: R7C4 = 2
d) 1 locked in 14(3) = 1{49/67} for R9

12. N7
a) 10(3) = {235} locked; 5 locked for R9
b) 11(3) = 2[54/81]
c) 6 locked in 21(3) -> 21(3) = {678} locked; 7 locked for R8
d) R9C3 = 9, R9C9 = 8
e) Hidden Single: R9C6 = 7 @ R9, R8C5 = 3 @ N8
f) R2C6 = 9, R2C5 = 7, R8C4 = 9, R8C1 = 2
g) Hidden Single: R7C9 = 9 @ N9

13. N9
a) 23(4) = {1589} -> R8C9 = 5, R8C8 = 1
b) 10(2) = {46} locked for R8
c) 2 locked in R9C78 -> 15(3) = {267}; 2,6 locked for R9+N9
d) R8C7 = 4, R8C6 = 6
e) Hidden Single: R7C2 = 6 @ R7
f) 18(3) = {378} -> R6C8 = 8
g) R6C3 = 5, R7C3 = 4, R7C1 = 1

14. N3
a) Hidden Single: R1C7 = 8 -> R1C6 = 4
b) Hidden Single: R1C8 = 9 @ C8, R4C7 = 9 @ C7, R5C5 = 9 @ C5
c) Hidden Single: R6C1 = 9 @ R6, R3C2 = 9 @ C2
d) R5C1 = 6, R4C2 = 3, R9C2 = 5, R9C1 = 3
e) 6(2) = {24} locked for C9

15. N6
a) 9(2) @ C9 = {36} -> R6C9 = 6, R5C9 = 3

16. Rest is clean-up and singles
Walkthrough by Andrew:
Nice puzzle Mike. I'll agree with your rating of 1.25.
CathyW wrote:
I didn't keep a WT but the outies of c123 and innies of N2356 proved quite helpful.
I used innies/outies for C123 (see comment about the difference between outies and innies/outies after step 28) but missed innies of N2356. At one stage I looked at multiple nonets but couldn't see any useful ones; don't know how I missed those ones which are probably the only useful multiple nonets. As it happens I had no problems with eliminations from R6C8 so it probably didn't have much affect on my solution path but it's annoying that I missed it; in a different puzzle it might have been important.

At one stage I wondered whether it would be possible to use "concentric" as a solving step. However there were also two outies from the outer ring so that didn't help. Still some puzzle creator might be able to use the idea of an outer concentric ring so that overlaps give the total for R19C9.

Here is my walkthrough. It took me until step 51, including prelims, to make the first placements but then it fell fairly quickly.

Didn't have time to check my walkthrough properly last night. I've now done that and made the necessary changes.

Thanks Mike for pointing out some more corrections to typos and extensions to a few of my steps.

1. R1C45 = {16/25/34}, no 7,8,9

2. R1C67 = {39/48/57}, no 1,2,6

3. R2C56 = {79}, locked for R2 and N2, clean-up: no 3,5 in R1C7

4. R34C2 = {39/48/57}, no 1,2,6

5. R34C9 = {15/24}

6. R45C3 = {18/27/36/45}, no 9

7. R45C8 = {18/27/36/45}, no 9

8. R56C2 = {14/23}

9. R56C9 = {18/27/36} (cannot be {45} which clashes with R34C9), no 4,5,9

10. R7C56 = {49/58/67}, no 1,2,3

11. R8C45 = {39/48/57}, no 1,2,6

12. R8C67 = {19/28/37/46}, no 5

13. R234C1 = {289/379/469/478/568}, no 1

14. 10(3) cage at R2C3 = {127/136/145/235}, no 8,9

15. 10(3) cage in N3 = {127/136/145/235}, no 8,9

16. 11(3) cage at R5C7 = {128/137/146/236/245}, no 9

17. 11(3) cage at R6C3 = {128/137/146/236/245}, no 9

18. 21(3) cage in N7 = {489/579/678}, no 1,2,3

19. 10(3) cage in N7 = {127/136/145/235}, no 8,9

[Mike commented "You could have gone one step further (with step 13) and eliminated the 3 from R34C1 due to {79} being unavailable in R2C1." I'm putting this comment here because it's not really part of a preliminary step.]

20. 45 rule on R1 2 outies R2C29 = 9 = {18/36/45}, no 2

21. 45 rule on R9 1 innie R9C9 – 6 = 1 outie R8C1, R8C1 = {123}, R9C9 = {789}

22. 45 rule on C1 1 innie R1C1 = 1 outie R9C2, no 8,9 in R1C1

23. 45 rule on C9 2 outies R18C8 = 10 = {19/28/37/46}, no 5

24. 45 rule on C12 2 outies R18C3 = 11 = [29/38]/{47/56}, no 1, no 8,9 in R1C3

25. 45 rule on R89 2 outies R7C29 = 15 = {69/78}
25a. R7C56 (step 10) = {49/58} (cannot be {67} which clashes with R7C29)
25b. Combined cage R7C2569 = 28 = 89{47/56}

26. Killer pair 8,9 in R7C29 and R7C56, locked for R7
26a. 18(3) cage at R6C8, 8,9 only in R6C8 -> no 1
[Mike added "Similarly, no {89} in R7C78 -> no 2,3,4 in R6C8".]

27. R8C45 (step 11) = {39/57} (cannot be {48} which clashes with R7C56), no 4,8
27a. R8C67 (step 12) = {19/28/46} (cannot be {37} which clashes with R8C45), no 3,7
27b. Combined cage R7C56 + R8C45 = {3589/4579} = 59{38/47}, 5,9 locked for N8, clean-up: no 1 in R8C7

28. 45 rule on C123 1 innie R9C3 – 4 = 2 outies R27C4, min R27C4 = 3 -> min R9C3 = 7, max R27C4 = 5, no 5,6,7
28a. Min R9C3 = 7 -> max R9C45 = 7, no 7,8
[I missed 45 rule on C123 4 outies R279C4 + R9C5 = 10. It's interesting to note the difference between using the two 45s. Innies/outies limits R27C4 to {1234} leaving R9C45 as {1..6}. Outies limits R2C4 and R9C5 to {1234} leaving R79C4 as {1..6}. If both are used then R27C4 and R9C5 are limited to {1234} leaving just R9C4 as {1..6}.]

29. Killer triple 7,8,9 in R9C3 and 21(3) cage, locked for N7, clean-up: no 7 in R1C1 (step 22)

30. 9 in C3 locked in R89C3, locked for N7, clean-up: no 6 in R7C9 (step 25)

31. 21(3) cage in N7 (step 18) = {489/579/678}
31a. 9 of {489/579} must be in R8C3 -> no 4,5 in R8C3, clean-up: no 6,7 in R1C3 (step 24)

32. 45 rule on R789 2 outies R6C38 – 12 = 1 innie R7C1, min R6C38 = 13, no 1,2,3, no 4 in R6C8, max R6C38 = 17 -> max R7C1 = 5
32a. Max R7C1 = 5 -> min R56C1 = 11, no 1

33. 18(3) cage at R6C8 = {279/369/378/567} (cannot be {459} which clashes with R7C56, cannot be {468} which clashes with R7C2569), no 4
33a. R7C78 cannot be {67} (clashes with R7C29) -> no 5 in R6C8
33b. R7C78 cannot be {57} (clashes with R7C2569) -> no 6 in R6C8
33c. R7C78 = {27/36/37/56}

34. Hidden killer quad 6,7,8,9 in R7C2, R7C56, R7C78 and R7C9 -> no 6 in R7C3

35. 45 rule on N3 3 innies R1C7 + R3C79 = 18, max R3C9 = 5 -> min R13C7 = 13, no 1,2,3
35a. 17(3) cage in N3 cannot contain both 8,9 -> R13C7 must contain at least one of 8,9

36. 45 rule on N9 3 outies R6C8 + R89C6 = 21, max R6C8 = 9 -> min R89C6 = 12, no 1,2,3, clean-up: no 8,9 in R8C7
[I could have done step 44 here but didn’t see it until later.]

37. 1,2 in N8 locked in R7C4 and R9C45
37a. R9C45 cannot contain both 1 and 2 (can’t make cage sum) -> R7C4 = {12}
37b. R9C345 = 7{16}/8{24}/9{14}/9{23} (cannot be 7{34} which doesn’t contain 1/2)

38. 45 rule on N5 2 innies R4C4 + R6C6 = 7 = {16/25/34}, no 7,8,9

39. 17(3) cage at R3C4 = {368/458} (only remaining combinations) = 8{36/45}, no 1,2, 8 locked in R3C45 for R3 and N2, clean-up: no 4 in R1C7, no 4 in R4C2, no 5,6, in R6C6

40. 45 rule on N6 3 innies R4C79 + R6C8 – 16 = 1 outie R6C6, min R4C79 + R6C8 = 17, max R4C9 + R6C8 = 14 -> min R4C7 = 3

41. 45 rule on C6789 4 outies R2457C5 = 27 = {3789/4689/5679}, no 1,2, 9 locked for C5, clean-up: no 3 in R8C4

42. 45 rule on R12 2 outies R3C38 = 1 innie R2C1, min R3C38 = 3 -> min R2C1 = 3

43. 23(4) cage in N9 = {1589/1679/2489/2579/2678/3479/3578} (cannot be {3569/4568} which only contain one of 7,8,9) -> R8C89 = {15/16/24/25/26/34/35}, no 7,8,9; clean-up: no 1,2,3 in R1C8 (step 23)

44. R6C8 + R89C6 = 21 (step 36) = [768/786/867] (cannot be [948/984] which clash with R7C56), no 9 in R6C8, no 4 in R89C6, clean-up: no 6 in R8C7
44a. 6 locked in R89C6, locked for C6 and N8

45. 18(3) cage at R6C8 (step 33) = {378/567}, no 2
45a. R7C78 = {37/56}
[Mike added "You could have also noted here that 7 is locked in the 18(3) cage -> no 7 in R9C8" (IOU).]

46. R8C89 (step 43) = {15/16/25/26/34} (cannot be {24} which clashes with R8C7, cannot be {35} which clashes with R7C78 and with R8C45)

47. R9C345 (step 37b) = 8{24}/9{14}/9{23} -> no 7 in R9C3
47a. 7 in N7 locked in 21(3) cage = {579/678}, no 4

48. 10(3) cage in N7 (step 19) = {136/145/235}
48a. 45 rule on N7 3 innies R7C13 + R9C3 = 14 = {149/239/248} (cannot be {158} which clashes with 10(3) cage), no 5
48b. Max R7C34 = 6 -> no 4 in R6C3

49. 21(3) cage in N7 (step 18) = {579/678}
49a. 7 must be in R8C23 (R8C23 = {59} clashes with R8C45, R8C23 = {68} clashes with R8C6) -> no 7 in R7C2

50. 7 in N7 locked in R8C23, locked for R8, clean-up: no 5 in R8C23

51. R8C45 = [93], clean-up: no 4 in R1C4, no 4 in R7C56, no 5 in R8C2 (step 49)

52. R9C3 = 9 (hidden single in C3), R9C45 = 5 = {14} (only remaining combination)
52a. R7C9 = 9 (hidden single in R7)
52b. R1C8 = 9 (hidden single in C8), clean-up: no 3 in R1C6
52b. R9C3 = 9 -> R7C13 = 5 (step 48a) = {14/23}
52c. R1C8 = 9 -> R12C9 = 8 = [26/35/53/71]

53. Naked pair {58} in R7C56, locked for R7 and N8 -> R7C2 = 6, R8C6 = 6, R8C7 = 4, R9C6 = 7, R9C9 = 8, R2C56 = [79], clean-up: no 1 in R56C9

54. Naked pair {14} in R9C45, locked for R9 and N8 -> R7C4 = 2, clean-up: no 5 in R1C5, no 3 in R7C13 (step 52b)
54a. R67C3 = 9 = [54/81]

55. 23(4) cage in N9 R79C9 = [98] -> R8C89 = 6 = [15] -> R8C1 = 2, clean-up: no 3 in R12C9 (step 52c), no 1 in R34C9, no 8 in R45C8

56. Naked pair {24} in R34C9, locked for C9, clean-up: no 7 in R56C9

57. Naked pair {36} in R56C9, locked for C9 and N6 -> R12C9 = [71], R1C67 = [48], R2C4 = 3

58. 10(3) cage in N3 = {235} (only remaining combination), locked for N3 -> R3C7 = 6, R3C8 = 3, R34C9 = [42], R7C78 = [37], R6C8 = 8, R67C3 = [54], R7C1 = 1, clean-up: no 7 in R3C2, no 8,9 in R4C2, no 7 in R5C3

59. Naked pair {58} in R3C45, locked for R3 and N2 -> R34C2 = [93], R3C1 = 7, R9C12 = [35], R9C78 = [26], R2C78 = [52], R2C3 = 6, R3C3 = 1, R1C123 = [523], R2C2 = 8, R2C1 = 4, R8C23 = [78], R4C1 = 8, R45C3 = [72]

60. R3C45 = {58} -> R4C4 = 4

and the rest is naked singles and a cage sum.
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Ed
PostPosted: Tue Jul 08, 2008 2:05 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 72 by Ruud (Oct 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:3328:4353:4353:4353:2820:3077:3077:3077:3592:3328:6154:2315:2315:2820:3854:3854:8208:3592:2578:6154:6154:2315:10006:3854:8208:8208:4378:2578:2578:6154:6154:10006:8208:8208:4378:4378:2852:2852:2852:10006:10006:10006:2602:2602:2602:4909:4909:5935:5935:10006:8242:8242:2612:2612:4909:5935:5935:4665:10006:2619:8242:8242:2612:3647:5935:4665:4665:2883:2619:2619:8242:1351:3647:3401:3401:3401:2883:4173:4173:4173:1351:
Solution: 
+-------+-------+-------+
| 7 2 9 | 6 4 8 | 3 1 5 |
| 6 3 5 | 1 7 2 | 4 8 9 |
| 1 4 8 | 3 5 9 | 2 7 6 |
+-------+-------+-------+
| 4 5 7 | 2 1 6 | 9 3 8 |
| 3 6 2 | 9 8 7 | 1 5 4 |
| 8 9 1 | 4 3 5 | 6 2 7 |
+-------+-------+-------+
| 2 7 3 | 5 6 4 | 8 9 1 |
| 9 8 6 | 7 2 1 | 5 4 3 |
| 5 1 4 | 8 9 3 | 7 6 2 |
+-------+-------+-------+
Quote:
frank: It is nice to claim victory for a change
Andrew: Fairly straightforward but I did need some hidden killers, expressed directly or indirectly so I can't rate it lower than 1.0
Afmob: I think the problem with this assassin was not to solve it but to make the solving path not too long so I tried to keep it as small as possible. Rating: 0.75 - 1.0 but more 1.0 than 0.75
CathyW: Enjoyed this one. I'd agree with the 1.0 rating
gary w: A nice easy one after last week's horror!
Walkthrough by Andrew:
Congratulations Frank for solving it so quickly! :D You finished it before I started.

Fairly straightforward but I did need some hidden killers, expressed directly or indirectly so I can't rate it lower than 1.0.

Here is my walkthrough for A72. There could be quicker routes near the end but it was getting late so I may have missed some.

Edit. I've now checked through my walkthrough (it was too late to do that on Thursday evening, well actually the "wee sma' hours" of Friday morning), tidying up the later steps and correcting an error near the end. Hope I've got it right now. :)

1. R12C1 = {49/58/67}, no 1,2,3

2. R12C5 = {29/38/47/56}, no 1

3. R12C9 = {59/68}

4. R89C1 = {59/68}

5. R89C5 = {{29/38/47/56}, no 1

6. R89C9 = {14/23}

7. 9(3) cage at R2C3 = {126/135/234}, no 7,8,9

8. 10(3) cage at R3C1 = {127/136/145/235}, no 8,9

9. R5C123 = {128/137/146/236/245}, no 9

10. R5C789 = {127/136/145/235}, no 8,9

11. 19(3) cage at R6C1 = {289/379/469/478/568}, no 1

12. 10(3) cage at R6C8 = {127/136/145/235}, no 8,9

13. 10(3) cage at R7C6 = {127/136/145/235}, no 8,9

14. 32(5) cage at R2C8 = {26789/35789/45689} = 89{267/357/456}, no 1

15. 32(5) cage at R6C6 = {26789/35789/45689} = 89{267/357/456}, no 1

16. 39(7) cage at R3C5 = {1356789/2346789} = 36789{15/24}

"It starts with a little present"
17. 45 rule on R5 and C5, R5C5 counts toward both row and column, total 82 -> R5C5 = 8, clean-up: no 3 in R12C5, no 3 in R89C5

"There are in fact two more little presents"
18. 45 rule on C1234 1 innie R5C4 = 9

19. 45 rule on C6789 1 innie R5C6 = 7

20. 9 in C5 locked in R12C5 or R89C5 -> one of these must be {29} -> 2 in C5 locked in R12C5 or R89C5

21. 45 rule on R1234 2 innies R34C5 = 6 = {15} (only remaining combination)
21a. Naked pair {15} in R34C5, locked for C5, clean-up: no 4 in 39(7) cage (step 16), no 6 in R12C5, no 6 in R89C5

22. 45 rule on R1 3 innies R1C159 = 16 = {259/268/457}
22a. 2 of {259} must be in R1C5 -> no 9 in R1C5, clean-up: no 2 in R2C5

23. 45 rule on R1 3 outies R2C159 = 22 = {589/679} = 9{58/67}, no 4, 9 locked for R2, clean-up: no 9 in R1C1, no 7 in R1C5

24. R1C159 (step 22) = {259/268/457}
24a. 4 of {457} must be in R1C5 -> no 4 in R1C1, clean-up: no 9 in R2C1

25. Killer pair 5,6 in R12C1 and R89C1, locked for C1
[I missed the clash between R12C1 and R89C1. However that allowed the interesting step 36.]

26. 1 in C1 locked in R345C1
26a. CPE no 1 in R4C2

27. 45 rule on R9 3 innies R9C159 = {169/178/259/349/367/457} (cannot be {268} because 6,8 only in R9C1, cannot be {358} because no 3,5,8 in R9C5)
27a. 2 of {259} must be in R9C9 -> no 2 in R9C5, clean-up: no 9 in R8C5

28. 9 in N8 locked in R9C56, locked for R9, clean-up: no 5 in R8C1

29. R9C159 (step 27) = {169/178/259/367/457} (cannot be {349} because no 3,4,9 in R9C1)
29a. 4 of {457} must be in R9C9 -> no 4 in R9C5, clean-up: no 7 in R8C5

30. R12C9 contains 8/9
30a. 32(5) cage at R2C8 (step 14) = 89{267/357/456} so must contain 8/9 in N3 -> R4C7 = {89}
30b. Killer pair 8,9 in R12C9 and 32(5) cage, locked for N3

31. Hidden killer pair 8,9 in R12C9 and R4C9 -> R4C9 = {89}

32. Killer pair 8,9 in R4C79, locked for R4 and N6

33. 32(5) cage at R6C6 (step 15) = 89{267/357/456}, 8,9 locked for N9

34. 10(3) cage at R3C1 (step 8) = {127/136/145/235}
34a. No 4 in R3C1 because R4C12 = [15] clashes with R4C5

35. 4 in C1 locked in R4567C1
35a. CPE no 4 in R6C2

36. R12C1 contains 7/8, R89C1 contains 8/9 -> R67C1 cannot contain more than one of 7,8,9 -> max R67C1 = 13 -> min R6C2 = 6
36a. Max R6C2 = 9 -> min R67C1 = 10 -> R67C1 must contain one of 7,8,9
36b. Killer triple 7,8,9 in R12C1, R67C1 and R89C1, locked for C1

37. 10(3) cage at R3C1 (step 8) = {127/136/145/235}
37a. 5,6,7 only in R4C2 -> R4C2 = {567}

38. 45 rule on C9 4 outies R4C8 + R5C78 + R6C8 = 11 = {1235}, locked for N6
38a. 7 in N6 locked in R6C79, locked for R6

39. R5C789 (step 10) = {136/145} (cannot be {127/235} because R5C9 only contains 4,6), no 2, 1 locked for R5 and N6
39a. 2 in N6 locked in R46C8, locked for C8

40. 2 in R5 locked in R5C123, locked for N4
40a. R5C123 (step 9) = {236/245}

41. 7 in R4 only in R4C23
41a. CPE no 7 in R23C2

42. 19(3) cage at R6C1 (step 11) = {289/379/469/478}
42a. 7 of {379} must be in R7C1 -> no 3 in R7C1
42b. 2 of {289} and 7 of {478} must be in R7C1 -> no 8 in R7C1

43. 10(3) cage at R6C8 (step 12) = {127/145} (cannot be {136} because R6C89 = [36] clashes with R6C5, cannot be {235} because R6C9 only contains 4,6,7) -> R7C9 = 1, R6C89 = [27/54], clean-up: no 4 in R89C9

44. Naked pair {23} in R89C9, locked for C9 and N9

45. 10(3) cage at R7C6 (step 13) = {127/136/145/235}
45a. Min R8C7 = 4 -> max R78C6 = 6, no 6
45b. 1 of {145} must be in R8C6
45c. 5 of {235} must be in R8C7
45d. -> no 4,5 in R8C6

46. R9C159 (step 29) = {259/367}, no 8, clean-up: no 6 in R8C1

47. R9C678 = {367/457} (cannot be {169/178/259/268/349/358} because 1,2,3,8,9 only in R9C6), no 1,2,8,9 in R9C6, 7 locked in R9C78 for R9 and N9
47a. 3 of {367} must be in R9C6 -> no 6 in R9C6

48. R9C5 = 9 (naked single), R8C5 = 2, R12C5 = [47], R89C9 = [32], R8C6 = 1, clean-up: no 6 in R1C1
[Could now have fixed R12C19 using step 22. This also applies after step 50.]

49. 1 in R9 locked in R9C23
49a. R9C234 = {148} (only remaining combination), locked for R9, clean-up: no 5 in R9C678 (step 47) -> R9C6 = 3, R67C6 = [36], clean-up: no 6 in R8C7 (step 13)

50. Naked pair {67} in R9C78, locked for R9 and N9 -> R9C1 = 5, R8C1 = 9, clean-up: no 8 in R12C1 -> R12C1 = [76], clean-up: no 8 in R1C9

51. 7 in N8 locked in R78C4, locked for 18(3) cage -> no 7 in R8C3
51a. 18(3) cage at R7C4 = {567} (only remaining combination), no 4,8 -> R8C3 = 6

52. Naked pair {57} in R78C4, locked for C4 and N8 -> R7C6 = 4, R7C1 = 2, R8C7 = 5, R78C4 = [57], R9C4 = 8
52a. Naked pair {14} in R9C23, locked for N7 -> R8C2 = 8, R8C8 = 4

53. R6C1 = 8 (hidden single in C1), R6C2 = 9

54. 4 in C1 locked in R45C1, locked for N4

55. 23(5) cage at R6C3, R7C23 = {37}, R8C2 = 8 -> R6C34 = 5 = [14]

I seem to have deleted too much when editing this walkthrough so the rest is straightforward but not quite down to naked singles
Walkthrough by Afmob:
I think the problem with this assassin was not to solve it but to make the solving path not too long so I tried to keep it as small as possible.

Assassin 72 Walkthrough:

1. Opening the present aka position of cells of 39(7)
a) Innies C1234 = R5C4 = 9
b) Innies C6789 = R5C6 = 7
c) Innies R5 = R5C5 = 8
d) Innies R1234 = 6(2) = {15/24}
e) Innies R6789 = 9(2) = {36/45}
f) Killer pair (45) of Innies R1234 = 6(2) blocks {45} of Innies R6789 = 9(2)
g) Innies R6789 = {36} locked for C5
h) Both 11(2) in C5 = {29/47} locked for C5

2. R1
a) Outies = 22(3) = 9{58/67} -> 9 locked for R2
b) 11(2) = [29/47]

3. C1
a) Killer pair (58) of 14(2) blocks {58} of 13(2)
b) 13(2) = [49/67/76]
c) Killer pair (69) in 13(2) + 14(2) locked

4. C89
a) Outies C9 = 11(4) = {1235} locked for N6
b) 10(3) @ R5 = 1{36/45} since R5C9 = (46) -> 1 locked for R5+N6
c) 2 locked in R46C8 for C8
d) 17(3): R3C9 <> 1,2 because R4C89 <= 14
e) 1,2 locked in R789C9 for N9

5. C1
a) 11(3) = 2{36/45} -> 2 locked for N4

6. R1
a) Outies = 22(3) = {679} locked for R2 because R2C15 <> 5,8
b) 14(2) = [59/86]
c) Innies = 16(3) = {268/457}: R1C1 <> 4 since Innies must have 2 xor 4
and R1C5 = (24)
d) 13(2) = {67} locked for C1+N1

7. C1
a) 14(2) = {59} locked for C1+N7
b) 8 locked in 19(3) = 8{29/47/56} -> no 3; R6C2 <> 8

8. C9
a) 10(3) @ R6: R7C9 <> 5,6,7 because R6C89 >= 6
b) 5 locked in R13C9 for N3

9. R3
a) 7 locked in R3C789 for N3

10. R67
a) 9 locked in R7C78 for N9
b) 9 locked in 32(5) in R7C78 -> R6C7 <> 9
c) 9 locked in R6C23 for N4
d) 10(3) = 1{27/36/45} since R6C9 = (467) -> R7C9 = 1

11. R89
a) 5(2) = {23} locked for C9+N9
b) Innies R9 = 16(3) = 9{25/34}; R9C5 <> 2 because R9C19 <= 12
c) 11(2): [29/74]
d) Hidden Single: R8C1 = 9 @ R8
e) R9C1 = 5
f) Innies R9 = 16(3) = {259} locked for R9
g) R9C9 = 2, R8C9 = 3, R9C5 = 9, R8C5 = 2
i) R1C5 = 4, R2C5 = 7, R2C1 = 6, R1C1 = 7, R2C9 = 9, R1C9 = 5
j) Hidden Single: R4C7 = 9 @ N6, R7C8 = 9 @ N9

12. R89
a) 13(3) = 4{18/36} -> 4 locked for R9
b) 10(3) = 1{36/45} -> R8C6 = 1
c) 16(3) = {367} locked for R9 -> R9C6 = 3; {67} locked for N9
d) R7C5 = 6, R6C5 = 3
e) 10(3) = {145}; R7C6+R8C7 = {45} -> R7C7 <> 4,5
f) R7C7 = 8
g) Hidden Single: R6C1 = 8 @ C1, R4C9 = 8 @ R4
h) Naked pair (45) locked in R8C78 for R8
i) 18(3) = 6[48/57] -> R8C4 = 6

13. C4
a) Hidden Single: R8C4 = 7
b) R8C2 = 8, R7C4 = 5, R7C6 = 4, R8C7 = 5, R8C8 = 4
c) R9C4 = 8, R7C1 = 2, R6C2 = 9

14. R6
a) 32(5) = {45689} -> R6C7 = 6, R6C6 = 5

15. Rest is clean-up and singles.

Rating: 0.75 - 1.0 but more 1.0 than 0.75
Walkthrough by CathyW:
I have 19 steps (excluding the prelims) in this WT before it gets to singles!

Prelims:

a) 13(2) r12c1 = {49/58/67}
b) 11(2) r12c5, r89c5 = {29/38/47/56}
c) 14(2) r12c9, r89c1 = {59/68} -> blocks {58} for 13(2) r12c1
d) 9(3) @ r2c3 = {126/135/234}
e) 32(5) @ r2c8, r6c6 = {26789/35789/45689} must have 8,9
f) 10(3) @ r3c1, r5c7, r6c8, r7c6 = {127/136/145/235}
g) 39(7) @ r3c5 = {1356789/2346789}
h) 11(3) r5c123: no 9
i) 19(3) @ r6c1: no 1
j) 5(2) r89c9 = {14/23}


1. Overlap r5, c5 -> r5c5 = 8 -> r12c5, r89c5 <> 3

2. Innies c1234 -> r5c4 = 9

3. Innies c6789 -> r5c6 = 7

4. Innies r1234 -> r34c5 = 6 -> r34c5 = {15/24}

5. Innies r6789 -> r67c5 = 9 -> r67c5 = {36} – only possible combo.
a) r12c5, r89c5 <> 5
b) 11(2) r12c5, r89c5 = {29/47}
c) r34c5 = {15}

6. Outies c9 -> r456c8 + r5c7 = 11 = {1235} n/e N6
a) 10(3) r5c789 = {13}6/{15}4 -> r5c123, r46c8 <> 1
b) 11(3) r5c123 = {236/245} -> r5c78 <> 2
c) 2 locked r46c8 n/e c8
d) 10(3) @ r6c8: r6c8 = (235), r6c9 = (467) -> r7c9 = 1 (only option)
e) 5(2) r89c9 = {23} n/e N9/c9
f) r6c89 = [27/54] ([36]) blocked by r6c5

7. Outies c1 -> r456c2 + r5c3 = 22
a) max from r4c2 + r5c23 = 5+6+7 = 18 -> r6c2 min 4
b) r4c2 <> 1 (can’t make 21 from other cells)

8. 5 locked r123c9 n/e N3

9. Innies r9: r9c159 = 16 = [592/943/673]
a) r8c1 <> 6
b) r8c5 <> 9

10. Outies r9: r8c159 = 14 = [923/842/572]

11. Innies r1: r1c159 = 16 = [925/628/475/745]
a) r2c5 <> 2
b) r2c9 <> 5,8


12. Outies r1: r2c159 = 22 = {679} – only possible combo, n/e r2
a) r1c1 <> 9
b) r1c5 <> 7
c) split 16(3) r1c159 = [628/745]
d) 13(2) r12c1 = {67} n/e N1/c1
e) 14(2) r89c1 = {59} n/e N7/c1

13. 8 locked r67c1 within 19(3) -> 19(3) = 8{29/47} -> r67c1 <> 3, r6c2 = (79)

14. Split 16(3) r9c159 = [592/943] -> r8c5 <> 4

15. 9 locked r9c56 in N7 n/e r9
a) r9c1 = 5, r8c1 = 9
b) r9c5 = 9, r8c5 = 2
c) r9c9 = 2, r8c9 = 3
d) r1c5 = 4, r2c5 = 7
e) r2c1 = 6, r1c1 = 7
f) r2c9 = 9, r1c9 = 5

16. HS r4c7 = 9, r7c8 = 9

17. 1 locked r6c34 within 23(5) -> r8c2 <> 1

18. 18(3) @ r7c4 <> 1
a) HS r8c6 = 1 -> r7c6 = (345), r8c7 = (456)
b) 1 locked r9c23 within 13(3) -> 13(3) = {148} -> 16(3) r9c678 = 3{67} -> {67} n/e N9
c) r7c5 = 6, r6c5 = 3
d) r7c6, r8c7 = {45} -> pointing cells r7c7, r8c4 <> 4,5 -> r7c7 = 8

19. 18(3) @ r7c4 = {468/567} -> r8c3 = 6
a) NP r8c78 = {45} n/e r8
b) r7c4 = (45) -> NP r7c46: 4,5 n/e r7/N8
c) NS: r9c4 = 8, r8c4 = 7, r8c2 = 8, r7c1 = 2
d) HS: r6c1 = 8
e) r7c23 = {37} -> r6c34 = {14} n/e r6

Singles and simple cage combos from here.

Enjoyed this one. :D
I'd agree with the 1.0 rating. Would be good to see more puzzles which make use of overlaps.
Early Steps by garyw:
A nice easy one after last week's horror!

Early steps after which it is pretty straightforward...


1. r5c5=8 overlap of c5 and r5
2. 45 rule c1234 -> r5c4=9 c6789 -> r5c6=7 r1234-> r34c5=6 r6789->r67c5=9 (must be3/6 as 4/5 blocks the 6/2 bit)..and r34c5={15}
3.O 0n r1 -> r2c159=22 so contains a 9 and also r2c5=7/9
4. O on c9 -> r5c7+ r456c8=11 {1235}
5. Thus r5c9= 4/6
6.So cannot be a 1 in the 11/2 cage N4 (remaining 10/2 then blocked )
7. Thus in r5 1 at c7/8
8. Thus r7c9 is a HS=1
9. r89c9={23}
10.9 cannot be in r9c6 now (blocked by 123 in N9)
11. So 11/2 cage N8 = {29}
12.Thus r2c5=7 (from 3. above) r1c5=4r2c9=9 r1c9=5 and r12c1=76

Mop up now


If there aren't any really fiendish variants I might get a chance to do something else this weekend!

Regards

Gary
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Ed
PostPosted: Tue Jul 08, 2008 2:11 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 72v2 by Para (Oct 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:3328:4353:4353:4353:2820:3077:3077:3077:3592:3328:6154:2315:2315:2820:3854:3854:8208:3592:2578:6154:6154:2315:5910:3854:8208:8208:4378:2578:2578:6154:6154:5910:8208:8208:4378:4378:5156:5156:5156:5156:5910:4393:4393:4393:4393:4909:4909:5935:5935:5910:8242:8242:2612:2612:4909:5935:5935:4665:5910:2619:8242:8242:2612:3647:5935:4665:4665:2883:2619:2619:8242:1351:3647:3401:3401:3401:2883:4173:4173:4173:1351:
Solution: 
+-------+-------+-------+
| 7 2 9 | 6 4 8 | 3 1 5 |
| 6 3 5 | 1 7 2 | 4 8 9 |
| 1 4 8 | 3 5 9 | 2 7 6 |
+-------+-------+-------+
| 4 5 7 | 2 1 6 | 9 3 8 |
| 3 6 2 | 9 8 7 | 1 5 4 |
| 8 9 1 | 4 3 5 | 6 2 7 |
+-------+-------+-------+
| 2 7 3 | 5 6 4 | 8 9 1 |
| 9 8 6 | 7 2 1 | 5 4 3 |
| 5 1 4 | 8 9 3 | 7 6 2 |
+-------+-------+-------+
Quote:
Para, lead-in: it opens up some interesting moves from the the original pattern that weren't needed to solve the puzzle before. It rates about a V1.75 the way i solved it
mhparker: In particular, there's one high-level (but simple and very powerful) move that effectively killed the puzzle off without having to resort to a lot of low-level combination crunching ..correspond to a rating of around 1.5
sudokuEd: Have given up...Wow. What a brute......Took me forever .... - but not for Mike!
sudokuEd: I really enjoyed the JSudoku log.... Really easy to follow and many, many clever things
Andrew: I don't think I'm going to be able to finish...
Andrew in 2011:..I have now... I agree with Mike's rating of 1.5. .... how JSudoku solved this puzzle....Mike gave a good summary at the end, showing that JSudoku's steps weren't particularly simple.
Andrew in 2022: This puzzle was considered too hard to meet the requirements for the Revisits started by Ed in 2020 but I decided to try it again. As with the other WTs the breakthrough was in the same area but I happened to find a different way to crack it.
A well crafted V2 by Para; this puzzle only reaches the difference from the original right at the end!
Walkthrough by mhparker:
Hi folks,

Is anyone else doing Para's "Old School" A72V2?

This is the way I did it, which is probably not the way Para intended it to be done. In particular, there's one high-level (but simple and very powerful) move (step 22) that effectively killed the puzzle off without having to resort to a lot of low-level combination crunching (which I personally prefer to leave to the automated solvers, which are better at it).

Would be interesting to see how others solve it. The way I did it would probably correspond to a rating of around 1.5.

Many thanks to Para for creating this variant. Fortunately, I didn't do the original this time, so it was like a brand new puzzle for me. Maybe this is a small advantage because I didn't have to suffer the negative psychological impact of seeing my V1 solving path brutally shut off? On the other hand, I didn't have the V1 as "training" and no doubt overlooked several moves you guys made in the original. So I guess it's a case of swings and roundabouts.


Assassin 72 V2 Walkthrough

Prelims:

13(2)n1 = {49/58/67}: no 1..3
11(2)n2 and 11(2)n8 = {29/38/47/56}: no 1
14(2)n3 and 14(2)n7 = {59/68}
9(3)n12 = {126/135/234}: no 7..9
32(5)n356 and 32(5)n569 = {(267/357/456)89}: no 1
10(3)n14, 10(3)n69 and 10(3)n89 = {127/136/145/235}: no 8,9
19(3)n47 = {289/379/469/478/568}: no 1
5(2)n9 = {14/23}

1. Innie r5: r5c5 = 8
1a. cleanup: no 3 in r1289c5
1b. 11(2)n2 and 11(2)n8 = {29/47/56}

2. Innies r1234: r34c5 = 6(2) = {15}, locked for c5
(Note: {24} blocked by 2 11(2) cages in c5 (step 1b))
2a. cleanup: no 6 in r1289c5

3. Hidden pair {36} in c5 at r67c5
3a. -> r67c5 = {36}

4. 8 of 32(5)n356 and 32(5)n569 now locked in c78
4a. -> 32(5)n356 and 32(5)n569 form generalized X-Wing on 8 in c78
4b. -> no 8 elsewhere in c78

5. 8 in n9 locked in 32(5)n569
5a. -> no 8 in r6c7

6. 8 in r6 locked in n4 -> not elsewhere in n4

7. {58} combo for 13(2)n1 blocked by 14(2)n7
7a. -> no 5,8 in r12c1

8. Outies r1: r2c159 = 22(3) = {679}, locked for r2
(Note: {589} blocked because {58} only in r2c9)
8a. cleanup: no 9 in r1c1; no 7,9 in r1c5; no 6,9 in r1c9

9. Innies r1: r1c159 = 16(3) = [628/745]
9a. -> no 4 in r1c1
9b. cleanup: no 9 in r2c1

10. Naked pair {67} at r12c1 -> no 6,7 elsewhere in c1 and n1
10a. cleanup: no 8 in r89c1

11. Naked pair {59} at r89c1 -> no 5,9 elsewhere in c1 and n7

12. 12(3)n23 = {129/138/147/345} (no 6)
(Note: {156/237/246} blocked by r1 innies (step 9))

13. 17(3)n12 = {179/269/359/368} (no 4)
(Note: {467} unplaceable, and {278/458} blocked by r1 innies (step 9))
13a. if {179} then 7 must go in r1c4; if {269/368} then 6 must go in r1c4
13b. -> no 1,2,8 in r1c4

14. 8 in c4 locked in n8 -> not elsewhere in n8

15. 8 in r9 locked in 13(3)n78 = {148/238} (no 5,6,7,9) = {(2/4)..}, {(3/4)..}

16. Innies r9: r9c159 = [574/592]
(Note: [943] blocked by 13(3)n78 (step 15))
16a. -> r8c1 = 9
16b. cleanup: no 7 in r8c5; no 2,4 in r8c9

17. 13(3)n78 and r9c9 form killer pair on {24} in r9 -> no 2,4 elsewhere in r9

18. 8 in c1 locked in 19(3)n47 = {289/478} (no 3,5,6)
(Note: {568} unplaceable)
18a. must have 1 of {79}, only available in r6c2
18b. -> r6c2 = {79}

19. 10(3)n14 can only contain 2 of {1..4}, which must go in r34c1
19a. -> no 1..4 in r4c2
19b. [415] permutation blocked by r4c5
19c. -> no 4 in r3c1

20. I/O diff. c1: r46c2 = r5c1 + 11
20a. r5c1 = 1..4 -> r46c2 = 12..15 = [57/59/67/69]
20b. -> no 7 in r4c2

21. I/O diff. c9: r46c8 = r5c9 + 1
21a. -> no 1 in r46c8 (IOU)
21b. max. r46c8 = 10
21c. -> no 9 in r4c8

--- this next move cracks it ---

22. no 1 in r8c9. Here's how.
22a. 1 in r8c9 -> no 1 in r8c3
22b. 1 in r8c9 -> no 1 in r6c9 -> 1 in r6c34 -> no 1 in r7c23+r8c2
22c. 1 in r8c9 -> 4 in r9c9 -> 13(3)n78 = {238} -> no 1 in r9c23
22d. but this leaves nowhere to place the 1 in n7
22e. -> no 1 in r8c9

--- just mop-up now ---

23. r89c9 = [32]
23a. -> r9c5 = 9 (step 16)
23b. -> r8c5 = 2

24. r12c5 = [47]
24a. -> r12c1 = [76]
24b. -> r12c9 = [59]

25. 16(3)n89 = {367} (only remaining combo)
25a. -> r9c6 = 3
25b. r9c78 = {67}, locked for n9

26. r67c5 = [36]

27. Hidden single (HS) in r1 at r1c4 = 6
27a. cleanup: no 1 in r1c23

28. 18(3)n78 = {468/567} (no 1)
28a. -> r8c3 = 6

29. 10(3)n89 = {145} (no 7) (only remaining combo)

30. Innies n8: r78c6+r9c4 = 13(3) = {148} (no 5) (only remaining combo), locked for n8
30a. -> r9c4 = 8

31. r8c7 = 5 (step 29)
31a. -> r78c4 = [57]

32. HS in n9 at r7c9 = 1
32a. -> r6c89 = [27/54]

33. r78c6 = [41]

34. HS in n9 at r8c8 = 4
34a. -> r8c2 = 8
34b. cleanup: no 3 in r1c3

35. Naked single (NS) at r7c1 = 2
35a. -> r6c12 = [89]

36. split 28(4) at r6c67+r7c78 = {5689} (only remaining combo)
36a. -> r6c67 = [56]

Now just naked singles and cage sums to end.
garyw & mhparker on step 22:
Hi Gary,
gary w wrote:
Just a question though,Mike.Step 22 was great but was it a hypothetical?
Thanks for raising this question. Don't know whether this is the answer you're looking for, but here's my response:

Possibly, but arguably no more so than a complex XY-Loop, aka. complex Nice Loop. If there were only 2 places (i.e., cells or cell groups) for the 1 in N7, with 2 inference chains emanating from the starting cell (r8c9), then there would have been a grouped strong link between the 2 cell groups in N7, allowing the logic to be re-expressed as a loop. The loop would have been a discontinuous complex Nice Loop, with two strong links at the discontinuity (r8c9), and would have shown that "if r8c9 is not 3, then it must be 3" (with the conclusion r8c9 = 3).

In this case (step 22 of my A72V2 WT), there are however three places for the 1 in N7, and three chains emanating from r8c9 instead of two, as in the theoretical example above. As such, it is no longer possible to create a simple loop. However, the chains themselves are bona fide bidirectional inference chains, just as in the former case. So it's arguably just an extension of the 2-way (loop) scenario to the 3-way case.

I hope this has answered your question and at least been partially comprehensible! :-)
JSudoku 24 step ALT ending by mhparker:
Hi folks,

Since no-one else has submitted a full WT for the Assassin 72 V2 :( , I decided to compare notes with JSudoku instead, in an attempt to see how much effort my high-level breakthrough move (step 22) saved compared to the traditional low(er)-level approach. The difference was huge, illustrating once again just how powerful some of these complex steps can be. In case some of you also might find the comparison interesting, I've decided to publish my results here.

Here's the grid state after my step 21, which I took as a common starting position:

Code:
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 67        | 123589      123589      35679     | 24        | 12345789    1234579     1234579   | 58        |
|           :-----------.-----------------------:           :-----------------------.-----------:           |
| 67        | 123458    | 12345       12345     | 79        | 123458      12345     | 23458     | 69        |
:-----------:           '-----------.           :-----------:           .-----------'           :-----------:
| 123       | 1234589     1234589   | 123456    | 15        | 123456789 | 23456789    23456789  | 123456789 |
|           '-----------.           '-----------:           :-----------'           .-----------'           |
| 1234        56        | 12345679    12345679  | 15        | 2345679     23456789  | 234567      123456789 |
:-----------------------'-----------------------:           :-----------------------'-----------------------:
| 1234        12345679    12345679    12345679  | 8         | 12345679    12345679    12345679    12345679  |
:-----------------------.-----------------------:           :-----------------------.-----------------------:
| 248         79        | 123456789   12345679  | 36        | 2345679     2345679   | 234567      1234567   |
|           .-----------'           .-----------:           :-----------.           '-----------.           |
| 248       | 1234678     1234678   | 123456789 | 36        | 123457    | 23456789    23456789  | 1234567   |
:-----------:           .-----------'           :-----------:           '-----------.           :-----------:
| 9         | 1234678   | 1234678     12345678  | 24        | 1234567     1234567   | 2345678   | 13        |
|           :-----------'-----------------------:           :-----------------------'-----------:           |
| 5         | 12348       12348       12348     | 79        | 13679       13679       13679     | 24        |
'-----------'-----------------------------------'-----------'-----------------------------------'-----------'


From this position, recall that I used the following single high-level move to crack the puzzle:

mhparker (in WT above) wrote:
22. no 1 in r8c9. Here's how.
22a. 1 in r8c9 -> no 1 in r8c3
22b. 1 in r8c9 -> no 1 in r6c9 -> 1 in r6c34 -> no 1 in r7c23+r8c2
22c. 1 in r8c9 -> 4 in r9c9 -> 13(3)n78 = {238} -> no 1 in r9c23
22d. but this leaves nowhere to place the 1 in n7
22e. -> no 1 in r8c9


From the same position, JSudoku required (with a bit of help from me) the following 24 (!) lower-level moves to crack the puzzle:
JSudoku (after heavy optimization and help from me, esp. with final step) wrote:
22. 18(3)n78 = {189/279/369/378/459/468/567}
22a. 9 only in r7c4
22b. -> no 1,2 in r7c4
22c. r78c4 cannot contain both of {79} due to r9c5
22d. -> no 2 in r8c3

23. 2 in r9 locked in 13(3)n78 or 5(2)n9
23a. if 13(3)n78 contains a 2, it must also contain a 3 ({238})
23b. if 5(2)n9 contains a 2, it must also contain a 3 ([32])
23c. -> 3 locked in r9c234 or r8c9
23d. -> no 3 in r9c78

24. 4 in r9 locked in 13(3)n78 or 5(2)n9
24a. if 13(3)n78 contains a 4, it must also contain a 1 ({148})
24b. if 5(2)n9 contains a 4, it must also contain a 1 ([14])
24c. -> 1 locked in r9c234 or r8c9
24d. -> no 1 in r9c78

25. Naked triple on {679} in r9 at r9c578 -> no 6,7,9 in r9c6

26. 6 in r9 locked in n9 -> not elsewhere in n9

27. Grouped XY-Chain eliminates 3 from r7c4. Here's how.
27a. Either r9c6 = 3, or...
27b. ...r9c6 <> 3 => r9c6 = 1 (strong link, bivalue cell)
27c. -> r9c78 <> {7..} (weak link, combinations 16(3))
27d. => r9c5 = 7 (strong link, r9)
27e. -> r9c5 <> 9 (weak link, r9c5)
27f. => r7c4 = 9 (strong link, n8)
27g. Either way, no 3 in r7c4

28. I/O difference n8: r7c5 + r9c46 = r8c37 + 6
28a. no 1,3 in r9c4. Here's how.
28b. min. r8c37 = 3 -> min. r7c5 + r9c46 = 9
28c. if r9c4 = 1/3, then r7c5 + r9c46 would be forced to {136} = 10
28d. -> r8c37 would have to sum to 4
28e. but this is impossible, since {13} in r8c37 blocked by r8c9
28f. -> no 1,3 in r9c4

29. (from steps 23c and 24c) {13} now locked in r9c23 or r8c9
29a. -> no 1,3 in r8c23

30. no 3 in r7c5. Here's how.
30a. min. r8c37 now 5 ([41])
30b. -> (from step 28) min. r7c5 + r9c46 = 11
30c. -> only possible permutation for r7c5 + r9c46 with a 3 in r7c5 is [381] = 12
30d. -> r8c37 = 6 = [42] (only possible permutation)
30e. but [42] for r8c37 is blocked by r8c5
30f. -> no 3 in r7c5

31. r67c5 = [36]

32. 10(3)n69 = {127/136/145/235} = {(1/3)..}
32a. {13} only available in c9
32b. -> 10(3)n69 and r8c9 form killer pair on {13} -> no 1,3 elsewhere in c9

33. 3 in c9 locked in n9 -> not elsewhere in n9

34. no 2 in r9c4. Here's how.
34a. I/O diff. n8: r9c46 = r8c37
34b. only possible permutation with 2 in r9c4 is r8c37+r9c46 = [4123]
34c. but this is blocked by r8c5
34d. -> no 2 in r9c4

35. CPE: r8c7 sees all 2's in n8 -> no 2 in r8c7

36. 10(3)n89 cannot have both of {27} within n8 due to 11(2)n8
36a. -> no 7 in r78c6
36b. 10(3)n89 = {127/145/235} = {(2/4)..}

37. 10(3)n89 (step 36b) and r8c5 form killer pair on {24}
37a. -> no 2,4 in r8c4

38. r78c4 and r9c5 form hidden killer pair on {79} within n8
38a. -> r78c4 = {(7/9)..}
38b. -> {468} combo blocked for 18(3)n78 = {189/369/378/459/567}
38c. -> possible permuations for 18(3)n78 = [981/963/783/945/567/765]
38d. -> r7c4 = {579}(no 4,8), r8c3 = {468}(no 7), r8c4 = {1357}(no 8)

39. Hidden single in c4 at r9c4 = 8

40. CPE: r8c7 sees all 4's in n8 -> no 4 in r8c7

41. I/O diff. n8: r8c37 = r9c6 + 8 = 9 or 11
41a. -> r8c37+r9c6 = [451/653/811]
(Note: [473] blocked by 13(3)n78)
41b. -> no 7 in r8c7

42. 7 now unavailable for 10(3)n89 (step 36b) = {145/235}
42a. 5 locked -> no 5 in r8c4
42b. -> no 4 in r8c3 (step 38c)

43. 2 in n8 locked in 10(3)n89 or split 14(3) at r8c159 (i.e., r9 outies)
43a. if 10(3)n89 contains a 2, it must also contain a 3 ({23}[5])
43b. if r8c159 contains a 2, it must also contain a 3 ([923])
43c. -> 3 locked in r78c6 or r8c9
43d. -> no 3 in r8c4

44. (from step 38c) 18(3)n78 = [981/567] (only remaining permutations)
44a. -> no 7 in r7c4

45. I/O diff. n8: r78c4 + r9c6 = r8c7 + 10 = 11 or 15
45a. -> r78c4 + r9c6 = [573] (only possible permutation)
45b. -> r8c7 = 5, r8c3 = 6 (18(3) cage sum)


Note that this is after a lot of analysis and simplification by me, with step 45 coming completely from me to shorten the solving path, otherwise JSudoku would have taken even longer to crack it! The fact that JSudoku didn't see step 45 is due to a known deficiency in innie/outie difference handling that I've already mentioned on this forum. Believe it or not, I've cut out the non-essential steps! I also took step 30 a bit further than JSudoku took it (where JSudoku for some reason failed to show that r7c5 cannot be a 3).

Note also that although I used the term "lower-level" for the JSudoku moves above, JSudoku still needed an AIC (step 27). It's also interesting to note that JSudoku made extensive use of Locked Cages moves (steps 23, 24 and 43).

BTW, for those who haven't yet tried interpreting JSudoku logs and converting them into standard walkthrough form, optimizing out the unnecessary steps, be warned that it is a very time-consuming process. The above analysis must have taken me at least 3 hours (!). However, I usually find the results very instructive, as is the case here.

I suppose that (to be fair) I should really do the same with Sudoku Solver. Trouble is, I've now run out of energy!
2011 Walkthrough by Andrew:
Thanks Para for a challenging variant. I tried this puzzle when it was first posted but got stuck and only came back to it this year.

Mike wrote:
In this case (step 22 of my A72V2 WT), there are however three places for the 1 in N7, and three chains emanating from r8c9 instead of two, as in the theoretical example above. As such, it is no longer possible to create a simple loop. However, the chains themselves are bona fide bidirectional inference chains, just as in the former case. So it's arguably just an extension of the 2-way (loop) scenario to the 3-way case.

I hope this has answered your question and at least been partially comprehensible! :-
In 2007 Andrew wrote:
I don't think I'm going to be able to finish...
I have now, but only with a bit of help. As I've commented before my step 29, I accidentally caught sight of a hint so then tried to find another way to continue but ended up finding a different way to do Mike's step. As he said, the chains are bidirectional and I found them from the other ends which, in my opinion, is a slightly cleaner way.

Mike wrote:
This is the way I did it, which is probably not the way Para intended it to be done.
I'm inclined to think the opposite. Para's rating isn't much too higher than Mike's so they may both have used the same, or similar, breakthrough.

Thanks Mike for presenting how JSudoku solved this puzzle, and for giving it help. Nice CPEs using locking cages in steps 23 and 24. Some of the other steps, including 37 and 43 were interesting; others were heavy going with some using contradiction. Mike gave a good summary at the end, showing that JSudoku's steps weren't particularly simple.


Here is my walkthrough for A72 V2.

I tried this puzzle back in 2007, when it first appeared, and managed about 50 steps (that included Prelims, which I’ve now listed first) before getting stuck. Those original steps have been re-written as I would write them if I’d started this puzzle now.

Prelims

a) R12C1 = {49/58/67}, no 1,2,3
b) R12C5 = {29/38/47/56}, no 1
c) R12C9 = {59/68}
d) R89C1 = {59/68}
e) R89C5 = {29/38/47/56}, no 1
f) R89C9 = {14/23}
g) 9(3) cage at R2C3 = {126/135/234}, no 7,8,9
h) 10(3) cage at R3C1 = {127/136/145/235}, no 8,9
i) 19(3) cage at R6C1 = {289/379/469/478/568}, no 1
j) 10(3) cage at R6C8 = {127/136/145/235}, no 8,9
k) 10(3) cage at R7C6 = {127/136/145/235}, no 8,9
l) 32(5) cage at R2C8 = {26789/35789/45689}, no 1
m) 32(5) cage at R6C6 = {26789/35789/45689}, no 1

It still starts with a little present; only one this time instead of the three in A72
1. 45 rule on R5 1 innie = 8, clean-up: no 3 in R12C5, no 3 in R89C5

2. R12C1 = {49/67} (cannot be {58} which clashes with R89C1)
2a. Killer pair 6,9 in R12C1 and R89C1, locked for C1

3. 45 rule on R1234 2 innies R34C5 = 6 = {15/24}

4. 3 in C5 only in R67C5
4a. 45 rule on R6789 2 innies R67C5 = 9 = {36}, locked for C5, clean-up: no 5 in R12C5, no 5 in R89C5

5. R34C5 = {15} (hidden pair in C5)

6. 45 rule on R1 3 outies R2C159 = 22 = {679} (cannot be {589} because 5,8 only in R2C9), locked for R2, clean-up: no 9 in R1C1, no 7,9 in R1C5, no 6,9 in R1C9

7. 45 rule on R1 3 innies R1C159 = 16 = {268/457}
7a. R1C5 = {24} -> no 4 in R1C1, clean-up: no 9 in R2C1

8. Naked pair {67} in R12C1, locked for C1 and N1, clean-up: no 8 in R89C1

9. Naked pair {59} in R89C1, locked for C1 and N7

10. 8 in C1 only in R67C1, locked for 19(3) cage at R6C1, no 8 in R6C2
10a. 19(3) cage at R6C1 = {289/478} (cannot be {568} because 5,6 only in R6C2), no 3,5,6
10b. 7,9 only in R6C2 -> R6C2 = {79}

11. 10(3) cage at R3C1 = {127/136/145/235}
11a. 5,6,7 only in R4C2 -> R4C2 = {567}
11b. 4 in {145} must be in R4C1 (R4C12 cannot be [15] which clashes with R4C5) -> no 4 in R3C1

12. 45 rule on C1 2 outies R46C2 = 1 outie R5C1 + 11
12a. R5C1 = {1234} -> R46C2 = 12,13,14,15 = [57/67/59/69] (cannot be [77]), no 7 in R4C2

13. 45 rule on R9 3 innies R9C159 = 16 = {259/349/457}, no 1, clean-up: no 4 in R8C9
13a. 2 of {259} must be in R9C9 -> no 2 in R9C5, clean-up: no 9 in R8C5

14. 45 rule on R9 3 outies R8C159 = 14 = {149/239/257} (cannot be {347} because R8C1 only contains 5,9)
14a. 10(3) cage at R7C6 = {127/136/145/235}
14b. 7 of {127} must be in R8C67 (R8C67 cannot be {12} which clashes with R8C159) -> no 7 in R7C6

15. 16(3) cage at R9C6 = {169/178/268/367} (cannot be {259/349/358/457} which clash with R9C159), no 4,5

16. 12(3) cage at R1C6 = {129/138/147/345} (cannot be {156/237/246} which clash with R1C159), no 6

17. 17(3) cage at R1C2 = {179/269/359/368} (cannot be {278/458/467} which clash with R1C159), no 4
17b. 6,7 of {179/269/368} must be in R1C4 -> no 1,2,8 in R1C4

18. Max R2C67 = 13 -> min R3C6 = 2

19. 9(3) cage at R2C3 = {126/135/234}
19a. 3 in {135/234} must be in R23C4 (R23C4 cannot be {15/24} which clash with R3C5 and R1C5) -> no 3 in R2C3

20. 10(3) cage at R6C8 = {127/136/145/235}
20a. 6,7 of {127/136} must be in R67C9 (R89C9 cannot be {12/13} which clash with R67C9) -> no 6,7 in R6C8

21. 32(5) cage at R2C8 and 32(5) cage at R6C6 must each contain both of 8,9
21a. Caged X-Wing for 8 in 32(5) cage at R2C8 and 32(5) cage at R6C6 for C78, no other 8 in C78
21b. 8 in N9 only in R7C78 + R8C8, locked for 32(5) cage at R6C6, no 8 in R6C7
21c. 8 in R6 only in R6C13, locked for N4

22. 8 in N2 only in R123C6, locked for C6
22a. 8 in R9 only in 13(3) cage at R9C2 = {148/238}, no 5,6,7,9
22b. Killer pair 2,4 in 13(3) cage and R9C159, locked for R9

23. R9C1 = 5 (hidden single in R9), R8C1 = 9
23a. R8C159 (step 14) = {149/239}, no 7, clean-up: no 4 in R9C5
23b. 1,3 of {149/239} only in R8C9 -> R8C9 = {13}, clean-up: no 3 in R9C9

24. 45 rule on C9 2 outies R46C8 = 1 innie R5C9 + 1, IOU no 1 in R46C8
24a. Min R46C8 = 5 -> min R5C9 = 4
24b. Max R5C9 = 9 -> max R46C8 = 10, no 9 in R4C8

25. 45 rule on C9 5(4+1) outies R4C8 + R5C678 + R6C8 = 18
25a. Min R4C8 + R5C78 + R6C8 = 10 -> no 9 in R5C6

26. 10(3) cage at R7C6 = {127/136/145/235}
26a. 6 of {136} must be in R8C67 (R8C67 cannot be {13} which clashes with R8C9) -> no 6 in R7C6

27. 18(3) cage at R7C4 = {189/279/369/378/459/468/567}
27a. 9 of {189/279} must be in R7C4 -> no 1,2 in R7C4
27b. 2 of {279} must be in R8C4 (R78C4 cannot be [97] which clashes with R9C5) -> no 2 in R8C3

28. 45 rule on C9 3 innies R345C9 = 1 outie R6C8 + 16
28a. Min R6C8 = 2 -> min R345C9 = 18 but cannot be {189} which clashes with R12C9 -> no 1 in R34C9
28b. Consider the candidates in R6C8
R6C8 = 2 => no 2 in R4C9
R6C8 = 3 => R345C9 = 19 but cannot be {289} which clashes with R12C9
R6C8 = 4,5 => R345C9 = 20,21 => no 2 in R4C9
28c. -> no 2 in R4C9

[At this stage I was again stuck for a bit. While looking for Mike’s “marks pic”, which I’d made a note of when I first tried to solve this puzzle, in order to check the eliminations I’d already made, I accidentally caught sight of his breakthrough step. I tried to find other ways to continue from that position but then spotted a different way to make that breakthrough, using a short forcing chain.]

29. Consider placements for 1 in N7
1 in R7C23 + R8C2, locked for 23(5) cage at R6C3 => no 1 in R6C34 => R6C9 = 1 (hidden single in R6) => no 1 in R8C9
1 in R8C3 => no 1 in R8C9
1 in R9C12 => 13(3) cage at R9C2 (step 22) = {148}, locked for R9 => R9C9 = 2 => R8C9 = 3
29a. -> no 1 in R8C9, clean-up: no 4 in R9C9

30. R89C9 = [32], R8C5 = 2 (step 23a), R9C5 = 9, R12C5 = [47], R12C1 = [76], R2C9 = 9, R1C9 = 5, clean-up: no 3 in 13(3) cage at R9C2 (step 22a)
30a. Naked triple {148} in 13(3) cage at R9C2, locked for R9
30b. Naked pair {67} in R9C78, locked for R9 and N9 -> R9C6 = 3, R67C5 = [36]

31. 12(3) cage at R1C6 (step 16) = {129/138}
31a. 8,9 only in R1C6 -> R1C6 = {89}
31b. R1C78 = {12/13}, 1 locked for R1 and N3

32. R1C4 = 6 (hidden single in R1)
32a. 17(3) cage at R1C2 (step 17) = {269/368}

33. 8 in C9 only in R34C9 -> 17(3) cage at R3C9 = {278/368/458}
33a. 2,3,5 only in R4C8 -> R4C8 = {235}

34. 1 in C9 only in R67C9 -> 10(3) cage at R6C8 = {127/145}, no 6
34a. 2,5 only in R6C8 -> R6C8 = {25}

35. 9 in 32(5) cage at R2C8 only in R4C67, locked for R4

36. 9 in N9 only in R7C78, locked for 32(5) cage at R6C6, no 9 in R6C67
36a. 32(5) cage at R6C6 = {45689} (only remaining combination, cannot be {26789} because 2,6,7 only in R6C67), no 2,7, 6 locked for R6

37. 10(3) cage at R7C6 = {145} (only remaining combination), no 7, CPE no 1,4,5 in R8C4

38. 7 in N8 only in R78C4, locked for C4 and 18(3) cage at R7C4, no 7 in R8C3
38a. 18(3) cage = {567} (only remaining combination) -> R8C3 = 6, R78C4 = [57]
38b. Naked pair {14} in R78C6, locked for C6, N8 and 10(3) cage at R7C6 -> R8C7 = 5, R9C4 = 8
38c. Naked pair {14} in R9C23, locked for N7 -> R8C2 = 8, R7C1 = 2

39. R8C8 = 4, R7C9 = 1, R6C7 = 6, R6C6 = 5, R6C8 = 2, R6C9 = 7 (step 34), R5C9 = 4, R34C5 = [51], R9C78 = [76]
39a. R6C2 = 9, R6C4 = 4, R6C3 = 1, R6C1 = 8, R4C4 = 2, R5C4 = 9
39b. R23C4 = {13}, R2C3 = 5 (step 19)

40. R5C1 = 3, R3C1 = 1, R4C1 = 4, R4C2 = 5 (cage sum)
40a. R5C78 = [15] = 6, R5C9 = 4 -> R5C6 = 7 (cage sum)

41. Naked pair {34} in R23C2, locked for C2 and N1 -> R1C2 = 2, R1C3 = 9 (step 32a)

and the rest is naked singles.


Rating Comment. I agree with Mike's rating of 1.5. I used two forcing chains, although the one in step 28b wasn't needed because step 29 makes the same elimination; however I've kept step 28b in because I saw it before I got stuck
2022 Walkthrough by Andrew:
Prelims

a) R12C1 = {49/58/67}, no 1,2,3
b) R12C5 = {29/38/47/56}, no 1
c) R12C9 = {59/68}
d) R89C1 = {59/68}
e) R89C5 = {29/38/47/56}, no 1
f) R89C9 = {14/23}
g) 9(3) cage at R2C3 = {126/135/234}, no 7,8,9
h) 10(3) cage at R3C1 = {127/136/145/235}, no 8,9
i) 19(3) cage at R6C1 = {289/379/469/478/568}, no 1
j) 10(3) cage at R6C8 = {127/136/145/235}, no 8,9
k) 10(3) cage at R7C6 = {127/136/145/235}, no 8,9
l) 32(5) cage at R2C8 = {26789/35789/45689}, no 1
m) 32(5) cage at R6C6 = {26789/35789/45689}, no 1

1a. R12C1 = {49/67} (cannot be {58} which clashes with R89C1), no 5,8
1b. Killer pair 6,9 in R12C1 and R89C1, locked for C1

2a. 45 rule on R5 1 innie R5C5 = 8, clean-up: no 3 in R12C5, no 3 in R89C5
2b. 45 rule on R1234 2 innies R34C5 = 6 = {15/24}
2c. 45 rule on R6789 2 innies R67C5 = 9 contains 3 for C5 = {36}, 6 locked for C5
2d. 1 in C5 only in R34C5 = {15}, 5 locked for C5
2e. Caged X-Wing for 8 in 32(5) cage at R2C8 and 32(5) cage at R6C6, no other 8 in C78
2f. 8 in N9 only in R7C78 + R8C8, locked for 32(5) cage at R6C6, no 8 in R6C7
2g. 8 in N6 only in R4C79, locked for R4

3a. 45 rule on R1 3 outies R2C159 = 22 = {679} (cannot be {589} because 5,8 only in R2C9), locked for R2, clean-up: no 9 in R1C1, no 7,9 in R1C5, no 6,9 in R1C9
3b. 45 rule on R1 3 innies R1C159 = 16 = {268/457} = [628/745], clean-up: no 9 in R2C1
3c. Naked pair {67} in R12C1, locked for C1 and N1, clean-up: no 8 in R89C1
3d. Naked pair {59} in R89C1, locked for N7, 5 locked for C1
3e. 8 in C1 only in R67C1 -> 19(3) cage at R6C1 = {289/478} (cannot be {568} because 5,6 only in R6C2), no 3,5,6
3f. 7,9 only in R6C2 -> R6C2 = {79}
3g. 10(3) cage at R3C1 = {136/145/235} (cannot be {127} which clashes with 19(3) cage), no 7
3h. 5,6 only in R4C2 -> R4C2 = {56}
3i. 4 of {145} must be in R4C1 (cannot be [415] which clashes with R4C5), no 4 in R3C1
3j. Consider combinations for 10(3) cage at R3C1 = {136/145/235}
10(3) cage = {136/145} => caged X-Wing for 1 in R34, no other 1 in R4
or 10(3) cage = {235} = {23}5 => R34C5 = [51]
-> 1 in R4C15, locked for R4
3k. 12(3) cage at R1C6 = {129/138/147/345} (cannot be {156/237/246} which clash with R1C159, no 6
3l. 9 of {129} must be in R1C6 (cannot be 1{29}/2{19} which clash with R1C159 and R12C9 = [59]), no 9 in R1C78
3m. 6 in R1 only in R1C159 = [628] or 17(3) cage at R1C2 = {269/368} (cannot be {467} which clashes with R1C159) -> 17(3) = {179/269/359/368} (cannot be {278/458}, blocking cages), no 4
3n. 6 of {368} must be in R1C4 -> no 8 in R1C4
3o. 8 in N2 only in R123C6, locked for C6
3p. 6,7 of {179/269} must be in R1C4 -> no 1,2 in R1C4
3q. 3 in 9(3) cage at R2C3 must be in R23C4 (R23C4 cannot be {15} which clash with R3C5, cannot be {24} which clash with R1C5), no 3 in R2C3
3r. Max R2C67 = 13 -> min R3C6 = 2

4a. 8 in R9 only in 13(3) cage at R9C2 = {148/238}, no 5,6,7,9
4b. 6 in R9 only in 16(3) cage at R9C6 = {169/367}, no 2,4,5
4c. Killer pair 1,3 in 13(3) cage and 16(3) cage, locked for R9, clean-up: no 2,4 in R8C1
4d. R9C1 = 5 (hidden single in R9) -> R8C1 = 9
4e. 45 rule on R9 2 innies R9C59 = 11 = [74/92], R8C59 = [23/41]
4f. Consider permutations for R9C59
R9C59 = [74] => R8C9 = 1 => 16(3) cage = 1{69} => 10(3) cage at R7C6 = {235}
or R9C59 = [92] => R8C59 = [23] => 16(3) cage = 3{67} => 10(3) cage {145}
-> R9C6 = {13}, R9C78 = {67/69}, 6 locked for N9, 10(3) cage = {145/235}, no 6,7
4g. 6,7 in N7 only in R89C23, CPE no 6,7 in R6C3

5a. 10(3) cage at R6C8 = {127/136/145/235}
5b. 6,7 of {127/136} must be in R67C9 (R67C9 cannot be {12/13}), no 6,7 in R6C8
5c. 45 rule on C9 2 outies R46C8 = 1 innie R5C9 + 1 -> no 1 in R6C8 (IOU)
5d. Min R46C8 = 5 -> min R5C9 = 4
5e. Max R46C8 = 10 -> no 9 in R4C8
5f. Consider combinations for R89C9 = [14/32]
R89C9 = [14] => 10(3) cage = {235}, CPE no 2,3,5 in R45C9
or R89C9 = [32] => 10(3) cage = {127/136/145} (cannot be {235} because then 2,3 only in R6C8), 1 locked for C9
-> 1 in R678C9, locked for C9, no 2,3 in R4C9

6a. 10(3) cage at R7C6 (step 4f) = {145/235}, R8C59 = [23/41]
6b. Combined cage 10(3) cage + R8C59 = {145}[23]/{235}[41], CPE no 1,2,3,4,5 in R8C4
6c. 18(3) cage at R7C4 = {369/378/468/567} (cannot be {189} = 9{18} which clashes with R8C59 + R9C5 = [239/417], cannot be {279} = [927] which clashes with R9C5, cannot be {459} because 5,9 only in R7C4), no 1,2
6d. Consider permutations for R8C59 = [23/41]
R8C59 = [23] => 18(3) cage = {378/468/567} (cannot be {369} = [936])
or R8C59 = [41] => 10(3) cage = {235} with 3 in R78C6 => R7C5 = 6 => 18(3) cage = {378/468/567} (cannot be {369} = [936])
or R8C59 = [41] => 10(3) cage = {235} with 3 in R8C7 => 18(3) cage = {378/468/567} (cannot be {369} = [936])
-> 18(3) cage = {378/468/567}, no 9
6e. R9C5 = 9 (hidden single in N8) -> R8C5 = 2, R89C9 = [32], R9C78 = {67}, 7 locked for N9, R9C6 = 3 (cage sum), R67C5 = [36], R12C5 = [47], R12C1 = [76] -> R12C9 = [59]
6f. R1C4 = 6 (hidden single in R1) -> R1C23 = 11 = {29/38}
6k. 12(3) cage at R1C6 = {129/138} -> R1C78 = {12/13}, 1 locked for N3, R1C6 = {89}
6l. 9 in N2 only in R13C6, locked for C6
6m. 32(5) cage at R2C8 = {26789/35789/45689} -> R4C7 = 9
6n. R7C8 = 9 (hidden single in N9)
6o. 32(5) cage at R6C6 = {45689} (only remaining combination, cannot be {26789 because 2,6,7 only in R6C67), no 2,7, 6 locked for R6
6p. 18(3) cage = {468/567} -> R8C3 = 6, R7C4 = {45}
6q. Naked triple {145} in R7C4 + R78C6, locked for N8, 1 locked for C6 and 10(3) cage
6r. R9C4 = 8, R8C4 = 7 -> R7C4 = 5
6s. Naked pair {14} in R78C6, 4 locked for C6 and 10(3) cage -> R8C7 = 5
6t. R7C9 = 1 (hidden single in N9) -> R78C6 = [41], R7C7 = 8, R8C8 = 4, R6C67 = [56], R8C2 = 8, R34C5 = [51], R6C8 = 2 -> R6C9 = 7 (cage sum), R5C9 = 4

7a. R23C9 = [68] -> R4C8 = 3 (cage sum)
7b. R5C789 = [154] -> R5C6 = 7 (cage sum)
7c. R23C8 = [87], R4C7 = 9 -> R3C7 + R4C6 = 8 = [26]
7d. R1C78 = [31] -> R1C6 = 8 (cage sum)

and the rest is naked singles.
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Ed
PostPosted: Tue Jul 08, 2008 2:15 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 73 by Ruud (Oct 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:2560:2560:4610:4610:2564:4357:4357:3335:3335:2560:5386:5386:4610:2564:4357:4623:4623:3335:3602:5386:5652:5652:2564:3351:3351:4623:4122:3602:3602:5652:5150:5151:5151:3351:4122:4122:2340:2340:5150:5150:5150:5151:5151:1067:1067:3117:3117:1583:7472:7472:7472:3891:4916:4916:3117:3639:1583:1583:7472:3891:3891:3645:4916:4927:3639:3639:3650:2371:4164:3645:3645:3399:4927:4927:3650:3650:2371:4164:4164:3399:3399:
Solution: 
+-------+-------+-------+
| 5 3 6 | 9 4 8 | 2 1 7 |
| 2 9 4 | 3 1 7 | 6 8 5 |
| 1 8 7 | 6 5 2 | 3 4 9 |
+-------+-------+-------+
| 7 6 9 | 4 3 1 | 8 5 2 |
| 4 5 8 | 2 6 9 | 7 3 1 |
| 3 1 2 | 7 8 5 | 4 9 6 |
+-------+-------+-------+
| 8 2 3 | 1 9 6 | 5 7 4 |
| 9 7 5 | 8 2 4 | 1 6 3 |
| 6 4 1 | 5 7 3 | 9 2 8 |
+-------+-------+-------+
Quote:
Andrew: Thank goodness for another fairly easy one. I liked the way that one theme kept repeating itself ... I'll rate this as 1.0
gary w: pretty straightforward one and no combo crunching required
Afmob: Although this Assassin was "just" of rating 1.0 it kept me busy
Walkthrough by Andrew:
Thank goodness for another fairly easy one. I needed a break from A72V2 where I seem to have ground to a halt and](*,) although I haven't yet given up.

I liked the way that one theme kept repeating itself in the way that I solved it.

Probably about the same level as A72 so I'll rate this as 1.0.

Here is my walkthrough for A73, posted in the "wee sma' hours" again so I'll check it as soon as I can and tidy it up then; I know that I've missed so clean-ups in the later stages so may be able to make it a bit shorter. I look forward to the clocks going back so that will give Frank and I an extra hour to try to solve Assassins on Thursdays. :wink:


1. R5C89 = {13}, locked for R5 and N6

2. R5C12 = {27/45}

3. R89C5 = {18/27/36/45}, no 9

4. 10(3) cage in N1 = {127/136/145/235}, no 8,9

5. 21(3) cage in N1 = {489/579/678}, no 1,2,3

6. R123C5 = {127/136/145/235}, no 8,9

7. 22(3) cage at R3C3 = 9{58/67}

8. 6(3) cage at R6C3 = {123}

9. 19(3) cage at R6C8 = {289/379/469/478/568}, no 1

10. 19(3) cage in N7 = {289/379/469/47/568}, no 1

11. 29(4) cage at R6C4 = {5789}
11a. CPE no 5,7,8,9 in R45C5

12. 45 rule on R5 3 outies R4C456 = 8 = 1{25/34}, 1 locked for R4

13. 45 rule on R89 2 outies R7C28 = 9 = {18/27/36/45}, no 9

14. 45 rule on C12 2 outies R28C3 = 9 = [45/54/63/72/81]

15. 45 rule on C89 2 outies R28C7 = 7 = {16/25/34}, no 7,8,9

16. 45 rule on C6789 1 innie R6C9 – 2 = 1 outie R4C5 -> R4C5 = 3, R6C9 = 5, clean-up: no 2 in R4C46 (step 12), no 6 in R89C5

17. Naked pair {14} in R4C46, locked for R4 and N5

18. R123C5 = {127/145} = 1{27/45}, no 6, 1 locked for C5 and N2, clean-up: no 8 in R89C5

19. R5C5 = 6 (hidden single in C5)

20. R67C5 = {89} (hidden pair in C5) -> R6C4 = 7

21. 20(4) cage at R4C5 = {1379/3458} (cannot be {2378} because R4C6 only contains 1,4), no 2
21a. 7 of {1379} must be in R5C7 -> no 9 in R5C7
21b. 4 of {3458} must be in R4C6 -> no 4 in R5C7
21c. 5 of {3458} must be in R5C7 -> no 8 in R5C7

22. R5C4 = 2 (hidden single in N5), clean-up: no 7 in R5C12

23. Naked pair {45} in R5C12, locked for R5 and N4 -> R5C7 = 7 -> R45C6 = [19] (step 21), R4C4 = 4, R5C3 = 8, R67C5 = [89] , clean-up: no 5 in R3C4 (step 7), no 1 in R8C3 (step 14)

24. 2 in 6(3) cage at R6C3 locked in R67C3, locked for C3, clean-up: no 7 in R2C3 (step 14)

25. 21(3) cage in N1 = {489/579/678}
25a. R2C3 only contains 4,5,6 -> no 4,5,6 in R23C2

26. 45 rule on N1 3 innies R1C3 + R3C13 = 14, min R3C3 = 5 -> max R1C3 + R3C1 = 9, no 9

27. 14(3) cage at R3C1 = {167/239/257} (cannot be {149/158/248/347/356} because R4C12 need two of 2,6,7,9), no 4,8
27a. 1,3,5 only in R3C1 -> R3C1 = {135}
27b. R4C12 = {27/29/67}

28. 2 in N1 locked in 10(3) cage = 2{17/35}, no 4,6

29. R1C3 + R3C13 (step 26) = {149/167/356} (cannot be {347} which clashes with 10(3) cage)
29a. 1 of {149/167} must be in R3C1 -> no 1 in R1C3

30. 8 in N1 locked in 21(3) cage = {489/678}, no 5, clean-up: no 4 in R8C3 (step 14)
30a. 8 locked in R23C2, locked for C2, clean-up: no 1 in R7C8 (step 13)
30b. 4 of N1 locked in R12C3, locked for C3

31. 14(3) cage at R8C4 = {158/167/356}, no 9

32. 9 in C3 locked in R34C3 -> no 9 in R3C4, clean-up: no 6 in R34C3 (step 7)

33. 9 in C4 locked in R12C4, 18(3) cage at R1C3 = 9{36/45}, no 7,8
33a. 4 of {459} must be in R1C3 -> no 5 in R1C3

34. R1C3 + R3C13 (step 29) = {149/167/356}
34a. 5 of {356} must be in R3C3 -> no 5 in R3C1
34b. 3 of {356} must be in R3C1 -> no 3 in R1C3

35. 18(3) cage at R1C3 = 9{36/45} (step 33)
35a. 6 of {369} must be in R1C3 -> no 6 in R12C4

36. Hidden killer pair 6,8 in R3C4 and R89C4 -> R89C4 must contain 6/8

37. 14(3) cage at R8C4 (step 31) = {158/167/356}
37a. R89C4 contains 6/8 (step 36) = {16/18/36/56/58}
37b. -> R9C3 = {1357}, no 6
[Alternatively naked pair {46} in R12C3, locked for C3.]

38. 19(3) cage at R6C8 = 19(3) cage at R6C8 = {289/469} (cannot be {379/478/568} because R6C89 needs two of 2,4,6,9) = 9{28/46}, no 3,5,7, 9 locked in R6C89 for R6 and N6
38a. 8 of {289} must be in R7C9 -> no 2 in R7C9

39. 16(3) cage at R3C9 = {259/268/358} (cannot be {169/178/349/367/457} because R4C89 needs two of 2,5,6,8}, no 1,4,7
39a. 9 of {259} and 3 of {358}must be in R3C9 -> no 5 in R3C9
39b. R4C89 cannot be {26} which clashes with R4C12 (step 27b) -> no 8 in R3C9

40. 12(3) cage at R6C1 = {138/156/237} {cannot be {147/345} because R6C12 needs two of 1,2,3,6, cannot be {246} because R6C12 = {26} clashes with R6C789), no 4
40a. 5,7,8 only in R7C1 -> R7C1 = {578}

41. 45 rule on N7 3 innies R7C13 + R9C3 = 12 = {138/237} (only remaining combinations) = 3{18/27}, no 5, 3 locked for N7, clean-up: no 6 in R7C8 (step 13)
41a. 7 of {237} must be in R7C1 -> no 7 in R9C3
41b. 3 locked in R79C3, locked for C3

42. R8C3 = 5 (naked single), R2C3 = 4 (step 14), R1C3 = 6, clean-up: no 2 in R2C7 (step 15), no 4 in R7C8 (step 13), no 3 in R8C7 (step 15), no 4 in R9C5, no 8 in R3C4 (step 7) -> R3C4 = 6
42a. R78C2 = 9 = {27} (only remaining combination), clean-up: R7C8 = {27} (step 13)
42b. Naked pair {27} in R78C2, locked for C2 and N7
42c. Naked pair {27} in R7C28, locked for R7

43. R7C1 = 8 (naked single) -> R6C12 = {13} -> R6C3 = 2

44. Naked pair {89} in R23C2, locked for C2 and N1 -> R4C2 = 6, R3C3 = 7, R4C3 = 9, R4C1 = 7, R3C1 = 1, R6C12 = [31], R9C2 = 4, R5C12 = [45], R1C2 = 3

45. R1C3 = 6 -> R12C4 = 12 = [93], R7C34 = [31], R9C3 = 1, R89C4 = [85], clean-up: no 4 in R8C5, no 4 in R8C7 (step 15)

46. Naked pair {27} in R89C5, locked for C5 and N8
46a. Naked triple {346} in R789C6, locked for C6
46b. Naked pair {27} in R8C25, locked for R8, clean-up: no 5 in R2C7 (step 15)

47. 7 in N2 locked in R12C6 -> 17(3) cage = {278} (only remaining permutation), no 1,4,5 in R1C7

48. Naked pair {16} in R28C7, locked for C7 -> R6C7 = 4, R7C67 = [65], R7C9 = 4, R89C6 = [43], R9C7 = 9, R89C1 = [96]

49. R3C7 = 3 (hidden single in C7)

50. 16(3) cage at R3C9 (step 39) = {259} (only remaining combination) -> R3C9 = 9, R23C2 = [98] , R3C6 = 2, R4C7 = 8, R1C7 = 2, R12C1 = [52], R6C89 = [96]

51. Naked pair {13} in R58C9, locked for C9

52. 14(3) cage at R7C8 = {167} (only remaining combination) -> R7C8 = 7, R78C2 = [27], R89C5 = [27]
51a. Naked pair {16} in R8C78 -> R8C9 = 3, R5C89 = [31]

52. 7 in N3 locked in R12C9, 13(3) cage = 7{15} -> R1C89 = [17], R2C9 = 5

and the rest is naked singles
Outline by gary w:
Yes,a pretty straightforward one and no combo crunching required.Here's an outline of how I solved it.



1. I-O c1-4 -> r2c4=r5c5+1
2. I-O c6-9 -> r6c6=r4c5+2
3. r4c456=8={134}

Together with 29(4) cage N5 can easily complete N5 as

431
269
785

4. Also r5c3=8 and r7c5=9 r5c12={45}
5. r9c3<>9
6. I N7=12
7. I c12-> r2378=26 <> 1... and r28c3=9 <> {18}
8. Thus 1 in N7 is at r7c13,r9c3.
9. But cannot be 1 at r7c1 as both options {129} and {138} blocked.
The clincher... both remaining options for the 1 in N7 -> R7C1=8.
as the [129} option is blocked and 1 at r7c3->2 at r6c3 and {13} r6c12
10. r7c1=8 -> r6c12={13} and r6c3=2
11. r3c1 now must=1 no other option so 10(3) cage N1={235} -> r4c12={67}
12.8 In n1 is at r23c2
13. The second clincher... the 58 pairs in both N1 and N4 (and not at r34c3)
-> the 22(3) cage N124 <> 9{58} therefore = {967}
14. From 11. above -> r4c3=9 r3c4=6 r3c3=7 r1c3=6

Now just a mop up.

Another very nice killer..thanks Ruud

Regards

Gary
Walkthrough by Afmob:
Although this Assassin was "just" of rating 1.0 it kept me busy all day long but it was more my fault since I messed up my walkthrough couple a times so that I hit a dead end.
I used pretty much the same trick as Cathy though without the outies of N5. And Ruud was right about the length of the walkthrough. After the first steps I thought the Assassin was finished but guess I was wrong or maybe my mop-up is not the shortest way.

Assassin 73 Walkthrough:

1. C1234 !
a) Outies C1 = 19(5) = 1{2349/2358/2367/2457/3456} -> 1 locked for C2
b) Outies C12 = 9(2) = [45/54/63/72/81]
c) 29(4) = {5789} -> locked for (N5+C2) -> R45C5 <> 5,7,8,9
d) ! Outies C1234 = 28(1+3) -> 3 of those 4 cells are in 29(4) = {5789}
-> 28(1+3) = R5C5 + R67C5+R6C6 = 28(4) = 89{47/56} (because every cell is related to the other 3 cells)
-> R5C5 = (46)
-> R6C4 <> 8,9 since it must be in R67C5+R6C6

2. C5
a) Innies = 26(4) = {3689/4589/4679/5678} because R5C5 = (46)
-> 26(4) <> {4589} since R45C5 <> 5,8,9
-> 6 locked in Innies = 6{389/479/578} for C5+N5
b) 9(2) <> 3
c) 10(3) = {127/145/235} -> Killer pair (57) blocks {5678} of Innies -> R67C5 <> 5
d) Innies = 26(4) = 69{38/47} -> 9 locked for C5 @ 29(4) -> R6C6 <> 9

3. C6789 !
a) Outies C89 = 7(2) = {16/25/34}
b) ! Outies C6789 = 27(1+3) -> 3 of those cells are in 29(4) = {5789}; R4C5 = (346)
-> 27(1+3) = R4C5 + R67C5+R6C6 = 27(4) = {3789/4689/5679} (every cell relates to the other 3 cells)
-> {4689} impossible because 29(4) <> 4,6 -> 27(4) = 79{38/56}
-> R4C5 = (36)
-> 7 locked @ 29(4) in Outies C6789 -> R6C6 <> 7

4. R456
a) 4(2) = {13} locked for R5+N6
b) 9(2) = {27/45}
c) 5 locked in R6C46 @ 29(4) for R6+N5
d) Outies R5 = 8(3) = {134} since R4C5 = (36)
-> {134} locked for R4+N5
-> R4C5 = 3, R5C5 = 6

5. C6789
a) Outies = 27(4) = {3789} -> R6C4 = 7, R6C6 = 5
b) 20(4) = 3{179/278/458} -> 3{278} impossible because R4C6 = (14)
-> 20(4) = 3{179/458}
-> R5C6 = (57) since it's the only position where they are possible
c) Hidden Single: R5C4 = 2 @ N5

6. N4
a) 20(4) N4 = {2468} -> R4C4 = 4, R5C3 = 8
b) R4C6 = 1, R5C6 = 9, R6C5 = 8, R7C5 = 9
c) 9(2) = {45} locked for R5+N4
d) R5C7 = 7
e) 6(3) = {123} -> 2 locked for C3
f) 14(3): R4C12 = {26/27/29/67} -> R3C1 = (1356)
-> R4C12 = {26} impossible because R3C1 would be 6
-> R3C1 <> 6

7. C245
a) 8 locked in R23C2 for C2 @ 21(3) = 8{49/67}
b) 9(2) @ C5 = {27/45}

8. C123
a) Outies = 9(2) = [45/63]
b) 21(3) must have 4 xor 6 and R2C3 = (46) -> R23C2 <> 4,6
c) 2 locked in 10(3) = 2{17/35}
d) 4,6 locked in R123C3 for C3

9. N6
a) 19(3): R6C89 = {29/46/49/69} -> R7C9 = (469) but R7C9 <> 9 since R7C5 = 9
-> 19(3) = {469}
-> 9 locked for R6+N6; 6 locked for R45C6

10. N47
a) 12(3): R6C12 = {12/13/16/23/26/36}
-> {12} impossible because R7C1 <> 9
-> {36} impossible because R7C1 would be 3
-> R7C1 = (4578)

11. R789
a) Outies R89 = 9(2) = {27/36/45}
b) 1 locked R79C3 for C3
c) 1 locked in 6(3) for R7

12. N4
a) 1 locked in 12(3) = 1{38/56} -> R7C1 = (58)
b) 19(3) <> 5 since {568} blocked by R7C1 = (58)

13. N7 !
a) ! Innies = 12(3) = {138} since they have no 4,6 and R7C1 = (58)
-> R7C1 = 8
-> R79C3 = {13} locked for C3+N7
b) R8C3 = 5, R6C3 = 2
c) 14(3) @ R7C2 = {257} -> {27} locked for C2+N7
d) 14(3) @ R8C4 = 5{18/36} -> R9C4 = 5
-> 14(3) must have 1 xor 3 and R9C3 = (13) -> R8C4 <> 1,3

14. N8
a) Hidden Single: R7C4 = 1
b) R7C3 = 3, R9C3 = 1
c) 14(3) = {158} -> R8C4 = 8

15. N1
a) 21(3) = {489} -> R2C3 = 4
b) 14(3) = {167} since R4C12 <> 3,5 -> R3C1 = 1, R4C1 = 7, R4C2 = 6
c) R6C1 = 3, R6C2 = 1, R4C3 = 9, R3C4 = 6, R3C3 = 7, R1C3 = 6
d) Hidden Single: R1C2 = 3
e) R1C4 = 9, R2C4 = 3
f) Hidden Single: R5C2 = 5 @ C2 -> R5C1 = 4, R9C2 = 4 @ C2

16. R789
a) 9(2) = {27} locked for C5+N8
b) 15(3) = {456} -> R7C7 = 5
c) 16(3) = {349} because R9C = (36) -> R9C7 = 9, R8C6 = 4, R9C6 = 3
d) R9C1 = 6, R8C1 = 9, R7C6 = 6, R6C7 = 4, R7C9 = 4
e) 14(3) @ R7C8 = {167} locked for N9 -> R7C8 = 7
f) R7C2 = 2, R8C2 = 7, R8C5 = 2, R9C5 = 7, R8C9 = 3, R5C9 = 1, R5C8 = 3

17. N3
a) Hidden Single: R3C7 = 3
b) 16(3) = {259} -> R3C9 = 9
c) 18(3) = {468} locked

18. Rest is clean-up and singles.
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