SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Thu Mar 28, 2024 8:25 pm

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
 Post subject: Assassin 382
PostPosted: Thu Aug 15, 2019 8:27 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Attachment:
a383.JPG
a383.JPG [ 59.02 KiB | Viewed 6504 times ]
Assassin 382

A very pleasent puzzle I thought. Nothing too strenuous with a neat way to finally crack it. SudokuSolver gives it 1.75 and JSudoku uses a chain.
triple click code:
3x3::k:4352:4352:9217:9217:4098:4098:4098:3331:3332:4352:2053:9217:9217:9217:9217:4098:3331:3332:6918:2053:3335:3335:9217:4872:4872:1289:1289:6918:5130:5130:3335:2571:2571:4872:1292:1292:6918:5130:3341:3341:3341:2574:2574:2574:4623:6918:0000:0000:0000:0000:0000:7186:7186:4623:6918:0000:0000:0000:0000:3856:7186:7186:4623:2577:0000:0000:0000:3856:3856:3856:7186:4371:2577:0000:0000:0000:4372:4372:4372:4371:4371:
solution:
Code:
+-------+-------+-------+
| 6 9 4 | 2 3 8 | 1 5 7 |
| 2 1 3 | 5 7 9 | 4 8 6 |
| 8 7 5 | 1 6 4 | 9 3 2 |
+-------+-------+-------+
| 3 5 9 | 7 8 2 | 6 1 4 |
| 7 6 8 | 4 1 3 | 5 2 9 |
| 4 2 1 | 6 9 5 | 3 7 8 |
+-------+-------+-------+
| 5 3 6 | 9 2 7 | 8 4 1 |
| 9 4 7 | 8 5 1 | 2 6 3 |
| 1 8 2 | 3 4 6 | 7 9 5 |
+-------+-------+-------+
Cheers
Ed


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 382
PostPosted: Thu Aug 22, 2019 1:27 am 
Offline
Grand Master
Grand Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Very pleasant indeed! Thanks Ed.
I can foresee a wide variety of solving paths on this one.
Here's mine...
Corrections & clarifications (first set) thanks to Ed.
Assassin 382 WT:
1. Outies c1 -> r1c2 = 9

2. Outies r12 -> r3c25 = +13(2) = [58], [67], or [76]
Whatever value goes in r3c5 goes in n1 in r12c1 -> cannot be 8.
-> r3c25 = {67}
-> 8(2)n1 = [17] or [26] and r12c1 = {26} or {17}
-> (12679) locked for n1

3. 9 must be in 36(7)r1c3 in n2 in r2c456
-> HS 9 in n3/r3 -> r3c7 = 9
-> r12c7 = +5(2), r1c56 = +11(2)
Also the two 13(2)s in n3 are {58} and {67}

4! (1267) in n1 prevents 5 in r34c4
-> 5 in r3 only in r3c13
Innies r1234 = r3c1 + r4c123 = +25(4)
20(3)n4 prevents r4c123 = +20(3)
-> 5 not in r3c1
-> HS 5 in r3 -> r3c3 = 5
-> 13(3)r3c3 = [517] or [526]

5. Whichever of (34) is in 5(2)r3c8 can only go in n1 in r12c3
-> can only go in n2 in r1c56
-> goes in n1 in r2c3
-> Either
(A) 16(4)r1c5 = [{38}{14}] or
(B) 16(4)r1c5 = [{47}{23}]
If the former -> r1c3,r3c1 = [48]
If the latter -> r1c3,r3c1 = {38}

6! r1c3 from (348) and r3c4 from (12) -> whichever of (348) is in r1c3 goes in n2 in r3c6
-> r1c3 + r4c7 = +10(2)
For Step 5(A) -> r1c3,r4c7 = [46]
For Step 5(B) -> r1c3,r4c7 cannot be [82] (since 2 in r12c7) must be [37]
-> r12c3 = {34} and r3c1 = 8
-> 19(3)r3c6 = [496] or [397]

7. 5 in r4 only in r4c12
Innies r5 -> r5c129 = +22(3)
Remaining innies r1234 = r4c123 = +17(3)
5 cannot be in r4c1 since that puts 20(3)n4 = [398] which leaves no solution for innies r5 = +22(3)
-> r4c2 = 5
-> r7c1 = 5

8. r4c123 only from [458] or [359]
But the former puts r5c2 = 7 and r12c1 = {17} which leaves no solution for 10(2)c1
-> r4c123 = [359]
-> r5c129 = [769]
-> 8(2)n1 = [17]
-> r12c1 = {26}
-> r3c5 = 6
Also HS 1 in r4 -> 5(2)r4 = {14} and 10(2)r4 = {28}
-> HS 8 in r5 -> 13(3)r5 = [8{14}] and 10(3)r5 = {235} with 2 in r5c78
Also 27(5)c1 = [83745] and 10(2)c1 = {19}

9! Consider Step 5(B) 16(4)r1c5 = [{47}{23}]
This puts r1c3 = 3 -> r1c7 = 2
and one of the 13(2)s in n3 = [67]
which prevents r12c1 from being {26}

-> 16(4)r1c5 = [{38}14]

10. Continuing
-> 5(2)r3c8 = {23}
Also -> r12c3 = [43]
-> 19(3)r3c6 = [496]
Also -> 13(3)r3c3 = [517]
-> 7 in n2 in r2c56
-> One of 13(2) in n3 = [76]
-> r12c1 = [62]
-> r1c4 = 2
-> r2c456 = {579}
-> One of 13(2) in n3 = [58]

11. Outies c9 = r349c8 = +13(3)
Possibilities are [319], [346], [247]
-> r9c8 from (679)
-> If 13(2)r1c8 is {67} -> r349c8 = [319]

12. Innies c89 = r5678c8 = +19(4)
Outies c89 = r5c6 + r567c7 = +19(4)
Given r34c7 = [96] -> r67c7 cannot be +14(2)
-> r5c67 cannot be +5(2)
-> 5 in r5c67

13! Consider r7c7.
That value cannot be from (1496) already in c7, or 5 already in r7.
The value cannot go in r5c8 since that would put (outies c89) r6c7 = 9
-> Since 2 in r5c78 -> r7c7 cannot be 2
-> r7c7 from (378) and that same value goes in n6 in r6c9
-> r67c9 from [36], [72], or [81]
The first two -> 13(2)r1c8 = [76] -> r349c8 = [319]
The last one -> 5(2)n6 = [14] -> r349c8 = [319]
In all cases r349c8 = [319]
-> r34c9 = [24]

14. -> NS r5c8 = 2
-> r678c8 = +17(3)
-> r67c7 = +11(2) can only be {38}
-> 10(3)r5 = [352]
Also NS 8 in c8 -> 13(2)r1c8 = [58]
-> r678c8 = [7{46}]
-> 13(2)r1c9 = [76]
-> 18(3)c9 = [981]
-> r67c7 = [38]
-> r89c9 = {35}
-> r89c7 = {27}

15. Finishing
r5c45 = [41]
r6c23 = [21]
10(2)c1 = [91]
r1c56 = [38]
->10(2)n5 = [82]
-> (38) in c4 in n8
-> 17(3)r9 can only be [467]
Also 15(4)n8 = [7512]
-> r89c9 = [35]
Also r7c5 = 2 and r789c4 = [983]
-> r789c3 = [672], r789c2 = [348], r78c8 = [46]
and r2c456 = [579]


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 382
PostPosted: Sun Aug 25, 2019 7:16 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Prediction true wellbeback! We worked in the same areas but very differently. Hopefully my final crucial step (27) is valid. [It is! Thanks Andrew for checking it for me]

WT:
Preliminaries by SudokuSolver
Cage 5(2) n3 - cells only uses 1234
Cage 5(2) n6 - cells only uses 1234
Cage 8(2) n1 - cells do not use 489
Cage 13(2) n3 - cells do not use 123
Cage 13(2) n3 - cells do not use 123
Cage 10(2) n7 - cells do not use 5
Cage 10(2) n5 - cells do not use 5
Cage 10(3) n56 - cells do not use 89
Cage 20(3) n4 - cells do not use 12
Cage 19(3) n236 - cells do not use 1

1. "45" on c1: 1 outie r1c2 = 9
1a. -> r12c1 = 8 (no 4,8)
1b. no 4 in r2c89

2. r3c5 sees all 8 in n1 -> no 8 in r3c5 (Common Peer Elimination, CPE)

3. "45" on r12: 1 innie r2c2 + 5 = 1 outie r3c5
3a. = [16/27]
3b. r3c2 = (67)

4. 36(7)r1c3 must have 9 which is only in r2c456: locked for r2, n2
4a. no 4 in r1c89

5. r3c7 = 9 (hsingle n3)
5a. -> r3c6 + r4c7 = 10 (no 5)

6. naked pair {67} in r3c25: both locked for r3 and no 6,7 in r12c3 since they see both those cells (CPE)

7. "45" on r1234: 1 outie r5c2 + 5 = 2 innies r34c1
7a. since r4c1 and r5c2 are in the same nonet, they can't be equal -> no 5 in r3c1 (innie-outie unequal; IOU)

8. 5 in r3 only in 13(2)r3c3 = {157/256}(no 3,4,8,9)
8a. r4c4 = (67)

9. hidden killer pair 6,7 in n1 -> sp8(2)r12c1 must have 6 or 7 = {17/26}(no 3,5)
[I could now have done naked quad {1267} in n1 -> r3c3 = 5: missed that. Thanks Andrew]

10. "45" on n3: 2 remaining innies r12c7 = 5 = {14/23}

11. 3 and 4 in r12 are only in two cages -> both must have both 3 & 4
11a. -> 36(7)r1c3 = {1345689/2345679}
11b. and 16(4)r1c5 = {1348/2347}(no 5,6)
11c. 16(4) splits into a 11(2) and 5(2) = {38/14}/{47/23}
11d. note: has 8 in n2 or 2 in n3
11e. r1c56 = {3478}

12. killer pair 1,2 in r3c3489: both locked for r3

13. r3c6 + r4c7 = 10: but [82] blocked by 16(4)r1c5 (step 11d)
13a. = [37/46]

14. r3c1 = 8 (hsingle r3)
14a. no 2 in 10(2)n7

15. hidden pair 3,4 in n1 -> r12c3 = {34}: both locked for c3 and locked for 36(7) cage

16. hidden pair 3,4 in r2 -> r2c7 = (34)
16a. r1c7 = (12) (h5(2))

17. naked pair {67} in r4c47: both locked for r4

18. 10(2)n5 = {19/28}(no 3,4)
18a. killer pair 1,2 with 5(2)n6: both locked for r4

19. 5 in r4 only in n4: locked for n4

20. "45" on r1234: 1 innie r4c1 + 3 = 1 outie r5c2 = [36/47/58]
20a. -> 20(3)r4c2 must have 5 in r4c23 or 8 in r5c2 since 5 locked in r4c123
20b. 20(3) = [398/596/{58}[7](no 4)
20c. note: must have 8 or 9 in r4c23 (alternatively, hidden killer pair 8,9 in r4 -> 8 or 9 in r4c23)

These next couple of steps are very Andrew like!
21. "45" on r5: 3 innies r5c129 = 22
21a. but [985] blocked by 8 or 9 in r4c23 (step 20c)
21b. = {679} only: all locked for r5
21c. no 5 in r4c1 (iodr1234=-3)
21d. 20(3)r4c2 = [596/{58}[7](no 3)
21e. -> r4c123 = [359]/[4]{58}

22. hidden single 5 in c1 -> r7c1 = 5
22a. -> r456c1 = 14
22b. can't be [473] because of 7 in r5c2 (iodr1234=-3]
22c. can't be [392] because of 9 also in r4c3 (step 21e)
22d = [491/374](no 2,6)
22e. must have 4: locked for c1 and n4

23. 2 in c1 only in sp8(2)r12c1 = {26}: both locked for n1
23a. r23c2 = [17]
23b. r5c2 = 6 -> r4c1 = 3 (iodr1234=-3)
23c. -> r4c123 = [359](step 21e)
23d. and r456c1 = [374](step 22d)
23e. r5c9 = 9

24. 5(2)n6 = {14} only: both locked for n6, 1 for r4
24a. 10(2)n5 = {28}: both locked for n5

25. 10(3)r5c6: {145} blocked by 1,4 only in r5c6
25a. = {235} only: all locked for r5: 2 locked for n5

26. 16(4)r1c5 = {1348/2347}: must have 3
26a. r12c7 = 5 = [14/23]
26b. -> r1c56 = 11 = {38/47}
26c. if [23] in r12c7 -> r5c67 = [35]
26d. or {38} in r1c67
26e. -> 3 locked in r1c67 or r5c6
26f. -> no 3 in r3c6
26g. r3c6 = 4 -> r4c7 = 6 (cage sum)
26h. r1c56 = {38}: 8 locked for r1, n2 and 3 locked for 16(4) cage
26i. r12c7 = [14](h5(2))

Now, a cool way to finally make decent progress. Very wellbeback like
27. "45" on c89: 1 innie r5c8 + 9 = 2 outies r67c7
27a. no 9 in r6c7 -> r5c8 <> r7c7 -> r5c8 repeats in c7 only in r89c7, ie, one of 2,3,5 - perhaps two of
27b."45" on n69: 1 outie r5c6 + 6 = 2 remaining innies r89c9
27c. r5c6 = (35) -> r89c7 = 9/11
27d. ie, can't have two of 2,3,5
27e. -> hidden killer triple 2,3,5 in c7 -> r67c7 must have one of 2,3,5
27f. -> r6c8 sees all of 2,3,5 in c7 (r7c7 is same cage) since r5c8 = (2,3,5) in r89c7
27g. -> r6c8 (78)

28. killer pair 7,8 in 13(2)r12c8 and r6c8: both locked for c8

29. "45" on c9: 3 outies r349c8 = 13
29a. must have 1 or 4 for r4c8 = [319/346], r9c8 = (69)
29b. r3c89 = [32]


30. 28(5)r6c7: any combo with both 6&9 blocked by r9c8 = (69)
30a. = {13789/24589/25678/34579/34678} = 6 or 9 in one of r78c8
30b. -> killer pair 6,9 with r9c8: both locked for n9, 6 for c8

31. r12c8 = [58], 13(2)r1c9 = {67}: 7 locked for c9
31a. r56c8 = [27]
31b. -> 2 must be in r89c7 (step 27a) and 7 in c7 only in r89c7 -> r89c7 = {27} only: 2 locked for n9
31c. r89c7 = 9 -> r5c6 = 3 (iodn69=+6)

32. 17(3)r9c5: {269} blocked by r9c8 = (69)
32a. {179} blocked by r9c1 = (19)
32b. and must have 2 or 7 for r9c7
32c. = {278/467}(no 1,3,5,9)
32d. must have 7, locked for r9

33. 5 in n9 only in 17(3)r8c9 = {359} only
33a. -> r9c8 = 9, 3,5 locked for c9
33b. -> r567c9 = [981]

34. r1c56 = [38], r4c56 = [82]

35. 17(3)r9c5 = [467] only permutation

much easier now
Cheers
Ed


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 382
PostPosted: Sun Aug 25, 2019 10:04 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. I got stuck and couldn't see how to continue until I realised that I'd overlooked some simple eliminations (step 9), after which it flowed fairly well as Assassins often do.

Here is my walkthrough for Assassin 382:
Prelims

a) R12C8 = {49/58/57}, no 1,2,3
b) R12C9 = {49/58/57}, no 1,2,3
c) R23C2 = {17/26/35}, no 4,8,9
d) R3C89 = {14/23}
e) R4C56 = {19/28/37/46}, no 5
f) R4C89 = {14/23}
g) R89C1 = {19/28/37/46}, no 5
h) 19(3) cage at R3C6 = {289/379/469/478/568}, no 1
i) 20(3) cage at R4C2 = {389/479/569/578}, no 1,2
j) 10(3) cage at R5C6 = {127/136/145/235}, no 8,9

1. 45 rule on C1 1 outie R1C2 = 9, clean-up: no 4 in R2C8, no 4 in R2C9
1a. R1C2 = 9 -> R12C1 = 8 = {17/26/35}, no 4,8
1b. 4,8 in N1 only in R12C3 + R3C13, CPE no 4,8 in R3C5
1c. 36(7) cage at R1C3 contains 9, locked for N2
1d. Max R45C2 = 15 -> min R4C3 = 5

2. 45 rule on R12 2 outies R3C25 = 13 = {67} (only possible combination), locked for R3
2a. R3C2 = {67} -> R2C2 = {12}
2b. R3C7 = 9 (hidden single in R3), clean-up: no 4 in R1C8, no 4 in R1C9
2c. R3C6 + R4C7 = 10 = {28}/[37/46], no 5, no 3,4 in R4C7
2d. Naked quad {5678} in R12C89, locked for N3
2e. 45 rule on N3 2 outies R1C56 = 11 = {38/47/56}, no 1,2
2f. Naked pair {67} in R3C25, CPE no 6,7 in R12C3
2g. Hidden killer pair 6,7 in R12C1 and R3C2 for N1, R3C2 = {67} -> R12C1 must contain one of 6,7 -> R12C1 (step 1a) = {17/26}, no 3,5
2h. Naked quad {1267} in R12C1 + R23C2, locked for N1

3. 45 rule on R5 3 innies R5C129 = 22 = {589/679}, 9 locked for R5
3a. 13(3) cage at R5C3 = {148/238/247/346} (cannot be {157/256} which clash with R5C129), no 5
3b. 10(3) cage at R5C6 = {127/136/145/235}
3c. {127/136} cannot be 7{12}/6{13} which clash with R4C89 -> no 6,7 in R5C6

4. 45 rule on C9 3 outies R349C8 = 13, max R34C8 = 7 -> min R9C8 = 6
[I won’t continue this step, because I don’t used Unique Rectangles, but I spotted that R3C89 and R4C89 must have different combinations because otherwise it would be impossible to determine their permutations; I think that’s right even though they’re in different nonets. If I’d used that, then R34C8 cannot total 5 so no 8 in R9C8.]

5. 45 rule on C89 2 outies R67C7 = 1 innie R5C8 + 9
5a. Max R67C7 = 15 -> max R5C8 = 6
5b. Min R5C8 = 1 -> min R67C7 = 10, no 1 in R67C7

6. 18(3) cage at R5C9 = {189/279/369/459} (cannot be {378/468/567} which clash with R12C9), 9 locked for C9

7. 45 rule on R12345 2 outies R67C1 = 1 innie R5C9
7a. Max R67C1 = 9, no 9 in R67C1

8. 13(3) cage at R3C3 = {148/157/238/256} (cannot be {139} = [319] which clashes with R3C89, cannot be {247} = [427] which clashes with R3C89, cannot be {346} = {34}6 which clashes with R3C89), no 9
8a. 5 of {157/256} must be in R3C3 -> no 5 in R34C4
8b. 5 in R3 only in R3C13, locked for N1

9. 5 in R4 only in R4C123, locked for N4
[First time through I’d overlooked this, so I’ve reworked from here using a few of my previous steps.]
9a. R5C129 (step 3) = [985]/{679}, R67C1 (step 7) = R5C9
9b. Consider combinations for 20(3) cage at R4C2 = {389/479/569/578}
20(3) cage = {389/479/569} => R4C3 = 9 => R5C9 = 9 (hidden single in R5) => R5C129 = {67}9
or 20(3) cage = {578}, locked for N4 => R345C1 cannot total 22 (because doesn’t contain 7 and 5,8 only in R3C1) => R67C1 cannot total 5 => no 5 in R5C9
-> R5C129 = {679}, locked for R5
9c. Naked pair {67} in R35C2, locked for C2
9d. 20(3) cage = {479/569/578} (cannot be {389} because R5C2 only contains 6,7), no 3
9e. 20(3) cage = {569/578} (cannot be {479} = [497] which clashes with R4C56 + R4C89, killer ALS block), no 4, 5 locked for N4
9f. R5C2 = {67} -> no 6,7 in R4C3

10. 45 rule on R12345 4(3+1/2+2) innies R34C1 + R5C19 = 27
10a. Max R5C19 = 16 -> min R34C1 = 11, no 9 in R3C1 -> no 1,2 in R4C1

11. 45 rule on R123 4 innies R3C1346 = 18 contains 5,8 for R3 = {1458/2358}
11a. 1 of {1458} must be in R3C4 -> no 4 in R3C4
11b. 13(3) cage at R3C3 (step 8) = {148/157/238/256}
11c. 1 of {148} must be in R3C4 (cannot form R3C1346 = {1458} with 4,8 in R3C34), 7 of {157} must be in R4C4 -> no 1 in R4C4
11d. 1 in R4 only in R4C56 = {19} or in R4C89 = {14} -> no 4 in R4C56 (locking-out cages), clean-up: no 6 in R4C56
[That was another step I overlooked first time through.]

12. R67C1 (step 7) = R5C9
12a. Consider permutations for R5C129 (step 9b) = {679}
6 in R5C12 => no 6 in R4C1
or R5C129 = [976] => R67C1 = 6, R567C1 = 15 => R34C1 = 12 cannot contain 6
-> no 6 in R4C1
12b. 45 rule on R123 2 outies R4C47 = 1 innie R3C1 + 5
12c. R3C1 = {3458} -> max R4C47 = 13 contains 6 for R4 -> no 8 in R4C47, clean-up: no 2 in R3C6 (step 2c)

13. R34C1 + R5C19 (step 10) = 27, R67C1 (step 7) = R5C9
13a. Consider combinations for R5C19 = {69/79}
R5C19 = {69} = 15, locked for R5 => R5C2 = 7, R567C1 = 15 => R34C1 = 12 but cannot be [57] which clashes with R5C2
or R5C19 = {79} = 16, R567C1 = 16 => R34C1 = 11 = {38}/[47]
-> no 5 in R3C1
13b. R3C3 = 5 (hidden single in R3) -> R34C4 = 8 = [17/26]
13c. R4C2 = 5 (hidden single in R4)
13d. R4C47 = R3C1 + 5 (step 12b)
13e. R4C47 = [62]/{67} = 8,13 -> R3C1 = {38}
13f. 4 in N1 only in R12C3, locked for C3 and 36(7) cage at R1C3
13g. 36(7) cage = {1245789/1345689/2345679}, 5 locked for N2, clean-up: no 6 in R1C56 (step 2e)

[After those additional steps, a different angle on my original first forcing chain.]
14. 36(7) cage at R1C3 contains 9 and three of {18}, {27}, {36} and {45}
14a. Consider placements for R3C4 = {12}
R3C4 = 1 => R3C89 = {23}, locked for R3 => R3C16 = [84]
or R3C4 = 2, R3C89 = {14}, locked for N3 => R12C7 = {23}, locked for C7, no 2 in R4C7 => no 8 in R3C6 (step 2c) => R3C6 = 3
-> R3C6 = {34}, clean-up: no 2 in R4C7 (step 2c)
14b. R3C1 = 8 (hidden single in R3), clean-up: no 2 in R89C1
14c. Naked pair {34} in R12C3, 3 locked for C3 and 36(7) cage at R1C3
14d. Killer pair 3,4 in R1C3 and R1C56, locked for R1
14e. Killer pair 1,2 in R1C7 and R3C89, locked for N3
14f. Naked pair {67} in R4C47, locked for R4, clean-up: no 3 in R4C56
14g. Killer pair 8,9 in R4C3 and R4C56, locked for R4

15. R5C129 (step 9b) = {679} cannot be [967] (which clashes with 20(3) cage at R4C2) -> no 7 in R5C9
15a. R67C1 (step 7) = R5C9, R7C1 = 5 (hidden single in C1) -> R5C9 = R6C1 + 5, R5C9 = {69} -> R6C1 = {14}
15b. 2 in C1 only in R12C1 = {26}, locked for N1, 6 also locked for C1 -> R23C1 = [17], R5C2 = 6, R4C3 = 9 (cage sum), R3C5 = 6, clean-up: no 1 in R4C56
15c. R357C1 = [875] = 20 -> R46C1 = 7 = [34], clean-up: no 2 in R4C89
15d. Naked pair {28} in R4C56, locked for N5
15e. 1 in N2 only in R13C4, locked for C4
15f. R5C3 = 8 (hidden single in R5) -> R6C23 = [21]
15g. R5C3 = 8 -> R5C45 = 5 = [41]
15h. R5C9 = 9 -> R67C9 = 9 = {36}/[54/72/81], no 7,8 in R7C9

16. 17(3) cage at R9C5 cannot be {179} (which clashes with R9C1) -> no 1 in R9C67
16a. 1 in N8 only in R78C6, locked for 15(4) cage at R7C6, no 1 in R8C7
16b. R1C7 = 1 (hidden single in C7), clean-up: no 4 in R3C89
[I’m happy that I didn’t have to assume that R3C89 and R4C89 had different combinations.]
16c. Naked pair {23} in R3C89, locked for R3, 3 locked for N3 -> R3C46 = [14], R2C7 = 4, R12C3 = [43], R4C4 = 7 (cage sum), R4C7 = 6, clean-up: no 7 in R1C56 (step 2e), no 3 in R7C9 (step 15h)
16d. Naked pair {38} in R1C56, 8 locked for R1 and N2, clean-up: no 5 in R2C8, no 5 in R2C9
16e. 5 in N3 only in R1C89, locked for R1 -> R1C4 = 2, R12C1 = [62], R2C89 = {68} (hidden pair in R2)
16f. 8 in C4 only in R789C4, locked for N8

[It ought to be getting easier now, but still some work to do.]
17. R67C7 = R5C8 + 9 (step 5)
17a. R67C7 cannot total 14 -> no 5 in R5C8
17b. Naked pair {23} in R35C8, locked for C8
17c. R5C8 = {23} -> R67C7 = 11,12 = {38}/[57], no 7 in R6C7, no 2 in R7C7

18. 45 rule on N69 2 remaining innies R89C7 = 1 outie R5C6 + 6
18a. R5C6 = {35} -> R89C7 = 9,11 = {27/38}, no 5
18b. 5 in C7 only in R56C7, locked for N6, clean-up: no 4 in R7C9 (step 15h)

19. 7 in C7 only in R789C7, locked for N9
19a. 17(3) cage at R8C9 = {269/359/368/458}, no 1
19b. R9C1 = 1 (hidden single in R9) -> R8C1 = 9

20. R349C8 (step 4) = 13 = [319/346], no 8 -> R3C89 = [32], clean-up: no 7 in R6C9 (step 15h)
20a. R5C8 = 2 -> R67C7 (step 5) = 11 = {38}, locked for C7, 8 locked 28(5) cage at R6C7 -> R5C67 = [35], R1C56 = [38], R4C56 = [82]
20b. R6C8 = 7, R1C8 = 5 -> R2C8 = 8, R12C9 = [76], R7C9 = 1 -> R6C9 = 8 (cage sum), R67C7 = [38], R4C89 = [14]
20c. R89C9 = {35} -> R9C8 = 9 (cage sum)
20d. 17(3) cage at R9C5 = [467] (only remaining permutation)
20e. R8C6 = 1 (hidden single in C6), R8C7 = 2 -> R7C6 + R8C5 = 12 = [75]

and the rest is naked singles.

Rating Comment:
I'll rate it at a full 1.5. I used several forcing chains; of course my rating doesn't take account of getting stuck.


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC


Who is online

Users browsing this forum: Google [Bot] and 7 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group