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 Post subject: Assassin 367
PostPosted: Tue Jan 01, 2019 7:05 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
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a367.JPG
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This is an x-killer. 1-9 cannot repeat on either diagonal

Assassin 367
code:
3x3:d:k:9984:9984:9984:9984:9984:9984:9984:2305:3586:9984:4867:4867:5892:5892:5892:2305:2305:3586:1029:4867:4867:5892:4614:2567:2567:2824:2824:1029:6409:6409:6409:4614:4614:4618:4618:3083:4108:4108:4108:6409:5645:6670:4618:4618:3083:3599:3599:5645:5645:5645:6670:6670:6670:3083:3599:2576:2576:5645:5905:6418:6418:6418:6418:3347:5905:5905:5905:5905:4884:4884:2069:6418:3347:3862:3862:3862:3095:3095:4884:2069:2069:
solution:
Code:
+-------+-------+-------+
| 8 9 3 | 4 2 1 | 7 6 5 |
| 5 4 6 | 7 8 3 | 2 1 9 |
| 1 7 2 | 5 9 6 | 4 3 8 |
+-------+-------+-------+
| 3 6 7 | 9 1 8 | 5 4 2 |
| 9 2 5 | 3 6 4 | 1 8 7 |
| 4 8 1 | 2 5 7 | 6 9 3 |
+-------+-------+-------+
| 2 1 9 | 8 4 5 | 3 7 6 |
| 6 3 8 | 1 7 2 | 9 5 4 |
| 7 5 4 | 6 3 9 | 8 2 1 |
+-------+-------+-------+
A very happy New Year's with a new Assassin! SudokuSolver has a shocker with this one (2.20) but JSudoku has no trouble (no chains). Caused me some trouble too but finally found an extension of a technique I use regularly. Used the one advanced step but some others were difficult to see. Perhaps I missed something.

Cheers
Ed


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 Post subject: Re: Assassin 367
PostPosted: Mon Jan 07, 2019 6:10 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Here's my WT. step 11 is my key. Many thanks to Andrew for some clarifications and corrections.

A367 WT:
Preliminaries courtesy of SudokuSolver
Cage 4(2) n14 - cells ={13}
Cage 14(2) n3 - cells only uses 5689
Cage 12(2) n8 - cells do not use 126
Cage 13(2) n7 - cells do not use 123
Cage 10(2) n23 - cells do not use 5
Cage 10(2) n7 - cells do not use 5
Cage 11(2) n3 - cells do not use 1
Cage 8(3) n9 - cells do not use 6789
Cage 9(3) n3 - cells do not use 789
Cage 19(3) n89 - cells do not use 1
Cage 26(4) n56 - cells do not use 1
Cage 39(8) n123 - cells ={12345789}

No routine cage clean-up done. Only if stated.
1. 4(2)r3c1 = {13}: both locked for c1

2. 39(8)r1c1: must have 1 & 3 which are only in r1: locked for r1

3. 6 in r1 only in n3: 6 locked for n3
3a. no 8 in r1c9

4. "45" on n3: 2 innies r13c7 = 11 (no 1,5)

5. 1 in n3 only in 9(3) = {126/135}(no 4)
5a. 1 only in r2: locked for r2
5b. 6 in {126} must be in r1c8 -> no 2 in r1c8
5c. 5 in {135} must be in r1c8 -> no 5 in r2c78

6. 2,4,7 & 8 in r1 only in 39(8) -> no 2,4,7,8 in r2c1

7. "45" on r1: 4 outies r2c1789 = 17 = {1259/1358} = 2 or 8
7a. must have 5 -> 5 locked for r2

8. "45" on r123: 2 innies r3c15 = 10 = [19/37]

9. 11(2)r3c8 = {29/38/47}(no 5) = 3 or 7 or 9 -> r3c15+r3c67 cannot be [19]+{37}
9a. -> {37} blocked from 10(2)r3c6
9b. = {28}/[64](no 1,3,7,9; no 4 in r3c6)
9c. -> r1c7 = (379)(h11(2)r13c7)

10. "45" on r12: 3 outies r3c234 = 14 and must have 5 for r3 = {158/257/356}(no 4,9)

An advanced level trick this one! Hope it is easy enough to follow
11. "45" on r1: 1 outie r2c1 + 6 = 2 innies r1c89
11a. since the number 6 must be in the innies r1 and it is also the Innie Outie Difference (iod) of 6 is -> 5 or 9 in r1c89 = (5/9) in r2c1. It must be the same number as r2c1
11b. -> the only place for that number in r3 is in n2
11c. -> 5 in r3c4 and/or 9 in r3c5
11d. -> 9 in 23(4)r2c4 must also have 5 or there would be no 5 or 9 for r3 (Locking-out cages)
11d. -> 23(4)n2: {1679/2489/3479} all blocked
11e. {1589} blocked since 1,5 only in r3c4
11f. {2678} blocked by r3c6 = (268)
11g. {2579} blocked by r3c5 = (79)
11h. 23(4) = {3569/3578/4568}(no 1,2)
11i. must have 5 -> r3c4 = 5
11j. note: could still have 9 in r3c5 so don't know for sure which pairs up with r2c1

12. "45" on r12: 2 remaining innies r2c23 = 10. But {28} blocked by h17(4)r2 = 2/8 (step 7.)
12a. = {37/46}(no 2,8,9) = [3/6..]
12b. r2c23 = 10 -> r3c23 = 9, but {36} blocked by r2c23
12c. r2c23 = {18/27}(no 3,6)

13. hidden single 6 in r3 -> r3c6 = 6, r3c7 = 4 (Placed for D/), r1c7 = 7 (h11(2)r13c7)

14. 2 in r2 only in 9(3)r1c8 = {126} only -> r1c8 = 6
14a. 2 locked for n3
14b. -> 14(2)n3 = {59} only: both locked for c9 and 9 for n3
14c. 11(2)n3 = {38} only: both locked for r3
14d. r3c15 = [19](h10(2)r3c15), r4c1 = 3
14e. r3c23 = {27}: both locked for n1
14f. h10(2)r2c23 = {46} only: 4 locked for r2 and n1

Time to move on.
15. 13(2)n7: {58} blocked by r12c1 = two of {589}
15a. 13(2) = [49]/{67}(no 5,8)

16. 8(3)n9 must have 1: 1 locked for n9

17. "45" on n9: 2 outies r78c6 = 7 = {25/34}(no 1,7,8,9)

18. "45" on r89: 1 outie r7c5 = 1 innie r8c9 (no 1,5)

19. "45" on r789: 2 innies r7c14 = 10
19a. no 5 in r7c1, no 7,9 in r7c4

20. 1 in r7 on 10(2)r7c2 or h10(2)r7c14 -> 9 locked for r7 (Locking cages)
20a. and 9 locked for n7

21. 13(2)n7 = {67}: both locked for c1 and n7
21a. no 3,4 in 10(2)n7
21b. no 3,4 in r7c4 (h10(2)r7c14)

22. r3c5 = 9 -> r4c56 = 9 = {18/27}/[45](no 5,6 in r4c5)

I took a long time to see this next step and it makes a very big difference.
23. 6 in n5 only in r56c5 or r456c4. r7c4 sees all these -> no 6 in r7c4 (CPE)
23a. -> no 4 in r7c1 (h10(2)r7c14)

24. naked quad {1289} in r7c1234: 2,8 locked for r7
24a. no 2,8 in r8c9 (IODr89=0)
24b. no 5 in r8c6 (h7(2)r78c6)

25. 19(3)r8c6 must have 2,3,4 for r8c6 = {289/469)(no 3,5)
25a. 2 must be in r8c6 -> no 2 in r89c7

26. 2 in n9 only in 8(3) = {125} only: 5 locked for n9 and c8

27. hsingle 5 in r7 -> r7c6 = 5, r8c6 = 2 (h7(2)r78c6), -> r89c7 = 17 = {89} only: both locked for c7

28. "45" on n6: 2 innies r6c78 = 15 = [69] only
28a. r7c7 = 3 (placed for D\)

29. 12(3)n6: {138} blocked by r3c1, {147} blocked by r78c9 = two of 4,6,7
29a. = {237} only: all locked for c9 and n6
29b. r9c9 = 1, r8c8 = 5 (both placed for D\), r9c8 = 2, r2c78 = [21] (1 placed for D/)

30. r6c78 = 15 -> r56c6 = 11 = [38]/{47}(no 8 in r5c6)

31. killer pair 7,8 with r4c6: both locked for c6 and n5
31a. r2c6 = 3
31b. -> r56c6 = {47} only: both locked for c6
31c. r4c56 = [18](8 placed for D/)

etc
Cheers
Ed


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 Post subject: Re: Assassin 367
PostPosted: Wed Jan 16, 2019 7:45 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks again Ed! Happy New Year :)
If you missed something then so did I. Our key steps are very similar though we expressed them in different ways.
Corrections & clarifications thanks to Ed. No matter how many times I proofread - errors still get through!
And more corrections from Andrew!
Assassin 367 WT:
1. 39(8) has no 6 -> 6 in r1 in r1c89
Whatever is in r2c1 (Call it 'X') is the other value in r1c89
Since 4(2)r3c1 = {13} -> X not from (13)
-> (13) in r1 in r1c2..7

2. Innies n3 = r13c7 = +11(2) (No 1)
-> 1 in n3 in 9(3).
-> 9(3)n3 = {6{12}] or [5{13}]
If the former -> 14(2)n3 = {59} -> r2c19 = {59}
If the latter -> 14(2)n3 = [68] -> r2c1 = 5
-> 5 locked in r2 in r2c19.

3. 6 in n1 only in 19(4)
-> Since r3c1 from (13) -> 9 in n1 only in 39(8)

4! Innies r123 -> r3c15 = +10(2) = [19] or [37]
6 in r1c89 prevents 11(2)n3 from being {56}
-> 5 in r3 in r3c234

Note that if 5 in 19(4)n1 -> 19(4)n1 = {1567}

Either:
(A) r3c15 = [19] -> 19(4)n1 cannot contain a 5 -> r3c4 = 5
or:
(B) r3c15 = [37] -> r1c7 = 7 -> 3 in 9(3)n3 must be [5{13}] -> r2c1 = 5 -> r3c4 = 5

Either way -> r3c4 = 5

5. Remaining Outies r12 -> r3c23 = +9(2) -> r2c23 = +10(2).
-> Either r2c23 = {46} or r3c23 = {36}
-> r2c23 cannot be {37}
-> 7 in r2 in n2 in r2c456
-> r3c15 = [19]
-> r4c1 = 3

6. Also since 9 already in 39(8) in n1 -> 9 in n3 only in 14(2)n3
-> 14(2)n3 = {59}
-> 9(3)n3 = [6{12}]
-> r2c19 = {59}
Also -> r2c23 = {46}
-> r2c456 = {378}
Also -> r3c23 = {27}
-> 11(2)r3c8 = {38}
-> r13c7 = [74]
-> r3c6 = 6
-> r1c456 = {124}

7. Innies r789 = r7c14 = +10(2)
-> 5 in r7 in r7c56789
Outies n9 -> r78c6 = +7(2) = {34} or {25}
Since 19(3) cannot contain two values = +7 -> whatever goes in r7c6 must go in 8(3) in n9.
-> Either
r78c6 = {25} -> 8(3)n9 = {125}
Or r78c6 = {34} -> 8(3)n9 = {134}
-> Whatever is in r8c6 goes in n9 in r9c89 and in n7 in r7c123
-> (Since 5 not in r7c123) r78c6 cannot be [25]


Also 7 in r1c7 -> r8c6 cannot be 3

-> Either r78c6 = [52] and 8(3)n9 = {125} and 19(3)r8c6 = [2{89}]
Or r78c6 = [34] and 8(3)n9 = {134} and 19(3)r8c6 = [4{69}]


8. Innies - Outies r89 -> r7c5 = r8c9
-> r7c5 cannot be 1
-> 1 in r7 in r7c234
-> Either 10(2)n7 = {19} or H+10(2)r7c14 = [91]
-> 9 in r7c123
-> 13(2)n7 cannot be {49}

9. 3 in r1c23
-> At least one of (58) in r12c1
-> 13(2)n7 = {67}

10. 6 cannot be in r7c4 (leaves no place for 6 in n5)
-> r7c1 cannot be 4 (since r7c14 = +10(2))
-> 4 in n7 in r89c23
-> (Since whatever is in r8c6 is in r9c89 and r7c123) r78c6 cannot be [34]
-> r78c6 = [52]
-> 19(3)r8c6 = [2{89}]
Also 8(3)n9 = {125}
-> 25(5)r7c6 = [5{3467}]

11. Outies n6 -> r56c6 = +11(2)
Must be {38} or {47}
-> r6c78 = [69]
11(2)n3 = {38} prevents 8 in 12(3)n6
-> 8 in n6 in r45c8
-> 11(2)n3 = [38]
5 in n6 in 18(4)
-> 7 in n6 in r456c9
-> r7c8 = 7

12. 9 cannot go in r5c4 (Leaves no place for 9 in n4)
-> HS 9 in n5 -> r4c4 = 9
-> HS 9 in n8 -> 12(2)n8 = [39]
-> r89c7 = [98]
-> Remaining Innies r9 -> r9c189 = [7{12}]
-> 15(3)r9 = [{45}6]
-> HS r7c9 = 6
Since whatever goes in r7c5 also goes in r8c9 it must also go in r9c123. This can only be a 4.
-> 25(5)r7c6 = [53764] and r7c5 = 4
-> 12(3)n6 = {237}
-> 8(3)n9 = [521]
-> r2c78 = [21]
-> r45c7 = {15} and r45c8 = {48}

13. HS 6 in n5 -> r5c5 = 6
-> 6 in n4 in r4c23
-> Since 9 in n4 in 16(3) -> 16(3)n4 = {259}
-> 25(4)r4c2 can only be [{67}93]
-> r6c123 = {148}
-> r56c6 = [47]
-> r4c56 = [18] (Since 1 already on D/)
-> r6c45 = [25]
-> 12(3)n6 = [273]
Also 18(4)n6 = [5418]

14. Also NT on D\ -> r1c1 = 8, r2c2 = 4, r3c3 = 2
HS 5 on D/ -> r1c9 = 5
-> r2c19 = [59]
Also NT on D/ -> r7c3 = 9, r8c2 = 3.
etc.


Last edited by wellbeback on Mon Jan 21, 2019 12:11 am, edited 2 times in total.

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 Post subject: Re: Assassin 367
PostPosted: Thu Jan 17, 2019 5:41 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
My solving path was quite a lot different from Ed's and wellbeback's.

I must have missed something as I found it difficult in the later stages. On going through Ed's walkthrough again, his step 20 was the most important thing that I missed; I've added a note about it at the end of my step 9.

I have no idea why SudokuSolver gave such a high score; I didn't see anything in Ed's or my walkthrough which it wouldn't be programmed to do.

Here's my walkthrough for Assassin 367:
Prelims

a) R12C9 = {59/68}
b) R34C1 = {13}
c) R3C67 = {19/28/37/46}, no 5
d) R3C89 = {29/38/47/56}, no 1
e) R7C23 = {19/28/37/46}, no 5
f) R89C1 = {49/58/67}, no 1,2,3
g) R9C56 = {39/48/57}, no 1,2,6
h) 9(3) cage at R1C8 = {126/135/234}, no 7,8,9
i) 19(3) cage at R8C6 = {289/379/469/478/568}, no 1
j) 8(3) cage at R8C8 = {125/134}
k) 26(4) cage at R5C6 = {2789/3689/4589/4679/5678}, no 1
l) 39(8) cage at R1C1 = {12345789}, no 6

Steps Resulting From Prelims
1a. Naked pair {13} in R34C1, locked for C1
1b. 1,3 in 39(8) cage at R1C1 only in R1C234567, locked for R1
1c. 8(3) cage at R8C8 = {125/134}, 1 locked for N9

2. 6 in R1 only in R1C89, locked for N3, clean-up: no 8 in R1C9, no 4 in R3C6, no 5 in R3C89
2a. 45 rule on N3 2 innies R13C7 = 11 = {29/38/47}, no 1,5, clean-up: no 9 in R3C6
2b. 1 in N3 only in 9(3) cage at R1C8 = {126/135}, no 4, 1 locked for R2
2c. 5 of {135} must be in R1C8 -> no 5 in R2C78
2d. 6 of {126} must be in R1C8 -> no 2 in R1C8
2e. 2,4,7,8 in R1 only in R1C1234567, locked for 39(8) cage at R1C1 -> no 2,4,7,8 in R2C1

3. 45 rule on N9 2 outies R78C6 = 7 = [16]/{25/34}, no 7,8,9, no 6 in R7C6
3a. 19(3) cage at R8C6 = {289/379/469/478/568}
3b. 2,3,4 of {289/379/469/478} must be in R8C6 -> no 2,3,4 in R89C7

4. 45 rule on R789 2 innies R7C14 = 10 = {28/46}/[73/91], no 5 in R7C1, no 5,7,9 in R7C4

5a. 45 rule on N6 2 innies R6C78 = 15 = {69/78}
5b. 45 rule on N6 2 outies R56C6 = 11 = {29/38/47/56}

6. 45 rule on R123 2 innies R3C15 = 10 = [19/37]
6a. 23(4) cage at R2C4 = {2489/2678/3569/3578/4568} (cannot be {1679/2579/3479} which clash with R3C5, cannot be {1589} = {589}1 which clashes with R2C1), no 1

7. 45 rule on R89 1 outie R7C5 = 1 innie R8C9, no 1 in R7C5

8. 5 in R3 only in R3C234
8a. 45 rule on R12 3 outies R3C234 = 14 = {158/257/356}, no 4,9
8b. 4 in R3 only in R3C789, locked for N3, clean-up: no 7 in R3C7 (step 2a), no 3 in R3C6
8c. R3C15 (step 6) = [19/37], R3C234 = {158/257/356} -> combined cage R3C15234 = [19]{257}/[19]{356}/[37]{158}
8d. R3C67 = [28/64/82] (cannot be [19/73] which clash with combined cage R3C15234), no 1,7 in R3C6, no 3,9 in R3C7 clean-up: no 2,8 in R1C7 (step 2a)
8e. 1 in R3 only in R3C123, locked for N1

9. Caged X-Wing for 5 in 39(8) cage at R1C1 (no 5 in R1C7) and R3C234 for N12 -> no 5 in R2C23456
9a. 23(4) cage at R2C4 = {2489/3569/3578/4568} (cannot be {2678} which clashes with R3C6)
9b. Consider placements for R2C1 = {59}
9c. R2C1 = 5 => R3C4 = 5 (hidden single in N2) => 23(4) cage = {3569/3578/4568}
or R2C1 = 9 => 23(4) cage = {3578/4568}
-> 23(4) cage = {3569/3578/4568}, no 2
-> R3C4 = 5
9d. R3C234 = 14 (step 8a) -> R3C23 = 9 -> R2C23 = 10 = {28/37/46}, no 9
9e. 9 in N1 only in R1C123 + R2C1, locked for 39(8) cage at R1C1, clean-up: no 2 in R3C7 (step 2a), no 8 in R3C6
9f. 19(4) cage at R2C2 = {1468/2368/2467} (cannot be {1378} which clashes with R3C1), with R2C23 = 10 and R3C23 = 9 = {28}{36}/{46}{18}/{46}{27} -> R2C23 = {28/46}, no 3,7
9g. 7 in R2 only in R2C456 -> 23(4) cage at R2C4 = {3578} (only remaining combination), locked for N2, 3,8 also locked for R2 -> R3C5 = 9, R3C1 = 1 (step 6), R4C1 = 3, clean-up: no 3 in R9C6
9h. Naked pair {12} in R2C78, 2 locked for R2 and N3, R1C8 = 6 (cage sum), clean-up: no 9 in R6C7 (step 5a)
9i. Naked pair {59} in R12C9, locked for C9, 9 also locked for N3, clean-up: no 5 in R7C5 (step 7)
9j. Naked pair {46} in R2C23, locked for N1 -> R3C23 = {27} (only remaining combination), locked for R3 and N1
9k. R3C6 = 6 -> R3C7 = 4, placed for D/, R1C7 = 7 (step 2a), clean-up: no 5 in R56C6 (step 5b), no 8 in R6C8 (step 5a), no 6 in R7C2, no 1 in R7C6 (step 3), no 9 in R8C1
9l. R89C1 = [49/67/76] (cannot be {58} which clashes with R12C1, ALS block), no 5,8
9m. R7C23 = {19/28/37} (cannot be [46] which clashes with R89C1), no 4,6
9n. R3C5 = 9 -> R4C56 = 9 = {18/27}/[45], no 5,6 in R4C5
[At this stage, after the elimination of 1 in step 9k, I missed Ed’s 1 in R7 only in R7C14 (step 4) = [91] or R7C23 = {19}, locking cages, 9 locked for R7 and even more important locked for N7. This would have avoided my difficult, but interesting, later steps.]

10. 19(3) cage at R8C6 (step 3a) = {289/469/568}, no 3, clean-up: no 4 in R7C6 (step 3)
10a. 5 of {568} must be in R8C6 -> no 5 in R89C7
10b. Naked triple {689} in R689C7, locked for C7, 9 also locked for N9
10c. 19(3) cage must contain 9 = {289/469}, no 5, clean-up: no 2 in R7C6 (step 3)
10d. 25(5) cage at R7C6 = {23578/34567} (cannot be {24568} because 2,4 in N9 clash with 8(3) cage at R8C8)
10e. 12(3) cage at R4C9 = {147/237/246} (cannot be {138} which clashes with R3C9), no 8
10f. Killer pair 6,7 in 12(3) cage and R6C78, locked for N6
10g. 5 in N6 only in 18(4) cage at R4C7 = {1359/1458/2358}
10h. 2 of {2358} must be in R45C7 (cannot be in R45C8 which would clash with R2C78 = [12] as 1 would be hidden single in C7), no 2 in R45C8

11. 25(4) cage at R4C2 and R4C56 ‘see’ each other so must form a 34(6) combined cage = {136789/145789/235789/245689/345679}
11a. 12(3) cage at R4C9 (step 10e) = {147/237/246}
11b. Variable hidden killer pair 2,6 for 34(6) combined cage and R4C79 for R4, R4C79 cannot be [26] which clashes with 12(3) cage = 6{24} -> 34(6) combined cage must contain at least one of 2,6 in R4C23456 -> 34(6) combined cage = {136789/235789/245689/345679} (cannot be {145789}
11c. Variable hidden killer pair 4,6 for 34(6) combined cage and R4C79 for R4, R4C89 cannot be [46] which clashes with 12(3) cage = 6{24} -> 34(6) combined cage must contain at least one of 4,6 in R4C23456 -> 34(6) combined cage = {136789/245689/345679} (cannot be {235789}
11d. R4C56 (step 9n) = {18}/[45] (cannot be {27} because 34(6) combined cage only contains one of 2,7), no 2,7
11e. 34(6) combined cage = {136789/245689/345679}, R4C56 = {18}/[45] -> 25(4) cage = {3679/2689}, no 1,4,5
11f. 3 of {3679} must be in R5C4 -> no 7 in R5C4

12. R56C6 (step 5b) = {29/38/47}, R4C56 (step 11d) = {18}/[45]
12a. Consider permutations for R78C6 (step 3) = [34/52]
12b. R78C6 = [34] => R56C6 = {29}
or R78C6 = [52] => R4C45 = {18}, locked for N5 => R56C6 = {47}
-> R56C6 = {29/47}, no 3,8
12c. Killer pair 2,4 in R56C6 and R8C6, locked for C6 -> R1C6 = 1, clean-up: no 8 in R4C5, no 8 in R9C5
12d. R56C6 contains one of 7,9, R6C8 = {79} -> 26(4) cage at R5C6 = {2789/4679}, CPE no 7,9 in R6C45
12e. 3 in N5 only in R56C45, CPE no 3 in R7C4, clean-up: no 7 in R7C1 (step 4)
12f. 6 in N5 only in R4C4 + R56C45, CPE no 6 in R7C4, clean-up: no 4 in R7C1 (step 4)
12g. Consider permutations for R4C56 = [18/45]
R4C56 = [18]
or R4C56 = [45], R56C6 = {29} => R6C78 = [87]
-> no 8 in R4C8
12h. 8 in R4 only in R4C2346, CPE no 8 in R5C4

13. 12(3) cage at R4C9 (step 10e) = {147/237/246}
13a. 45 rule on R1234 3 innies R4C789 = 1 outie R5C4 + 8
13b. Apart from 3, which is already placed for R4, the value in R5C4 must be in R4C789
13c. R5C4 cannot be 2 because R4C789 = 10 cannot be {127} = [217] clashes with 12(3) cage = 7{14}/7{23} (alternatively 18(4) cage at R4C7 (step 10g) cannot contain both of 1,2)
R5C4 cannot be 6 because 6 in R4C9 and R4C78 cannot total 8
R5C4 cannot be 9 because 9 in R4C8 and R4C79 cannot total 8 = [17] because 18(4) cage at R4C7 = [19]{35} clashes with 12(3) cage = 7{14}/7{23})
-> R5C4 = 3
[It took me a long time to spot the power of that 45, which cracks the puzzle.]
13d. R5C4 = 3 -> R4C789 = 11 = {245} (only remaining combination, cannot be {146} which clashes with R4C5), locked for R4 and N6, R4C5 = 1, R4C6 = 8, placed for D/, R5C7 = 1, clean-up: no 2 in R7C2, no 4 in R9C5
13e. R6C9 = 3 (hidden single in N6) -> R45C9 = 9 = [27], R4C78 = [54], R5C8 = 8 (cage sum), R3C89 = [38]
13f. R6C78 = [69] = 15 -> R56C6 = 11 = [47], 7 placed for D\, R3C3 = 2, placed for D\, R6C4 = 2, placed for D/, R6C5 = 5, R5C5 = 6, placed for both diagonals, R4C4 = 9, placed for D\, R2C8 = 1, placed for D/, R2C2 = 4, placed for D\, R8C8 = 5, placed for D\, R9C8 = 2, R7C678 = [537], R7C3 = 9 placed for D/, R7C2 = 1 (cage sum)

and the rest is naked singles, without using the diagonals.

Rating comment:
I'll rate my walkthrough for A367 at 1.5.


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 Post subject: Re: Assassin 367
PostPosted: Sun Jan 20, 2019 5:21 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Thought of another way to express/see three of the techniques used in this puzzle thread. First, wellbeback's step 7. See pic below at that spot.
Attachment:
a367WellStep7.JPG
a367WellStep7.JPG [ 40.29 KiB | Viewed 6524 times ]
Quote:
Outies n9 -> r78c6 = +7(2) = {34} or {25}
Since 19(3) cannot contain two values = +7 -> whatever goes in r7c6 must go in 8(3) in n9
Like the way this excludes the clone of r7c6 from r89c7 using just the cage total. I reached the same conclusion but used the innie/outie difference of r7c6 + 12 = r89c7 -> r7c6 cannot equal either cell in r89c7 since the remaining innie can't be 12. Never thought about wellbeback's way before!
Quote:
7...whatever goes in r7c6 must go in 8(3) in n9.
-> Either r78c8 = {25} -> 8(3)n9 = {125}
Or r78c6 = {34} -> 8(3)n9 = {134}
Another way to see this is that an 8(3) must have 1 -> the other two cells = 7. Since one cell of the h7(2)r78c6 must be in one of the remaining two cells that total 7 in the 8(3)n9 -> both numbers from the h7(2)r78c6 must be in those two cells in n9 that total 7. I think this pattern of 7s in both places means I'm more likely to see this type of step.

Now, Andrew's step 9a. See pic below (this excludes the first part of step 9. to show the power of this alternative way. Note: 5 in r3 is only in r3c234
Attachment:
a367AndrewStep9b.JPG
a367AndrewStep9b.JPG [ 81.24 KiB | Viewed 6524 times ]
Quote:
9a. 23(4) cage at R2C4 = {2489/3569/3578/4568} (cannot be {2678} which clashes with R3C6)
9b. Consider placements for R2C1 = {59}
9c. R2C1 = 5 => R3C4 = 5 (hidden single in N2) => 23(4) cage = {3569/3578/4568}
or R2C1 = 9 => 23(4) cage = {3578/4568}
-> 23(4) cage = {3569/3578/4568}, no 2
An alternative
9c. R2C1 = 5 => R3C4 = 5 (hidden single in R3) which is in the 23(4)r2c4
or R2c1 = 9
-> 9 in 23(4)r2c4 must have 5 in r3c4
-> {2489} blocked
-> 23(4) cage = {3569/3578/4568}, no 2

Gives the same result in a slightly simpler way I think. wellbeback and I did a more complicated version of Andrew's step 9 to place r3c4.

Cheers
Ed


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