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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: 781
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: 781
Location: Sydney, Australia
Here's my WT. step 11 is my key.

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) -> r3c1567 <> {19/37}
9a. -> 10(2)r3c6 = {28}/[64](no 1,3,7,9; no 4 in r3c6)
9b. -> 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 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 r67c5 = 1 innie r8c9 (no 1,5,9)

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 6, no 5 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: 174
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!
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 r2c345
-> 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 r78c8 = {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(3)
-> 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 Thu Jan 17, 2019 6:31 pm, edited 1 time 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: 1606
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 R3C2 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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