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Assassin 403
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Author:  Ed [ Thu Jul 01, 2021 6:52 am ]
Post subject:  Assassin 403
Attachment:
a403.JPG
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NOTE: it has four 'zero' cells in n14 that do not form a cage and can have repeats.

Assassin 403

As much as I've enjoyed the Revisits, have been missing big cages. I found this one hard - probably because of the big cages! SudokuSolver gives it 2.00 but it has a known weakness with zero killers so don't worry about that score too much. JSudoku gets it in a snap with 4 'complex intersections'.

With so much sport around, happy to hang off any more puzzles for a while. Let me know.

code: triple click:
3x3::k:8704:8704:8961:8961:8961:770:770:3075:3075:8704:0000:0000:4869:8961:2566:7175:7175:3592:8704:8704:0000:4869:8961:2566:7175:7177:3592:2058:8704:0000:4869:5387:2566:7175:7177:3592:2058:8704:9228:5387:5387:7177:7177:7177:4365:9228:9228:9228:9228:5387:5387:2830:2830:4365:9228:3599:5648:5648:1553:1553:3090:4365:4365:3599:1811:5648:3860:3860:5653:4365:3090:4356:1811:3599:5648:5648:5653:5653:5653:4356:3090:
solution: 
+-------+-------+-------+
| 7 4 8 | 6 5 2 | 1 3 9 |
| 1 6 2 | 4 9 3 | 8 5 7 |
| 9 3 5 | 8 7 1 | 6 4 2 |
+-------+-------+-------+
| 3 8 1 | 7 4 6 | 9 2 5 |
| 5 2 4 | 3 8 9 | 7 6 1 |
| 6 7 9 | 2 1 5 | 3 8 4 |
+-------+-------+-------+
| 8 9 6 | 1 2 4 | 5 7 3 |
| 4 5 3 | 9 6 7 | 2 1 8 |
| 2 1 7 | 5 3 8 | 4 9 6 |
+-------+-------+-------+
Cheers
Ed
Author:  Ed [ Sat Jul 10, 2021 6:54 pm ]
Post subject:  Re: Assassin 403
Here's how I did it. Really enjoyed this solution in the end but had a lot of trouble getting it optimised because it's long (31 steps for a 'start'!). Hope you guys found one of your clever shortcuts. [Thanks to Andrew for some suggestions on alternative steps. Love how there are so many ways to see things.]
a403 start:
Preliminaries from SudokuSolver
Cage 3(2) n23 - cells ={12}
Cage 17(2) n9 - cells ={89}
Cage 6(2) n8 - cells only uses 1245
Cage 15(2) n8 - cells only uses 6789
Cage 7(2) n7 - cells do not use 789
Cage 8(2) n4 - cells do not use 489
Cage 12(2) n3 - cells do not use 126
Cage 11(2) n6 - cells do not use 1
Cage 10(3) n25 - cells do not use 89
Cage 19(3) n25 - cells do not use 1
Cage 28(4) n36 - cells do not use 123
Cage 35(5) n12 - cells ={56789}
Cage 17(5) n69 - cells do not use 89

1. 10(3)r2c6: {127} blocked by r1c6 = (12)
1a. = {136/145/235}(no 7) = 1 or 2

2. killer pair 1,2 in r1c6 and 10(3): both locked for c6
2a. r7c5 = (12)

3. 10(3)r2c6: {145} blocked by r7c6 = (45)
3a. = {136/235}(no 4)
3b. 3 locked for c6

4. 4 in n2 only in 19(3) = {469/478}(no 2,3,5)
4a. 4 locked for c4

5. hidden killer pair 1,2 in n2 -> the one 1 or 2 in 10(3) must be in r23c6
5a. -> no 1,2 in r4c6
[Andrew noticed that steps 5 & 6 could be combined as "Hidden triple {123} in r123c6"]

6. 3 in n2 only in 10(3) -> no 3 in r4c6
6a. 5 or 6 for 10(3) must be in r4c6 -> no 5,6 in r23c6

7. 5 in n2 only in 35(5)r1c3 -> no 5 in r1c3

8. "45" on c789: 2 innies r19c7 + 4 = 1 outie r5c6
8a. max. two innies = 5 (r9c7 = (1..4))
8b. r5c6 = (789)

9. "45" on n89: 1 innie r7c1 - 4 = 1 outie r9c7
9a. r7c1 = (5..8)

10. "45" on n7: 1 innie r7c1 - 2 = 2 outies r79c4
10a. two outies can't be {12} since r7c5 = (12) -> min. two outies = {13} = 4 -> min r7c1 = 6
10b. max. 2 outies = 6 (no 6,7,8,9)
10c. no 1 in r9c7 (iodn78=+4)

11. "45" on n2356789: 1 outie r1c3 + 2 = 2 innies r6c4 + r7c1
11a. max 2 innies = 11 -> max. r6c4 = 5
11b. min. 2 innies = 8 -> no 1 in r6c4

12. 1 & 4 in n5 only in 21(5) = 14{259/268/358/367}

13. hidden killer pair 2,3 in n5 since 21(5) can only have one of them
13a. -> r6c4 = (23)

14. from step 10, r7c1 - 2 = r79c4
14a. r79c4 can't be {23} because r6c4 = (23)
14b. = {13/15}(no 2)
14c. 1 locked for c4 and n8 and 22(5)r7c3
14d. r79c4 = 4 or 6 -> r7c1 = (68)
14e. r9c7 = (24)(iodn78=+4)

15. r7c56 = [24]

16. from step 8, r19c7 + 4 = r5c6
16a. r9c7 from (24) -> no 2 in r1c7
16b. r1c67 = [21]
16c. r1c7 = 1 -> r5c6 - 5 = r9c7
16d. r5c6 = (79)

17. 10(3)r2c6 = {13}[6]

18. 36(6)r5c3 must have 2 or 3 for r6c4 = {26/35}4789
18a. must have 4,7,9: all locked for n4
18b. 2 or 3 must be in r6c4 -> neither in n4
18c. no 1 in r45c1

19. 8(2)n4 = {26/35} = 5 or 6
19a. -> killer single 5 with 36(6): 5 locked for n4

20. "45" on c12: 4 innies r2c2 + r6c12 + r7c1 = 27
20a. max. r6c12 + r7c1 = 24 -> min. r2c3 = 3

SudokuSolver can't do this step, hence, the very high SS score
21. "45" on the whole grid -> 4 zero cells in n14 = 14
and must have 1 for c3
21a. since min. r2c2 = 3 -> min. any 3 of those = 6
21b. -> no 9 in r2c2 nor r23c3

very slow seeing this next step
22. 34(7)r1c1: {1345678} blocked by r2c2 = (345678)
22a. = 129{3478/3568/4567}
22b. 9 locked for n1

23. from step 11, r1c3 + 2 = r6c4 + r7c1
23a. -> max. r6c4 + r7c1 = 10
23a. -> {345789} blocked from 36(6) since r6c4 + r7c1 = [38] = 11
23b. -> = {246789} only (no 3, 5)
23c. r6c4 = 2
23d. must have 6 -> no 6 in r45c1 (Common Peer Elimination)

24. 8(2)n4 = {35} only: both locked for c1, 3 for n4

Loved this next cascade.
25. from step 22a. 34(7)r1c1 = 129{3478/3568/4567} = 3 or 5 in n1 -> r23c3 cannot have both 3 & 5
25a. also, r789c3 cannot have both 3 & 5 since r79c4 must have one
25b. -> both r23c3 and r789c3 must have one each
25c. -> 34(7)n1 can't have both 3,5
25d. = 12479{38/56} = 3 or 5
25e. 4 & 7 both locked for n1
[Can also see step 25 through 22(5)n78 = 135{49/67}]

26. killer pair 3,5 in n1 in 34(7) and r23c3: locked for n1

27. naked pair {68} in r1c3 + r2c2: both locked for n1

28. 4 zeros cells = 14 = [6]{125}/[8]{123}
28a. 2 locked for c3
28b. no 8 in r4c3

29. from step 11, r1c3 = r7c1
29a. from step 20, r2c2 + r6c12 + r7c1 = 27
29a. -> r2c2 + r7c1 = {68} only (not {66/88}) = 14 -> r6c12 = 13 = {49/67}(no 8)

30. from 25b. r789c3 has one of 3 or 5
30a. -> {158}{34} blocked from combined cages 14(3) & 7(2)n7
30b. -> no 8 in r8c1 since no 2,4 are in r7c2

31. hidden single 8 in c1 -> r7c1 = 8
31a. -> r9c7 = 4 (iodn78=+4)
31b. and r79c4 = 6 (iodn7=+2) = {15} only: 5 locked for c4 and n8 and 22(5) cage
31c. and r1c3 = 8 (r1c3=r7c1)
31d. and r5c6 = 9 (iodc789=-4)

Much more straight forward now
Cheers
Ed
Author:  wellbeback [ Sun Jul 11, 2021 6:59 pm ]
Post subject:  Re: Assassin 403
Great puzzle Ed! I have never used as many IODs in a solution before. Here's an optimized version of how I did it. It is a bit rushed and I haven't looked at Ed's yet since England vs Italy is about to start, so will try and look it later.
Assassin 403 WT:
1. 35(5)r1c3 = {56789}
Of the numbers (1234) in n2 - at most one is in 19(3)r2c4
-> r123c6 all from (1234)
Since r1c6 from (12) -> 10(3)r2c6 cannot be {127}
Both 3(2)r1c6 and 10(3)r2c6 each contain one of (12) in c6
-> 6(2)r7c5 from [15] or [24]
-> 10(3)r2c6 cannot be {145}
-> Either 3(2)r1c6 = [12] and 10(3)r2c6 = [{23}5]
or 3(2)r1c6 = [21] and 10(3)r2c6 = [{13}6]

2. 4 in n2 in 19(3) in r23c4
-> 19(3)r2c4 = {469} or {478}
-> One of (6789) in r1c3 and in r23c4
If (69) in 19(3)r2c4 -> 15(2)n8 = {78} and vice versa
-> Whichever of (6789) is in r8c5 is also in 35(5) in r1c4
-> Remaining cells in c4 -> r5679c4 = {1235}

3. IOD c789 -> r5c6 = r1c7 + r9c7 + 4
-> r5c6 is Min 7, r19c7 is max +5, r9c7 is max 4.

4. IOD n8 -> r79c4 = r9c7 + 2
Since r79c4 is Min +4 (One of (12) is in 6(2)n8) -> r9c7 is min 2.
I.e., r9c7 from (234)

5. IOD n4 -> Innies n4 (r4c2 + r4c3 + r5c2) = Outies n4 (r6c4 + r7c1) + 1
Since value in outie r7c1 must also be in one of the innies in n4 -> remaining innies n4 = r6c4 + 1 (Where have I seen that before? :))
Since remaining two innies n4 are min +3 -> min r6c4 = 2

6. Innies n5. r4c4 is min 6, r4c6 from (56), r5c6 is min 7
Since (1234) cannot all go in 21(5)n5 -> r6c4 must be one of them.
-> r6c4 from (23)

7! Since 4 in r23c4 and r6c4 from (23) -> Innies n8 = r7c4 + r9c4 cannot be +5
-> IOD n8 -> r9c7 cannot be 3
-> r9c7 from (24)
Since r1c7 from (12) and r9c7 from (24) and max r1c7 + r9c7 = +5 -> r1c7 cannot be 2!
-> 3(2)r1c6 = [21] and 10(3)r2c6 = [{13}6]

8. -> Innies n5 from {2679} or {3678}
Either r9c7 = 2 -> Innies n8 = r79c4 = {13} -> r6c4 = 2
Or r9c7 = 4 -> r5c6 = 9 -> Innies n5 = [7692]
Either way -> r6c4 = 2
r4c4,r5c6 = {79} and 21(5)n5 = {13458}

9. -> Remaining Innies n4 (Step 5 Line 2) = +3(2) = {12}
-> 8(2)n4 = {35}
-> 36(6)r5c3 = {246789} with 2 in r6c4

10. IOD n8 -> r9c7,{r7c4,r9c4} = 2,{13} or 4,{15}
-> 6(2)n8 = [24]

11! This one finally breaks it I think
IOD n78 -> [r7c1,r9c7] = [62] or [84]
-> (479) in n4 in 36(6)
-> 11(2)n6 cannot be {47}
Also 2 in r6c4 -> 11(2)n6 cannot be {29}
-> 11(2)n6 from {38} or {56}

IOD n9 -> r56c9 is max +5(2)
-> Whichever of (79) is in r4c4 goes in n6 in r5c78
i.e., 28(5) contains both (79)

2 in n3 in r3c8 or r23c9
If the latter this puts 2 in n6 in the 28(5)
I.e., 28(5) must contain all of (279)
-> 28(5) = {24679}

IOD n3 -> r4c79 = r3c8 + 10
Since r4c79 cannot be {79} -> r3c8 cannot be 6
-> 6 in 28(5) in r5c78

12. Easier now. For example,
-> 11(2)n6 = {38}
6 in n4 in r6c123
-> [r7c1,r9c7] = [84]
-> r5c6 = 9 and r4c4 = 7
-> 7 in n6 in r5c78
-> 36(6)r5c3 = [4{679}28]
-> r56c9 = [14]
-> r34c8 = [42]
Also r79c4 = {15}
-> r5c4 = 3
Also (HS) r9c5 = 3
-> r89c6 = {78}
-> 15(2)n8 = {69}
Also 17(5)r5c9 = [14{37}2]
Also 17(2)n9 = [89]
etc.
Author:  Andrew [ Sun Jul 11, 2021 8:57 pm ]
Post subject:  Re: Assassin 403
Thanks Ed for a completely new Assassin. First time through I made some false assumptions so had to rework from step 8 looking in different areas.

It's a long time since I was last at Wembley, for the 1966 World Cup Final.

While I'm waiting until the game is almost over before watching the recording with fast-forwarding and just looking at any interesting bits.

Ed and I used many of the same 45s but some in different ways; my way is possibly harder, I used two forcing chains.

I haven't been rating the Revisits so I won't rate this one either; just to add that the SS score of 2.0 is far too high, because of what SS isn't programmed to do.

Here's how I solved Assassin 403:
Prelims

a) R1C67 = {12}
b) R1C89 = {39/48/57}, no 1,2,6
c) R45C1 = {17/26/35}, no 4,8,9
d) R6C78 = {29/38/47/56}, no 1
e) R7C56 = {15/24}
f) 7(2) cage at R8C2 = {16/25/34}, no 7,8,9
g) R8C45 = {69/78}
h) 17(2) cage at R8C9 = {89}
i) 19(3) cage at R2C4 = {289/379/469/478/568}, no 1
j) 10(3) cage at R2C6 = {127/136/145/235}, no 8,9
k) 28(4) cage at R2C7 = {4789/5689}, no 1,2,3
l) 35(5) cage at R1C3 = {56789}
m) 17(5) cage at R5C9 = {12347/12356}, no 8,9

1a. Naked pair {12} in R1C67, locked for R1
1b. Naked pair {89} in 17(2) cage at R8C9, locked for N9
1c. Killer pair 8,9 in R8C45 and R8C9, locked for R8
1d. 10(3) cage at R2C6 = {136/145/235} (cannot be {127} which clashes with R1C6), no 7
1e. Killer pair 1,2 in R1C6 and 10(3) cage, locked for C6, clean-up: no 4,5 in R7C5
1f. 10(3) cage = {136/235} (cannot be {145} which clashes with R7C6), no 4, 3 locked for C6
1g. 4 in N2 only in R23C4, locked for C4
1h. 19(3) cage at R2C4 = {469/478}, no 2,3,5
1i. R123C6 = {123} (hidden triple in N2), locked for C6
1j. 5 in N2 only in R1C45 + R23C5, locked for 35(7) cage at R1C3, no 5 in R1C3
1k. 17(5) cage at R5C9 = {12347/12356}, CPE no 1,2,3 in R9C9
1l. 34(7) cage at R1C1 must contain 1, CPE no 1 in R2C2

2a. 45 rule on complete grid 4(1+3) innies R2C2 + R234C3 = 14
2b. Min R234C3 = 6 -> max R2C2 = 8
2c. Min R2C2 + R23C3 = 6 -> max R4C2 = 8

3a. 45 rule on N7 1 innie R7C1 = 2 outies R79C4 + 2
3b. Min R79C4 = 4 (cannot be {12} which clashes with R7C5) -> min R7C1 = 6
3c. Max R7C1 = 9 -> max R79C4 = 7, no 7,8,9 in R79C4
3d. Max R79C4 + R7C56 = 13 must contain 1, locked for N8
[Ed pointed out CPE no 1 in R7C3. He’s better at spotting those less obvious CPEs than I am!]
3e. 45 rule on N78 1 innie R7C1 = 1 outie R9C7 + 4, R7C1 = {6789} -> R9C7 = {2345}
3f. 45 rule on N9 2 outies R56C9 = 1 innie R9C7 + 1, max R9C7 = 5 -> max R56C9 = 6, no 6,7 in R56C9
3g. 8,9 in R7 only in R7C123, locked for N7

4a. 45 rule on C6789 2 innies R67C6 = 1 outie R9C5 + 6, IOU no 6 in R6C6
4b. Min R67C6 = 9 -> min R9C5 = 3
4c. Max R67C6 = 14 -> min R9C5 = 8
4d. R67C6 cannot total 10 -> no 4 in R9C5
4e. 4 in C5 only in R456C5, locked for N5

5a. 22(4) cage at R8C6 = {2569/2578/3469/3478} (cannot be {2389/2479/3568/4567} which clash with R8C45)
5b. 2 of {2569/2578} must be in R9C7 -> no 5 in R9C7, clean-up: no 9 in R7C1 (step 3e)
5c. Killer quad 6,7,8,9 in 22(4) cage and R8C45, locked for N8
5d. Max R9C7 = 4 -> max R56C9 (step 3f) = 5, no 5 in R56C9

6a. 45 rule on N2356789 2 innies R6C4 + R7C1 = 1 outie R1C3 + 2
[Or 45 rule on N14 using step 2a for the 4 zero cells.]
6b. Max R1C3 = 9 -> max R6C4 + R7C1 = 11, min R7C1 = 6 -> max R6C4 = 5
6c. 36(6) cage at R5C3 must contain 9, locked for N4

7a. 45 rule on N3 2 outies R4C79 = 2 innies R1C7 + R3C8 + 9
7b. Max R4C79 = 17 -> max R1C7 + R3C8 = 8 -> max R3C8 = 7
7c. Min R1C7 + R3C8 = 3 -> min R4C79 = 12, no 1,2 in R4C9

8a. 45 rule on N36 1 outie R5C6 = 3(1+2) innies R1C7 + R56C9 + 3
8b. Min R1C7 + R56C9 = 1 + 3 -> min R5C6 = 7
8c. 45 rule on N5 4 innies R4C46 + R5C6 + R6C4 = 24 = {1689/2589/2679/3579/3678} -> R6C4 = {123}
8d. R4C6 = {56} -> no 6 in R4C4
8e. 19(3) cage at R2C4 (step 1h) = {469/478}
8f. 9 of {469} only in R4C4 -> no 9 in R23C4
8g. 9 in N2 only in R1C45 + R23C5, locked for 35(7) cage at R1C3, no 9 in R1C3

9a. R7C1 = R9C7 + 4 (step 3e), R6C4 + R7C1 = R1C3 + 2 (step 6a), 22(4) cage at R8C6 (step 5a) = {2569/2578/3469/3478}, R4C46 + R5C6 + R6C4 = 24 (step 8c)
9b. Consider placements for R9C7 = {234}
R9C7 = 2 => R1C67 = [21], R23C6 = {13} => R4C6 = 6 (cage sum)
or R9C7 = 3 => 22(4) cage must contain 4, locked for N8 => R7C6 = 5, R4C6 = 6
or R9C7 = 4, 22(4) cage must contain 3, R7C6 = 4 (hidden single in N8) => R7C5 = 2, R79C4 = {15}, 1 locked for C4, R7C1 = 8 => R1C3 + R6C4 = [82] (only possibility) => R23C4 = {48} (hidden pair in N2), R4C4 = 7 (cage sum), R46C4 = [72] = 9 => R46C6 = 15 = [69]
-> R4C6 = 6, R23C6 = 4 = {13}, R1C67 = [21], clean-up: no 2 in R5C1
9c. 1 in C9 only in R567C9, locked for 17(5) cage at R5C9, no 1 in R7C8
9d. 14(3) cage at R2C9 = {239/248/257/347/356}
9e. R56C9 = R9C7 + 1 (step 3f)
9f. R9C7 = {234} -> R56C9 = 3,4,5 = {12/13/14} (cannot be {23} which clashes with 14(3) cage), 1 locked for C9 and N6
9g. R8C8 = 1 (hidden single in N9) -> R7C7 + R9C9 = 11 = {47/56}, no 1,2, clean-up: no 6 in R9C1

10a. R6C4 + R7C1 = R1C3 + 2 (step 6a), R7C1 = R79C4 + 2 (step 3a)
10b. Consider placements for R7C1 = {678}
R7C1 = 6 => R79C4 = 4 = {13}, 3 locked for C4
or R7C1 = 7 => R79C4 = 5 = {23}, 3 locked for C4
or R7C1 = 8 => R1C3 + R6C4 = [71/82]
-> R6C4 = {12}
[The final key step. The rest is long but fairly straightforward.]
10c. R4C46 + R5C6 + R6C4 (step 8c) = {1689/2589/2679}, 9 locked for N5
10d. 36(6) at R5C3 = {156789/246789} (cannot be {345789} because R6C4 only contains 1,2)
10e. R6C4 = {12} -> 1,2 in R5C3 + R6C123
10f. 36(6) = {156789/246789}, CPE no 6,7 in R45C1, clean-up: no 1,2 in R45C1
10g. Naked pair {35} in R45C1, locked for N4, clean-up: no 2,4 in R8C2
10h. 36(6) = {246789} -> R6C4 = 2, clean-up: no 9 in R6C78
10i. R4C46 + R5C6 + R6C4 = 24 (step 8c), R4C6 + R6C4 = [62] = 8 -> R4C4 + R5C6 = 16 = {79}, 7 locked for N5
10j. 19(3) cage at R2C4 = {469/478}, R4C4 = {79} -> no 7 in R23C4
10k. R7C5 = 2 (hidden single in N8) -> R7C6 = 4, clean-up: no 7 in R9C9 (step 9g)
10l. 1 in N8 only in R79C4, locked for C4
10m. R79C4 = {13/15} = 4,6 -> R7C1 = {68}, clean-up: no 3 in R9C7 (step 3e)
10n. R6C4 = 2 -> R1C3 = R7C1 = {68}
10o. 2 in N9 only in R89C7, locked for C7
10p. 3 in N9 only in R7C89 + R8C7, locked for 17(5) cage at R5C9, no 3 in R56C9
10q. 9 in R6 only in R6C123, locked for 36(6) cage, no 9 in R5C3
10r. 9 in R5 only in R5C567, locked for 28(5) cage at R3C8, no 9 in R4C8
10s. 14(3) cage at R23C9 = {239/257/347/356} (cannot be {248} which clashes with R56C9, ALS block), no 8

11a. R67C6 = R9C5 + 6 (step 4a), R7C6 = 4 -> R6C6 + R9C5 = [53/86]
11b. Hidden killer pair 8,9 in R8C45 and R9C6, R8C45 contains only of 8,9 -> R9C6 = {89}
11c. 7 in R9 only in R9C23, locked for N7
11d. 7 in R7 only in R7C789, locked for N9

12a. R4C79 = R1C7 + R3C8 + 9 (step 7a)
12b. R4C79 cannot be {79} which clashes with R4C4 -> R1C7 + R3C8 cannot total 7, no 6 in R3C8
12c. 28(5) cage at R3C8 contains 2 for C8 and 9 for R5 = {23689/24589/24679}
12d. {23689} must have one of 2,3 in R3C8, {24589/24679} must have one of 2,4 in R3C8 (both of 2,4 in R4C8 + R5C78 would clash with R56C9, ALS block) -> R3C8 = {234}
12e. R6C78 = {38/47} (cannot be {56} which clashes with 28(5) cage), no 5,6
12f. 6 in R6 only in R6C123, locked for 36(6) cage -> R7C1 = 8
12g. R6C78 = {38} (cannot be {47} which clashes with R6C123, ALS block), locked for R6 and N6
12h. R5689C6 = [9578], naked pair {69} in R8C45, locked for R8, 6 locked for N8, R9C5 = 3 -> R9C7 = 4 (cage sum), clean-up: no 7 in R7C7 (step 9g), no 3 in R8C2, no 1 in R9C1
12i. R8C7 = 2 (hidden single in N9), R7C89 = {37} (hidden pair in N9), naked pair {14} in R56C9, 4 locked for C9 and N6
12j. R4C4 = 7 -> naked pair {59} in R4C79, 5 locked for R4 and N6, R4C8 = 2
12k. R4C79 = {59} = 14 -> R1C7 + R3C8 = 5 = [14]
12l. Naked pair {67} in R5C78, locked for R5 -> R5C3 = 4
12m. R4C4 = 7 -> R23C4 = 12 = [48]

13a. R8C3 = 3 (hidden single in R8), R79C4 = {15} = 6 -> R79C3 = 13 = [67]
13b. R8C1 = 4, 7(2) cage at R8C2 = [52] -> R79C2 = [91], R45C2 = [82]
13c. R16C3 = [89], R4C3 = 1, R23C3 = {25} = 7 -> R2C2 = 6 (cage sum for zero cages, step 2a)
13d. Killer pair 7,9 in R1C1 and R1C89, locked for R1
13e. Naked pair {56} in R1C45, locked for N2, 5 locked for R1, clean-up: no 7 in R1C89

and the rest is naked singles.
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