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 Post subject: Assassin 366
PostPosted: Wed Dec 19, 2018 6:53 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
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Assassin 366
Merry Christmas! Sorry, not a themed puzzle. Very pleasant solution but quite long. All standard vanilla Assassin type steps. Hopefully you guys can find a shortcut. SS has a ridiculously hard time with with this (3.25) but JSudoku (6 chains) had no more trouble with it than the original 1.25 score puzzle I started with.

code:
3x3::k:0000:0000:10241:2306:2306:10241:771:2820:2820:0000:4357:10241:10241:10241:10241:771:9222:3335:0000:4357:10241:3849:3850:3850:9222:9222:3335:0000:4357:10241:3849:3850:3850:6923:9222:2060:3085:3085:7694:3849:6923:6923:6923:9222:2060:2831:2831:7694:7694:2832:2832:9222:9222:7697:4370:7694:7694:3859:4628:4628:4628:4628:7697:4370:4373:7694:3859:3859:3080:2816:7697:7697:4370:4373:4373:4373:3859:3080:2816:7697:7697:
solution:
Code:
+-------+-------+-------+
| 9 7 6 | 4 5 1 | 2 3 8 |
| 2 5 3 | 9 8 7 | 1 4 6 |
| 1 8 4 | 2 3 6 | 5 9 7 |
+-------+-------+-------+
| 8 4 2 | 6 1 5 | 9 7 3 |
| 3 9 1 | 7 4 8 | 6 2 5 |
| 5 6 7 | 3 9 2 | 8 1 4 |
+-------+-------+-------+
| 7 2 8 | 1 6 4 | 3 5 9 |
| 6 1 9 | 5 7 3 | 4 8 2 |
| 4 3 5 | 8 2 9 | 7 6 1 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 366
PostPosted: Thu Dec 27, 2018 5:22 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Felt a lot harder than this WT suggests. Step 18 was hard to see even though it is very basic. Thanks to Andrew for some suggestions and corrections. Also, Andrew wanted to know about how I did the 'overlap' in step 17 - wrongly it turns out!!
A366:
Preliminaries courtesy of SudokuSolver
Cage 3(2) n3 - cells ={12}
Cage 8(2) n6 - cells do not use 489
Cage 12(2) n8 - cells do not use 126
Cage 12(2) n4 - cells do not use 126
Cage 13(2) n3 - cells do not use 123
Cage 9(2) n2 - cells do not use 9
Cage 11(2) n4 - cells do not use 1
Cage 11(2) n5 - cells do not use 1
Cage 11(2) n9 - cells do not use 1
Cage 11(2) n3 - cells do not use 1
Cage 27(4) n56 - cells do not use 12
Cage 40(8) n124 - cells ={12346789}

No routine clean-up done unless stated.

1. 3(2)n3 = {12}: both locked for n3 and c7

2. "45" on c3-9: 3 outies r789c2 = 6 = {123} only: all locked for c2 and n7

3. 17(3)n7 = {458/467}(no 9)
3a. must have 4: 4 locked for n7 and c1

4. 17(3)r2c2 = {458/467}(no 9)
4a. must have 4: 4 locked for c2

5. combined cage 12(2)+11(2)n4: 12(2 ) = [39]/{57}(no 8); 11(2) = [29/38]{56}(no 7);
5a. if 12(2) = {39} -> 11(2) = {56}
5b. or 12(2) = {57}
5c. must have 5: 5 locked for n4

6. 5 in c3 only in n7: 5 locked for n7

7. 17(3)n7 = {467} only: 6&7 locked for c1 and n7
7a. no 5 in r5c2, no 5 in r6c2

8. "45" on c2-9: 3 innies r156c2 = 22
8a. but {589} as [598] only blocked by [33] in r56c1
8b. = {679} only: 6,7 locked for c2
8c. but [976] blocked by [55] in r56c1 -> no 9 in r1c2
8d. no 3 in r6c1

9. naked triple {589} in r789c3: 8 & 9 locked for c3

10. 40(8)r1c3 must have 8 & 9 which are only in n2: locked for n2
10a. no 1 in 9(2)n2

11. 11(2)n3 = {38/47/56}
11a. -> {36} blocked from 9(2)n2 since [7}{36} in r1c245 clashes with 11(2)n3
11b. -> 9(2)n2 = {27/45}(no 3,6) = 4/7
11c. ->{47} blocked from 11(2)n3
11d. & {56} blocked from 11(2) by r1c2 and 9(2)n2
11e. 11(2)n3 = {38}: both locked for r1 and n3
11f. no 5 in 13(2)n3

12. 5 in n4 only in r56c1: locked for c1

13. 5 in r1 only in 9(2)n2 = {45}: both locked for n2

14. r3c3 sees all 1,2,3,6,7 in n2 -> can't have any of these (CPE)
14a. r3c3 = 4

15. naked pair {58} in r23c2: 8 locked for n1
15a. r4c2 = 4

16. r4c1 = 8 (hsingle n4)
16a. r3c2 = 8 (hsingle r3), r2c2 = 5

At this spot, can place 9 in r1c1 using overlap of r1 and c1 but doesn't add anything to the solution. Was fun finding it though! But flawed arithmetic! So ignore this.

17. "45" on n3: 3 innies r2c8 + r3c78 = 18
17a. must have 5 for n3 = {459/567}
17b. 5 locked for 36(7) -> no 5 in r456c8 nor r6c7
17c. note: split 18(4) in n6

This is the step that took the longest to find and makes a big difference.
18. "45" on r6789: remembering sp18(4)n6, 3 outies r5c3 + r45c8 = 10
18a. -> no 9 in r45c8

19. 9 in r4 only in r4c4567
19a. r5c56 see all of those -> no 9 in r5c56 (CPE)

20. 27(4)r4c7 = {3789/4689/5679}: must have 9 -> 9 locked for n6 and c7

21. 36(7)r2c8 must have 9 -> h18(3)n3 = {459} only (step 17a)
21a. = [459] only permutation

22. 13(2)n3 = {67}: both locked for c9
22a. -> 8(2)n6 = {35} only: both locked for c9 and n6
22b. r1c89 = [38]

23. "45" on c9: two remaining outies r89c8 = 14 = {68} only: both locked for c8 and n9
23a. -> 11(2)n9 = {47} only: both locked for n9 and c7
23b. r7c7 = 3

24. "45" on c89: 1 remaining outie r6c7 - 3 = 1 innie r7c8 = [85] only permutation
(Andrew suggested an alternative: r456c8 = {127} = 10 -> r6c7 = 8 (cage sum); hidden single 5 in n9 -> r7c8 = 5)

25. r6c9 = 4 (hsingle n6)

26. "45" on n89: 1 innie r9c4 = 8, r89c8 = [86]
26a. r9c4 = 8 -> 17(4)r8c2 = {13}[58]: 1 locked n7
26b. r7c23 =[28], r8c3 = 9

27. r7c78 = 8 -> r7c56 = 10
27a. but {19} blocked by r7c9 = (19)
27b. r7c56 = {46} only: both locked for r7 and n8
27c. r789c1 = [764], r89c7 = [47]

28. r89c6 = [39] (only permutation)
28a. r7c4 = 1, r9c5 = 2

29. naked pair {35} in r5c19: both locked for r5

30. r45c7 = 15 -> r5c56 = 12 = {48} only: -> no 4 in r5c4

31. 15(3)r3c4: {357} blocked by r8c4 = (57)
31a. -> = {267} only: all locked for c4
31b. r8c45 = [57], r6c4 = 3

32. 2 & 3 in c3 only in 40(8)r1c3 -> no 2 or 3 in r1c6 nor r2c56

33. hsingle 3 in n2 -> r3c5 = 3

34. 2 in n2 only in r3: locked for r3

35. r6c4 + r7c23 + r8c3 = 22 -> r56c3 = 8 = {17} only: both locked for n4 and c3
easy now
Cheers
Ed


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 Post subject: Re: Assassin 366
PostPosted: Tue Jan 01, 2019 5:23 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Gradually getting back into Assassin practice. Just managed to finish this for 2018. In my case I was staring at my worksheet for quite a long time before I spotted the fairly obvious step 8 which cracked it for me.

Don't be put off by the SS score; that's because it's not programmed for step 2.

Happy New Year to all Assassins! Some of you are already into 2019 as I post this. ;)

Thanks Ed for your comments and corrections; I hope that the explanation for step 5a is now clearer.
Here is my walkthrough for Assassin 366:
Prelims

a) R1C45 = {18/27/36/45}, no 9
b) R12C7 = {12}
c) R1C89 = {29/38/47/56}, no 1
d) R23C9 = {49/58/67}, no 1,2,3
e) R45C9 = {17/26/35}, no 4,8,9
f) R5C12 = {39/48/57}, no 1,2,6
g) R6C12 = {29/38/47/56}, no 1
h) R6C56 = {29/38/47/56}, no 1
i) R89C6 = {39/48/57}, no 1,2,6
j) R89C7 = {29/38/47/56}, no 1
k) 27(4) cage at R4C7 = {3789/4689/5679}, no 1,2
l) 40(8) cage at R1C3 = {12346789}, no 5

Steps Resulting From Prelims
1a. Naked pair {12} in R12C7, locked for C7 and N3, clean-up: no 9 in R1C89, no 9 in R89C7
1b. 27(4) cage at R4C7 = {3789/4689/5679}, CPE no 9 in R5C8

2. 45 rule on C3456789 3 outies R789C2 = 6 = {123}, locked for C2 and N7, clean-up: no 9 in R5C1, no 8,9 in R6C1
2a. 17(3) cage at R2C2 = {458/467}, no 9, 4 locked for C2, clean-up: no 8 in R5C1, no 7 in R6C1
2b. 17(3) cage at R7C1 = {458/467}, no 9, 4 locked for C1 and N7, clean-up: no 8 in R5C2, no 7 in R6C2
2c. 9 in N7 only in R789C3, locked for C3
2d. 9 in 40(8) cage at R1C3 only in R1C6 + R2C456, locked for N2
2e. 17(4) cage at R8C2 can only contain two of 1,2,3 in R89C2 -> no 1,2,3 in R9C4
2f. R5C12 = [39]/{57}, R6C12 = [29/38]/{56} -> combined cage R56C12 = [39]{56}/[29]{57}/[38]{57}, 5 locked for N4
2g. 5 in C3 only in R789C3, locked for N7
2h. 17(3) cage at R7C1 = {467} (only remaining combination), locked for C1 and N7, clean-up: no 5 in R5C2, no 5 in R6C2
2i. 5 in N4 only in R56C1, locked for C1
2j. 8 in N7 only in R789C7, locked for C7
2k. 8 in 40(8) cage at R1C3 only in R1C6 + R2C456, locked for N2, clean-up: no 1 in R1C45

3a. 45 rule on R789 3 innies R7C23 + R8C3 = 1 outie R6C9 + 15
3b. Max R7C23 + R8C3 = 20 -> max R6C9 = 5
3c. 45 rule on N36 3 innies R45C7 + R6C9 = 19 = {289/379/469/478/568} -> R6C9 = {2345}, no 3,4,5 in R45C7
3d. 45 rule on N89 1 innie R9C4 = 1 outie R6C9 + 4, R6C9 = {2345} -> R9C4 = {6789}
3e. Min R89C2 + R9C4 = 9 -> max R9C3 = 8
3f. 9 in N7 only in R78C3, locked for 30(6) cage at R5C3

4. 45 rule on N9 2 innies R7C78 = 1 outie R6C9 + 4, max R6C9 = 5 -> max R7C78 = 9, no 9 in R7C7, no 7,8,9 in R7C8
4a. R6C9 + R7C78 cannot be [581] which clashes with R45C7 + R6C9 (step 3a) = {68}5 -> no 8 in R7C7
4b. R6C9 + R7C78 cannot be [536] because R89C7 = {38/47} and there’s no place for 5 in N9 -> no 6 in R7C8
4c. 9 in C7 only in R3456C7, CPE no 9 in R46C8
4d. 9 in N6 only in R456C7, locked for C7
4e. R45C7 + R6C9 (step 3c) = {289/379/469/478/568}
4f. Consider combinations for R89C7 = {38/47/56}
R89C7 = {38/56} => R45C7 + R6C9 cannot be {568} = {68}5
or R89C7 = {47} => R7C78 = [63] => R45C7 + R6C9 cannot be {568}
or R89C7 = {47} and R7C78 totals less than 9 => no 5 in R6C9
-> no 5 in R6C9, clean-up: no 9 in R9C4 (step 3d)
4g. Consider placements for R6C9 = {234}
R6C9 = {23} => R45C7 = {89/79}
or R6C9 = 4 => R45C7 = {69}
or R6C9 = 4 => R7C78 = 8 = [62] => R89C7 = {38/47} => R45C7 cannot be {78}
or R6C9 = 4 => R7C78 = 8 = {35} => R89C7 = {47} => R45C7 cannot be {78}
or R6C9 = 4 => R7C78 = 8 = [71] => R45C7 cannot be {78}
-> R45C7 + R6C9 = {289/379/469} (cannot be {478}, 9 locked for N6 and 27(4) cage at R4C7
[Ed pointed out that the eliminations in steps 4f and 4g could be done in the same way as step 4b.]
4h. 36(7) cage at R2C8 must contain 9 in R23C8, locked for C8 and N3, clean-up: no 4 in R23C9
4i. R1C89 = {38/47}, cannot be {56} which clashes with R23C9
4j. Killer pair 7,8 in R1C89 + R23C9, locked for N3

5. The two digits missing from 36(7) cage at R2C8 must total 9
5a. 36(7) cage must contain one or both of 1,8 and 2,7 which aren’t in R2C8 + R3C78 so must be in R456C8 + R6C7 -> R45C9 = {26/35} (cannot be {17} which clashes with 1,8 and/or 2,7), no 1,7
5b. Killer pair 5,6 in R23C9 and R45C9, locked for C9
5c. 1 in N6 only in R456C8, locked for C8 -> R456C8 + R6C7 must contain 8, locked for N6
5d. R45C7 + R6C9 (step 4g) = {379/469}, no 2, clean-up: no 6 in R9C4 (step 3d)
5e. 17(4) cage at R8C2 = {1358/2357} -> R9C3 = 5, 3 locked for N7, clean-up: no 7 in R8C6, no 6 in R8C7
5f. Naked pair {89} in R78C7, 8 locked for 30(6) cage at R5C3
5g. 45 rule on N3 3 outies R456C8 + R6C7 = 18 and contains 1,8 = {1278/1458} (cannot be {1368} which clashes with R45C9), no 3,6
5h. 3 in N6 only in R456C9, locked for C9, clean-up: no 8 in R1C8
5i. 8 in N3 only in R123C9, locked for C9
5j. Max R6C9 = 4 -> max R7C78 (step 4) = 8, no 7 in R7C7
5k. Consider placements for R6C9
R6C9 = 3 => max R7C78 = 7 => no 6 in R7C7
or R6C9 = 4 => R45C7 = {69}, locked for C7
-> no 6 in R7C7

6. 30(6) cage at R5C3 contains 8,9 in R78C3 = {123789/124689} (cannot be {134589} with R56C3 = {34}, R6C4 = 5, CPE no 3,5 in R6C1 so R6C14 = [25] + R6C39 = {34} clash with R6C56), no 5 in R6C4
[I’ve replaced my original step 6a, which was done when my mind was still focussed on getting more from the 30(6) cage, but Ed pointed out that it hadn’t eliminated R56C12 = [5738]. The new step 6a is simpler.]
6a. R5C12 = [39/57], R6C12 = [29/38/56] -> combined cage R56C12 = [3956/5729] (cannot be [5738] clashes with 17(3) cage at R2C2)
-> 9 in R56C2, locked for C2 and N4
[Ed had found 45 rule on C23456789 3 remaining innies R156C2 = 22 = {679} (cannot be [598] which would make R56C1 both 3), then cannot be [976] which would make R56C1 both 5) -> 9 locked in R56C2 for C2 and N4.]
6b. R6C12 = [29/56], no 3 in R6C1, no 8 in R6C2
6c. 8 in N4 only in R4C12, locked for R4
6d. 17(3) cage at R2C2 (step 2a) = {458} (only remaining combination, cannot be {467} which clashes with R56C2, ALS block), locked for C2
6e. 9 in R6 only in R6C12 = [29] or R6C56 = {29}, 2 locked for R6 (locking cages)

7. 45 rule on N8 3 innies R7C56 + R9C4 = 18 = {189/279/468/567} (cannot be {369} because R9C4 only contains 7,8, cannot be {378} which clashes with R89C6, cannot be {459} which doesn’t contain 7,8), no 3
7a. 7,8 of each combination must be in R9C4 -> no 7,8 in R7C56
7b. Consider combinations for R7C56 = {19/29/46/56}
R7C56 = {19/46} = 10 => R7C78 = 8 = {35}
or R7C56 = {29/56} = 11 => R7C78 = 7 = {34} (cannot be [52] which clashes with R7C56)
-> R7C78 = {34/35}, no 2, 3 locked for R7 and N9, clean-up: no 8 in R89C7
7c. Killer pair 4,5 in R7C78 and R89C7, locked for N9
7d. Killer pair 6,7 in R45C7 and R89C7, locked for C7
7e. R6C7 = 8 (hidden single in C7), clean-up: no 3 in R6C56

8. 45 rule on R6789 1 outie R5C3 = 1 remaining innie R6C8, no 2,3,6 in R5C3, no 5 in R6C8
8a. 30(6) cage at R5C3 (step 6) = {123789/124689} -> R7C2 = 2
[Cracked, the rest is straightforward.]
8b. R89C2 = {13}, R9C3 = 5 -> R9C4 = 8 (cage sum), R6C9 = 4 (step 3d), clean-up: no 7 in R1C8, no 4 in R5C3, no 7 in R6C56, no 4 in R89C6
8c. 30(6) cage = {123789} (only remaining combination), no 6
8d. R45C7 + R6C9 = 19 (step 3c), R6C9 = 4 -> R45C7 = 15 = {69}, 6 locked for C7, N6 and 27(4) cage at R5C7, clean-up: no 2 in R45C9, no 5 in R8C7
8e. Naked pair {35} in R45C9, locked for C9 and N6, clean-up: no 8 in R23C9
8f. Naked pair {67} in R23C9, locked for C9 and N3, clean-up: no 4 in R1C8
8g. Naked pair {47} in R89C7, locked for C7 and N9
8h. R1C89 = [38] -> R3C7 = 5, R7C78 = [35], clean-up: no 6 in R1C45
8i. R2C2 = 5 (hidden single in C2)
8j. R89C8 = [86] (hidden pair in C8) -> R78C3 = [89], clean-up: no 3 in R9C6
8k. Naked pair {35} in R5C19, locked for R5
8l. R45C7 = 15 -> R5C56 = 12 = {48}, 4 locked for N5
8m. R7C78 = [35] = 8 -> R7C56 = 10 = {46} (cannot be {19} which clashes with R7C9), locked for R7 and N8 -> R7C1 = 7, R89C1 = [64], R89C7 = [47], R9C6 = 9 -> R8C6 = 3, R7C49 = [19], R9C5 = 2, clean-up: no 7 in R1C4
8n. 30(6) cage = {123789}, 1 locked for C3 and N4
8o. 40(8) cage at R1C3 = {12346789}, 1 locked for N2
8p. 5 in R1 only in R1C45 = {45}, locked for R1 and N2
8q. R3C12 = [18] (hidden pair in R3), R4C2 = 4, R4C1 = 8 (hidden single in R4)
8r. R3C8 = 9 (hidden single in R3), R2C8 = 4, R3C3 = 4 (hidden single in C3)

8s. 2,6 in C3 only in R124C3, locked for 40(8) cage
8t. 3 in R3 only in R3C456, locked for N2
8u. 3 in 40(8) cage only in R24C3, locked for C3
8v. Naked pair {17} in R56C3, locked for C3, N4 and 30(6) cage at R5C3 -> R6C4 = 3, R5C2 = 9 -> R5C1 = 3, R6C2 = 6 -> R6C1 = 5
8w. 15(3) cage at R3C4 = {267} (only remaining combination), locked for C4

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A366 at 1.5. I used a few forcing chains.


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 Post subject: Re: Assassin 366
PostPosted: Fri Jan 04, 2019 7:45 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Happy New Year everyone, and thanks again Ed! Here is my WT. We all started in almost exactly the same way, but after that although our conclusions followed much the same path - our reasoning to reach those conclusions differed a little.

Assassin 366 WT:
1. Outies c3456789 = r789c2 = +6(3) = {123}
-> 4 in 17(3)n7 and 4 in 17(3)r2c2
-> 12(2)n4 from {57} or [39]
If the latter -> 11(2)n4 = {56}
-> 5 in n4 in r56c12
-> 5 in c3 in r789c3
-> 17(3)n7 = {467}
-> r789c3 = {589}

2. 12(2)n4 = [57] or [39]
If the latter -> 11(2)n4 = [56]
-> Remaining Innies c2 = r156c2 = +22(3)
and since either r5c2 is 7 or r6c2 is 6 -> r156c2 = {679}
-> 17(3)r2c2 = {458} with 5 in r23c2 (since 5 previously shown to be in n4 in r56c12)

3. Innies n36 = r45c7 + r6c9 = +19(3)
-> Min r6c9 = 2
Innies n3689 = r45c7 + r9c4 = +23(3)
-> Min r45c7 = +14(2)
-> Max r6c9 = 5

4. 3(2)n3 = {12}
-> 11(2)n9 cannot be {29}

5. r6c9 cannot be 5 since that puts r45c6 = {68)
which puts 11(2)n9 = {47}
which leaves no place for 5 in n9 (since remaining Innies n9 = r7c78 = +9(2)).
-> Max r6c9 = 4

6. Possibilities for Innies n36
r6c9 = 2 -> r45c7 = {89}
r6c9 = 3 -> r45c7 = {79}
r6c9 = 4 -> Innies n9 = r7c78 = +8(2) (No 4) -> 11(2)n9 = {47} -> r45c7 = {69}
-> 9 locked in r45c7

7! -> 9 in 36(7) in r23c8
-> Innies n3 = r2c8 + r3c78 = +18(3) must be {369} or {459}
-> 11(2)n3 cannot be {56}
-> HS 5 in r1 -> 9(2)n2 = {45}

8! Whatever goes in r3c3 (not 5!) can only go in 9(2) in n2
-> r3c3 = 4
-> 17(3)r2c2 = [{58}4]
Also HS 4 in r2 -> r2c8 = 4
-> Remaining Innies n3 -> r3c78 = [59]
-> r23c2 = [58]
Also 13(2)n3 = {67}
-> 8(2)n6 = {35}
-> 11(2)n3 = [38]
Also HS 4 in n6 -> r6c9 = 4
-> 11(2)n9 = {47} (Since remaining Innies n9 = +8(2))
-> Remaining Innies n36 -> r45c7 = {69}
Also r789c9 = {129}
-> r89c8 = {68}
-> r7c78 = [35]
-> r6c7 = 8 and r456c8 = {127}

9. Also remaining Innie n3689 = r9c4 = 8
-> 17(4)n7 = [{13}58]
-> r7c2 = 2 and r78c3 = {89}

10. 11(2)r6c1 and 11(2)r6c5 must be between them {29} and {56}
-> r6c348 = {137}
-> 30(6)r5c3 = [{137}2{89}]
Since 12(2)n4 from [39] or [57] -> r6c4 from (37) and 1 in r56c3

11. HS 8 in n4 -> r4c1 = 8
-> Both (48) in n5 in r5c456
-> Since r5c56 = +12(2) -> r5c56 = {48}

12. Given previous placements only solution for 15(3)r3c4 = {267}
(Cannot be {357} since r6c4 from (37))
(Cannot be [1{59}] since 1 in n2 in 40(8))
-> r6c4 = 3
-> r56c3 = {17}
-> 12(2)n4 = [39]
-> 11(2)n4 = [56]
-> 11(2)n5 = {29}
Also r45c7 = [96]
-> 15(3)r3c4 = [267]
Also NS r4c3 = 2
-> r4c56 = {15}
-> r456c8 = [721]
-> r56c3 = [17]
Also r3c56 = {36}

13. r7c9 from (19)
-> r7c56 = +10(2) = {46}
-> 9(2)n2 = [45]
-> r4c56 = [15]
-> 12(2)n8 = {39}
etc.


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