SudokuSolver Forum

3x3::k:5632:5632:3330:3330:3076:3589:3589:3335:3335:5632:3850:3850:3330:3076:3589:4623:4623:3335:2066:3850:5396:5396:3076:3095:3095:4623:3866:2066:2066:5396:4894:5919:5919:3095:3866:3866:3108:3108:4894:4894:4894:5919:5919:2091:2091:4909:4909:3119:3888:3888:3888:5427:4660:4660:4909:2359:3119:3119:3888:5427:5427:1597:4660:3647:2359:2359:4930:2883:4420:1597:1597:4935:3647:3647:4930:4930:2883:4420:4420:4935:4935:

1. 6(3)n9 = {123}
Innies r89 = r8c2378 = +10(4) = {1234}
-> 4 in r8c23
-> 9(3)n7 = {234}
-> r7c28 = {23} and 1 in r8c78

2. 21(3)r3c3 -> r3c3 is min 4
Innies n1 = +8(3)
-> r3c3 is max 5
-> Outies c12 = r28c3 = +9(2) is not {45}
-> r8c2 = 4 and r28c3 either [72] or [63]
-> 22(3)n1 = {589}
-> 15(3)n1 = {267} with 2 in c2
-> 9(3)n7 = [342]
-> r2c3 = 7 and r23c2 = {26}
Also 21(3)r3c3 = [4{89}]
Also r1c3,r3c1 = {13}
Also 6(3)n9 = [2{13}]

3. Innies n7 = r7c13,r9c3 = +22(3) = {589} or [7{69}]
But that latter case leaves no place for 6 in n4.
-> Innies n7 = {589}
-> 14(3)n7 = {167} with 1 in r9 and 6 in c1
-> 6 in n4 in r56c3
Also -> 1 in n8 in r7c45

4. Whichever of (89) is in r4c3 goes in n7 in r7c1 and in n1 in r1c2
-> 5 in n1 in r12c1 and in n7 in r79c3
Also since 6 not in 19(3)r6c1 -> 5 not in r6c2 -> 5 in r45c2

Since 2 not in 12(3)r6c3 -> 9 not in r7c3
Also Innies n7 cannot be [985] since that puts r89c4 = {68} contradicting 21(3)r3c3 = [4{89}]
-> r7c3 = 5

This is pretty much where I hit a wall for a long time! Eventually I found the following:

5. Part 1.
Outies c89 = r28c7 = +7(2) = [61] or [43]
In the former case since 4 already in r3 and r8c8 = 3 -> 18(3)n3 = [648], [657], or [675]
In the latter case -> 18(3)n3 = [4{59}] or [4{68}]
-> r3c8 from (56789)

5. Part 2.
Outies r12 = r3c258 = +17(3)
Since r3c2 from (26), 4 already in r3, and r3c8 from (56789)...
-> r3c258 from [2{78}], [2{69}], [629], or [638]
-> 5 not in r3c8
-> r3c8 from (6789)

5. Part 3.
r3c258 cannot be [287] since that puts both r2c2 and r2c7 = 6
-> r3c8 from (689)

5. Part 4. (Here's a small chain)
5 in r3 only in r3c679
Trying r3c258 = [{26}9] and therefore 18(3)n3 = [459] ...
... puts 12(3)r3c6 = [516] which leaves no solution for remaining innies n3 (= r1c7,r3c9 = +13(2)).
-> r3c8 from (68)
-> 18(3)n3 = {468} with 4 in r2c78
-> r3c258 = [638] or [296]

5! Part 5. Finally!
Both (46) in r1 in n2 (r1c456)
Whether r1c3 is 1 or 3 -> 6 not in r1c4
Whether r3c5 is 3 or 9 -> 6 not in r1c5
-> r1c6 = 6
-> 2 not in r1c7
Also given 18(3)n3 = {468} -> 2 not in 13(3)n3
-> 2 in n3 in r3c79
-> r3c258 = [638]
-> 15(3)n1 = [276] and 18(3)n3 = [{64}8]

Essentially cracked now. Simple continuation. E.g.,

6. 21(3)r3c3 = [498]
-> Innies n7 = [859]
-> r1c2 = 8 and r12c1 = {59}
Also r3c1 = 1 and r1c3 = 3
-> 13(3)n3 = [{19}3]
Also r3c12 = [25]
-> 12(2)n4 = [39]
-> 19(3)r6c1 = [478]
Also r1c5 = 4 -> r2c5 = 5
Also r12c4 = [28]
Also 14(3)r1c6 = [671]
-> 12(3)r3c6 = [723]
etc.

Prelims

a) R5C12 = {39/48/57}, no 1,2,6
b) R5C89 = {17/26/35}, no 4,8,9
c) R89C5 = {29/38/47/56}, no 1
d) 22(3) cage at R1C1 = {589/679}
e) 8(3) cage at R3C1 = {125/134}
f) 21(3) cage at R3C3 = {489/579/678}, no 1,2,3
g) 19(3) cage at R6C1 = {289/379/469/478/568}, no 1
h) 21(3) cage at R6C7 = {489/579/678}, no 1,2,3
i) 9(3) cage at R7C2 = {126/135/234}, no 7,8,9
j) 6(3) cage at R7C8 = {123}
k) 19(3) cage at R8C4 = {289/379/469/478/568}, no 1
l) 19(3) cage at R8C9 = {289/379/469/478/568}, no 1

1a. 22(3) cage at R1C1 = {589/679}, 9 locked for N1
1b. Naked triple {123} in 6(3) cage at R7C8, locked for N9
1c. 45 rule on N1 3 innies R1C3 + R3C13 = 8 = {125/134} -> R3C3 = {45}, R1C3 + R3C1 = {12/13}, 1 locked for N1
1d. 21(3) cage at R3C3 = {489/579} (cannot be {678} because R3C3 only contains 4,5) -> R3C4 + R4C3 = {79/89}
1e. 45 rule on N7 3 innies R7C13 + R9C3 = 22 = {589/679}, 9 locked for N7
1f. Min R7C3 = 5 -> max R6C3 + R7C4 = 7, no 7,8,9 in R6C3 + R7C4
1g. 45 rule on R89 4 innies R8C2378 = 10 = {1234}, locked for R8, 4 locked for N7, clean-up: no 7,8,9 in R9C5
1h. 45 rule on R89 2 outies R8C28 = 5 = {23}
1i. 9(3) cage at R7C2 = {234} (only remaining combination), locked for N7
1j. 45 rule on C12 4 innies R2378C2 = 15 = {2346}, locked for C2, 6 locked for N1, clean-up: no 7 in 22(3) cage at R1C1, no 8,9 in R5C1
1k. R2C3 = 7 (hidden single in N1) -> R23C2 = 8 = {26}, 2 locked for C2 and N1
1l. R78C2 = [34] -> R8C3 = 2, R7C8 = 2, clean-up: no 6 in R5C9
1m. R3C3 = 4 (hidden single in N1) -> R3C4 + R4C3 = {89}, CPE no 8,9 in R4C4
1n. 45 rule on C89 2 outies R28C7 = 7 = [43/61]
1o. Max R2C7 = 6 -> min R23C8 = 12, no 1 in R23C8
1p. 45 rule on C89 3 remaining innies R238C8 = 15 = {159/168/348/357} (cannot be {456} because R8C8 only contains 1,3
1q. R8C8 = {13} -> no 3 in R23C8
1r. 1 in N7 only in R9C12, locked for R9
1s. 45 rule on C1234 1 outie R5C5 = 1 innie R6C4 + 3, no 1,2,3 in R5C5, no 7,8,9 in R6C4
1t. 45 rule on C6789 1 outie R4C5 = 1 innie R6C6, no 1,2,3,4 in R4C5, no 6,7,8,9 in R6C6
1u. 45 rule on R12 3 outies R3C258 = 17 = {269/278/368} (cannot be {cannot be {179/359} because R3C2 only contains 2,6), no 1,5
1v. Killer pair 8,9 in R3C258 and R3C4, locked for R3

2a. Consider permutations for 8(3) cage at R3C1 = [125/341] -> R4C2 = {24}
8(3) cage = [125]
or 8(3) cage = [341] => R5C12 = {57}
-> 5 in R4C2 + R5C12, locked for N4
2b. 5 in C3 only in R79C3, locked for N7
2c. R7C13 + R9C3 (step 1e) = {589}, 8 locked for N7
2d. 14(3) cage at R8C1 = {167}, 6 locked for C1
2e. Naked triple {589} in R127C1, 5 locked for C1 and N1, clean-up: no 7 in R5C2
2f. Naked triple {589} in R479C3, 8,9 locked for C3
2g. 19(3) cage at R6C1 = {289/379/478} -> R6C1 = {234}
2h. 12(3) cage at R6C3 = {138/156/345}, no 9
2i. R7C3 = {58} -> no 5 in R7C4

3a. 18(3) cage at R2C7 = {459/468/567} = [459/468/486/648/657], no 9 in R2C8
3b. R3C258 (step 1u) = {269/368} (cannot be [278] because R2C2 = 6 clashes with R2C7 = {46}cannot be [287] because R2C2 = 6 clashes with R2C78 = [65]), no 7, 6 locked for R3
3c. 3 of {368} must be in R3C5 -> no 8 in R3C5
3d. 18(3) cage at R2C7 = {459/468}, 4 locked for R2 and N3
3e. Hidden killer pair 5,7 in R3C67 and R3C9 for R3, R3C67 cannot contain both of 5,7 -> 12(3) cage at R3C6 must contain one of 5,7, R3C9 = {57}
3f. 12(3) cage at R3C6 = {147/156/237/345} -> R4C7 = {2346}
3g. 45 rule on N3 3 innies R1C7 + R3C79 = 14 = {158/167/257} (cannot be {239} because R3C9 only contains 5,7, cannot be {356} which clashes with 18(3) cage), no 3,9
3h. 6,8 of {158/167} only in R1C7 -> no 1 in R1C7
3i. Killer pair 5,6 in R1C7 + R3C79 and 18(3) cage, locked for N3
3j. 13(3) cage at R1C8 = {139/238}, no 7
3k. 7 in N3 only in R1C7 + R3C79 = {167/257}, no 8
3l. 12(3) cage = {147} can only be [174] (cannot be [714] which clashes with R28C7), 12(3) cage = {156} can only be [156] (cannot be [516] which clashes with R1C7 + R3C79 = [617] -> no 1 in R3C7
3m. R1C7 + R3C79 = {257} (only remaining combination), 2,5 locked for N3, 2 locked for C7
3n. 18(3) cage = {468} (only remaining combination), 8 locked for C8 and N3
3o. 12(3) cage = {156/345} must be [156/354] -> no 5 in R3C6
3p. 5 in R3 only in R3C79, locked for N3
3q. 9 in R3 only in R3C45, locked for N2
3r. R3C258 = {269/368} must be [296/638], no 2,6 in R3C5
3s. R3C5 = {39} -> no 6 in 12(3) cage at R1C5
3t. 6 in R1 only in R1C46, locked for N2
3u. R1C3 = {13} -> 13(3) cage at R1C3 cannot contain 6 (because no 4 in R2C4)
3v. R1C6 = 6 (hidden single in R1) -> R1C7 + R2C6 = 8 = [71], R3C79 = [25], R3C28 = [68] -> R3C1456 = [1937], R4C7 = 3 (cage sum), R4C2 = 5 -> R4C1 = 2 (cage sum), R8C7 = 1 -> R2C7 = 6 (step 1n), R2C8 = 4
[Cracked; clean-ups omitted from here.]

4a. R3C5 = 3 -> R12C5 = 9 = [45]
4b. R2C2 = 2 -> R12C4 = [28], R2C1 = 9, R1C12 = [58], R5C2 = 9 -> R5C1 = 3, R6C12 = [47], R7C1 = 8 -> R79C3 = [59]
4c. R89C5 = [92] (only remaining permutation) -> 21(3) cage at R6C7 = [849] (only remaining permutation)
4d. R9C3 = 9 -> R89C4 = 10 = [73], R89C1 = [67]
4e. R8C9 = 8 -> R9C89 = 11 = [56]
4f. R7C9 = 7 -> R6C89 = 11 = [92], R5C9 = 1 -> R5C8 = 7

and the rest is naked singles.

From that spot (I will have missed some clean-up), SS finds the clean-ups that Andrew does at 3s and 3u (but with many more candidates still in that 12(3) cage so hard to see) -> no 6 in r1c45. This locks 6 in r1 in the 14(3)r1c6. This is a big help (gets rid of 2 from r1c7).

But the interesting one is that from "45" on n3: 2 innies r1c7 + r3c9 = 2 outies r3c6 + r4c7 +2
Since r4c7 sees r1c7, they cannot be equal -> r3c6 cannot be 2 less than r3c9
-> 5 blocked from r3c6
-> 5 locked in r3 in the h14(3)n3 = {257} only
-> r3c7 = 2