SudokuSolver Forum

3x3:d:k:3328:3328:1794:1794:2564:3077:3077:4359:4359:3328:5642:2819:2819:2564:3598:3598:3856:4359:5642:5642:3348:3348:4374:4102:4102:3856:3856:5642:4124:3348:3348:4374:1824:1824:4898:3856:4124:4124:3366:3366:4374:2601:2601:4898:4898:4124:4910:3631:3631:2353:1842:1842:5428:4898:4910:4910:5688:5688:2353:2611:2611:5428:5428:4910:5184:5688:5688:2371:2884:2884:4678:5428:5184:5184:2634:2634:2371:769:769:4678:4678:

Prelims

a) R1C34 = {16/25/34}, no 7,8,9
b) R12C5 = {19/28/37/46}, no 5
c) R1C67 = {39/48/57}, no 1,2,6
d) R2C45 = {29/38/47/56}, no 1
e) R2C67 = {59/68}
f) R3C67 = {79}
g) R4C67 = {16/25/34}, no 7,8,9
h) R5C34 = {49/58/67}, no 1,2,3
i) R5C67 = {19/28/37/46}, no 5
j) R6C34 = {59/68}
k) R67C5 = {18/27/36/45}, no 9
l) R6C78 = {16/25/34}, no 7,8,9
m) R7C67 = {19/28/37/46}, no 5
n) R89C5 = {18/27/36/45}, no 9
o) R8C67 = {29/38/47/56}, no 1
p) R9C34 = {19/28/37/46}, no 5
q) R9C67 = {12}
r) 20(3) cage at R8C2 = {389/479/569/578}, no 1,2
s) 13(4) cage at R3C3 = {1237/1246/1345}, no 8,9

1a. Naked pair {79} in R3C67, locked for R3
1b. Naked pair {12} in R9C67, locked for R9, clean-up: no 7,8 in R8C5, no 8,9 in R9C34
1c. R2C45 = {29/38/47} (cannot be {56} which clashes with R2C67), no 5,6

2a. 45 rule on R6789 2 innies R6C19 = 7 = {16/25/34}, no 7,8,9
2b. Killer triple 4,5,6 in R6C19, R6C34 and R6C67, locked for R6, clean-up: no 3,4,5 in R7C5
2c. 4 in R6 only in R6C19 = {34} or R6C67 = {34}, 3 locked for R6 (locking cages), clean-up: no 6 in R7C5
[Alternatively 45 rule on R789 3 outies R6C258 = 17 contains 7 for R6 = {179/278}, no 3]
2d. R12C5 = {19/37/46} (cannot be {28} which clashes with R67C5), no 2,8
2e. 17(3) cage at R3C5 = {269/359/368/458} (cannot be {179/278} which clash with R67C5, cannot be {467} which clashes with R6789C5 which must contain at least one of 4,6,7), no 1,7

3a. 45 rule on N3 4(3+1) outies R123C6 + R4C9 = 29
3b. Max R123C6 = 24 -> min R4C9 = 5

[I missed 45 rule on R12 2 innies R2C28 = 6 which would have made this step a lot simpler, particular after R2C4 = {24} giving R2C28 = {15}.]
4a. 45 rule on R1 3 outies R2C159 = 14 = {149/167/239/248/257/347} (cannot be {158/356} which clash with R2C67)
4b. R2C34 (step 1c) = {29/38/47}, R2C67 = {59/68} -> combined cage R2C3467 = {29}{68}/{38}{59}/{47}{59}/{47}{68}
4c. R2C159 = {167/239/347} (cannot be {149/248/257} which clash with R2C3467), no 5,8
4d. R2C3467 = {29}{68}/{38}{59}/{47}{68} (cannot be {47}{59} which clashes with R2C159), 8 locked for R2
4e. Killer pair 7,9 in R2C159 and R2C3467, locked for R2
4f. Hidden killer pair 7,9 in 13(3) cage at R1C1 and R2C3, 13(3) cage cannot contain both of 7,9 -> R2C3 = {79}, R2C4 = {24}
4g. 8 in R2 only in R2C67 = {68}, 6 locked for R2, clean-up: no 4 in R1C5
4h. R2C159 = {239/347}, no 1, 3 locked for R2, clean-up: no 9 in R1C5
4i. R2C28 = {15} (hidden pair in R2), CPE no 5 in R5C5 + R8C2, no 1,5 in R8C8 using both diagonals
4j. 13(3) cage = {139/247} (cannot be {157} which clashes with R2C2, cannot be {148/238/256/346} because 13(3) cage must contain one of 7,9), no 5,6,8
4k. 8 in N1 only in R3C12, locked for R3 and 22(4) cage at R2C2
= {1489/1678/2578/3568} (cannot be {2389/3478} because R2C2 only contains 1,5)
4l. R2C2 = {15} -> no 1,5 in R3C12, R4C1 = {3679}
4m. 8 in N2 only in R12C6, locked for C6, clean-up: no 2 in R5C7, no 2 in R7C7, no 3 in R8C7
4n. Consider placement for R2C3 = {79}
R2C3 = 7 => 13(3) cage = {139}, 1 locked for R1
or R2C3 = 9
-> R12C5 = {37}/[64], no 1,9
4o. R89C5 = [18/27]/{45} (cannot be {36} which clashes with R12C5)
4p. 9 in C5 only in R45C5, locked for N5, clean-up: no 4 in R5C3, no 1 in R4C7, no 5 in R6C3
4q. 17(3) cage at R3C5 (step 2e) = {269/359}, no 4,8
4r. 13(3) cage cannot be {47}2 which clashes with R2C34 -> no 2 in R2C1, clean-up: no 9 in R2C9
4s. 13(3) cage + R2C34 cannot be {19}[374] which clashes with R2C5 -> no 3 in R2C1
4t. 13(3) cage = {13}9/{247}, no 9 in R1C12

5a. Hidden killer pair 8,9 in R1C67 and R1C89, neither can contain both of 8,9 -> R1C67 = {39/48}, no 5,7,
17(3) cage at R1C8 = {269/278/359/458} (cannot be {179} which clashes with R3C7, cannot be {368} which clashes with R2C7, cannot be {467} which doesn’t contain one of 8,9), no 1
5b. 3,4 of {359/458} must be in R2C9 -> no 3,4 in R1C89
5c. R1C34 = {16/25} (cannot be {34} which clashes with R1C67), no 3,4
5d. Killer pair 1,2 in 13(3) cage and R1C34, locked for R1
5e. 17(3) cage must contain one of 2,3,4 -> R2C9 = {234}
5f. 9 in N2 only in R13C6, locked for C6, clean-up: no 1 in R7C7, no 2 in R8C7
5g. R2C159 (step 4h) = {239/347} = {347}/[932]
5h. 17(3) cage = {269/359/458} (cannot be {78}2 which clashes with R12C5 + R2C19 = [7392]), no 7
5i. R3C7 = 7 (hidden single in N3), placed for D/, R3C6 = 9, clean-up: no 3 in R1C7, no 3 in R5C6, no 3 in R7C6, no 4 in R8C6
5j. Killer pair {56} in R1C34 and 17(3) cage, locked for R1, clean-up: no 4 in R2C5
5k. Naked pair {37} in R12C5, locked for C5, 3 locked for N2, clean-up: no 9 in R1C7, no 2 in R67C5, no 2 in R8C5
5l. Naked pair {18} in R67C5, locked for C5
5m. Naked pair {45} in R89C5, locked for N8, 5 locked for C5, clean-up: no 6 in R7C7, no 6 in R8C7, no 6 in R9C3
5n. Naked pair {48} in R1C67, locked for R1
5o. 8 in N3 only in R12C7, locked for C7, clean-up: no 2 in R5C6, no 2 in R7C6, no 3 in R8C6
5p. 3 in C6 only in R46C6, locked for N5
5q. 3 in R4C67 = [34] or R6C67 = [34] -> 4 in R46C7 (locking cages), locked for C7 and N6 -> R1C67 = [48], R2C67 = [86], R2C4 = 2 -> R2C3 = 9, R3C5 = 6
5r. Naked pair {59} in R1C89, 5 locked for R1 and N3, R2C8 = 1, placed for D/, R2C2 = 5, placed for D\
5s. R1C89 = {59} -> R2C9 = 3 (cage total), R12C5 = [37], R2C1 = 4, R1C4 = 1 -> R1C3 = 6, R6C3 = 8 -> R6C4 = 6, placed for D/
5t. Naked pair {24} in R3C89, 2 locked for R3
5u. R2C2 = 5, R3C12 = {38} = 11 -> R4C1 = 6 (cage sum)
5v. R2C8 = 1, R3C89 = {24} = 6 -> R4C9 = 8 (cage sum)
5w. R3C3 = 1 (hidden single in R3), placed for D\, R3C4 = 5 -> R4C34 = 7 = [34], 4 placed for D\
5x. R67C5 = [18], R5C6 = 7 -> R5C7 = 3, R5C34 = [58], R7C7 = 9, placed for D\ -> R7C6 = 1, R5C5 = 2, placed for both diagonals, R4C6 = 5, placed for D/, R1C9 = 9, placed for D/, R1C1 = 7, placed for D\, R6C6 = 3, placed for D\
5y. R8C8 + R9C9= [86] -> R9C8 = 4 (cage sum)

and the rest is naked singles.

1. 16(2)r3c6 = {79}
Innies r12 = r2c28 = +6(2)
-> For (79) in n1, one must go in 13(3)n1 and the other must go in r2c3
-> 11(2)r2c3 = [92] or [74]
-> Innies r12 = r2c28 = +6(2) = {15}
-> 14(2)r2c6 = {68}
In n1 either 13(3)n1 = {139} and 11(2)r2c3 = [74]
or 13(3)n1 = {247} and 11(2)r2c3 = [92]
Also 8 in n1 only in r3c12

2! For the numbers (6789) in c67:
One is in r1c67
Two are in r2c67
Two are in r3c67
One is in r5c67
One is in r7c67
One is in r8c67
-> No (6789) elsewhere in c67
-> Neither of the 7(2)s in c67 can be {16}
-> For the number 1 in c67:
One is in 3(2)r9c6
The other can only be in one of the 10(2)s in c67
i.e., Either 10(2)r5c6 = {19} or 10(2)r7c6 = {19}

3! -> 9 in n2 only in 10(2)r1c5 or r3c6
3 in r2 only in r2c1, r2c5, or r2c9
3 in r2c1 puts r1c12 = {19}, puts r3c6 = 9
3 in r2c5 puts 10(2)r1c5 = [73], puts r3c6 = 9
3 in r2c9 puts r1c89 = {59}, puts r3c67 = [97]
-> In all cases 16(2)r3c6 = [97]
-> 9 in c5 in r45c5
Also one of 10(2)r5c6 or 10(2)r7c6 = [19]
-> 3(2)r9c6 = [21]
(Simpler is:
1 in r2 in r2c28.
9 in n1 either in 13(3) = {139} -> 1 nowhere else in r12 -> 10(2)n2 not {19}
or 9 in n1 in r2c3 -> 10(2)n2 not {19}
Either way 9 in n2 only in r3c6)

4! Whether 10(2)r5c6 = [19] or 10(2)r7c6 = [19] -> r5c5 cannot be 9
-> r4c5 = 9
Since r3c7 = 7 -> neither r3c5 or r5c5 can be 7
-> r35c5 cannot be {17}
-> (1 in c5) One of the 9(2)s in c5 must be {18}
-> 10(2)n2 cannot be {28}

5. 9 in n3 only in 17(3)n3
-> 8 in n3 only in r12c7
-> Either r1c6 = 4 or r2c6 = 6
-> 10(2)n2 cannot be {46}
-> 10(2)n2 = {37}
-> the two 9(2)s in c5 are {18} and {45}
-> 17(3)r3c5 = <296>

6! 12(2)r1c6 can only be {48}
-> For the two 7(2)s in c67 - since:
neither can be {16}
and the numbers (24) are used once each already in c67
-> one of the 7(2)s in c67 = {34} and the other = {25}
-> (since both 4s already placed in c67) 7 in c7 only in one of the 10(2)s in c67
i.e., either 10(2)r5c6 = [73] or 10(2)r7c6 = [73]
-> One of the 7(2)s in c67 must be [34]
-> 12(2)r1c6 = [48]
-> 14(2)r2c6 = [86]
-> 17(3)n3 = {359} with 5 in r1c89
-> 15(4)r2c8 = [1{24}8]
-> 17(3)r3c5 = [692]
-> 7(2)r1c3 = [61]
-> 11(2)r2c3 = [92]
-> r3c4 = 5
Also r2c2 = 5
Also 13(3)n1 = [724]
-> 22(4)r2c2 = [5{38}6]
Also 10(2)r1c5 = [37]
-> 17(3)n3 = [{59}3]

7. Since r3c4 = 5 -> 13(4)r3c3 can only be {1345}
Since at least one of (34) in r4c34 -> 7(2)r4c6 = {25} and 7(2)r6c6 = [34]
-> 13(4)r3c3 = [1534]

etc.

for pairs of rows or columns, but not for as many as four digits. It was good that he continued with that approach with other digits. Clearly that approach is the intended feature of this puzzle.