SudokuSolver Forum

3x3:d:k:4608:4608:4610:4610:3076:5381:5381:4615:4615:4608:3338:4610:4610:3076:5381:5381:3088:4615:5394:3338:3338:4885:3076:5399:3088:3088:3354:5394:5394:4885:4885:5399:5399:5399:3354:3354:5412:5394:5394:6183:6183:6183:6698:6698:3354:5412:5412:3631:3631:3631:3634:3634:6698:6698:5412:5431:3384:3631:3898:3634:2364:3901:6698:831:5431:3384:4162:3898:4164:2364:3901:3143:831:5431:4162:4162:3898:4164:4164:3901:3143:

Prelims

a) R78C3 = {49/58/67}, no 1,2,3
b) R78C7 = {18/27/36/45}, no 9
c) R89C1 = {12}
d) R89C9 = {39/48/57}, no 1,2,6
e) 19(3) cage at R3C4 = {289/379/469/478/568}, no 1
f) 24(3) cage at R5C5 = {789}
g) 21(3) cage at R7C2 = {489/579/678}, no 1,2,3
h) 13(4) cage at R3C9 = {1237/1246/1345}, no 8,9
i) 14(4) cage at R6C3 = {1238/1247/1256/1346/2345}, no 9

1a. Naked pair {12} in R89C1, locked for C1 and N7
1b. 45 rule on N7 2 innies R7C1 + R9C3 = 8 = {35}, 5 locked for N7
1c. 21(3) cage at R7C2 = {489/678}, 8 locked for C2 and N7
1d. Max R9C3 = 5 -> min R89C4 = 11, no 1 in R89C4
1e. 45 rule on N9 2 innies R7C9 + R9C7 = 9 = {18/27/36/45}, no 9
1f. 13(4) cage at R3C9 = {1237/1246/1345}, CPE no 1 in R6C9

2a. Naked triple {789} in 24(3) cage at R5C5, locked for R5 and N5
2b. 19(3) cage at R3C4 = {289/379/469/478/568}
2c. 2,3 of {289/379} must be in R4C4 -> no 2,3 in R3C4 + R4C3
2d. 45 rule on R1234 3 outies R5C239 = 6 = {123}, locked for R5
2e. 45 rule on C12 2 outies R35C3 = 7 = [43/52/61]
2f. Min R3C3 = 4 -> max R23C2 = 9, no 9 in R23C2
2g. 45 rule on C89 2 outies R35C7 = 6 = [15/24]
2h. R78C7 = {18/27/36} (cannot be {45} which clashes with R5C7), no 4,5
2i. Naked pair {12} in R3C7 + R9C1, locked for D/
2j. Naked pair {12} in R3C7 + R9C1, CPE no 1,2 in R9C7 using D/, clean-up: no 7,8 in R7C9 (step 1e)
2k. Min R2C8 + R3C7 = 4 -> max R3C8 = 8
2l. 45 rule on C1234 1 innie R5C4 = 1 outie R6C5 + 3
2m. R5C4 = {789} -> R6C5 = {456}
2n. 45 rule on C6789 1 innie R5C6 = 1 outie R4C5 + 3
2o. R5C6 = {789} -> R4C5 = {456}

3a. R6C6 = 1 (hidden single in N5), placed for D\, clean-up: no 8 in R8C7
3b. R6C6 = 1 -> R6C7 + R7C6 = 13 = {49/58/67}, no 2,3
3c. R4C4 = 2 (hidden single in N5), placed for D\, clean-up: no 7 in R8C7
3d. R4C4 = 2 -> R3C4 + R4C3 = 17 = {89}
3e. 3 in N5 only in R4C6 + R6C4, locked for D/
3f. Min R2C8 + R3C7 = 5 -> max R3C8 = 7
3g. Min R6C45 = 7 -> max R6C3 + R7C4 = 7, no 7,8 in R6C3, no 6,7,8 in R7C4
3h. 2 in N6 only in R5C9 + R6C89, CPE no 2 in R7C9, clean-up: no 7 in R9C7 (step 1e)
3i. 16(3) cage at R8C4 = {349/358/367/457}
3j. Killer triple 7,8,9 in R35C4 and 16(3) cage, locked for C4
3k. 45 rule on N1 2 outies R12C4 = 1 innie R3C1 + 4
3l. Max R12C4 = 11 -> max R3C1 = 7

4a. 45 rule on R789 4 innies R7C1469 = 15 = {1347/1356}, no 8,9, 1,3 locked for R7, clean-up: no 4,5 in R6C7 (step 3b), no 6 in R8C7
4b. 7 of {1347} must be in R7C6 -> no 4 in R7C6, clean-up: no 9 in R6C7 (step 3b)

[I ought to have spotted this sooner …]
5a. R7C9 + R9C7 (step 1e) = [18]/{36} (cannot be {45} which clashes with R5C7 using 26(5) cage at R5C7), no 4,5
5b. Hidden killer pair 4,5 in 15(3) cage at R7C8 and R89C9 for N9, 15(3) cage cannot contain both of 4,5 (because {456} clashes with R5C8) -> 15(3) cage and R89C9 must each contain one of 4,5 -> R89C9 = {48/57}, no 3,9
5c. 9 in N9 only in 15(3) cage = {159/249}, no 3,6,7,8, 9 locked for C8
5d. 1 of {159} must be in R9C8 -> no 5 in R9C8

6a. Combined cage R78C7 + R7C9 + R9C7 (step 1e) = [72]{36}/[81]{36} (cannot be [63][18]/[72][18] which clash with R6C7 + R7C6 -> R7C9 + R9C7 = {36}, locked for N9
[Alternatively a forcing chain using R6C7 + R7C6 gives the same result.]
6b. Naked pair {12} in R38C7, locked for C7
6c. 3 in R8 only in R8C456, locked for N8
6d. 1 in N6 only in R4C89 + R5C9, locked for 13(4) cage at R3C9, no 1 in R3C9
6e. R7C1469 (step 4a) = {1347/1356} -> R7C4 = 1
6f. 14(4) cage at R6C3 = {1256/1346}, 6 locked for R6, clean-up: no 7 in R7C6 (step 3b)
6g. 2 of {1256} must be in R6C3 -> no 5 in R6C3
6h. Naked triple {356} in R7C169, locked for R7, clean-up: no 7 in R8C3
6i. Naked pair {78} in R67C7, locked for C7
6j. 1 in N2 only in 12(3) cage at R1C5 = {129/138/147} (cannot be {156} which clashes with R46C5, ALS block), no 5,6
6k. 3,6 on D\ only in R1C1 + R2C2 + R3C3, locked for N1
6l. 13(3) cage at R2C2 must contain at least one of 3,6 = {256/346} (cannot be {157/247} which don’t contain 3,6), no 1,7, 6 locked for N1
6m. 2 of {256} must be in R3C2 -> no 5 in R3C2
6n. 4 of {346} must be in R3C2 -> no 4 in R2C2 + R3C3, clean-up: no 3 in R5C3 (step 2e)
6o. 18(3) cage at R1C1 = {189/279/378} (cannot be {459} which clashes with 13(3) cage), no 4,5
6p. 4 on D\ only in R8C8 + R9C9, clean-up: no 8 in R9C9

7a. 45 rule on N3 3 innies R12C7 + R3C9 = 15 = {249/357/456} (cannot be {267} because 2,7 only in R3C9)
7b. 7 of {357} must be in R3C9 -> no 3 in R3C9
7c. 6 of {456} must be in R12C7 (R12C7 cannot be {45} which clashes with R5C7) -> no 6 in R3C9

8a. 26(5) cage at R5C7 = {24569/34568} (cannot be {24578} because R7C9 only contains 3,6, cannot be {23489/23579/23678} because R5C78 require two of 4,5,6), no 7, 4 locked for N6
8b. 3 of {34568} must be in R7C9 (cannot be 6 which clashes with R6C7 + R7C6 = [76] when 8 in R6C89) -> no 3 in R6C89

9a. 45 rule on N1 3 innies R12C3 + R3C1 = 14 = {149/158/257} (cannot be {248} which clashes with R3C2)
9b. 4 of {149} must be in R3C1 -> no 4 in R12C3
9c. 4 in N1 only in R3C12, locked for R3
9d. 13(4) cage at R3C9 = {1237} (only remaining combination), no 5,6, 2 locked for C9
9e. R12C7 + R3C9 (step 7a) = {249/357} (cannot be {456} because R3C9 only contains 2,7), no 6
9f. Killer pair 4,5 in R12C7 and R5C7, locked for C7
9g. 12(3) cage at R2C8 = {138/156/246} (cannot be {147/237/345} which clash with R12C7 + R3C9), no 7
9h. R3C7 = {12} -> no 1,2 in R3C8
9i. 15(3) cage at R7C8 (step 5c) = {159/249} with 4,5 in R8C8 -> R8C78 = [14/25] -> 12(3) cage = {138} (cannot be {156/246} which clash with R8C78) -> R3C7 = 1, placed for D/, R23C8 = [83], 8 placed for D/, R5C7 = 5 (step 2g)
9j. R8C7 = 2 -> R7C7 = 7, placed for D\, R5C5 = 9, placed for both diagonals
9k. R79C8 = [91] -> R8C8 = 5 (cage sum), placed for D\, R89C9 = [84], R7C3 = 4, placed for D/, R8C3 = 9, R4C3 = 8 -> R3C4 = 9
9l. Naked pair {49} in R12C7, locked for N3 and 21(4) cage at R1C6, 9 locked for C7
9m. R12C7 = {49} -> R3C9 = 2 (step 7a)
9n. R6C7 = 8 -> R7C6 = 5 (cage sum)

10a. R9C3 = 5 (hidden single in R9), R3C3 = 6
10b. Naked pair {78} in R3C56, locked for N2, 7 locked for R3
10c. R12C7 = {49} = 13 -> R12C6 = 8 = {26}, locked for C6 and N2 -> R4C6 = 3
10d. 5 in N2 only in R12C4, locked for C4 -> R6C4 = 6, placed for D/ -> R8C2 = 7, placed for D/
10e. R8C6 = 4, R8C4 = 3, R9C3 = 5 -> R9C4 = 8 (cage sum)

and the rest is naked singles.

Preliminaries from SudokuSolver
Cage 3(2) n7 - cells ={12}
Cage 12(2) n9 - cells do not use 126
Cage 13(2) n7 - cells do not use 123
Cage 9(2) n9 - cells do not use 9
Cage 24(3) n5 - cells ={789}
Cage 21(3) n7 - cells do not use 123
Cage 19(3) n245 - cells do not use 1
Cage 13(4) n36 - cells do not use 89
Cage 14(4) n458 - cells do not use 9

No clean-up done unless stated
1. 24(3)n5 = {789}: all locked for n5 and r5

2. "45" on c5: 3 innies r456c5 = 18 and can only have one of 7,8,9
2a. = {369/459/468/567}(no 1,2)

3. "45" on r6789: 3 outies r5c178 = 15 = {456} only: all locked for r5

4. "45" on c89: 2 outies r35c7 = 6 = [15/24]

5. naked pair 1,2 in r3c7 and r9c1: both locked for D/
5a. no 1,2 in r3c1 nor r9c7 since they also see that nakedness (Common Peer Elimination CPE)

6. r4c4 + r6c6 = [21] (hidden singles n5): both placed for D\
6a. -> r3c4 + r4c3 = 17 = {89} only
6b. -> no 8,9 in r3c3 (CPE)
6c. and r6c7 + r7c6 = 13 (no 2,3)

This step got me hooked on this puzzle. The simple pleasures of life!
7. "45" on n9: 2 innies r7c9 + r9c7 = 9 (no 9)
7a. but {45} in both the hidden cage and 9(2)r7c7 are blocked by r5c7 = (45) (no 4,5 in both cages; no 7,8 in r7c9)

8. hidden killer triple 4,5,9 in n9. 12(2)n9 = {39/48/57}, has one of them -> 15(3) must have two
8a. but {456} blocked by r5c8 = (456)
8b. 15(3) = {159/249}(no 3,6,7,8)
8c. 9 locked for n9 and c8
8d. 12(2) = {48/57}(no 3)

9. 2 in n6 only in 26(5) or r5c9
9a. -> no 2 in r7c9 (CPE)

10. 26(5)r5c7 must have two of {456} for r5c78 + (136) for r7c9
10a. = {14579/14678/24569/34568}
10b. must have 4: locked for n5

11. h9(2)n9 = [18]/{36}
11a. and 8 in n6 only in c7 or in 26(5)
11b. -> 1 in 26(5) must also have 8 for n6
11c. -> {14579} blocked from 26(5)
11d. = {14678/24569/34568}
11e. must have 6 -> no 6 in r4c9 (CPE)

The key step. Not easy.
12. 5 in n9 in 15(3) = {159} or in 12(2) = {57} = 1 or 7
12a. 13(4)r3c9 = {1237/1246/1345}
12b. {1237} must have 3,7 in r4c8 to avoid no 5 in n9
12c. {1246} must have [61] in r4c89 since r4c9 from (1357)
12d. {1345} must have 5 in r4c8 to avoid clashing with 12(2)n9
12e. -> r4c8 = (3,5,6,7)
12f. -> 1 which must be in 13(4) only in c9: locked for c9

13. 26(5)r5c7 = {24569/34568}(no 7)
13a. must have 5: locked for n6

14. 13(4)r3c9: can't have both 4,5 since they are only in r3c9
14a. = {1237/1246}(no 5)
14b. must have 2: locked for c9

15. 7 in c8 only in r1234c7 -> no 7 in r3c9 (CPE)

16. h9(2)n9 = {36} only: both locked for n9
16a. 9(2)r7c7 = [72/81]

17. naked pair {12} in r38c7: both locked for c7

18. 1 in n5 only in 13(4) -> no 1 in r3c9
18a. 13(4) = {1237/1246}
18b. 4 in {1246} must be in r3c9 -> no 6 in r3c9

19. "45" on c12: 2 outies r35c3 = 7 = [61/52/43]
19a. r3c3 = (456)

20. 3 and 6 on D\ only in n1: both locked for n1
20a. hidden killer pair 3,6 in n1 -> 13(3) must have at least one of them and one of 4,5,6 for r3c3
20b. = {256/346}(no 1,7,8,9)
20c. = {56}[2]{56}/[346]
20d. r3c2 = (24), r2c2 = (356), r3c3 = (56)
20e. 6 locked for n1
20f. -> r3c23 = [25/26/46]
20g. note: r3c29 = {234}: no elims yet

Final cracker
21. 12(3)r2c8 must have 1 or 2 for r2c7
21a. but r3c7 + r23c8 as [2]{37} blocked by [44] in r3c29 (Almost Locked Set ALS)
21b. = {138/147/156/246}
21c. but r3c78 as [24/26] blocked by r3c23 (step 20f)
21d. -> no 2 in r3c7
21e. -> r3c7 = 1 (placed for D/)
21f. -> r5c7 = 5 (h6(2)r35c7)

22. r78c7 = [72] (7 place for d\)

23. 12(2)n9 = {48}: both locked for n9 and c9
23a. -> 15(3) = {159} only: 5 locked for c8

24. 4 in n6 only in r56c8: locked for c8

25. r23c8 = 11 (cage sum) = {38} only: both locked for c8 and n3
25a. -> r3c29 = [42]
25b. -> r2c2 + r3c3 = [36]
25c. -> r5c3 = 1 (h7(2)r35c3)

Much easier now. Don't forget the Ds!

1. 24(3)n5 = {789}
Outies c1234 = r5c56 + r6c5 = +21(3)
-> r6c5 from (456)
Outies c6789 = r5c45 + r4c5 = +21(3)
-> r4c5 from (456)
-> (123) in corners of n5

2. Outies r1234 = r5c239 = +6(3) = {123}
-> Remaining cells in r5 = r5c178 = {456}

3. Outies c89 = r35c7 = +6(2) = [15] or [24]
3(2)n7 = {12}
-> {12} locked in D/ in r3c7,r9c1
-> (HS 1 in n5) r6c6 = 1
-> (HS 2 in n5) r4c4 = 2
-> 19(3)r3c4 = [{89}2]

4. Innies n7 = r7c1,r9c3 = {35}
-> 8 in n7 in 21(3)

5. 14(4)r6c3 either contains a 2 in r6c3 or is [{346}1]

6. Outies c12 = r35c3 = +7(2) = [61], [52], or [43]
Consider case where that is [52]
That puts 14(4)r6c3 = [{346}1]
It also puts r7c1 = 5 which puts r4c2 = 5
This leaves no place for 5 in n5
-> r35c3 from [61] or [43]

7! 13(2)n7 from {49} or {67}
-> (46) in c3 locked in r378c3
-> 14(4)r6c3 either has 2 in r6c3 or it is [3{46}1]
I.e., r6c3 only from (23)
-> (123) in n4 locked in r5c23 and r6c3

8. All combinations for 21(4) contain at least one number from (123)
-> Innies n7 = [r7c1,r9c3] = [35]
-> Remaining cells in 21(4)r5c1 = +18(3)
-> r4c123 = +21(3)

9!! Since r4c3 from (89) and r6c3 from (23) -> (IOD n4) r3c1 from (456)
Whatever that value is can only go in n4 in r6c2
-> Whatever values are in r4c23 can only go in c1 in r12c1

Whatever value is in r4c3 (8 or 9) goes in n7 in the 21(3)n7
21(3)n7 must be different than the H+21(3)r4c123
-> The other two values in 21(3)n7 can only go in c1 in r56c1
-> r56c1 + r4c3 = +21(3)
Since r56c1 + r6c2 = +18 -> r6c2 = r4c3 - 3
Since r3c1 = r6c2 -> r3c1 = r6c3 - 3

-> r3c1 + r4c12 = +18(3)
-> r5c23 = +3(2) = [21]

Straightforward from here

6. 3 on D\ only in n1: locked for n1

7. from 3k. iod n1 = -4
7a. min. r3c1 = 4 -> min. r12c4 = 8 (no 1)

8. r7c4 = 1 (hsingle c4)

This gets rid of the 1 from r7c4 so makes the h9(2) in n9 = {36} only.

SS does the ending differently as well but not up to optimising that.

My start showing r7c1,r9c3 = [35] (I have a better way to do this now) with your starts showing one of the 9(2)s in n9 is {36}
-> 3 in n9 in r89c7
-> (HS 3 in D\) r2c2 = 3
-> 13(3)n1 = [3{46}]
-> r3c1 = 5 (No other place for 5 in n1)