SudokuSolver Forum

Prelims

a) R1C45 = {18/27/36/45}, no 9
b) R12C1 = {19/28/37/46}, no 5
c) R12C6 = {16/25/34}, no 7,8,9
d) R34C7 = {19/28/37/46}, no 5
e) R34C9 = {17/26/35}, no 4,8,9
f) R45C5 = {19/28/37/46}, no 5
g) R45C8 = {59/68}
h) R5C12 = {14/23}
i) R56C9 = {19/28/37/46}, no 5
j) R6C12 = {14/23}
k) R6C56 = {19/28/37/46}, no 5
l) R7C23 = {12}
m) R78C4 = {18/27/36/45}, no 9
n) R78C9 = {19/28/37/46}, no 5
o) R89C3 = {49/58/67}, no 1,2,3
p) R9C12 = {69/78}
q) R9C45 = {12}
r) 10(3) cage at R8C8 = {127/136/145/235}, no 8,9
s) 27(4) cage at R2C3 = {3789/4689/5679}, no 1,2
t) 29(4) cage at R7C5 = {5789}

Steps resulting from Prelims
1a. Naked pair {12} in R7C23, locked for R7 and N7
1b. Naked pair {12} in R9C45, locked for R9 and N8
1c. Naked quad {1234} in R56C12, locked for N4
1d. Naked quad {5789} in 29(4) cage at R7C5, locked for N8
1e. 3 in C3 only in R123C3, locked for N1
1f. Clean-up: no 7 in R23C1, no 8,9 in R8C9 (other clean-ups deliberately omitted)

2. 45 rule on N8 1 innie R9C6 = 4, clean-up: no 3 in R23C6, no 6 in R6C5, no 9 in R8C3
2a. R89C7 = 12 = {39/57}

3. R89C3 = [49]/{58} (cannot be {67} which clashes with R9C12), no 6,7
3a. Killer pair 8,9 in R89C3 and R9C12, locked for N7
3b. 9 in N7 only in R9C123, locked for R9, clean-up: no 3 in R8C7 (step 2a)

4. Naked pair {36} in R78C4, locked for C4, clean-up: no 3,6 in R1C5

5. R8C89 = {12} (hidden pair in R8), R7C9 = {89}
5a. 10(3) cage at R8C8 = {136/235} (cannot be {127} because 1,2 only in R8C8), no 7, 3 locked for N9, clean-up: no 9 in R8C7 (step 2a)
5b. Naked pair {57} in R89C7, locked for C7 and N9, clean-up: no 3 in R34C7
5c. 10(3) cage = {136} (only remaining combination) -> R8C8 = 1, R8C9 = 2, R7C9 = 8, clean-up: no 6 in R34C9

6. Naked pair {36} in R9C89, locked for R9 and N9, clean-up: no 9 in R9C12
6a. Naked pair {49} in R7C78, locked for R7
6b. Naked pair {57} in R7C56, locked for R7 and N8
6c. Naked pair {78} in R9C12, locked for R9 and N9 -> R89C7 = [75], R9C3 = 9, R8C3 = 4
6d. 9 in N4 only in R4C12, locked for R4, clean-up: no 1 in R3C7, no 1 in R5C5, no 5 in R5C8

7. R56C9 = {19/46} (cannot be {37} which clashes with R34C9), no 3,7
7a. Killer pair 6,9 in R45C8 and R56C9, locked for N6, clean-up: no 4 in R3C7
7b. 6 in C7 only in R123C7, locked for N3

8. Killer triple 1,3,6 in R34C9, R56C9 and R9C9, locked for C9

9. 27(4) cage at R2C3 = {3789/4689/5679}, 9 locked for N2

10. 14(3) cage at R4C6 = {149/158/239/248/347} (cannot be {257} which clashes with R7C6, cannot be {356} which clashes with R23C6, cannot be {167} which clashes with R23C6 + R7C6, killer ALS block), no 6
10a. 8 of {158} must be in R45C6 (R45C6 cannot be {15} which clashes with R23C6), 4 of {248} must be in R5C7 -> no 8 in R5C7
10b. 9 of {149} must be in R5C6, 1 of {158} must be in R5C7 -> no 1 in R5C6

11. 6 in N5 only in R45C5 = {46} or R6C56 = [46] -> 4 must be in R456C5, locked for C5 and N5 (locking cages), clean-up: no 5 in R1C4

12. 45 rule on N5 3 innies R456C4 = 1 outie R5C7 + 11
12a. R456C4 cannot total 12 (because {129} clashes with R9C4 and R456C4 doesn’t contain any of 3,4,6) -> min R456C4 = 13, min R5C7 = 2

13. 14(3) cage at R4C6 (step 10) = {149/239/248/347} (cannot be {158} because R5C7 only contains 2,3,4), no 5
13a. 5 in N5 only in R456C4, locked for C4

14. R456C4 = 1 outie R5C7 + 11 (step 12)
14a. R5C7 = {234} -> R456C4 = 13,14,15 and must contain 5 for N5 = {157/158/257/159/258}
14b. Killer pair 1,2 in R456C4 and R9C4, locked for C4, clean-up: no 7,8 in R1C5

15. 27(4) cage at R2C3 = {3789/4689/5679}
15a. 3 of {3789} must be in R23C5 (cannot be 3{789} which clashes with R13C4, ALS block), no 3 in R2C3

[I was finding it hard to make further progress. Then I spotted a contradiction move (14(3) cage at R4C6 cannot be {239} because of a contradiction in N6 using R5C12) and, after a bit more thought, managed to replace it with a short forcing chain which cracked this puzzle.]
16. 14(3) cage at R4C6 (step 13) = {149/239/248/347}
16a. Consider combinations for R5C12
R5C12 = {14}, locked for R5 => R5C9 = {69}, R6C9 = {14}, 9 in N6 only in R5C89, locked for R5 => 14(3) cage = {248/347}
or R5C12 = {23}, locked for R5 => R5C7 = 4 => 14(3) cage = {149/248/347}
-> 14(3) cage = {149/248/347} -> R5C7 = 4, R7C78 = [94], clean-up: no 6 in R3C7, no 6 in R4C5, no 1 in R4C7, no 1 in R5C12, no 6 in R56C9
[Cracked.]

17. Naked pair {23} in R5C12, locked for R5 and N4, clean-up: no 7,8 in R4C5
17a. Naked pair {14} in R6C12, locked for R6 -> R56C9 = [19], clean-up: no 7 in R34C9, no 5 in R4C8, no 6 in R6C6

18. Naked pair {68} in R45C8, locked for C8 and N6 -> R34C7 = [82], R6C7 = 3, R34C9 = [35], R6C8 = 7, R9C89 = [36], clean-up: no 2 in R2C1, no 8 in R5C5

19. Naked pair {28} in R6C56, locked for R6 and N5 -> R6C34 = [65]

20. R49C4 = [12] (hidden pair in C4), R9C5 = 1, clean-up: no 8 in R1C4, no 9 in R5C5

21. Naked pair {47} in R13C4, locked for C4 and N2 -> R5C4 = 9, R2C4 = 8, R45C6 = [37], R45C5 = [46], R45C8 = [68], R5C3 = 5, R2C3 = 7, R4C3 = 8, clean-up: no 2 in R3C1

22. 27(4) cage at R2C3 = {3789} (only remaining combination) -> R23C5 = [39], R8C56 = [89], R6C56 = [28], R1C5 = 5, R7C56 = [75], R12C9 = [74], R13C4 = [47], clean-up: no 2 in R23C6, no 1 in R2C1, no 6 in R3C1

23. Naked pair {16} in R2C67, locked for R2 -> R2C1 = 9, R3C1 = 1

and the rest is naked singles.

I'll rate my walkthrough for A238 at 1.5. I used a fairly short forcing chain but one path felt too much for Easy 1.5.

I hope that means that you found a simpler way to crack this puzzle.

1. Max. Lr9c12 = 17 -> max. Hr9c45 = 17/5; ie must be less than 4 -> H = 3

2. Ar7c4 must be 3 x Hr9c4 -> A = 9
2a. min. Cr2c3 = 10 since 4-cell cage -> can't be 1/3 of Ar1c45 -> Cr2c3 must be 3 x Ar1c4
2b. Cr2c3 = 27

3. Kr8c7 & Br8c8: neither can be 5 since min. 3 cell-cages = 6 -> must be 16/10 or 24/15
3a. r9c45 = 3 = {12} only -> min. r9c6 = 3
3b. ->from "45" on n8: r78c56 max. = 30 -> B+B at r78c56 Max. 21
3c. -> from substep 3. above, B = 10
3d. -> Kr8c7 = 16
3e. Br2c1 = 10 -> Gr5c1 = 5 (r5c12 can't be 20)
3f. Br5c9 = 10 -> Er3c9 must be 8 (can't have 5/4 of 10)

4. Jr89c3 can't be {89} or {79} since these would block 15(2)r9c12 = {69/78} = [8/9,7/9..]
4a. -> max. r89c3 = 15 -> from "45" on n7, min. Fr7c1 = 13
4b. Dr23c6: max. 17 -> can't be double Fr4c6: must be 1/2 of Fr4c6: ie, F must be even (and greater than 13)
4c. Max. Fr45c3 = 17 since 2-cell cage -> F must be 14/16
4c. F can't be 16 since Kr8c7 = 16: must be 14 only
4d. -> Dr2c6 = 7

Summary:
A = 9
B = 10
C = 27
D = 7
E = 8
F = 14
G = 5
H = 3
I
J = 13
K = 16
L = 15

Candidates to end of Andrew's step 13a.

.-------------------------------.-------------------------------.-------------------------------.
| 12456789  12456789  1235678   | 12478     12578     1235678   | 1234689   2345789   4579      |
| 124689    12456789  35678     | 4789      356789    1256      | 1234689   2345789   4579      |
| 124689    12456789  1235678   | 12478     356789    1256      | 2689      2345789   1357      |
:-------------------------------+-------------------------------+-------------------------------:
| 56789     56789     5678      | 12578     1234678   12378     | 1248      568       1357      |
| 1234      1234      5678      | 125789    2346789   23789     | 234       689       1469      |
| 1234      1234      5678      | 125789    1234789   1236789   | 12348     234578    1469      |
:-------------------------------+-------------------------------+-------------------------------:
| 36        12        12        | 36        57        57        | 49        49        8         |
| 356       356       4         | 36        89        89        | 7         1         2         |
| 78        78        9         | 12        12        4         | 5         36        36        |
'-------------------------------.-------------------------------.-------------------------------'

13b. 14(3)r4c6 = [4/9] = [4 in r5c7 or 9 in r5c6]

14. Combined cage 24(4)r45c8+r56c9: but [59]{46} clashes with 14(3)r4c6 (step 13b)
14b. -> Combined cage 24(4)r45c8+r56c9 = {68}{19} only
14c. 14(2)r4c8 = {68} only: both locked for C8 and n6
14d. 10(2)r5c9 = {19} only: both locked for c9 and n6

etc. Cracked

1. L = 5*H, min H = 3, max L = 17 but must be a multiple of 5 -> L = 15, H = 3
1a. A:H = 1:3, H = 3 -> A = 9
1b. A:C = 1:3, min C = 10 (for 4 cell cage) -> C = 27

2. Min B = 9, only way to fit B(2) and B(3) cages in the same nonet without repeating numbers, max B = 15, only way to fit two B(2) cages in same column
2a. B:K ratio 5:8, no multiple of 8 in range 9 to 15, cannot be 15 because L is already 15 -> B = 10, K = 16
2b. A + B + B = 29 for cage at R8C5
[Note. A + B + B could also have been used to set maximum value of B, since max for 4-cell cage = 30.]

3. B:G = 1:2, B = 10, max G = 17 (for 2 cell cage) -> G = 5

4. E:B = 4:5, B = 10 isn’t a multiple of 4 -> E = 4/5*B = 8

5. This only leaves D, F and J to be determined, D:F = 1:2 with min F = 6 (for 3 cell cage) while F + J = 27 (remaining cages in N7)
5a. Only remaining values for D and F are 6 and 12 or 7 and 14 (others would repeat one of the cage totals already determined), F cannot be 12 because that would make J equal to L -> F = 14, D = 7, J = 13

No need to apologise. That and L=5*H were what made the Kenyan part easy to solve.


My walkthrough for the killer part of this puzzle is posted earlier in this thread.