SudokuSolver Forum

1. Call all the cells except the ones in the faux "Windoku" nonets 'Z'.
Counting all those cages gets: 27+30+26+31+30+26+27+25+27 = 249.

Call the corner cells + the center cell 'X'. They must all be different and therefore sum at least 15.
Call the four cells at r19c5 + r5c19 'Y' They have a min total of 6. ({12} + {12}).

Now, r1+r5+r9+c1+c5+c9 = 6*45 = 270.
The cells counted twice in that total are r159c159 - i.e., the same as 'X' + 'Y'.
Taking those away gives 'Z' - i.e., 249.
-> 270 - (X + Y) = Z = 249
-> X + Y = 21.
-> (Since X is Min 15) -> Max Y = 6
But since Min Y is also 6
-> r19c5 = {12} and r5c19 = {12} and X = {12345}

2. -> (12) in n5 in the corners
-> r5c5 from (345)
-> r19c19 (i.e., the corners) are from (12) plus two of (345).
Since all of r19c5 and r5c19 are from (12) -> at most one more of (12) can be in any of r1, r9, c1, c9
-> (12) in the corners are on the same diagonal.
Also -> (12) nowhere else in r1, r9, c1, c9 other than one of the corners and the centre cell.

Also since (12) are in the corners of n5 - they must be on the other diagonal in the corners in n5.
-> (12) nowhere else on the diagonals other than the corners of the puzzle and the corners of n5.

3. 3(2)r2 = {12}
-> 8(2)r2 = {35}
-> Only places available for (12) in n3 are r1c9 + r3c8
-> r1c9 + r9c1 = {12}
-> r4c4 + r6c6 = {12}
-> r1c1,r5c5,r9c9 = {345}

Also HP r2c3,r3c2 = {12}
-> r3c28 = {12}
and since r4c2 and r4c8 can't both be 4
-> 6(2)c2 = [15] and 5(2)c8 = [23]

-> HP r7c8,r8c7 = [12]
etc.

There's similarity to HS22, where I found the HS bit fairly quickly. Maybe it needs a mental leap to go from 4 double-counting corner cells, which I'd seen before, to 9 double-counting cells R159C159?!

Prelims

a) R2C34 = {12}
b) R2C67 = {17/26/35}, no 4,8,9
c) R34C2 = {15/24}
d) R34C8 = {14/23}

Steps resulting from Prelims
1a. Naked pair {12} in R2C34, locked for R2, clean-up: no 6,7 in R2C67
1b. Naked pair {35} in R2C67, locked for R2

2. 31(5) cage at R4C1 must contain 9, locked for N4

3. 45 rule on R1234 3 innies R4C159 = 22 = {589/679}, 9 locked for R4

4. 45 rule on R6789 3 innies R6C159 = 20 = {389/479/569/578}, no 1,2

5. 45 rule on C1234 3 innies R159C4 = 21 = {489/579/678}, no 1,2,3

6. 45 rule on C6789 3 innies R159C6 = 19 = {289/379/469/478/568}, no 1

7. 45 rule on N2 4 innies R23C46 = 15 = {1239/1257/1347/1356/2346} (cannot be {1248} because R2C6 only contains 3,5), no 8

8. 45 rule on N1 4 innies R23C23 = 18 = {1269/1278/1359/1458/1467/2349/2358/2457} (cannot be {3456} because R2C3 only contains 1,2, cannot be {1368/2367} because 3,6,7,8 only in R2C2 + R3C3)
8a. 1,2 of {1269/1278} must be in R2C3 + R3C2 -> no 1,2 in R3C3

9. 45 rule on N1 5(3+2) outies R23C4 + R4C234 = 15, min R4C234 = 6 -> max R23C4 = 9, no 9 in R3C4

[Since I’m not making any significant progress, it was time to try to find something similar to the HS step used in HS22 Sixes X 2.]

10. Sum of the four L-shaped cages, the four T-shaped cages and the centre cage = 249, sum of the three rows R159 and the three columns C159 = 270 but these include nine cells R159C159 which are in both rows and columns; the difference between these two totals is the ‘double counting’ of R159C159 -> R159C159 = 21
10a. R19C19 and R5C5 ‘see’ each other so must be different -> min R19C19 + R5C5 = (12345} = 15, min R19C5 = {12} = 3, min R5C19 = {12} = 3 -> min R159C159 = 21 but R159C159 = 21 -> R19C19 + R5C5 = {12345}, R19C5 = {12}, locked for C5, R5C19 = {12}, locked for R5
10b. 30(5) cage at R4C5 = {34689/35679/45678}, 6 locked for N5

11. R19C19 + R5C5 (step 10a) = {12345}
11a. 1,2 cannot be in the same row because they would clash with R1C5 or R9C5, they cannot be in the same column because they would clash with R5C1 or R5C9 -> either R1C1 + R9C9 = {12}, locked for D\ or R1C9 + R9C1 = {12}, locked for D/
11b. 1,2 in N5 only in R46C46 and must be on the other diagonal than the one containing 1,2 in the corner cells -> no 1,2 in R3C7, R7C37 and R8C28
11c. R1C5 and one of R1C19 contain 1,2, locked for R1
11d. R9C5 and one of R9C19 contain 1,2, locked for R9
11e. R5C1 and one of R19C1 contain 1,2, locked for C1
11f. R5C9 and one of R19C9 contain 1,2, locked for C9
11g. Naked pair 1,2 in R1C5 + R2C4, locked for N2
11h. R23C46 (step 7) = {1347/1356/2346} (cannot be {1239/1257} because 1,2 only in R2C4), no 9, 3 locked for N2

12. 31(5) cage at R4C1 = {16789/25789} (because R5C1 only contains 1,2), no 3,4, 7,8 locked for N4

13. R1C9 + R3C8 = {12} (hidden pair in N3) -> R1C9 + R9C1 = {12} (step 11b), locked for D/, clean-up: no 1,2 in R4C8
13a. Naked pair {12} in R1C59, locked for R1
13b. Naked pair {12} in R9C15, locked for R9
13c. Naked triple {345} in R1C1 + R5C5 + R9C9, locked for D\
13d. R4C4 + R6C6 = {12} (hidden pair in N5)
13e. Naked pair {12} in R24C4, locked for C4

14. R3C28 = {12} (hidden pair in R3), clean-up: no 1,2 in R4C2
14a. R34C2 = 6, R34C8 = 5 -> R3C2 cannot be 1 more than R3C8 (because that would make R4C2 and R4C8 equal) -> R34C2 = [15], R34C8 = [23], R2C34 = [21], R1C59 = [21], R5C19 = [12], R4C4 + R6C6 = [21], R9C15 = [21]
[Alternatively R34C2 = 6, R34C8 = 5, R3C28 = {12} = 3 -> R4C28 = 8 = [53] …]
14b. Naked quad {6789} in R46C1 + R5C23, locked for N4 -> R4C3 = 4, R6C23 = [23]
14c. R4C7 = 1 (hidden single in R4)
14d. R7C8 = 1 (hidden single in R7)
14e. R8C3 = 1 (hidden single in R8)
14f. R7C6 = 2 (hidden single in R7)
14g. R8C7 = 2 (hidden single in R8)

15. 45 rule on N6 2 remaining innies R6C78 = 15 = {69/78}
15a. 4 in N6 only in R5C78, locked for R5

16. R23C4 + R4C234 (step 9) = 15, R2C4 = 1, R4C234 = [542] = 11 -> R3C4 = 3, R2C67 = [53]

17. R67C6 = [12] -> 25(5) cage at R6C6 = {12679}, no 8
17a. 8 on D\ only in R2C2 + R3C3, locked for N1
17b. 24(5) cage at R2C2 contains 2,3,4,8 = {23478}, 7 locked for N1 and D\
17c. Naked pair {69} in R7C7 + R8C8, locked for N9 and 25(5) cage at R6C6 -> R6C7 = 7 -> R6C8 = 8 (step 15), R8C6 = 4 (cage sum)

18. R4C159 (step 3) = {679} (only remaining combination), locked for R4 -> R4C6 = 8, placed for D/
18a. R6C159 (step 4) = {569} (only remaining combination), locked for R6 -> R6C4 = 4, placed for D/

19. 45 rule on N3 2 remaining innies R2C8 + R3C7 = 14 = [95], placed for D/, R3C6 = 6 (cage sum)
19a. R5C5 = 3, placed for D\
19b. R7C7 + R8C8 = [96] -> R7C3 + R8C2 = [67], R7C4 = 5 (cage sum)

20. 45 rule on N8 1 remaining innie R8C4 = 9 -> R7C2 = 4 (cage sum)

and the rest is naked singles, without using the diagonals.