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1. Outies n4 -> r7c2 = 2

2. Innies - Outies n1236 = 0.

-> Innies  r1c148  (Min +6)
         + r3c14   (Min +3)
         + r26c9   (Min +3)
         + r6c7    (Min +5) = Total Min +17
= Outies = r7c79   (Max +17)

->        r1c148 = {123}
and       r3c14 = {12}
and       r26c9 = {12}
and       r6c7  = 5
and       r7c79 = {89}

-> Innies n9 r9c79 = [13]
Also r6c6 = 9
-> 10/2@r3c4 = [28]
-> Innies n5 r6c5 = 6
Also r3c1 = 1.
Also HS 8 in n6 r5c8 = 8

3. r1c4 must be 1 or 3 and 8/2@r5c4 = {17} or [53]
-> (13) locked in c4 in r156c4
-> HS 1 in n8 r8c5 = 1
-> HS 1 in n7 r7c3 = 1

4. 26/5@r5c2 cannot contain both a 6 and a 9 (already in r6)
-> 1 not in 26/5@r5c2
-> HS 1 in n4 r4c2 = 1
-> 1 in n6 in r6c89

Also 2 in r6 can only go in r6c89
-> r6c89 = {12}
-> r6c1234 = {3478}
-> r5c24 = +10. Cannot be two 5s
-> Only possibility r6c4 = 7
-> r5c24 = [91]
and r6c123 = {348}

-> 5/2@r5c5 = {23}
-> 9/2@r4c5 = {45}

5. 4 in n6 in r5c79
-> Remaining two cells in 30/5@r5c7 = +9. Must be [72]
-> r6c9 = 1
-> r2c9 = 2
-> r1c148 = [231]
etc. etc.

use of minmax on innies-outies for N1236 was a very powerful way into this puzzle. I'll have to use it when I try the harder version. I used innies for N123 and innies-outies for N6 but didn't think of directly combining the two; I ought to have spotted that there are six 30(5) cages in, or partly in, N1236 which total 180.

Prelims

a) R34C4 = {19/28/37/46}, no 5
b) R4C56 = {18/27/36/45}, no 9
c) R56C4 = {17/26/35}, no 4,8,9
d) R5C56 = {14/23}
e) R6C67 = {59/68}
f) 6(3) disjoint cage at R1C1 = {123}

1. Naked triple {123} in 6(3) disjoint cage at R1C1, locked for C1

2. 45 rule on N4 1 outie R7C2 = 2
2a. 2 in C1 only in R13C1, locked for N1

3. 45 rule on N5 3 innies R4C4 + R6C56 = 23 = {689}, locked for N5, clean-up: R3C4 = {124}, no 1,3 in R4C56, no 2 in R56C4, no 9 in R6C7
3a. Variable combined cage R6C5 = {689} + R6C67 = {59/68} must contain 9, locked for R6 and N5, clean-up: no 1 in R3C4
3b. 2 in R6 only in R6C89, locked for N6

4. 45 rule on N7 3 innies R8C13 + R9C2 = 16 can only contain one of 1,3, R8C1 = {13} -> no 1,3 in R8C3 + R9C2

5. 45 rule on N8 2(1+1) outies R8C3 + R9C7 = 1 innie R9C6 + 1
5a. Min R8C3 + R9C7 = 5 -> min R9C6 = 4
5b. Max R8C3 + R9C7 = 10, min R8C3 = 4 -> max R9C7 = 6

[And now to try what I first spotted when I was setting up the worksheet …]
6. 45 rule on R123 7(3+1+3 or 2+2+3) innies R1C148 + R2C9 + R3C148 = 15
6a. Min R1C148 = 6, min R3C148 = 6 -> max R2C9 = 3
6b. Min R1C148 + R2C9 = 7 -> max R3C148 = 8 = {123/124/125/134}, 1 locked for R3, no 6,7,8,9 in R3C8
6c. Min R2C9 + R3C148 = 7 -> max R1C148 = 8 = {123/124/125/134}, 1 locked for R1, no 6,7,8,9 in R1C48

7. There are four 30(5) cages in R123 but only three 9s -> one of the 30(5) cages must be {45678} and the other three cages must contain 9 and must also each contain one of 1,2,3
7a. R2C9 = {123} -> one of the 1,2,3 in the 30(5) cages must be in R1 or R3 -> R1C148 and R3C148 cannot both total 6 -> max R2C9 = 2
[I wondered whether it would be possible to determine the combinations for the four 30(5) cages, using the fact that they require three each of 6,7,8, with the {45678} containing all of 6,7,8. No such luck at this stage.]

8. 45 rule on N6 3(1+2) outies R3C8 + R7C79 = 2 innies R6C79 + 15
8a. Min R6C79 = 6 -> min R3C8 + R7C79 = 21, max R7C79 = 17 -> min R3C8 = 4
8b. R3C8 + R7C79 = 4{89}/5{79}/5{89}, R7C79 = {789}, 9 locked for R7 and 30(5) cage at R5C7, no 9 in R5C79
8c. R3C8 + R7C79 = 21,22 -> R6C79 = 6,7 = [51/52/61], clean-up: no 6 in R6C6
8d. Naked pair {12} in R26C9, locked for C9

9. R3C1 = 1 (hidden single in R3), R8C1 = 3, R1C1 = 2
9a. R3C148 (step 6c) = {124/125} -> R3C4 = 2, R4C4 = 8, R6C6 = 9, R6C57 = [65], clean-up: no 3 in R5C4
9b. 45 rule on N1 2 remaining innies R13C3 = 12 = {39/48/57}, no 6

10. R1C148 + R2C9 + R3C148 = 15 (step 6), min R2C9 + R3C148 = 1 + 7 = 8 -> max R1C148 = 7, no 5 in R1C48

11. 45 rule on N3 4 innies R1C8 + R2C79 + R3C8 = 15 = {1248/1257/1347/1356/2346} (cannot be {1239} because R3C8 only contains 4,5), no 9
11a. 6,7,8 only in R2C7 -> R2C7 = {678}
11b. 2 in C7 only in R89C7, locked for N9
11c. 2 in R9 only in R9C57, locked for 21(5) cage at R7C5, no 2 in R8C6
11d. 2 in N8 only in R89C5, locked for C5, clean-up: no 7 in R4C6, no 3 in R5C6

12. 24(5) cage at R7C8 = {14568/23568/24567} (cannot be {12678/13578/23478} which clash with R7C79, ALS block), 5,6 locked for N9
12a. Killer pair 7,8 in R7C79 and 24(5) cage, locked for N9
12b. Killer pair 3,4 in 24(5) cage and R9C9, locked for N9

13. 30(5) cage at R5C7 contains 9 in R7C79 = {24789/34689}, no 1, 4 locked for N6
13a. 2 of {24789} must be in R6C8, 3,4,6 of {34689} must be in R5C79 + R6C8 -> no 7,8 in R6C8

14. 8 in R6 only in R6C123, locked for N4
14a. 26(5) cage at R5C2 contains 2,8 = {23489/23678/24578} (cannot be {12689} because 6,9 only in R5C2), no 1
14b. 5,6,9 only in R5C2 -> R5C2 = {569}

15. 9 in N6 only in 30(5) cage at R3C8 = {15789/34689/35679}
15a. 8 only in R5C8 -> no 1 in R5C8

16. 45 rule on N12 1 outie R2C7 = 1 remaining innie R1C4 + 5, R1C4 = {13}, R2C7 = {68}
16a. Killer pair 1,3 in R1C4 and R56C4, locked for C4

17. R1C8 + R2C79 + R3C8 (step 11) = {1248/1356/2346}
17a. R3C8 = {45} -> no 4 in R1C8
17b. Naked pair {13} in R1C48, locked for R1, clean-up: no 9 in R3C3 (step 9b)

18. 30(5) cage at R1C6 must either be {45678} or contain 9 in R2C5 -> no 1,3 in R2C5

19. R8C3 + R9C7 = R9C6 + 1 (step 5)
19a. Max R9C6 = 8 -> max R8C3 + R9C7 = 9, no 9 in R8C3
19b. R8C13 + R9C2 = 16 (step 4) = 3[49]/3{58}/3{67}, no 4 in R9C2

20. Hidden killer triple 4,6,9 in R2C4 and R789C4 for C4 -> R789C4 must contain at least two of 4,6,9
20a. 25(5) cage at R7C4 = {12679/14569/24568} (other combinations don’t contain at least two of 4,6,9)
20b. 25(5) cage ={14569/24568} (cannot be {12679} because 1,2 only in R8C5), no 7, clean-up: no 6 in R9C2 (step 19b)
20c. 1,2 only in R8C5 -> R8C5 = {12}
20d. 2 in C5 only in R89C5, 2 in R9 only in R9C57 -> R8C5 and R9C7 must be “clones”, both 1 or both 2
20e. Max R8C3 + R9C7 = 9 (step 19a) -> max R8C3 + R8C5 = 9 -> 25(5) cage ={14569} (only remaining combination, cannot be {24568} because R8C35 cannot be [82]) -> R8C5 = 1, R9C5 = 2 (hidden single in C5), R9C7 = 1, clean-up: no 4 in R5C6
20f. 25(5) cage = {14569}, no 8, 9 locked for C4, clean-up: no 5 in R9C2 (step 19b)
20g. 25(5) cage = {14569}, CPE no 4,5,6 in R8C6

21. R7C3 = 1 (hidden single in R7), R8C7 = 2 (hidden single in R8), R4C2 = 1 (hidden single in C2), R6C89 = [21] (hidden pair in N6), R2C9 = 2, clean-up: no 7 in R5C4

22. R1C8 + R2C79 + R3C8 (step 17) contains 2 = {1248/2346} -> R3C8 = 4, clean-up: no 8 in R1C3 (step 9b)

23. 30(5) cage at R3C8 (step 15) = {34689} (only remaining combination) -> R5C8 = 8, R4C789 = {369}, locked for R4 and N5

24. Naked pair {47} in R5C79, locked for R5 and 30(5) cage at R5C7, no 7 in R7C79 -> R5C5 = 3, R5C6 = 2, R6C4 = 7, R5C4 = 1, R1C4 = 3, R1C8 = 1
24a. Naked pair {45} in R4C56, locked for R4 -> R4C13 = [72]

25. R6C123 = {348} = 15, R7C2 = 2 -> R5C2 = 9 (cage sum), clean-up: no 4 in R8C3 (step 19b)

26. Naked quad {4569} in R789C4 + R8C3, 4 locked for C4 and N8

27. R7C6 = 3 (hidden single in N8), R9C57 = [21] -> R7C4 + R8C5 = 15 = {78}, locked for N8

28. R2C6 = 1 (hidden single in N2) -> 30(5) cage at R1C6 = {15789} (only remaining combination) -> R2C57 = [98], R13C6 = {57}, locked for C6 and N2 -> R13C5 = [48], R2C4 = 6, R7C5 + R8C6 = [78], R9C6 = 6

29. R1C3 = 9 (hidden single in N1), R3C3 = 3 (step 9b)

30. R1C2 = 8 (hidden single in N1), R9C2 = 7, R8C3 = 6 (step 19b), R5C13 = [65]

31. Naked pair {45} in R2C12, locked for R2 and N1 -> R2C3 = 7, R3C2 = 6

and the rest is naked singles.

I'll rate my walkthrough for A247 at 1.5. I used a "clone" step as a shortcut for my final breakthrough. I wasn't sure how to rate steps 6 and 7; making that decision wasn't necessary after using the "clone".

While it's possible that SudokuSolver may be able to do step 6, I doubt that it can do step 7.

Prelims

a) R34C4 = {19/28/37/46}, no 5
b) R4C56 = {18/27/36/45}, no 9
c) R56C4 = {17/26/35}, no 4,8,9
d) R5C56 = {14/23}
e) R6C67 = {59/68}

1. 45 rule on N4 1 outie R7C2 = 2

2. 45 rule on N5 3 innies R4C4 + R6C56 = 23 = {689}, locked for N5, clean-up: R3C4 = {124}, no 1,3 in R4C56, no 2 in R56C4, no 9 in R6C7
2a. Combined cage R6C5 = {689} + R6C67 = {59/68} must contain 9, locked for R6 and N5, clean-up: no 1 in R3C4
2b. 2 in R6 only in R6C89, locked for N6

[Instead of using what I first spotted when I was setting up the worksheet, I’ll use wellbeback’s much more powerful one …]
3. 45 rule on N1234 8 innies R1C148 + R26C9 + R3C14 + R6C7 = 2 outies R7C79
Min R1C148 = {123} = 6, min R26C9 = {12} = 3, min R3C14 = 3, min R6C7 = 5 -> min R1C148 + R26C9 + R3C14 + R6C7 = 17
Max R7C79 = {89} = 17
-> R1C148 = {123}, locked for R1, R26C9 = {12}, locked for C9, R3C14 = 3 = [12], R6C7 = 5, R4C4 = 8, R6C6 = 9, R6C5 = 6, R7C79 = {89}, locked for R7, N9 and 30(5) cage at R5C7, no 8,9 in R5C79 + R6C8, clean-up: no 3 in R5C4
3a. R5C8 = 8 (hidden single in N6)
3b. Killer pair 1,3 in R1C4 and R56C4, locked for C4

4. 45 rule on N9 R7C79 = 17 -> 2 remaining innies R9C79 = 4 = [13]
4a. R8C5 = 1 (hidden single in N8), clean-up: no 4 in R5C6
4b. R7C3 = 1 (hidden single in N7)

5. 45 rule on R123 5(3+2) remaining innies R1C148 + R2C9 + R3C8 = 12, R1C148 = {123} = 6 -> R2C9 + R3C8 = 6 = [15/24]

6. 45 rule on N8 1 innie R9C6 = 1 remaining outie R8C3, no 3,9 in R8C3, no 2 in R9C6

7. 21(5) cage at R7C5 contains 1,2,3 = {12369/12378}, no 4,5
7a. {12369} must be [36291], {12378} must be {37}{28}1 -> 3 locked for R7, no 3,6,7 in R8C6, no 7 in R9C5

8. 26(5) cage at R5C2 contains 2,8 = {23489/23678/24578} (cannot be {12689} because 6,9 only in R5C2), no 1
8a. 5,6,9 only in R5C2 -> R5C2 = {569}
8b. R4C2 = 1 (hidden single in N4)

9. 30(5) cage at R5C7 contains 8,9 = {24789/34689}, no 1
9a. R6C89 = [21] (hidden pair in N6), R2C9 = 2
9b. 30(5) cage = {24789} (only remaining combination), R5C79 = {47}, locked for R5 and N6, clean-up: no 1 in R5C6, no 1 in R6C4
9c. Naked pair {23} in R5C56, locked for R5 and N5 -> R6C4 = 7, R5C4 = 1, R1C4 = 3, R1C18 = [21]
9d. R4C13 = [72] (hidden pair in N4)

10. R4C789 = {369} = 18, R5C8 = 8 -> R3C8 = 4 (cage sum)
10a. R6C123 = {348} = 15, R7C2 = 2 -> R5C2 = 9 (cage sum)

11. 45 rule on N3 1 remaining innie R2C7 = 8, R7C79 = [98]
11a. R2C6 = 1 (hidden single in N2) -> 30(5) cage at R1C6 = {15789} (only remaining combination) -> R2C5 = 9, R13C6 = {57}, locked for C6 and N2, R13C5 = [48], R2C4 = 6

12. Naked triple {459} in R789C4, locked for N8, R8C5 = 1, R8C3 = 6 (cage sum), R5C13 = [65]
12a. R9C5 = 2, R8C6 = 8, R9C6 = 6, R7C56 = [73]

13. Naked pair {57} in R89C8, locked for C8 and N9 -> R7C8 = 6, R24C8 = [39], R4C79 = [36]

14. Naked triple {457} in R2C123, locked for N1 -> R13C3 = 12 = [93], R13C2 = [86]

15. 45 rule on N7 2 remaining innies R8C1 + R9C2 = 10 = [37]

and the rest is naked singles.

The rating for A247 Harder Version should be at least as high as for the original A247. The question is how to rate the key step. I’ll go for 1.5, because it’s a combination of minmax on two outies and a large number of innies. At a human solvable level, some might think this rating is a bit too high but it isn’t an easy step to spot.