SudokuSolver Forum

1. The puzzle is one of those "Rotational Inverse" ones. I.e., where every cage has a corresponding cage of the same size and shape when rotated by 180 degrees and with a value of 10 times the number of cells in the cage - the original cage total.

As we have seen in previous examples (I am not going to explain it here) this means:
a) The cell 180 degrees rotated from any given cell has a value of 10 - the original cell value.
b) r5c5 = 5.

2. 7/3@r1c7 = {124}
23/3@r7c1 = {689}

3. 39/7@r6c9 is missing two values which total 6. One of them must be in r9c9.
-> r9c9 from (124)
-> (124) locked in c9 in r139c9.

4. The other missing value must be in r7c7, r8c9, or r9c8.
If it were in r7c7 this would imply:
a) [r7c7,r9c9] = [42] (only combination of +6/2 that works given r5c5 = 5)
b) -> 6/2@r9c4 = [51]
c) -> r8c4 = 4
But that leaves no place for 4 in n7/r9

-> Other missing value in r8c9 or r9c8.
-> 15/3@r8c9 = { (15|24) 9 }

5. 9 must be in the 39/7@r6c9 somewhere - but not in n9
-> 9 in r9c6 and r8c9 -or- r9c8 and r6c9. I.e., 9 locked in r9 in r9c68.
-> r7c1 = 9

6. -> r1c1 not = 9
-> (Rotational Inverse rule) r9c9 not = 1
-> r9c9 from (24)
-> 15/3@r8c9 = {249}

But 2 and 4 in there can't both go in c9
-> r8c9 = 9 and r9c89 = {24}

-> 12/2@r4c9 = [57]
-> 8/2@r5c8 = {26}
-> r9c89 = [42]
-> r1c12 = [86]
-> r9c13 = [68]
Also 6/2@r9c4 = [51]

7. Whatever goes in r6c9 must go in n9 in r7c7.
Only candidate that works in both places is a 3.
-> 7/2@r6c6 = [43]
Also HS 3 in r9 -> r9c2 = 3.
-> r9 = [638519742]
-> r1 = [863195274] (Rotational Inverse rule)

etc., etc.

My initial thoughts were that the Human Solvable element might be innies-outies for N3 and N7, but I couldn’t get anything out of that. When wellbeback posted, I wondered whether he’d got more out of these innies-outies but he’d found something completely different. He’d spotted that the cage pattern is “Rotationally Inverse”. I’d forgotten about it since A123 and A123V2, which were easier using it but it wasn’t really essential for those puzzles. I’ll try to solve this puzzle with that knowledge, although I only started this week.

I found that I needed to use some steps which were like one’s in wellbeback’s walkthrough, but steps 6 and 7 were very different. I hope I’ve got those steps right now; originally I’d thought that I could eliminate more combinations.

Prelims

a) R1C56 = {59/68}
b) R2C45 = {29/38/47/56}, no 1
c) 13(2) cage at R3C3 = {49/58/67}, no 1,2,3
d) R45C2 = {39/48/57}, no 1,2,6
e) R45C9 = {39/48/57}, no 1,2,6
f) R56C1 = {17/26/35}, no 4,8,9
g) R56C8 = {17/26/35}, no 4,8,9
h) 7(2) cage at R6C6 = {16/25/34}, no 7,8,9
i) R8C56 = {18/27/36/45}, no 9
j) R9C45 = {15/24}
k) 7(3) disjoint cage at R1C7 = {124}
l) 23(3) disjoint cage at R7C1 = {689}

Steps resulting from Prelims
1a. R2C45 = {29/38/47} (cannot be {56} which clashes with R1C56), no 5,6
1b. R8C56 = {18/27/36} (cannot be {45} which clashes with R9C45), no 4,5
1c. Naked triple {124} in 7(3) disjoint cage at R1C7, locked for N3
1d. Naked triple {689} in 23(3) disjoint cage at R7C1, locked for N7

[HATMAN had said that once one gets the Human Solvable element, it’s a bit below Assassin level. My initial thoughts were that the Human Solvable element might be innies-outies for N3 and N7, but I couldn’t get anything out of that. When wellbeback posted, I wondered whether he’d got more out of these innies-outies but he’d found something completely different. He’d spotted that the cage pattern is “Rotationally Inverse”. I’d forgotten about it since A123 and A123V2, which were easier using it but it wasn’t really essential for those puzzles. I’ll try to solve this puzzle with that knowledge.]

[Using “Rotationally Inverse, R5C5 must be 5 and all pairs RmCn + R(10-m)C(10-n) must total 10 …, although I’ve used corresponding steps rather than Rotationally Inverse clean-ups.]

2. R5C5 = 5, placed for both diagonals, clean-up: no 9 in R1C6, no 8 in 13(2) cage at R3C3, no 7 in R4C2, no 7 in R4C9, no 3 in R6C1, no 2 in 7(2) cage at R6C6, no 3 in R6C8, no 1 in R9C4

3. Killer pair 4,6 in 13(2) cage at R3C3 and 7(2) cage at R6C6, locked for D\

4. 5 in N3 only in R1C8 + R2C79 + R3C8, locked for 40(7) cage at R1C8, no 5 in R2C6 + R4C8

5. 5 in N7 only in R7C2 + R8C13 + R9C2, locked for 30(7) cage at R6C2, no 5 in R6C2 + R8C4

6. 45 rule on N1 3(2+1) outies R1C4 + R4C14 = 14 = {149/167}/{25}7/{34}7 (other combinations blocked because R4C4 “sees” R1C1 + R1C4 + R4C1, R1C4 + R4C1 “see” all cells in N1 except for R2C1 + R1C2 respectively and no 5 or the value of R3C3 in R1C1), no 8
6a. R1C4 + R4C14 = {149/167}, 1 locked for 31(7) cage at R1C3, CPE no 1 in R1C1, 1 in N1 only in 15(3) cage at R1C1 = {159/168} => R1C1 = {89}
or R1C4 + R4C14 = {25}7/{34}7 => R1C2 + R2C1 = {25/34} => R1C1 = 8
-> R1C1 = {89}
6b. 15(3) cage = {159/168/258/348}, no 7
6c. Killer pair 8,9 in R1C1 and R1C56, locked for R1

[Similarly …]
7. 45 rule on N9 3(2+1) outies R6C69 + R9C6 = 16 = {169/349}/3{58}/3{67} (other combinations blocked for the “mirror image” reasons corresponding to those in step 6), no 2
7a. R6C69 + R9C6 = 16 = {169/349}, 9 locked for 39(7) cage at R6C9, CPE no 9 in R9C9, 9 in N1 only in 15(3) cage at R8C9 = {159/249} => R9C9 = {12}
or R6C69 + R9C6 = 3{58}/3{67} => R8C9 + R9C8 = {58/67} => R9C9 = 2
-> R9C9 = {12}
7b. 15(3) cage = {159/249/258/267}, no 3
7c. Killer pair 1,2 in R9C45 and R9C9, locked for R9

8. Naked triple {689} in R179C1, locked for C1, 6 also locked for N7, clean-up: no 2 in R56C1
8a. 12(2) cage at R4C2 = {39/48} (cannot be [57] which clashes with R56C1), no 5,7

9. Naked triple {124} in R139C9, locked for C9, 4 also locked for N3, clean-up: no 8 in R45C9
9a. 8(2) cage at R5C8 = {17/26} (cannot be [35] which clashes with 12(2) cage at R4C9)

10. R1C4 + R4C14 (step 6) = {149/167}/{25}7/{34}7
10a. 7,9 of {149}/{34}7 must be in R4C4 -> no 4 in R4C4, clean-up: no 9 in R3C3

11. R6C69 + R9C6 (step 7) = {169/349}/3{58}/3{67}
11a. 1,3 of {169}/3{67} must be in R6C6 -> no 6 in R6C6, clean-up: no 1 in R7C7

12. 31(7) cage at R1C3 = {1234579/1234678}
12a. 31(7) cage = {1234579} -> R1C1 = 8 -> R1C56 = [95] -> no 5 in R1C34
12b. 31(7) = {1234678} -> R1C1 = 9 -> R1C56 = {68} -> no 6 in R1C34

13. 39(7) cage at R6C9 = {1356789/2346789}
13a. 39(7) cage = {1356789} -> R9C9 = 2 -> R9C45 = [51] -> no 5 in R9C67
13b. 39(7) cage = {2346789} -> R9C9 = 1 -> R9C45 = {24} -> no 4 in R9C67

[I’ve now reached a position where I can use one of wellbeback’s breakthrough steps, but I’ll look at it the other way round.]
14. Consider placements for 4 in R9
R9C2 = 4 => 4 in R8 only in R8C7 => no 4 in R7C7
or 4 in R9C45 = {24}, locked for R9 => R9C9 = 1
or 4 in R9C8 => no 4 in R7C7
-> R7C7 + R9C9 cannot be [42]
-> R7C7 + R9C9 cannot contain both of the two numbers 1,5 or 2,4 missing from 39(7) cage at R6C9
-> 15(3) cage at R8C9 must contain the two numbers 1,5 or 2,4 missing from 39(7) cage
-> 15(3) cage = {159/249}, 9 locked for N9

[Similarly, or one can use the “Rotational Inverse” property]
15. Consider placements for 6 in R1
6 in R1C2 => no 6 in R3C3
or 6 in R1C56 = {68}, locked for R1 => R1C1 = 9
or 6 in R1C8 => 6 in R2 only in R2C3 => no 6 in R3C3
-> R1C1 + R3C3 cannot be [86]
-> R1C1 + R3C3 cannot contain both of the two numbers 5,9 or 6,8 missing from 31(7) cage at R1C3
-> 15(3) cage must contain the two numbers 5,9 or 6,8 missing from the 31(7) cage
-> 15(3) cage = {159/168}, 1 locked for N1

16. 15(3) cage at R1C1 and 31(7) cage at R1C3 both contain 1 -> 1 in R1C2 + R4C1 or R1C4 + R2C1 -> there must be 1 in R1C24, locked for R1 -> R1C7 = 2, R1C9 = 4, placed for D/, R3C9 = 1

17. 39(7) cage at R6C9 and 15(3) cage at R8C9 both contain 9 -> 9 in R6C9 + R9C8 or R8C9 + R9C6 -> there must be 9 in R9C68, locked for R9 -> R9C3 = 8, R9C1 = 6, placed for D/, R7C1 = 9

18. R1C1 = 8, placed for D\
18a. 15(3) cage at R1C1 = {168} (only remaining combination) -> R1C2 = 6, R2C1, R1C56 = [95], clean-up: no 2 in R4C45, no 7 in R4C4, no 7 in R56C1
18b. R56C1 = [35], clean-up: no 9 in R45C2, no 9 in R4C9
18c. Naked pair {48} in R45C2, locked for C2 and N4

19. R9C9 = 2, placed for D\
19a. 15(2) cage at R9C9 = {249} (only remaining combination) -> R9C8 = 4, R8C9 = 9, R5C9 = 7, R4C9 = 5, R9C45 = [51], clean-up: no 1 in R56C8, no 3 in R6C6, no 8 in R8C56

20. R1C4 + R4C14 (step 6) = {167} (only remaining combination) -> R1C4 = 1, R4C1 = 7, R4C4 = 6, placed for D\, R3C3 = 7, placed for D\, R7C7 = 3, placed for D\, R6C6 = 4

21. R1C3 = 3, R1C8 = 7, R2C2 = 9, R8C8 = 1, R9C7 = 7, R9C2 = 3, R9C6 = 9

22. R6C69 + R9C6 = 16 (step 7), R6C6 = 4, R9C6 = 9 -> R6C9 = 3

23. R2C45 = {47} (only remaining combination, cannot be {38} which clashes with R2C8), locked for R2 and N2

24. R8C56 = {36} (only remaining combination, cannot be {27} which clashes with R8C2), locked for R8 and N8

25. R3C1 = 4 (hidden single in N1), R8C1 = 2, R8C2 = 7, placed for D/, R7C3 = 1, placed for D/

26. R7C9 = 6 (hidden single in N9, R2C9 = 8, R2C8 = 3, placed for D/, R3C7 = 9, placed for D/

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

The SS score for this puzzle is 8.65, because it hasn’t been programmed to recognise “Rotationally Inverse”; this isn’t surprising since so few puzzles have this property. After placing R5C5 = 5, the SS score is 2.60.

I won’t attempt to rate my walkthrough for two reasons. I’ve no idea what rating to give for “Rotationally Inverse”. Also I wouldn’t know what rating to give for my steps 6 and 7. I’ll assume that SS can’t do these steps.

Prelims

a) R1C56 = {59/68}
b) R2C45 = {29/38/47/56}, no 1
c) R45C2 = {39/48/57}, no 1,2,6
d) R45C9 = {39/48/57}, no 1,2,6
e) R56C1 = {17/26/35}, no 4,8,9
f) R56C8 = {17/26/35}, no 4,8,9
g) R8C56 = {18/27/36/45}, no 9
h) R9C45 = {15/24}
i) 7(3) disjoint cage at R1C7 = {124}
j) 23(3) disjoint cage at R7C1 = {689}

Steps resulting from Prelims
1a. R2C45 = {29/38/47} (cannot be {56} which clashes with R1C56), no 5,6
1b. R8C56 = {18/27/36} (cannot be {45} which clashes with R9C45), no 4,5
1c. Naked triple {124} in 7(3) disjoint cage at R1C7, locked for N3
1d. Naked triple {689} in 23(3) disjoint cage at R7C1, locked for N7
[Note that the eliminations in steps 1a and 1b can also be done because of the non-consecutive condition.]

[As with the other version, where I failed to spot that the cage pattern is “Rotationally Inverse”, I’ll work from that starting point.

Using “Rotationally Inverse, R5C5 must be 5 and all pairs RmCn + R(10-m)C(10-n) must total 10, although I’ve used corresponding steps rather than Rotationally Inverse clean-ups.]

2. R5C5 = 5, placed for both diagonals, non consecutive (NC): no 4,6 in R46C5 + R6C46, clean-up: no 9 in R1C6, no 7 in R4C2, no 7 in R4C9, no 3 in R6C1, no 3 in R6C8, no 1 in R9C4

3. 45 rule on N1 2 outies R1C4 + R4C1 = 1 innie R3C3 + 1
3a. Min R1C4 + R4C1 = 3 -> min R3C3 = 2
3b. 31(7) cage at R1C3 = {1234579/1234678}, CPE no 1,2,3,4,7 in R1C1

4. Naked triple {689} in R179C1, locked for C1, clean-up: no 2 in R56C1
4a. R45C2 = {39/48} (cannot be [57] which clashes with R56C1), no 5,7

5. 45 rule on N9 2 outies R6C9 + R9C6 = 1 innie R7C7 + 9
5a. Max R6C9 + R9C6 = 17 -> max R7C7 = 8
5b. 39(7) cage at R6C9 = {1356789/2346789}, CPE no 3,6,7,8,9 in R9C9

6. Naked triple {124} in R139C9, locked for C9, clean-up: no 8 in R45C9
6a. R56C8 = {17/26} (cannot be [53] which clashes with R45C9), no 3,5

7. R1C1 = {689}, R1C56 = {68}/[95] -> variable combined cage R1C1 + R1C56 = 6[95]/8[95]/9{68}, 9 locked for R1

8. R9C56 = {24}/[51], R9C9 = {124} -> variable combined cage R9C56 + R9C9 = {24}1/[51]2/[51]4, 1 locked for R9

9. R1C56 = {68}/[95] -> NC no 8 in R2C5, clean-up: no 3 in R2C4

10. R9C56 = {24}/[51] -> NC no 2 in R8C5, clean-up: no 7 in R8C6

11. 15(3) cage at R1C1 = {159/168/249/258/348} (cannot be {267/456} because of NC, cannot be {357} because R1C1 only contains 6,8,9), no 7

12. 15(3) cage at R9C9 = {159/168/249/258/267} (cannot be {348/456} because of NC, cannot be {357} because R9C9 only contains 1,2,3), no 3

13. 5 in N3 only in R1C8 + R2C79 + R3C8, locked for 40(7) cage at R1C8, no 5 in R1C6 + R4C8

14. 5 in N7 only in R7C2 + R8C13 + R9C2, locked for 30(7) cage at R6C2, no 5 in R6C2 + R8C4

15. 31(7) cage at R1C3 = {1234579/1234678} contains 5,9 or 6,8
15a. 31(7) cage = {1234579} -> R1C1 = {68} -> R1C56 = [95] -> no 5 in R1C34
15b. 31(7) cage = {1234678} -> R1C1 = 9 -> R1C56 = {68} -> no 6,8 in R1C34

16. 39(7) cage at R6C9 = {1356789/2346789} contains 1,5 or 2,4
16a. 39(7) cage = {1356789} -> R9C9 = {24} -> R9C45 = [51] -> no 5 in R9C67
16b. 39(7) cage = {2346789} -> R9C9 = 1 -> R9C45 = {24} -> no 2,4 in R9C67

17. R1C4 + R4C1 = R3C3 + 1 (step 3)
17a. R1C4 + R4C1 cannot total 9 (cannot be [27/45/72] because 15(3) cage at R1C1 cannot contain both of 2,7 or both of 4,5 and 8 would be in R3C3 if R1C4 + R4C1 totalled 9) -> no 8 in R3C3
17b. R1C4 + R4C1 cannot total 10 (cannot be {37} because 15(3) cage at R1C1 cannot contain both of 3,7 and 9 would be in R3C3 if R1C4 + R4C1 totalled 10) -> no 9 in R3C3

18. R6C9 + R9C6 = R7C7 + 9 (step 5)
18a. R6C9 + R9C6 cannot total 10 (cannot be {37} because 15(3) cage at R8C9 cannot contain both of 3,7 and 1 would be in R7C7 if R6C9 + R9C6 totalled 10) -> no 1 in R7C7
18b. R6C9 + R9C6 cannot total 11 (cannot be [38/56/83] because 15(3) cage at R8C9 cannot contain both of 3,8 or both of 5,6 and 2 would be in R7C7 if R6C9 + R9C6 totalled 11) -> no 2 in R7C7

[I’ve now reached a position where I can use one of wellbeback’s breakthrough steps, but I’ll look at it the other way round.]
19. Consider placements for 4 in R9
R9C2 = 4 => 4 in R8 only in R8C78, locked for N9 => no 4 in R7C7
or 4 in R9C45 = {24}, locked for R9 => R9C9 = 1
or 4 in R9C89, locked for N9 => no 4 in R7C7
-> R7C7 + R9C9 cannot be [42]
-> R7C7 + R9C9 cannot contain both of the two numbers 1,5 or 2,4 missing from 39(7) cage at R6C9
-> 15(3) cage at R8C9 must contain the two numbers 1,5 or 2,4 missing from 39(7) cage
-> 15(3) cage = {159/249}, 9 locked for N9

[Similarly, or one can use the “Rotational Inverse” property]
20. Consider placements for 6 in R1
6 in R1C12, locked for N1 => no 6 in R3C3
or 6 in R1C56 = {68}, locked for R1 => R1C1 = 9
or 6 in R1C8 => 6 in R2 only in R2C23, locked for R2 => no 6 in R3C3
-> R1C1 + R3C3 cannot be [86]
-> R1C1 + R3C3 cannot contain both of the two numbers 5,9 or 6,8 missing from 31(7) cage at R1C3
-> 15(3) cage must contain the two numbers 5,9 or 6,8 missing from the 31(7) cage
-> 15(3) cage = {159/168}, 1 locked for N1

21. 15(3) cage at R1C1 and 31(7) cage at R1C3 both contain 1 -> 1 in R1C2 + R4C1 or R1C4 + R2C1 -> there must be 1 in R1C24, locked for R1

22. 39(7) cage at R6C9 and 15(3) cage at R8C9 both contain 9 -> 9 in R6C9 + R9C8 or R8C9 + R9C6 -> there must be 9 in R9C68, locked for R9

[Non-consecutive needed for a unique solution but the puzzle is cracked. Non-consecutive also makes the final stages a lot shorter.]

23. Naked pair {24} in R1C79, locked for R1 and N3 -> R3C9 = 1, NC no 3,5 in R1C8
23a. R9C5 = 1 (hidden single in R9), R9C4 = 5, NC: no 4,6 in R8C4, no 6 in R9C3, clean-up: no 8 in R8C56

24. R9C3 = 8, R9C1 = 6, placed for D/, R7C1 = 9, R1C1 = 8, placed for D\, NC no 5,7 in R8C1, no 7 in R8C3 + R9C2
24a. R1C5 = 9 (hidden single in R1), R1C6 = 5, NC: no 4 in R1C7, no 4,6 in R2C6, clean-up: no 2 in R2C45

25. R1C7 = 2, R1C9 = 4, placed for D/, R9C9 = 2, placed for D\, NC: no 3,5 in R2C9

26. R4C4 + R6C6 = {46} (hidden pair in N5), locked for D\
[This may not seem the natural next step, but I’d been looking for some time to see which diagonal the {46} pair in N5 would be on. However it proves to be helpful and leads to the natural steps.]
26a. Naked pair {37} in R3C3 + R7C7, locked for D\ -> R2C2 = 9, R8C8 = 1, clean-up: no 3 in R45C2, no 7 in R56C8
26b. Naked pair {37} in R3C3 + R7C7, CPE no 3,7 in R3C7 + R7C3
26c. Naked pair {48} in R45C2, locked for C2 and N4 -> R9C2 = 3, R9C67 = [97], R9C8 = 4, R8C9 = 9 (cage sum), NC: no 8, in R7C9, clean-up: no 3 in R45C9

27. R45C9 = [57], NC no 6 in R4C8, no 6 in R5C8, no 6,8 in R6C9
27a. R56C8 = [26], R6C9 = 3, R7C9 = 6, R1C8 = 7, R2C9 = 8, R2C8 = 3, placed for D/

28. R1C34 = [31], R1C2 = 6, R2C1 = 1 (cage sum), R5C1 = 3, R6C1 = 5, NC no 2 in R3C1, no 4 in R5C2
28a. R45C2 = [48]

29. R4C8 = 9, R3C8 = 5, R3C3 = 7, R3C2 = 2, R8C2 = 7, placed for D/

and the rest is naked singles, without using diagonals or non-consecutive.

The SS score for this puzzle is 4.25, because it hasn’t been programmed to recognise “Rotationally Inverse”; this isn’t surprising since so few puzzles have this property. After placing R5C5 = 5, the SS score is 1.85.

I won’t attempt to rate my walkthrough. I’ve no idea what rating to give for “Rotationally Inverse”.