Thanks HATMAN for another nice puzzle. I solved it first using Paper Solvable steps; some of them are fairly long so maybe “moderate Paper Solvable” is about the limit of what I can manage that way. Afterward, I then solved it using normal elimination solving to check my solution.
Here is my Paper Solvable walkthrough:
There are four extra groups on the outer circumference, which I’ll call W1, W2, W3 and W4.
There are diagonals at R1C1, R1C4, R1C6, R1C9, R4C1 and R4C9 where numbers cannot be repeated; the shorter diagonals only contain six of the nine numbers.
1. 22(3) cage at R1C1 contains 9, locked for D\ at R1C1, no 8,9 in 10(3) cages in N5 -> 9 in N5 must be in R4C6 + R6C4, locked for D/ at R1C9 -> 21(3) cage at R7C3 = {678}, locked for D/ at R1C9 -> 8 in N5 must be in R4C4 + R6C6, locked for D\ at R1C1 -> 22(3) cage at R1C1 = {679} 45 rule on D\ at R1C1 R4C4 + R5C5 + R6C6 = 13 contains 8 = {148/238} 45 rule on D/ at R1C9 R4C6 + R5C5 + R6C4 = 15 contains 9 = {159/249} 7 in N5 must be in one of the 10(3) cages = {127} 5 on D\ at R1C1 only in 10(3) cage at R7C7 = {145/235} 8(3) cage at R7C9 = {134} (cannot be {125} which clashes with 10(3) cage) -> R8C8 = 3, placed for D\ at R1C1 10(3) cage = {235}, locked for D\ at R1C1 -> R4C4 + R5C5 + R6C6 = 13 = {148} -> R5C5 = 1 (because there cannot be two 10(3) cages in N5 which both don’t contain 1), placed for both diagonals
2. 9(3) cage at R1C9 = {234} -> R2C8 = {24}, 12(3) cage at R1C7 = {129/147} (cannot be {138/156} because R2C8 only contains 2,4, cannot be {237/246/345} which clash with 9(3) cage), 1 locked for N3 8(3) cage at R7C9 = {134}, 1 locked for N9 1 in C8 only in R46C8 -> 10(3) cage at R4C7 = {235}, 21(3) cage at R4C9 cannot contain 2 -> R5C8 = 5, placed for D\ at R1C4 and D/ at R4C9
3. R5C8 = 5 -> 21(3) cage at R4C9 = {579}, locked for D/ at R4C9 -> 18(3) cage at R7C6 = {468} 8(3) cage at R7C9 = {134}, 1 locked for N9 21(3) cage at R7C3 = {678} (deduced in step 1) -> 10(3) cage at R7C1 = {127/136}, 1 locked for N7 1 in R8 only in R8C46 12(3) cage at R7C4 = {345} (cannot be {237} because R8C5 only contains 4,6,8, cannot be {246} which clashes with 18(3) cage) -> R8C5 = 4, placed for D\ at R4C1 and D/ at R4C9
4. R4C4 + R5C5 + R6C6 = {148} (deduced in step 1), 8 locked for N5 18(3) cage at R7C6 = {468} (deduced in step 3) -> 8 in C5 only in R123C5 21(3) cage at R1C4 = {489/678} -> R2C5 = 8, placed for D/ at R1C6
5. R4C6 + R6C4 contain 9 (deduced in step 1), locked for N5 R5C8 = 5 -> 21(3) cage at R4C9 = {579}, 9 locked for N6 -> 9 in R5 only in R5C123 21(3) cage at R4C3 = {579} -> R5C2 = 9, placed for D/ at R1C6
6. 21(3) cage at R4C3 = {579} (deduced in step 5), 5,7 locked for N4 12(3) cage at R7C4 = {345} (deduced in step 3), locked for D\ at R4C1 R5C2 = 9 -> 17(3) cage at R4C1 = {269}, 2,6 locked for N4 21(3) cage at R7C3 = {678} (deduced in step 1), 10(3) cage at R7C1 = {127/136} (deduced in step 3) -> 4 in N7 only in R79C2, locked for C2 -> 4 in N5 only in R5C13, locked for R5 10(3) cage at R4C7 = {235} and 21(3) cage at R4C9 = {579} (deduced in step 3) -> 4 in N6 only in R46C8, locked for C8 9(3) cage at R1C9 = {234} (deduced in step 2) -> R2C8 = 2
7. 22(3) cage at R1C1 = {679} (deduced in step 1) with R2C2 = {67} 15(3) cage at R1C3 = {168/357/456} (cannot be {159/249/258/348} because R2C2 only contains 6,7, cannot be {267} which clashes with 22(3) cage) -> 2 in N1 must be in R13C2, locked for C2 21(3) cage at R7C3 = {678} (deduced in step 1), 10(3) cage at R7C1 cannot contain 9 -> 9 in N7 must be in R8C13 12(3) cage at R7C4 = {345} (deduced in step 3), 5 locked for N8, 10(3) cage at R7C7 = {235} (deduced in step 1), 5 locked for N9 -> 5 in N7 must be in R8C13 -> R8C13 = {59} -> 2 in N7 only in 10(3) cage at R7C1, 10(3) cage also contains 1 (deduced in step 3) -> 10(3) cage = {127} -> R8C2 = 7, R2C2 = 6 [So, as I’d half expected, all the crossover cells in the snowflakes have different values. Also, as I’d anticipated, all the crossover cells are placed before any other cells.]
8. Using the remaining pairs of 3-cell cages already deduced R7C1 + R9C3 = {12}, R7C3 + R9C1 = {68}, R8C13 = {59} -> R79C2 = {34}, locked for C2 R4C1 + R6C3 = {26}, R4C3 + R6C1 = {57} -> R46C2 = {18}, locked for C2 and N4 -> R5C13 = {34}, locked for R5 R1C1 + R3C3 = {79}, only remaining place for 8 in 15(3) cage at R1C3 = {168} -> R1C3 + R3C1 = {18}, R13C2 = {25} -> R2C13 = {34} R4C4 + R6C6 = {48}, R4C6 + R6C4 = {59}, 10(3) cage at R5C4 = {127} (cannot be {136} which clashes with R5C13) -> R46C5 = {36}, locked for C5 R7C4 + R9C6 = {35}, R7C6 + R9C4 = {68} -> R79C5 = {79} (hidden pair in N8), locked for C5 -> R13C5 = {25} 12(3) cage at R1C7 contains 1 (deduced in step 2), R2C8 = 2 -> R1C7 + R3C9 = {19} [Now to use an outer circumference window for the first time, using more remaining pairs which have already been deduced …] R1C14 = {79} (hidden pair in window at R1C1), locked for R1 -> R1C7 = 1, R3C9 = 9, R1C3 = 8, R3C1 = 1, R7C1 = 2, R9C3 = 1, R7C3 = 6, R9C1 = 8, R6C3 = 2, R4C1 = 6, R46C5 = [36], R3C3 = 7, R1C1 = 9, R8C13 = [59], R6C1 = 7, R4C3 = 5, R4C6 = 9, R6C4 = 5, R7C4 = 3, R9C6 = 5, R79C2 = [43], R4C7 = 2, R6C9 = 3, R1C9 = 4, R3C7 = 3, R7C7 = 5, R9C9 = 2, R7C6 = 8, R9C4 = 6, R6C6 = 4, R4C4 = 8, R46C2 = [18], R4C9 = 7, R6C7 = 9, R46C8 = [41], R1C4 = 7, R3C6 = 6 (cage sum), R5C46 = [27], R7C9 = 1, R9C7 = 4
I also solved it using elimination solving, to be able to give a rating for this puzzle.
There are four extra groups on the outer circumference, which I’ll call W1, W2, W3 and W4.
There are diagonals at R1C1, R1C4, R1C6, R1C9, R4C1 and R4C9 where numbers cannot be repeated; the shorter diagonals only contain six of the nine numbers.
Prelims
a) 22(3) cage at R1C1 = {589/679} b) 21(3) cage at R1C4 = {489/579/678}, no 1,2,3 c) 9(3) cage at R1C9 = {126/135/234}, no 7,8,9 d) 21(3) cage at R4C3 = {489/579/678}, no 1,2,3 e) 10(3) cage at R4C5 = {127/136/145/235}, no 8,9 f) 10(3) cage at R4C7 = {127/136/145/235}, no 8,9 g) 21(3) cage at R4C9 = {489/579/678}, no 1,2,3 h) 10(3) cage at R5C4 = {127/136/145/235}, no 8,9 i) 10(3) cage at R7C1 = {127/136/145/235}, no 8,9 j) 21(3) cage at R7C3 = {489/579/678}, no 1,2,3 k) 10(3) cage at R7C7 = {127/136/145/235}, no 8,9 l) 8(3) cage at R7C9 = {125/134}
Steps resulting from Prelims 1a. 22(3) cage at R1C1 = {589/679}, 9 locked for N1 and D\ 1b. 8(3) cage at R7C9 = {125/134}, 1 locked for N9
2. 9 in N5 only in R4C6 + R6C4, locked for D/ 2a. 21(3) cage at R7C3 = {678}, locked for N7 and D/ 2b. 9(3) cage at R1C9 = {135/234}, 3 locked for N3 and D/ 2c. 10(3) cage at R7C1 = {127/136}, 1 locked for N7 and W3
3. 8 in N5 only in R4C4 + R6C6, locked for D\ 3a. 22(3) cage at R1C1 = {679} (only remaining combination), locked for N1 and D\ 3b. 10(3) cage at R7C7 = {145/235}, 5 locked for N9 and D\ 3c. 8(3) cage at R7C9 = {134} (only remaining combination, cannot be {125} which clashes with 10(3) cage), locked for N9 3d. Naked pair {25} in R7C7 + R9C9, locked for N9 and D\, R8C8 = 3 (cage sum), placed for D\ 3e. Naked pair {14} in R7C9 + R9C7, 1 locked for W4 3f. 1 in R8 only in R8C456, locked for N8
4. Naked triple {148} in R4C4 + R5C5 + R6C6, locked for N5 4a. 10(3) cages at R4C5 and R5C4 = {127/136} (cannot be {145} because 1,4 only in R5C5, cannot be {235} because R5C5 only contains 1,4) -> R5C5 = 1, placed for D/ 4b. 9(3) cage at R1C9 (step 2b) = {234} (only remaining combination), locked for N3 and D/ 4c. Naked pair {59} in R4C6 + R6C4 = {59}, locked for N5
5. 12(3) cage at R1C7 = {129/147} (cannot be {138/156} because R2C8 only contains 2,4, cannot be {237/246/345} because 2,3,4 only in 9(3) cage), 1 locked for N3 and W2
6. 1 in C8 only in R46C8, locked for N5 6a. 10(3) cage at R4C7 = {235} (only remaining combination) -> R5C8 = 5, placed for D\ at R1C4 and D/ at R4C9, R4C7 + R6C9 = {23}, locked for N6 6b. 21(3) cage at R4C9 = {579} (only remaining combination), locked for N6 and D/ at R4C9 6c. 5 in N3 only in R2C79, locked for R2
7. 18(4) cage at R7C6 = {468} (only remaining combination), locked for N8 7a. 12(3) cage at R7C4= {345} (only remaining combination, cannot be {237} because R8C5 only contains 4,6,8, cannot be {246} because 4,6 only in R8C5) -> R8C5 = 4, R7C4 + R9C6 = {35}, locked for N8 and D\ at R4C1 7b. 4 in N7 only in R79C2, locked for C2
8. 21(3) cage at R1C4 = {489/678}, 8 locked for N2 8a. R2C5 = 8 (hidden single in C5), placed for D/ at R1C6 8b. 8 in N3 only in R13C8, locked for C8 8c. 8 in N6 only in R5C79, locked for R5
9. 21(3) cage at R4C3 = {579} (only remaining combination), locked for N4 and D/ at R1C6 9a. 15(3) cage at R1C6 = [384/483/681] 9b. Killer pair 4,6 in 15(3) cage and 21(3) cage at R1C4, locked for N2
10. 4 in N4 only in R5C13, locked for R5 10a. 4 in N6 only in R46C8, locked for C8 -> R2C8 = 2 [I ought to have seen this earlier as a hidden single in C8] 10b. 12(3) cage at R1C7 (step 5) = {129} (only remaining combination), 9 locked for N3 and W2 -> R4C9 = 7, placed for W2, R6C7 = 9, R1C7 = 1, R3C9 = 9, R9C7 = 4, R7C9 = 1, R3C7 = 3, R1C9 = 4, R4C7 = 2, R6C9 = 3, R7C7 = 5, R9C9 = 2, R6C4 = 5, R4C6 = 9, R4C3 = 5, R6C1 = 7, placed for W3, R5C2 = 9, R7C4 = 3, R9C6 = 5
12. 15(3) cage at R1C3 = {168} (only remaining combination, cannot be {456} because 4,5 only in R3C1) -> R3C1 = 1, R1C3 = 8, R7C3 = 6, R1C8 = 6, placed for W2, R1C4 = 7, R3C6 = 6 (cage sum)
13. R4C5 = 3 (hidden single in R4), R6C5 = 6 (cage sum)
14. R2C9 = 5, placed for W2 -> R1C5 = 2
and the rest is naked singles, without using the outer circumference windows.
Rating Comment:
I’ll rate this puzzle at Easy 1.0 using elimination solving, because that’s the lowest rating I give when using elimination solving, although solving this way it felt easier than that.
As Paper Solvable possibly in the 0.75 range; I won’t go lower because some of the early placements needed fairly long series of steps.
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