Don’t Explain

Here’s the solution and a walkthrough. Corrections and queries are welcome!

Solution:

6 3 8 7 9 4 5 2 1
5 9 1 6 2 8 3 4 7
7 4 2 1 5 3 9 8 6
3 2 5 9 4 6 7 1 8
4 1 9 8 7 2 6 5 3
8 7 6 5 3 1 4 9 2
9 5 3 2 1 7 8 6 4
1 8 4 3 6 9 2 7 5
2 6 7 4 8 5 1 3 9

Walkthrough:

Step 1. 29(4) cage in N47 = {5789}. 45 rule on N789 => R7C127 = R6C6 + 21 => R6C6 = {123}, R7C127 = 22, 23 or 24 = some combo of {56789}, with 9 an obligatory inclusion & thus locked in those cells for R7.

Step 2. 45 rule on N9 => R789C7 = 11(3) => R7C7 = {5678}, R89C7 = {1234}, and the 9 is locked in R7C12 within N7 and R7. They are also locked in R7C12 for the 29(4) cage, so R6C12 = {578}.

Step 3. Let’s examine row 7 more closely. First, look at the 19(4) cage. R78C6 must contain {5..9}, because of the {1..4} in R6C6 and R8C7 (if this isn’t obvious to you, note that if a four-cell cage contained three cells less than 5 then the max. cage value would be 18 (2 + 3 + 4 + 9)). Secondly, note that R7C89 must contain one cell of value 5 or greater (because otherwise the max of R7C89 = 3 + 4 which is impossible in a 17(3) cage). So in R7 we actually have a hidden quintuplet on {56789} in R7C1267 + one of R7C89! Therefore R7C345 = {1234}.

Step 4. Since 12(4) must contain {56}, R8C5 = {56}, and {12} are locked in R7C345 within R7. 11(2) cage in N2 = {29|38|47} ({56} blocked by R8C5) => the 9(3) cage in N2 cannot be {234} => it is {1(26|35)}.

Step 5. R2C3 = 1 (the only spot for it now in N1). 45 rule on R3 => R3C789 = R4C3 + 18 => R4C3 = {2..6}, R3C789 = 20..24.

Step 6. In N4, the only place for the 9 is the three cells of the 33(7) cage => the 33(7) cage = {12369(48|57)}, i.e. it must contain {123}. Note that the {123} in R6C6 forces one of these candidates into the three cells R5C23+R6C3 within N4. But the 9(3) cage in N4 must also contain two candidates from {123}, so in conjunction with the 3 cells of the 33(7) (R5C23+R6C3) there is a hidden triplet on {123} in N4! => R4C3 = {456}.

Step 7. This in turn means (by 45 rule on R3) that R3C789 = 22, 23 or 24, i.e. {5..9}. Therefore the 4 in R3 is locked in R3C123 => R4C3 = {56}, R12C2 = {39|57}. 45 rule on N1 => R1C3 = {89}.

Step 8. If R4C3 = 5 then R3C123 = {247} (the only possible combo, since {346} is blocked by the 9(3) cage in N2). If R4C3 = 6 then R3C123 = {345} (the only possible combo). Therefore R3C123 must contain either 5 or 7 => R12C2 = {39}, R1C3 = 8, R12C1 = {56}, R3C123 = {247}, R4C3 = 5, R3C456 = {135}, R3C789 = {689}, R6C12 = {78}, R7C12 = [95].

Step 9. It’s straightforward mop-up from here on out. Our old plus-minus deduction from step 1 says that R7C127 = 22, 23 or 24 and R6C6 = {123}, but with R7C12 = [95], clearly R7C7 = 8, R6C6 = 1. Therefore (since R789C7 = 11: see step 2) R89C7 = [21], R78C6 = [79], R9C56 = {58}, R8C5 = 6, R7C345 = {123}, R7C89 = {46}, R8C8 = 7, R8C9 = 5, R9C89 = {39}. 45 rule on N7 => R79C3 = 10 = [37], R89C4 = [34], R12C4 = {67} (because {489} is blocked), R12C5 = {29}, R7C45 = [21], R3C4 = 1, R12C6 = [48], R9C56 = [85], R3C56 = [53], R23C7 = [39], R12C2 = [39], R12C5 = [92].

Step 10. Mop-up (cont’d). Since we know the 33(7) cage in N45 must have a 1, it must go in the portion within N4 => R5C2 = 1, R8C1 = 1, R9C12 = [26] (because 6 is blocked from the rest of N7), R8C23 = [84], R3C3 = 2, R4C2 = 2, R45C1 = {34}, R3C12 = [74], R6C12 = [87], R45C6 = [62], R56C3 = {69}. The 3 remaining cells of the 33(7) cage (R5C5+R6C45) must be {357} since {348} is blocked => R6C4 = 5, R56C5 = [73], R4C5 = 4, R45C1 = [34], R45C4 = {89}, R4C78 = [71] (only possibility remaining for the cage), and at this point the puzzle more or less solves itself.

This entry was posted on Tuesday, January 9th, 2007 at 3:23 am and is filed under sudoku. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.