Don’t Explain

Here are solutions to #5 and #6 (with step-by-step instructions for how to get to it!).

#5:

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

#6:

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

#6 is a very tricky one. I initially posted a partial walkthrough to get solvers past the hard bit at the start; since then I’ve written up a complete walkthrough. I’ll include both as the differences in approach (combinations vs. min/max) and notation style may be instructive. 

 

OLD WALKTHROUGH: 

Step 1: 45 rule on nonet 2 gets you a differential between the 3 internal cells & 1 external cell of 23. There is only one possibility for this: r2c3 is 1, and r1c4 and c6 and r3c4 are {7,8,9}.

Step 2: Since r1c6 = {7,8,9} then the remaining four cells of the rightmost 19(5) cage must total 10, 11 or 12. Note that all possible combinations for those sums involve the numbers 1 and 2, and contain no numbers greater than 6.

Step 3: The 12(4) cage immediately below must be 1-2-3-6 or 1-2-4-5. (Again, note that there are no numbers over 6.) The fact that the 19(5) cage above must have 1 and 2 in it leads to two conclusions:

1. cell r3c9 must be {3,4,5,6}
2. the three remaining cells of the 12(4) cage must include a 1 and 2.

Step 4: This means that the three cells of the 16(4) cage in nonet 6 cannot contain a 1 or 2. So the minimum sum of these three cells = 3 + 4 + 5 = 12. But that wouldn’t work (as the 4th cell, r7c9, would have to be a 4, which would be a duplicate). Nor would the next possibility, 3 + 4 + 6 = 13, as r7c9 would then be 3, another duplicate. Thus the three cells must add to either 14 or 15 (don’t worry about the combinations right now), and r7c9 must be either 1 or 2. Note that column 9 now contains only 3 cells where {7,8,9} are a possibility. So r[6,8,9]c9 must be {7,8,9}

Step 5: The 19(5) cage in nonet 8 must contain one of the following combos:

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

Note that all these combinations contain a 1, ruling out the possibility of a 1 for r7c9. So r7c9 = 2.

These steps should be enough to “crack” the puzzle; it’s more straightforward from here on out, so I won’t list all the subsequent steps. If you want more help, though, read on…

 

NEW (COMPLETE) WALKTHROUGH:

Step 1: 45 rule on N2 -> R1C46 + R3C4 - R2C3 = 23. There is only one possibility: R2C3 = 1, R1C46 + R3C4 = {789}.

Step 2: Let’s focus on the 19(5) cage in N23. We have already determined that R1C6 = {789}. The four remaining cells of the cage must have a 1, because otherwise the minimum sum would be 2 + 3 + 4 + 5 + 7 = 21. They must also have a 2, since otherwise the minimum would be 1 + 3 + 4 + 5 + 7 = 20. Thus {12} in N3 are locked in the 19(5) cage. Lastly, those four cells cannot have {789} because this would make the minimum sum of the cage 1 + 2 + 3 + 7 + 8 = 21.  So R1C789 + R2C9 = {123456}, with {12} locked in those cells within N3. 

Step 3: The 12(4) cage in N36 = {12(36)|(45)} -> R3C9 = {3456} and {12} is locked within N6 inside the three cells of the 12(4) cage. Now consider the 16(4) cage in N69. This cage must contain either 1 or 2 in it, because otherwise it would have a minimum value of 3 + 4 + 5 + 6 = 18. But {12} is blocked from the three cells of the 16(4) cage in N6 -> R7C9 = {12}.

Step 4: The 19(5) cage in N89 must contain a 1 (because otherwise the minimum sum = 2 + 3 + 4 + 5 + 6 = 20) -> it blocks 1 from R7C9! So R7C9 = 2, R4C8 = 2, R1C7 = 2, and the 1 in N6 is trapped in R45C9 -> R1C8 = 1. The only possible combo for the 19(5) cage in N89 = {13456}, since the 2 is blocked from it. At this point there are now only three cells in C9 in which {789} can go, so R689C9 = {789}.

Step 5: The minimum value of R56C8 = {34} (because {12} are blocked), so the maximum value of R6C9 = 16 - 4 - 3 - 2 = 7. But as we just showed, that is also the minimum value of R6C9!  So R6C9 = 7, the 21(3) cage at R6C567 = {489} (with the 4 locked in N5), R56C8 = [43], R45C9 = {156} (with 1 locked in those cells), R3C9 = {34}, R89C9 = {89}.

Step 6: {89} in N6 are locked in R456C7, so R23C8 = {89}, R4C7 = {56}, R56C7 = {89}. The 19(5) cage in N89 blocks {56} from R7C8, so R7C8 = 7, R89C8 = {56} (and {56} are locked in N8/R7 within R7C456!!), R789C7 = {134}, R234C7 = {567} (with 7 locked in N3). R4C6 = 36 - 9 - 8 - 7 - 6 - 5 = 1. R5C9 = 1, R4C9 = {56}, R3C5 = 1.

Step 7: 45 rule on N789 -> R7C123 = 20 = {389}, R8C7 = 3. With 3 now totally blocked from the 19(5) cage in N78, there is only one possible combination: {12457}. This leaves only one spot in R8 for the 6 (because we’ve already locked it in N8 within R7C456) -> R89C89 = [68/59], R9C67 = [34] (because [61] is blocked), R7C7 = 1 (which locks {456} in N8 and R7 within R7C456!), R23C6 = {26}.

Step 8: At this point we know that R12C9 = either {45} or {36} (i.e. the two numbers excluded from the 12(4) cage below), i.e. in either case they add to 9. So R1C6 = 19 - 2 - 1 - 9 = 7, R13C4 = {89}, R9C5 = 8, R8C6 = 9 (because all other numbers are now blocked from that cell!).

Step 9: In R4 the 8 cannot go in R4C45 so it is locked in R4C123, and 4 is also locked in R4C123 (because it’s already locked in R56 within R5C8 and R6C56). This means that the 22(5) cage has {48} in it -> the remaining three cells add to 10 -> it cannot have 9 in it (obviously) or 7 in it (because the three remaining cells would be {127} which is blocked) -> R4C45 = [79], R6C567 = [489], R5C67 = [58], R5C5 = 2, R12C5 = {35}, R2C4 = 4, R78C5 = [67], R7C46 = [54], R9C1 = 7.

Step 10: Mop-up from here on in. R4C123 = {348}, R3C12 = {25}, R56C4 = [36], R3C34 = 13 = [49], R23C6 = [26], R1C4 = 8, R23C78 = [69/78], R34C9 = [36], R12C9 = [45], R4C7 = 5, R1C23 = {39}, R1C1 = 6, R2C12 = [87].  In R6 {12} cannot go in the 27(4) cage so must go in R6C12 -> R6C3 = 5 -> 27(4) cage contains {5679} -> R7C3 = 9, R5C123 = [967], R1C23 = [93], R4C3 = 8, R89C3 = [26], R89C4 = [12], and the rest of the puzzle is clear sailing with only naked pairs to contend with.

This entry was posted on Monday, December 26th, 2005 at 10:55 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.