Don’t Explain

Here you go…

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

& udosuk has kindly sent me a walkthrough, as follows –

1. In n5, r4c5+r456c6=45-31=14, while r56c6 is at least 3, so r4c56 cannot exceed 14-3=11. Also, r3c7+r4c567=27, while r34c7 cannot exceed 17, so r4c56 is at least 27-17=10. Therefore r4c56=10..11, r34c7=16..17={(78)9}, r56c6=3..4={1(23)}.

2. In n6, the 22(3)-cage must be {(58|67)9}, so eliminate 9 from rest of n6, and r3c7=9, r4c7=7|8. Now r4c7 & the 22(3)-cage form a complex naked pair of {78}, so eliminate 7,8 from rest of n6.

3. 45-rule on r789 gives r7c156=22={(58|67)9}, so eliminate 9 from rest of r7. Also, r1c23 cannot exceed 17, so r3c12 is at least 45-22-17=6, r4c23 cannot exceed 20-6=14, r6c12 is at least 45-23-14=8. Therefore r7c1 cannot exceed 16-8=8={5678}. Therefore r7c56={(5678)9}, 9 is eliminated from rest of n8.

4. Now the 25(4)-cage on r9 cannot contain a 9, so it must be {4678}, while the 16(5)-cage in n7/8 must be {12346}. Hence r9c123={123}, r9c78={59}, r8c34={46}. Now eliminate the relevant naked subsets from n7,n9,r8. Now we have a naked quad of {4678} in r8c4+r9c456, so eliminate {4678} from rest of n8, making r7c56={59}, so r7c1=8. Eliminate {59} from rest of r7/n8. Now we have a hidden pair of {59} in r8c12, so r7c234=30-5-9=16={(12346)7)}={367}. Now r7c23={67}, r7c4=3, r8c3456=[4612], and we have naked triples r8c789={378} & r9c456={478}, making r9c7=6.

5. Looking at the 22(5)-cage in n5/6/8 now, r56c6={13}, r7c56={59}, so r6c7=22-1-3-5-9=4, r4c56=14-1-3=10, r4c7=27-10-9=8. So the 22(3)-cage in n6={679} now, and the remaining 4 cells of n6 are {1235} totalling 11, so r3c9=19-11=8. Also, r4c56=10 must be the naked pair {46} now, while r8c8 is the hidden
single 8. Now n56789 are all settled…

6. r6c12=16-8=8={17|26|35}, r4c23=45-23-8=14={59}, naked pair on r4/n4/cage, producing the naked pair r48c2={59} on c2. Then r3c12=20-14=6={24}, naked pair on r3/n1, and so r1c23=45-22-6=17=[89], so r4c23=[95], r8c12=[95]. Now, r12c4=24-17=7={25}, naked pair on c4/n2, making r34c4=[17], and r6c5=5 (hidden single), r7c56=[95]. With the naked pair r56c4={89}, r5c5=2, r9c4=4. Also, r12c6=45-23-7=15=[(67)(89)], giving us a naked pair r13c6={67}. So r123c5=[483], r123c6=[697], r4c56=[64], r9c56=[78], r3c38=[65], r7c23=[67], r9c89=[95], r8c7=23-1-2-8-9=3, r8c9=7. Finally, r1c78=20-15=5=[23] and the rest are almost all naked singles…

This entry was posted on Friday, February 3rd, 2006 at 7:20 pm 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.