AnotherLife wrote:Can anyone offer a good solution to this not very hard sudoku? I mean, it should be human but not artificial computer-based (adieu Berthier, c'est fini).
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.1..5.628..6.........89..4....41.26.1.......5.62.85....7..28.........7..245.6..1.
Nobody on this forum decides who may answer a question or not. If you don't accept this policy, you'd better open your own private club, where only people venerating you are allowed to speak.
Almost everybody on this forum uses computer assistance in a way or another. Some say it, some prefer not to say it.
Computer analysis of a puzzle allows to see things that a manual solver is unlikely to find.
The puzzle (SER 6.3) is quite easy.
18 Singles and three whips[1] (pointing/claiming....):
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whip[1]: c7n5{r3 .} ==> r2c8 ≠ 5
whip[1]: r3n7{c3 .} ==> r2c1 ≠ 7, r1c1 ≠ 7, r1c3 ≠ 7
whip[1]: r1n9{c3 .} ==> r2c2 ≠ 9, r2c1 ≠ 9
lead to the following resolution state, which will be the starting point for the forthcoming solutions.
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349 1 39 37 5 347 6 2 8
3458 358 6 2 34 1 359 379 379
357 2 37 8 9 6 35 4 1
35789 3589 3789 4 1 39 2 6 379
1 39 4 6 7 2 39 8 5
379 6 2 39 8 5 1 379 4
39 7 139 1359 2 8 4 359 6
6 389 1389 1359 34 349 7 359 2
2 4 5 379 6 379 8 1 39
1) Simplest-first solution, using only reversible patterns of size no more than 3: Show hidden-pairs-in-a-column: c4{n1 n5}{r7 r8} ==> r8c4 ≠ 9, r8c4 ≠ 3, r7c4 ≠ 9, r7c4 ≠ 3
z-chain[2]: c4n9{r6 r9} - b9n9{r9c9 .} ==> r6c8 ≠ 9
z-chain[2]: r5n3{c2 c7} - r3n3{c7 .} ==> r2c2 ≠ 3
biv-chain[3]: r4c6{n3 n9} - c4n9{r6 r9} - b8n7{r9c4 r9c6} ==> r9c6 ≠ 3
biv-chain[2]: r9n3{c9 c4} - b5n3{r6c4 r4c6} ==> r4c9 ≠ 3
biv-chain[2]: c9n3{r9 r2} - c5n3{r2 r8} ==> r8c8 ≠ 3, r9c4 ≠ 3
hidden-single-in-a-row ==> r9c9 = 3
whip[1]: r7n3{c3 .} ==> r8c2 ≠ 3, r8c3 ≠ 3
whip[1]: c2n3{r5 .} ==> r4c1 ≠ 3, r4c3 ≠ 3, r6c1 ≠ 3
whip[1]: r9n9{c6 .} ==> r8c6 ≠ 9
whip[1]: b9n9{r8c8 .} ==> r2c8 ≠ 9
biv-chain[2]: c8n3{r2 r6} - c4n3{r6 r1} ==> r2c5 ≠ 3
singles ==> r2c5 = 4, r8c5 = 3, r8c6 = 4, r1c1 = 4, r1c3 = 9
biv-chain[3]: r6c8{n3 n7} - r4c9{n7 n9} - b5n9{r4c6 r6c4} ==> r6c4 ≠ 3
stte
It is likely that some of these steps can be skipped.
2) 1-step solutionThere is no W1-anti-backdoor and therefore no solution with only one elimination.
(This doesn't mean there can't be any 1-step solution with a pattern making several eliminations at a time, but this restricts the possibilities.)
3) 2-step solutionsThere are many 2-step solutions with whips, but they use whips that are absurdly long for such a simple puzzle. Two of the simplest ones are:
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whip[5]: r5n3{c2 c7} - r3n3{c7 c3} - c2n3{r2 r8} - c5n3{r8 r2} - b3n3{r2c7 .} ==> r6c1 ≠ 3
whip[7]: r1c4{n3 n7} - c6n7{r1 r9} - r9n3{c6 c9} - c8n3{r8 r2} - r2n7{c8 c9} - r4c9{n7 n9} - r4c6{n9 .} ==> r6c4 ≠ 3
stte
whip[5]: r5n3{c2 c7} - r3n3{c7 c3} - c2n3{r2 r8} - c5n3{r8 r2} - b3n3{r2c7 .} ==> r6c1 ≠ 3
whip[7]: r6c4{n9 n3} - r1c4{n3 n7} - c6n7{r1 r9} - r9n3{c6 c9} - c8n3{r8 r2} - r2n7{c8 c9} - c9n9{r2 .} ==> r4c6 ≠ 9
stte
Hajime wrote:I could not solve this manually in 1 or 2 steps
Considering the above results, it shouldn't be too surprising.
This puzzle is one more example that a 1- or 2- step requirement often leads to absurdly complex solutions. Puzzles with this requirement have to be designed specially for it, as has been the case for most of those proposed in this section of the forum.
