## OK, here is how far I got

Advanced methods and approaches for solving Sudoku puzzles

### OK, here is how far I got

Code: Select all
`Solution So Far564 32- 98-938 --- -2-217 849 6--89- 472 51-1-- 968 ----42 1-- 89--21 59- 34848- --- --93-9 -84 1--Pencilmarks So Far-         -         -         -         -         17        -         -         17       -         -         -         67        15        1567      47        -         147      -         -         -         -         -         -         -         35        35       -         -         36        -         -         -         -         -         36       -         57        35        -         -         -         24        37        24       67        -         -         -         35        35        -         -         67       67        -         -         -         -         67        -         -         -        -         -         56        267       13        13        27        567       -        -         57        -         267       -         -         -         567       257      `

Unable to solve without guessing.

Mac
QBasicMac

Posts: 441
Joined: 13 July 2005

Code: Select all
`Colouring 7: double exclusion found, eliminating 7 from r2c9, (r1c9 => r6c9 => r6c1 => r5c2 => r5c8 => r6c9)Colouring 7: double exclusion found, eliminating 7 from r9c9, (r1c9 => r6c9 => r6c1 => r5c2 => r5c8 => r6c9)Colouring 7: double exclusion found, eliminating 7 from r2c6, (r1c6 => r1c9 => r6c9 => r6c1 => r7c1 => r7c6)r8c7 = 7: forced from either candidate of r6c1r2c7 = 4: only possible value for this cellr5c7 = 2: only possible value for this celletc`
simes

Posts: 324
Joined: 11 March 2005
Location: UK

The analysis from simes is spot-on but the following alternative to the fourth line uses shorter chains, so might be regarded simpler:

Consider the chains r5c8-3-r3c8~5~r8c8 and r5c8-3-r5c3-5-r8c3.
Whichever of the 2 candidates in Row 8 contains the value 5, the cell r5c8 does not contain the value 3.
Therefore r5c8=7.

after which all is straightforward.
Sue De Coq

Posts: 93
Joined: 01 April 2005

simes wrote:Coloring...

No way I could follow that!

It looks like guessing is the way to go for me. Well, T&E is a better term ("serial hacking" LOL).

But thanks a lot for the info, anyway. It verifies that there is no easy stuff I overlooked, even including X-Wing and Swordfish.

Mac

P.S. I see Sue De Coq wrote something. Will study that. Looks easier on the surface.
QBasicMac

Posts: 441
Joined: 13 July 2005

Sue De Coq wrote:might be regarded simpler

Heh - well, shorter, at least. Total gibberish to me, though.

Well, have fun, guys, you are WAY beyond where I ever want to be in SuDoku solving. T&E - quick and simple!

Mac
QBasicMac

Posts: 441
Joined: 13 July 2005

### Re: OK, here is how far I got

QBasicMac wrote:Unable to solve without guessing.

Try a Swordfish on 7s for a start.
Then after 2 xy-wings it solves easily.
angusj

Posts: 306
Joined: 12 June 2005

Here's the full extract from my solver log for the first move, after which, as angusj points out, the puzzle is straightforward to complete. You might find it useful because it attempts to describe the logic from first principles and avoids specialist terms (apart from the chain notation, of course) such as 'Colouring' and 'Swordfish'. The downside is that more experienced users might find it frustratingly verbose. When I write out a chain, I use the symbol '-' to indicate a strong link (i.e. exactly two candidates for that cell/row/column/box) or '~' for a weak link (i.e. more than two candidates). I believe wolfgang introduced this notation to the forum:

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`Consider the chain r9c2-7-r5c2-7-r5c8-7-r6c9~7~r9c9.When the cell r9c9 contains the value 7, so does the cell r9c2 - a contradiction.Therefore, the cell r9c9 cannot contain the value 7.- The move r9c9:=7 has been eliminated.Consider the chain r1c9-7-r1c6~7~r7c6-7-r7c1-7-r6c1-7-r6c9.The cell r6c9 must contain the value 7 if the cell r1c9 doesn't.Therefore, these two cells are the only candidates for the value 7 in Column 9.- The move r2c9:=7 has been eliminated.Consider the chain r1c6-7-r1c9-7-r6c9-7-r6c1-7-r7c1-7-r7c6.The cell r7c6 must contain the value 7 if the cell r1c6 doesn't.Therefore, these two cells are the only candidates for the value 7 in Column 6.- The move r2c6:=7 has been eliminated.Consider the chains r5c8-3-r3c8~5~r8c8 and r5c8-3-r5c3-5-r8c3.Whichever of the 2 candidates in Row 8 contains the value 5, the cell r5c8 does not contain the value 3.- The move r5c8:=3 has been eliminated.The value 7 is the only candidate for the cell r5c8.`
Sue De Coq

Posts: 93
Joined: 01 April 2005