Looking for a LOGICAL next step

Advanced methods and approaches for solving Sudoku puzzles

Postby tso » Wed Aug 17, 2005 10:25 pm


The partial puzzle doesn't have enough information as posted. Even when considering the cells with 3 candidates, it has 3 solutions, two of which have different numbers in each cell, so no forcing chain or any other proof is possible.


1) r1c9 -- the 4 doesn't connect to anything so this loop in the upper right corner cannot form a forcing chain. (If r1c9=4, the rest of the loop can exist in two states.) These five cells are a dead end and can be ignored.

2) Exactly one of r89c3 must be 9
Exactly one of r79c6 must be 9
Therefore, these four cells are in a binary group. The line from r9c3-r9c6 is changed from RED to BLUE
Last edited by tso on Wed Aug 17, 2005 7:06 pm, edited 1 time in total.
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Postby Karyobin » Wed Aug 17, 2005 10:53 pm

Fantastic!! From a maths/logic point of view, you always know you're in the presence of some pretty impressive stuff when it reads like absolute b****cks!

I don't have clue what he's talking about but I'm gonna learn it and quote it in conversation. Especially the bit about "the rest of the loop can exit in two states" and "The line from r9c3-r9c6 is changed from RED to BLUE". I mean, that's Dan Brown stuff, that is.

I'm definitely gonna read the rest of the thread now. It's a kind of Damascus-thing.
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Postby tso » Wed Aug 17, 2005 11:09 pm

"Exit" was a typo, it should have read "exist" -- if that helps.
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Postby tso » Wed Aug 17, 2005 11:33 pm


Re: edge labelling -- Cool! Works even without bothering to differenciate between blue and red lines or worry about binary groups -- just connect all cells with numbered lines and look for the nice loop. However, when combined with the blue/red coloring, I can't see how the "it's all T&E" crowd could see this as having the slightest bit of "look-ahead". The deduction will be more obvious than naked or hidden trips, etc.

Can this be extended somehow to include cells with 3 candidates?
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Postby tso » Fri Aug 19, 2005 11:03 pm

Scott H wrote:The first flavor of nice loop (all adjacent pairs different) is a consistent loop where all nodes in the loop can have one of two values. You can change all red edges in such loops to blue edges. Also, for each unit (row/column/block) containing such a red edge, you can exclude the label of the of the edge from all other cells in the unit (this deduction deserves to be more widely known).

I'm not sure I understand what this. Could you explain further?
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Postby Scott H » Sat Aug 20, 2005 6:37 am

The edge labelling idea can be extended, but extending it too far could risk confusion or complication. I'm still exploring the best threshold for how far to extend it. I'll discuss that and the last two questions in a new thread in the next couple of days.
Scott H
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