Coloring a Double Implication Chain

Steve · Posted: Sun Sep 17, 2006 7:47 pm Post subject:

Group colouring is discussed at http://www.sudoku.com/forums/viewtopic.php?t=2954

Steve

Ruud · Site Admin

This type of coloring has become very popular lately.

Check this recent topic for detailed information.

Names used by different people are:
- Weak coloring
- X-Colors
- Equivalence Marking

It looks like a double Nishio trick, where you ask yourself the question:

"Are there any candidates that would be eliminated when I place either of these 2 conjugate candidates in this house?"

When you backtrack from the eliminated candidate, you can use it as a single Nishio starting location.

Ruud
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rkral

daj95376 · Posted: Mon Sep 18, 2006 4:37 pm Post subject:

Thanks steve, Ruud, and rkral for replying to my query!

rkral: Yes, the elimination I described can be obtained in several ways. I'm not comfortable with nice loops and need to rectify that shortcoming. My solver uses Templates and it's always popping out eliminations that I quickly recognize as simple DICs, so that's how I reference them. Since my Templates are based on a single value, in this case <5>, I started viewing the DICs as an extended form of Coloring. That's why I started this thread.

BTW, the DIC that I spotted for this puzzle is the one I showed as colored:

[r1c7]=5 => [r1c1]!=5
[r5c7]=5 => [r6c3]=5 => [r9c1]=5 => [r1c1]!=5

TheSpaniard · Posted: Fri Sep 22, 2006 9:22 am Post subject:

To daj95376:

Ruud had mentioned that you can eliminate the "5" in r1c1 using X-Colors.
Steve and rkral have mention different ways of eliminating it too.

I will complete Ruud's comment, showing how X-Colors easily solves this situation:

Step 1: r1c7-Blue and r5c7-Green.

Step 2 (the "Single colors" stuff): No new cells coloured, as they are not other conjugate cells in "5's".

Step 3 (the "X-Ray" stuff):
a) r6c3-Green (is the only cell in its box not been a peer of r5c7, that is Green, so r6c3 is Green).
b) r9c1-Green (is the only cell in its box not been a peer of r6c3, that is Green, so r9c1 is Green).

Step 4.1) You can eliminate r1c1's candidate 5, due to the fact that it is peer of two cells (r1c7 and r9c1) coloured with two different colors.

Note that X-Colors overrides in a single and well defined technique both "Single Colors" and "Multicolors", and solves some other situations (like the one you proposed) that cannot be solved nor with Colors neither with Multicolors, needing other kind of more complicated techniques (Weak Coloring, Grouped Coloring, MC+Hinge, Implication chains, Nishio subsets, etc).
It is not clear until now (at least it is not clear for me) if X-Colors can completely override some or all of this different techniques; I have never found any situation that can be solved with any of these techniques and not with X-Colors (the contradiction rule), but I have not proven that rule yet (for instance, that WeakColoring(x,z) implies X-Colors(x,z)) in a more mathematical way.

Best Regards.

Carcul · Posted: Mon Sep 25, 2006 5:46 pm Post subject: