dailysudoku.com

[ravel]

TTHsieh posted this puzzle on the other forum.

[Earl]

I got this far, but I do not see the next step.

Any help out there?

Earl

[Earl]

I have made some progress.
But still need help.

Earl

[Marty R.]

Ravel, this was pretty interesting, very similar to the one you posted two weeks ago, in that a lot of techniques were required and each gave up only a very little ground. I used seven different techniques, one of them twice and another three times, for a total of 10 moves needed in order to bust the puzzle.

[jLo]

[jLo]

[ravel]

This is, what i had:

[Mogulmeister]

Very nice puzzle.

[nataraj]

ravel,

could you please explain how your second W-wing works? I get the one (@) all right, but spent a long time to make something out of the 57s in the lower right corner, without success. I am referring to your diagram:

[nataraj]

on second thought, I think I see how it works:

it is sufficient to show that r5c3 and r7c9 cannot BOTH be 7.

so lets assume both are 7.
Then r9c3=5, r9c7=3 => r9c6=2 => r9c4=1 and no possible value for r9c8.
Therefore at least one of the two pincers must be 5 and we can remove 5 from r5c9.

Neat.

[Asellus]

Many different routes. Mine:

1. XY Wing (same as Ravel's)
2. Sashimi Swordfish: -7 in R7C2
3. Color Wing (or ER or XY Chain) in C3,R5,B6: -5 in R9C7
4. Finned X-Wing: -5 in R1C7
5. 29 UR in R37C46: -2 in R7C6 (involves 9 X-Wing)
6. Finned X-Wing: -2 in R1C7 ("copy" of step 4!)
7. XY Chain from R1C7 to R9C3 (via C7,R8,B7): -7 in R9C7

I didn't determine whether or not every step was necessary.

Note: I use "Color Wing" as defined by Ruud. What Ravel calls a "Color Wing" above, Ruud calls a "Color Trap." More naming confusion!

[TKiel]

I reached this point, at which Simple Sudoku had no more hints, which means there are no locked candidates, naked/hidden subsets, fish of any size, XY-wings or simple/multi coloring chains. (It verifies that the puzzle has only one solution, so it must use techniques other than what I've listed, but it doesn't give them as hints.)

[TKiel]

After posting the above and looking at it some more, it seems that the letters I used to distinguish the different chains can be junked and the whole thing considered one grouped coloring chain, which basically makes this what Asellus would call a 'Color Wrap', in that C1 has two (-) cells. Therefore, no cells marked with (-) contain 5.

[Asellus]

Tracy,

As I see it, there are only two color clusters on 5, which, following your approach, I've labelled A and B:

[Steve R]

Tracy

I’m afraid I couldn’t follow the eliminations you made in colouring terms. They are certainly right enough. Perhaps someone with a keener eye will do better.

In the meantime it may be helpful to mention one elimination derived from (group) colouring on 5. What I have done is start at r7c9, marked “A.” The conjugate group in box 9 is marked “b” and so on: the alphabetical sequence enumerates the chain and polarity of each group is shown by whether it is upper or lower case.

[Asellus]

Steve,

Take a look at my "ER Loop" posting on the Techiques board. [Edit: It's actually on the "Other Puzzles" board.] It may look familiar.

[TKiel]

I think I recognize the flaw in my reasoning. I assumed that since r5c3 & r6c1 were peers, they would end up having opposite polarity, which is true if one of them were +. But both of them could also be -, which throws off the polarity of my chain and negates the exclusion.

Steve,

Thanks. Something like that is exactly what I was looking for.

[Asellus]

[keith]

[Asellus]

After giving it a bit of thought, I'm pretty sure a "Color Wing Wrap" is not logically possible. Here's the reasoning:

We have two color clusters, A and B, with polarities within each denoted as Aa and Bb. We then posit a weak link somewhere between two cells with polarity "A" and "B" that forms a "bridge" between the clusters.

Assuming all eliminations/contradictions due to either individual cluster alone have already been exploited, the only additional eliminations/contradictions possible are between cells of polarity "a" and "b".

Now, assume that somewhere in the grid an "a" cell and "b" cell share a house. They thus have a weak link (since the clusters are separate there is no strong link). Because they cause eliminations, all other candidates for the subject digit within that house are eliminated. As a result, they now acquire a strong link! There is no "Wrap" contradiction here. Instead, the two separate clusters combine into a single cluster and all "b" cells become "A" cells and all "B" cells become "a" cells.

However, if, as a result of this merger into a single cluster, we discover two "A" cells (or two "a" cells) within a single house, then there is a resulting "Color Wrap." However, we are back to a single strongly-linked cluster.

I don't know if that will make the situation clear; however, I'm fairly confident that the reasoning is sound.