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Marty R. · Posted: Mon Apr 24, 2006 5:23 am Post subject: Non-unique puzzle?

This is from the L.A. Times, April 23. Note how a "6" is forced into r6c4, resulting in the "deadly pattern" in boxes 4 and 5. Would they have done this deliberately or is it a mistake?

Ruud · Joined: 18 Jan 2006 Posts: 31

Hi Marty R.,

What you have found is an "unavoidable set". These form the basis for most uniqueness techniques.

However, you can only use these patterns if one of the following is true:

a) All clues for the 2 digits are part of the pattern;
b) No clues are part of the pattern;

In this case, there are 3 clues in the pattern, and 4 additional 6's and 9's outside the pattern. It is not deadly.

cheers,
Ruud.

ravel · Joined: 21 Apr 2006 Posts: 536

Hi Marty,

the puzzle is unique. To be deadly, none of the 4 numbers of such a pattern must be a given. In this case 3 are givens, i.e. a second solution with 9-6-9-6 is not possible.

But see here for recent examples, how to use such a deadly pattern to eliminate a number (first 2 chains in the post).

Marty R. · Posted: Mon Apr 24, 2006 4:35 pm Post subject:

Thank you for that illuminating information. Just a couple of days ago I read something in an old thread on some forum and started eliminating these "deadly patterns" based on already-filled cells, rather than just pairs in a unique rectangle.

However, almost every time, I neared an apparent solution, only to end up with an invalid result i.e., duplicates in a group. In fact, that happened seven times on one of the "Nightmare" puzzles. I now realize that I must've been using some cells that were givens.

Oh well, I thought it was a little too good to be true to stumble on such a simple technique that so effectively eliminated candidates. I'm glad I found that puzzle I posted here and asked, otherwise, who knows how much longer I would have continued to frustrate myself based on an invalid technique.

Thanks again.