## 27 Jan 2006 Nightmare

If you invented that new way to solve these little puzzles, tell us about it
David Bryant
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### 27 Jan 2006 Nightmare

I'm not sure this is a new technique. But in the course of solving today's "Nightmare" I hit an interesting formation -- I'm not sure I've ever encountered one exactly like it before.

With 45 cells resolved and 36 still to go the puzzle looked like this (I had only made the obvious moves up to this point, plus I had used some simple coloring to eliminate one "6" [at r8c5] and one "8" [at r5c8] from the matrix).

Code: Select all

``````  6    8    2    9    5    7    4    9    1
1    4    3    8   26    9    5   26    7
5    7    9   124 1246  146  68   268   3
3    9   78    6   78    5    2    1    4
278  12    6   147   3   14    9   57   58
4    5   178  179 1789   2   78    3    6
278  26    5  2479 24679  3    1   478  89
27    3    4  1279 1279   8   67   567  59
9   16   178   5  1467  146   3  4678   2``````
I wasn't seeing anything obvious and was focusing on the simple pairs, looking for a regular XY-Wing, when I spotted an unusual pattern near the bottom right corner. It's sort of a "crippled" XY-Wing.

r7c9 = 8 ==> r5c9 = 5 ==> r5c8 = 7 ==> no "7"s in rest of col 8
r7c9 = 9 ==> r8c9 = 5 ==> {6, 7} pair in row 8 ==> r7c8, r9c8 <> 7

So either way there can't be a "7" at r7c8 or r9c8 -- this leaves a naked quad and a "hidden" pair {5, 7} in column 8, and the "6" in r8c7 is revealed as unique in row 8 -- the rest of the puzzle just falls apart!

I'm sure there are other ways to solve this puzzle -- I don't think it's as hard as a typical "Nightmare." But I sort of like this way. It's cute. dcb
Myth Jellies
Hooked
Posts: 42
Joined: Tue Apr 04, 2006 7:07 am
You can get this solution out of a nice grouped candidate chain as follows...

Code: Select all

``````  6    8    2    9    5    7    4     9     1
1    4    3    8   26    9    5    26     7
5    7    9   124 1246  146  68    268    3
3    9   78    6   78    5    2     1     4
278  12    6   147   3   14    9   *57   *58
4    5   178  179 1789   2   78     3     6
278  26    5  2479 24679  3    1    478  *89
27    3    4  1279 1279   8  *67   *567  *59
9   16   178   5  1467  146   3   4678    2

&#91;7==5&#93;-&#91;5==8&#93;-&#91;8==9&#93;-&#91;9==5&#93;-&#91;5=&#40;6&7&#41;&#93;
&#91;r5c8&#93; &#91;r5c9&#93; &#91;r7c9&#93; &#91;r8c9&#93; &#91; r8c78 &#93;
chain implies that all cells seeing r5c8, r8c7, & r8c8 cannot be 7
``````
Ron Moore
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Joined: Sun Aug 13, 2006 3:34 am
Location: New Mexico

### An XYZ wing with two eliminations

I just joined the forum, so I'm aware that this is very old business. I wouldn't have commented except that the solving guide states that no XYZ wing resulting in two eliminations had been discovered in practice. Well, here is one. r8c8 (pivot), r5c8, and r8c7 form the XYZ wing pattern (no need to consider column 9), which, as the previous posts have concluded, eliminates 7 from r79c8.
David Bryant
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Joined: Fri Jan 20, 2006 6:21 pm
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### Astute observation

That's an astute observation, Ron. I'm not too good with names for some of these newer "techniques" -- I tend to rely on logic & first principles. Thanks for pointing out that this is an "XYZ-Wing" pattern.

Maybe we can get Ruud to update the solving guide. Here's an example of the double elimination XYZ-Wing in one of the early "nightmares" he published! dcb
Ruud
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Joined: Fri Dec 30, 2005 10:21 pm
I've updated the guide. A pointer to this Nightmare has been added to the XYZ-Wing topic.

Thanks, Ron. This is the second time in a row that others have found examples I was looking for in my own puzzles.

Ruud
Ron Moore