dailysudoku.com

[Steve R]

This is Claudia Bach’s puzzle for 31 May 2008:

[Marty R.]

I couldn't find a thing other than an XYZ-Wing. I turned to Medusa, quickly trapped a couple of numbers and the puzzle fell apart. When it falls apart so easily, it makes me wonder what I missed.

Who is Claudia Bach? I've been here over two years and never saw that name before.

[ravel]

When i need chains, its always the question, where to start with. I like to start from UR's, useless skyscraper/kites, xy- or w-wings, seldom xyz-wings, or simply bivalue cells or strong links (preferable on multi-candidate cells). Steve R once mentioned, that you can use the either/or of two digits, which share a cell and both have a strong link in the same unit.
E.g. in row 8 2 and 8 both have a strong link and the common cell r8c1. Since not both can be in this cell, either r8c7 must be 2 or r8c9 must be 8 (or both).

I tried to find them systematically in this puzzle (this is quickly done, when you can highlight numbers, but harder on paper) and looked, how i could use them.

[wapati]

[Asellus]

Two Medusa clusters, bridged on the <6>s in r1c25 (or in r28c3) quickly lead to the elimination of <6> from r7c8 and <8> from r8c9 and the puzzle falls apart.

Obviously, there are lots of AICs lurking in this grid.

I like ravel's "28" based AIC with the strong link discontinuity. The chain can be shortened:

(2)r8c7=(2-8)r8c1=(8)r8c9-(8)r2c9=(8-6)r2c5=(6)r2c3-ALS,r8c357[(6)r8c3=(2)r8c7]; r8c7=2

Essentially, there are 3 important strong inference links here:
(2)r8c7=(8)r8c9
(6)r2c3=(8)r2c9
both of which ravel pointed out, and
(6)r8c3=(2)r8c7
within the {1267} ALS of r8c357.

Because the <6>s lined up in c3 are weakly linked and the <8>s lined up in c9 are weakly linked, we have our short chain fixing r8c7 as <2>. Or, visually: