dailysudoku.com

Asellus · Posted: Tue Jul 17, 2007 4:50 pm Post subject:

jLo,

I am quite sure that the <4> in the lower left must be red IF the <7> in @ is red based only on the cells shown since the lower right cell must be <2>. It is the same logic as an XY-Chain.

I believe I have erred however in saying that all the red values can be eliminated. It is only certain that the <7> in @ can be eliminated since the implications vanish if @ is <1>. (That brings those other cells in the grid into play.)

As it happens, none of the red values in this case are in the final solution, so my error was a "lucky" one.

There is Sue de Coq-like structure in the region involved in those XY-Chains and George's comment. However, I couldn't work it out in any way that made sense to me. It did lead me to seeing that implication that I posted, however. As you may know, Sue de Coq often involves solutions based on avoiding the elimination of both candidates of a bivalue cell.

Steve R · Posted: Tue Jul 17, 2007 5:34 pm Post subject:

George,

It is an ALS chain:

-7- r3c9 -6- r9c9 -2- {r8c3, r9c3} -7-

which eliminates 7 from any common associate of r3c9, r8c3 and r9c3.

As Asellus pointed out, these chains have exactly the same logic as XY-chains. The only difference is that, in the XY case, the nodes are single cells with two candidates while, in the general case, the nodes may be two cells sharing three candidates, three cells sharing four candidates and so on.

Steve

Glassman · Joined: 21 Oct 2005 Posts: 50 Location: England

I'm completely stuck on this — I haven't even got as far as Captain Pete. How did he do the eliminations in c7?

This is my feeble effort. Enlightenment please!

TKiel · Joined: 22 Feb 2006 Posts: 292 Location: Kalamazoo, MI

X-wing, followed by naked triple.

Glassman · Joined: 21 Oct 2005 Posts: 50 Location: England

Tracy — Thanks, I've found it.

Glassman Cool