dailysudoku.com

chriswarren · Joined: 29 Sep 2010 Posts: 4

Trying to solve the puzzle below via XY chains appears to give two answers to R1C4??

Starting from R1C6 -> R4C6-> R5C6->R6C4 -> suggests you can remove the 5 out of R1C4...

...and R1C6->R9C6->R7C4 -> suggests you can remove the 6 from R1C4!

Am I interpreting XY chains wrong?

http://www.dailysudoku.com/sudoku/play.shtml?p=89:4:1:56:2:56:7:89:3:5678:567:578:9:3:4:1:268:2 568:2:569:3:8:1:7:56:69:4:5689:569:4:2:7:58:3:1:68 9:1789:2:789:3:6:18:4:5:89:1568:3:58:15:4:9:268:7: 268:3:159:59:16:89:2:568:4:7:1579:1579:6:4:89:3:25 8:28:1258:4:8:2:7:5:16:9:3:16:

Please copy the complete link above and paste into browser.

Marty R. · Posted: Thu Sep 30, 2010 3:40 pm Post subject:

Welcome to the forum. I'm unable to see anything from the link other than boxes 123.

What puzzle is this?

Marty R. · Posted: Thu Sep 30, 2010 4:40 pm Post subject:

OK, I figured out the link.

chriswarren · Joined: 29 Sep 2010 Posts: 4

Hi Marty - thanks for the response - but I'm still confused!

Maybe I've got the name of the strategy wrong, but I thought that if you follow a chain of bi-value cells where they always share a digit, if you find a cell that is seen by both ends of the chain, and shares a digit with these two ends - then you can remove it.

That being said, depending on which chain route you follow in my puzzle, you get a different answer for R1C4. Somethings wrong somewhere!

Marty R. · Posted: Thu Sep 30, 2010 7:14 pm Post subject:

Chris, if you look at the first chain, a 5 in r1c6 proves a 5 in r6c4. Now suppose that r1c6 is not 5, i.e., it's a 6. The 6 proves a 1 in r6c4. There's nothing there that says either r1c6 or r6c4 must be a 5. That is the essence of pincers, that at least one or the other must be true.

Let's use a stupid example because I don't see any valid chains there. A 6 in r1c6 proves a 5 in r4c6. So if r1c6=6, then r4c6 must=5 and if r1c6<>6, then it must be=5, therefore one of r1c6 or r4c6 must be=5.

Of course one or the other must be 5 because there are only two 5s there, but I hope this can illustrate the principle of the XY-Chain.

Keep asking if this isn't clear.

daj95376 · Joined: 23 Aug 2008 Posts: 3854

Hello Chris,

Your description of an XY-Chain is incorrect. You want to start with the assumption that a condition is true. This is a common mistake when first encountering XY-Chains. You need to read the following article, and then ask questions if you are still confused.

http://www.sudopedia.org/wiki/XY-Chain

Regards, Danny

chriswarren · Joined: 29 Sep 2010 Posts: 4

Hi Marty and Danny,

Many thanks for your help.

I'm obviously missing something. I've reread the sudopedia article - and I still feel that in the example I've given, there are two XY chains i.e. bivalue cells, strongly linked with at least one value. Depending on the route of the chain, you either have a start and end cell with a common 5 (R1C6 and R6C4) or a start and end with a common 6 (R1C6 and R7C4).

The point of the strategy I thought is that you can then remove the common number found in the pincered cell R1C4. But obviously my two chains (if they really are XY chains) result in wanting to remove both the 5 and 6 from R1C4!

I'm sure I'm misinterpreting something (I have to be!!) - but it's frustrating not knowing exactly what. Thanks for any more help you can give.

Marty R. · Posted: Fri Oct 01, 2010 2:59 pm Post subject:

Chris,

Try this and see if it's any clearer than what you've read thus far.

http://www.sudokuonline.us/xy-chains.php

chriswarren · Joined: 29 Sep 2010 Posts: 4

Thanks Marty!

Have finally twigged. The key point in your last link was that the number you elimate from the pincered cell has to be the one that is "unused" in the end of the chain (i.e. not used to make the link to the previous cell).

Hope that's the explanation - can sleep easy now!