Strategies for Number Puzzles of all kinds
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The Logic of Sudoku
  XY-Chains

The Y-Wing Chains are infact part of a more encompassing strategy called XY-Chains. The commonality is the same pincer-like attack on candidates that both ends can see and that the chain is made of bi-value cells. With Y-Chains the hinge was expanded to a chain of identical bi-value cells but in an XY-Chain these can be different - as long as there is one candidate to make all the links. The "X" and the "Y" in the name represent these two values in each chain link.

The example here is a very simple XY-Chain of length 4 which removed all 5's in the pink cells. The chain ends are A7 and C2 - so all cells that can see both of these are under fire. It's possible to start at either end but lets follow the example from A7. We can reason as follows

  • If A7 is 5 then A3/C7/C9 cannot be.
  • if A7 is NOT 5 then it's 9, so A5 must be 2, which forces A1 to be 6. If A1 is 6 then C2 is 5.

 XY-Chain 1
XY-Chain 1: Load Example or : From the Start
Which ever choice in A7 the 5's in A3/C7/C9 cannot be 5. The same logic can be traced from C2 to A7 so the strategy is bi-directional, in the jargon.
Here is a much longer seemingly more complex XY-Chain of length 10 - attacking 6's in the pink cells. I've shaded the two chain ends in green. I've also left the arrow head's off so you can trace the XY-Chain from either end. Doing so you'll quickly release that 6 must be at one end or the other, so there is no chance of other 6's which both ends can see.

When you've got a good spread of bi-values this is a useful trick. Remote Pairs, described below are also a special sub-set of XY-Chains - they merely have a double-chain of inference through both values in each chain link because they contain the same values from start to end. Remote Pairs were discovered/described first before this more general approach.
 XY-Chain 2
XY-Chain 2: Load Example or : From the Start

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Comments...

Monday 16-Nov-2009

... by: Matt Lala

I read something on a different site (if I understand this correctly) is that when you reach the end of the chain, you are using the "leftover" value to make your elimination. In the first example, the last green arrow is to a 6, and the leftover in that cell is a 5, which is what gets eliminated. If the last link had been to the 5, then the leftover is 6, and you cannot use that to end the chain and eliminate the 5's. Not sure what conditions are needed for the start. I may be wrong on this.

Francois, I don't think you need a solver, but it does help to have a program that highlights all of them. I glance at the 'busiest' groupings of bivalue cells and then pick one and just start driving. If a fork presents itself I'll choose whichever option seems to steer me back towards the start of the chain.

Anton, I don't think they have to be locked, they just need to be bivalue. Try plugging in a 2 at the start of that chain in the 2nd example, and follow the links, and you'll see how the green cell at the end becomes a 6 (despite the unlocked cells used). And plugging in a 6 at the start makes the eliminations pretty obvious.

Wednesday 28-Oct-2009

... by: Francois Tremblay

I understand the logic but my problem is how do you spot this without the help of a solver? Visually, the start and end of such a chain are tough to find. Do you go through all the grid to start the chain at all possible bi-value cells? In the "simple" cases above, you have respectively 21 & 22 bi-value cells (even your book's example on page 62, figure 22.1 shows 21 of them) which would represent a lot of permutations that only a computer could run through. As a human solver, is there a trick to find those chains?


Thursday 18-Jun-2009

... by: Anton Delprado

Maybe I am missing something but it looks like 4E and 8E are not locked for the value 6 because 9E is potentially 6 as well.

An XY-Chain is possible from this although it would have the pivot chain below:
3A - 3C - 3H - 6H - 6F - 5F - 9F

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Article created on 11-April-2008. Views: 21643
This page was last modified on 16-July-2008, at 14:14.
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Copyright Andrew Stuart @ Scanraid Ltd, 2008