Berthier’s xyt-chains in THLS

Next we review the introduction of xyt-chains in The Hidden Logic of Sudoku. These are modified XY-chains which allow extra candidates in the chain cells, and unlike regular XY-chains, make eliminations regardless of whether the chain starting candidate is true or false. This post displays several of the shorter xyt-chain examples in THLS.

Denis Berthier invented something worthwhile, but dangerous, with his modification of the XY -chain into the xyt-chain.  The xyt chain is an XY chain with “extra” candidates embedded in its node cells.  Unlike the XY-chain, the xyt-chain has a direction, and one “starting” candidate.  The embedded chain is built with an assumption that the starting candidate is false. Under this assumption, every “right” candidate with the exit wink it true, and every “left” candidate with an incoming wink is also false.

Extra candidates do not stop the embedded XY chain AIC action as long as they “see” a prior right candidate in the chain. The “full” xyt-chain ends with a right candidate matching the starting candidate.  Any outside candidate seeing both terminal candidates is false, because either the starting candidate is true, or it forces the ending candidate to be true via the xyt-chain.

In THLS the xyt-chain is described as a more general XY-chain, of which the “pure” xy-chain is a special case. That is incorrect, because the designated direction of the xyt-chain, allowing branching of the chain, is a special condition applied nowhere else in the AIC family.  The mistaken idea that AIC are somehow not lot logical may come from similar confusion with branched forcing chains.

After we actually look at some of these chains on the grid, perhaps you will agree with me that, while innovative, and something to be aware of, a comprehensive search for them should be among last resort measures.

17-2769-ur-gridLet’s check it out the UR you were to skip in the Royle 17-2769 homework. Here is the unique rectangle grid, with the very easy decision force a 9 into the rectangle.



Following the THLS policy to save the UR for a last resort, SudoRules comes across the following xyt chain.

17-2769-xytLook at 9r9c7 as a possible starting candidate. Assuming it false, we see an XY-chain moving to r8c2, but for the extra candidate 7. That assumption would make 7r8c8 true, removing that value from r8c2 (red wink) and allowing the red slink.  So, not 9r9c7 => not 9r9c2 by the chain. But clearly,  9r9c7 => not 9r9c2 as well. 9r9c7 has to be true or false, making 9r9c7 false. Same result, fatal in both cases.

The start of an xyt chain does look like an arbitrary guess, but its construction is logical, if unpredictable. In a comment on this post, Mittleman points out the ALS-XZ in this chain. The r8c23 ALS 379 and the r89c78 ALS679 have a 7 restricted common, and the chain’s victim sees all 9’s in both ALS. It’s worth noting that the starting bv pair of an xyt-chain usually form a promising first ALS with two singles. Not always, as we see next.

17-5105-x-chainThe next example, Royle-5105 requires a 5-chain to prepare the grid for the xyt-chain. Berthier does not include it in the trace and the elaboration, though the xyt example requires it.






17-5105-xytAn XY-chain starting with 9r1c3 is completed in r1c4 by erasing 4r1c3 by means of the assumption value 4r1c3.


17-5105-xyHowever, the Sysudoku OB gives a more easily spotted alternative.

The simple XY chain to the right removes three 4-candidates, setting up a regular XYZ-wing below with two ER victims, and 17-5101 is done.








Our final example is an illustration of an xyt-chain that might be spotted as a needed modification of a normal XY chain.

It is a coloring scenario in Royle 17-1365. Again, an abundance of bv encourages us to build two clusters.

17-1365-no-bridgeBridging logic applies with

 not(orange and green) => red or blue,

but 9 is the only number common to both clusters, and no red 9 means no bridge.

Then we see an “almost” XY loop spoiled by an extra 9 in one cell. 9r2c4 would erase it. Take 3 3r2c4 as the starting candidate. If false, the XY-chain moves around to 23r2c7 and 3r2c2 is removed.

17-1365-color-xyzt-color-trapThe extended cluster places a red 2 in r3c8, where two 2 candidates see red and blue. One of them wraps orange, and shortly thereafter, red forces blue.







17-1365-293-wingOn the other hand, I might have followed the normal SOB course, and pulled out this i293 wing, which confirms 3r4c7 and permits the remote pair that confirms 3r2c4 and collapses 17-1365.

My conclusion,  after these few examples, is that xyt-chains are harder to spot than regular Sysudoku alternatives.


While they are constructed straightforwardly enough, a comprehensive search for them would take a too many unproductive constructions, and should be deferred at least until the advanced SOB methods are given attention.

royle-17-20565Next we look for examples of  longer xyt chains in THLS XVII, and encounter surprises. The first one is an even stronger tie-in with coloring, perfect for Valentine’s Day. If you  would like to be there with your own slides and have mislaid your THLS again, here’s my next target, Royle 17-20565.  You’re welcome.

About Sudent

I'm John Welch, a retired engineering professor, father of 3 wonderful daughters and granddad to 7 fabulous grandchildren. Sudoku analysis and illustration is a great hobby and a healthy mental challenge.
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4 Responses to Berthier’s xyt-chains in THLS

  1. dovmittelman says:

    Just a follow up on last week’s conversation. I tend to call a puzzle with one trilocal candidate in one tri-value square simply a BUG, partly for brevity and partly because I conceive of the thing that the trilocal candidate is the grave that ‘buries’ (ie solves) the bi-value network. I understand that you call it a near BUG, apparently with the notion that the bi-value network is the grave that ‘buries’ (ie cooks) the puzzle. But whatever we call it, consider a puzzle with all bi-value squares except one tri-value one, which has two bilocal candidates and one trilocal one. Now I start coloring. I will perforce give some color to every candidate but that one, which means two colored candidates in the tri-value square. The whole network is a black box, so maybe there’s a color clash and maybe there isn’t, and maybe the tri-value square has two the same color (and at least a wrap) and maybe it has two different colors (and at least a trap), but in any event that tri-value square must resolve somehow, by eliminating at least one candidate. So where does solution #3 come from?

    • Sudent says:

      Dov, please comment on the post you’re commenting on. This discussion belongs on “Mashing the Bug” of Jan 1, 2013.

      Look at the diagram again. Take the third candidate off, and the completed coloring defines two solutions consistently. That is the “grave”. Put the third candidate on and you destroy the coloring done up to the near BUG. That coloring is invalid. Starting over from the third candidate produces the third solution. That solution mixes the invalid colors, because the slink network the colors represent is damaged.

  2. dovmittelman says:

    And one on the topic of the week: it looks like the xyt formula might not even be necessary for these eliminations. In the first example, the ALS means the 9 in (8,2) slinks directly to a 3/7 locked pair in (8,2+3). The 7 of that pair winks at 7(8,8) and off we go. The others don’t look obvious to me but all the xyt red links connect squares that make an ALS. It may be possible to rework the structure to ALS loops and remove the ad hoc assumption.

    • Sudent says:

      You’re right on the ALS in the 17-2769 example. Except that slinks and winks are between candidates, not cells. In the ALS-XZ, 7’s are a restricted common, and the 9’s are a toxic set. Starting the xyt-chain, we can’t anticipate the ALS, but we can say that the starting cells are an ALS with two single candidates, a promising start in looking for that companion ALS. I’ll be adding that to the post, to your credit.

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