This post concludes my reading of Denis Berthier’s *The Hidden Logic of Sudoku*, with two example puzzles to illustrate his xyzt-chain. While it is good to be aware of Berthier’s “t” versions of classic XY chains, I don’t recommend a comprehensive search for them. I’ll explain why.

Here is the basic trace of the homework puzzle, Royle 17-9373. It was selected to introduce the xyzt-chain in *THLS*.

To the requirements for the xyt, Denis allows one exception for the xyzt. In only one internal chain cell, there can be an additional candidate of the starting cell number. This makes the construction of the embedded XY-chain dependent on the victim, along with the directional assumption of the chain starting with a false candidate. A new thing in itself. Before using the cell, you need to make sure there is a potential victim seeing the starting candidate. Then the victim must see the ending candidate as well.

Does this work? Yes, if the added candidate is false and the starting candidate is false, the chain continues and the end candidate is true. A true candidate cannot see all three. Having a 3-candidate toxic set is a disadvantage shared with the xyz wing.

Is this a good deal for the human solver? No. The xyzt allowance adds very little to the chain construction budget, in return for the victim’s having to see a third candidate. Berthier has reported many of these xyzt-chains being built, in this a previous chapters of THLS, but I have yet to see one not cooked by a more easily spotted technique.

In Royle 17-9373, Berthier’s “type 2” xyzt-chain is found on the line marked grid. Starting on r8c5 with assumption 5r8c5 is false. Selection of r8c6 as the third cell anticipates a victim on r8 or c6, and gets the chain to r8c3 for the final slink. So 5r8c7 if false, whether 5r8c5 is true or false, as it allows the chain to create the embedded XY-chain that removes it.

The only reasonable way to look for this is to start an xyt chain on every possible pair, go as far as you can with the xyt, then look for a linking cell that would be next, except for a candidate of the starting number, and a corresponding victim.

The xyzt removal enables a 5-chain, leading to a collapse. But here is a simple kraken 5-wing bringing the same collapse. One of the wing’s victims sees the fin, 5r5c7. So the victim, like xyt and xyzt victims, is false if the fin is false (clean fish), but also false if the fin is true. The kraken logic is very similar to the logic of the xyt-chain.

Pardon the side lecture, but if the fish’s victim is in the same box as the fin, it’s just the finned fish we saw earlier in the THLS review. If its not, and forcing chain “seeing” is required, it’s a kraken fish.

For another of Berthier’s “simplest” xyzt-chain examples, here is the first one in Royle-4601. The victim see’s the candidate that is preventing its removal by the embedded XY chain. Strangely enough, it is the terminal 9r9c2 that provides the rational for the removal. It removes the victim directly if true, and by XY chain with the victim’s cooperation, if false. You could call it a suicide xyt chain.

The removal is indecisive, but another victim is strapping on the vest. The starting candidate partner 9r2c4 removes one barrier 9r2c9, but spares 9r1c1 for the assumption chain, with the victim stepping in to make its removal possible.

Next in the *THLS* trace, but __after__ the removal of 8r1c8 by a hidden hxy-chain, a third xyzt-chain, starting in r3c1 like the first chain.

This time the solver somehow sees the enabled terminal cell waiting and that a victim will allow the chain through r2c9. In the defiance this flabbergasting spotting, 17-4601 demands another xyt-chain and an xy-chain for its collapse.

But wait. There are much more reasonable alternatives.

The first one illustrates how badly Berthier’s logical complexity fails as an order of battle for human solvers. There is a simple XY-chain of five cells that wipes out 17-4601, all by itself. Sorry for displaying it, after having you laboriously trace out all these extreme four cell chains, but here it is, fresh out of line marking.

Or, with 7 of 9 boxes containing bv only, you might take an even easier path, with your crayons. Stay within the lines, please.

Coloring the line marked grid, I have two small clusters connected by the 5’s. The bridging logic is easy:

Row r3 says,

not red and green.

Column 1 says,

Not orange and green.

But one of these is true. We must have red or orange, so it can’t be green.

With blue candidates, 17-4601 folds up like an accordian.

We’ve gone far enough with xyt and xyzt chains. For human solvers, it makes no sense to work through the uncertainties of the “t” chains when there is certain progress to be made with techniques Berthier ignores or rejects. This review has shown why and how these chains work. I advise trying to find them yourself to appreciate just how speculative they are. Then forget’em, until you are really desperate.

The review of *The Hidden Logic of Sudoku* closes with the next post, with a return to XY chains for a reminder on ALS toxic sets, *a.k.a.* ALZ-XZ. This is another widely known advanced method ignored in the design of SudoRules, a human emulating solver. The puzzle for this demonstration is one of those generated and selected by Denis Berthier to compare regular and hidden XY resolution paths, Sudogen 17-3403.