Are there infinitely many 4th row single-piece minimal edge templates?

I've been reading some of the hex literature. In Cameron Browne's book "connection games" he lists all the single-cell minimal edge templates that he knew of in 2005 ("minimal" means that if you remove one of the empty hexes it's not a template any more). There were 12 (this includes reflections, so, for example, there were three single-piece third row templates, one symmetric and the other two reflections of each other) and he even talks about his list as "the known single-piece minimal edge templates", implicitly raising the question as to whether there are more. Since then a couple more have been discovered; there is a single-cell 4th row template on drking's site that's not in Browne's book, and here on this site a single-cell 6th row template was discovered. I was re-reading that thread, and thinking about "Tom's move" which Arek taught me in this thread, and inspired by a comment made by Art Duval in the 6th row template thread, the following construction occurred to me.

I think that the following position

is a win for black (it's white to play) and furthermore I think that it can be extended to the left as far as you like -- basically it's four clear rows for as long as you want, and then three cells on the 5th row and a bit more space at the end.

The reason that this is interesting from a theoretical point of view is that *if I am right* (and I am a hex amateur so I might well not be) then this proves that there are infinitely many minimal edge templates and as far as I know this result is not in the literature. It might well be known to the experts of course, but I've not seen anyone formalise it this way.

Before I go on, let me explain the main line for black. Black can create both a 4th row and a 2nd row ladder at the same time. He first ladders along the 4th row up until the area where the 5th row free cells are. White can't let him drop to the 3rd row because then this gives black a second row ladder escape. Black then ladders along the second row and finally plays Tom's move (and the usual continuation), which works because of the three 5th row empty cells. Note that Tom's move (o17) is the unique winning move in this position. More generally, computer analysis seems to indicate that the only winning lines for black involving laddering along the 4th and 2nd rows, making sure that the 4th row ladder "stays ahead" (if the second row ladder catches up or overtakes the 4th row ladder then white can counter a 4th row move with a second row move completely blocking the second row ladder and then black's in trouble)

The line I show here is optimal for black and I think there's some chance of proving that this is a minimal template. This might make a nice Masters thesis project; one would have to rigorously prove that removing one cell breaks the template but I am cautiously optimistic that this can be done; experts might instantly know this to be true of course. The reason I'm optimistic is that computer analysis indicates that basically the only winning line is the one I have tried to explain above, so probably this can be proved theoretically and then similarly one can start removing stones and constuct proofs that white can break the connection; in the arbitrarily-large-sized strip of length 4 one would probably only have to check that removing a cell from any row breaks the connection, and then just deal with cells at the beginning and end separately. This sketch proof could perhaps be made rigorous.

Another way of phrasing what I am trying to say here is that the area on the right of the template (with the three 5th row cells) is a ladder escape for two incoming ladders, one on the 4th and one on the 2nd row (black has to have both), and the escape *has no cells at all*, which makes it completely different to all the ladder escapes that I've seen so far. Again of course I will stress that I have very limited hex experience and that perhaps several such ladder escapes are already well-known. I am just trying to get these ideas down because I've not seen them in the literature; if I am wrong then so be it, but if I am right then we have what seems to me to be a great candidate for an infinite family of single-piece minimal templates and I don't think I've ever seen the existence of such a thing explicitly stated in the literature.

I can't edit my posts in this (rather primitive, in this day and age) forum and on re-reading I'd like to change/clarify some things, so it has to be another post. Sorry for the spam.

A correction: when I said " this proves that there are infinitely many minimal edge templates and as far as I know this result is not in the literature" I should have said "this proves that there are infinitely many *single piece* minimal edge templates...". It's easy to produce large multi-cell templates.

I should *definitely* have credited Art Duval for his comment on 31/12/08 on this thread; my post above is basically my attempt to understand some of the things he said here.

I should clarify that for this thread, when I say "n'th row template" I am allowing the possibility that there are cells in the template which are higher than the n'th row. There was some talk on the thread linked to above about whether this should be allowed in the definition.

I should note that shalev has a conjectural single piece 7th row template, which has not (as far as I know, at the time of writing) been verified by computer (the template is 23 columns long and is symmetric so you have to work out the missing four columns for yourself) but perhaps humans are confident about this.

Finally I should note that the question of whether there exists a single-cell n'th row template for arbitrarily large n is still, as far as I know, open.

Grr. Another typo -- "the escape has no cells at all" should be "the escape has no stones at all".

I'm convinced that your conjecture, that this is a sequence of minimal edge templates, is true, and that a formal proof is possible but rather tedious. I like how you see it as a ladder escape without black pieces.

Looks like an interesting result! As a beginner, I'm very intrigued by your recent Hex-related posts. :) Unfortunately, as a beginner, I can't really say much gameplay-wise; but most likely a formal proof will be fairly tedious as urmaul pointed out.

Yes I agree a formal proof would be tedious -- but I believe it would be a finite process (i.e. I don't believe that the proof that it's minimal would involve constructions of ever-increasing complexity as the area got bigger).

Dr King pointed out to me that I need to say more about white's defence during the 4th row ladder phase (I think he might be reluctant to expand his edge template web page to a page of infinite length). At any stage white does not have to play on the 3rd row directly blocking the ladder -- there are six other moves he can play which stop black from connecting instantly via a standard 3rd row edge template and each of these has to be refuted, so in fact my argument above is incomplete. Here's an example of an attempt to block the double-ladder with a second row intrusion, plus black's counter: link to trmph [trmph.com]. I checked all six.

This pattern shows that the only candidates for white are j17 and i19. If white plays i19, black can play j17.

Yes, the analysis here is correct. There's a similar pattern for 5th row, and probably 6th row as well, though it starts getting tricky to check.

It might start getting tricky to check, but that's what computers are there for; it only needs to be done once. Like analysing all openings on small boards -- hard, but once it's done, it's done.