HEX double ladder tactics

I posted about how to win when you have two attacking ladders.

http://boardgamegeek.com/thread/1299269/how-win-double-ladder

Please shoot holes in my analysis. Oh, and please set my house on fire and kick my dog.

I can't find any mistakes there although you haven't examined several side lines (they most interesting ones are alternatives to j5). Then again I guess everyone could easily figure out the solutions by himself.

Anyway, thanks for your effort to teach this cool technique.

I'd also be interested in an analysis about how to escape using a 3rd- and a 5th-row ladder (as opposed to a 2nd- and a 4th-row ladder) as I think this should be possible given enough room on the edge.

What about other combinations of attacking ladders?  For example 4+6, 5+7, or even something like 2+5 or 3+5+7?

This is what the Hex wiki called "Tom's move". It does work on higher up double ladders like 3rd and 5th. This trick is the main contestant for a solution for the "n-th row template problem".

There are numerous different ways of making escapes in various positions. Probably it's best to always look for an escape yourselves, but surely it could be fun to make a kind of collection. Some escapes are very rare and unobvious, just look at this one:

http://www.trmph.com/hex/board#13,h9h9j7h10f9e11g10f12g11g13g12f13k11j12k8

(it's actually the best way for black there)

Nice

Very cool escape Arek!

Regarding the 3+5-ladder escape: I can't figure it out by myself. I guess this position should be won for black: http://www.trmph.com/hex/board#13,c9c10b10a12b9d8a9e7a8f7a7g7a6h7a5i7a4j7a3k7a2l7a1m7.

But I am stuck, can't find a win in the following positions: http://www.trmph.com/hex/board#13,c9c10b10a12b9d8a9e7a8f7a7g7a6h7a5i7a4j7a3k7a2l7a1m7d9d11 and http://www.trmph.com/hex/board#13,c9c10b10a12b9d8a9e7a8f7a7g7a6h7a5i7a4j7a3k7a2l7a1m7d9d10f10f12.

For 4+6 it's certainly too complex for me to analyze. Is it known if it works for all combinations of attacking ladders in rows n and n+2 (n>1)? Also, if this is indeed true, how would it imply a guaranteed escape to the edge with only one piece at n-th row given a sufficiently large edge (at least this is what I assume you mean by "n-th row template problem)?

Loony: for the 3-5 ladder, try this and this.

This type of escape is suspected to work for all ladders in rows n and n+2 (with n>1), but this isn't proven. If it is true, then it is also probably true that templates like this work from any height, because the natural way of trying to stop such a stone from connecting is by forming a double ladder.

Actually, the n-th row template might need to look like this instead. Anyway, if such an asymmetric template always works, then I strongly suspect that more vanilla n-th row templates like this always work as well, because the natural ways to defend against them just reduce to the asymmetric template ( here's an example).

hmm, shalev, how to play this http://www.trmph.com/hex/board#13,c9c10b10a12b9d8a9e7a8f7a7g7a6h7a5i7a4j7a3k7a2l7a1m7d9d10f10e11 ?

I've always thought that the only double ladder escape is on lines 2+4 in a closed position like you're showing.

@Arek: I already analyzed the line you are questioning.

http://www.trmph.com/hex/board#13,c9c10b10a12b9d8a9e7a8f7a7g7a6h7a5i7a4j7a3k7a2l7a1m7d9d10f10e11b11b12c11c12d11d12g9g11h10h11f11

http://www.trmph.com/hex/board#13,c9c10b10a12b9d8a9e7a8f7a7g7a6h7a5i7a4j7a3k7a2l7a1m7d9d10f10e11b11b12c11c12d11d13d12c13g11f12e12e13h10f11g9

@shalev: Thanks, I will try using your suggestions!

The n-th row template is very interesting, but I suspect it's hard to prove anything because with increasing n the number of options (others than running into the double ladder immediately) to defend increases too.

I forgot to mention obligatory http://www.trmph.com/hex/board#13,c9c10b10a12b9d8a9e7a8f7a7g7a6h7a5i7a4j7a3k7a2l7a1m7d9d10f10e11b11b12c11c13e9e10f9f11d12.

yes Loony, you're right. But for lines 4+6 the 13x13 board is surely too small. Interesting what size is big enough then!

Thanks to your suggested moves I finally managed to prove the 3+5 ladder escape is working!

I'm interested in doing the exercise for 4+6 but will probably take me months..

Following only the standard pattern I get: http://www.trmph.com/hex/board#19,d14d15c15b17c14e13b14f12a14g12a13h12a12i12a11j12a10k12a9l12a8m12a7n12a6o12a5p12a4q12a3r12a2s12a1d13e14e15g15f16c16c17d16d17e16e17h14f15g13g16i15i16h16g18k16j17h17h18i17i18l15l17m16m17k17j19k18k19l18l19o17n18m18m19p16o18q17p19r18

So I guess the minimum of required edge cells is 18. I can imagine there is a defense that makes black require even more.

Loony, I believe that the standard pattern is the best white can do... here is the proper starting position: http://www.trmph.com/hex/board#19,f14f15e15d17f13g12f12h12g11i12h11j12i11k12j11l12k11m12l11n12m11o12n11p12o11q12p11r12q11s12r11b18h15g16e16e17f16f17i14i16j15j16h16g18h17h18i17i18l16k17m15j17l14m17n16n17l17k19l18l19m18m19p17o18q16

(hence minimum I would say 19x19)

in a position that you have shown there is another factor that lets white do this and probably win:

http://www.trmph.com/hex/board#19,d14d15c15b17c14e13b14f12a14g12a13h12a12i12a11j12a10k12a9l12a8m12a7n12a6o12a5p12a4q12a3r12a2s12a1d13e14f17

I'd be interested in verifying this version of the template, though it probably requires a lot of work. I suspect it is correct and minimal, but I'm not 100% sure.

I think a long time ago I verified this 5th row template, which is analogous, and of course there's this 4th row template.

Another question is coming up with any 7th-row template. The one analogous to these would have width 24, so it won't fit on trmph (but it would still fit on jhex I think). I'm not sure if it works or not. I briefly tried thinking about how wide a symmetric template would have to be (assuming it is analogous to the standard 6th-row template), and I got width 25, which is even worse.

I'll have a look at it on the weekend. If 38.f17 wins for white the ladder escape is less worthy because a main idea is that you are able to extend the upper ladder as far as necessary until the escape can be applied. For 2+4 and 3+5 an adjusted starting position is not mandatory.

Shalev, for that 6th row template, what about F16

David: I think you're right, I don't see how to connect after that move. Nice! I guess this means the 6th-row template needs another space on the left, for 20 in total... maybe something like this (imagine there is an extra column at the bottom right corner, with 3 empty cells). I haven't tried to verify this though, so it might still be wrong.

David, Looney, Shalev, Arek Thank you for all of the above. This greatly adds to my understanding of the n-th row problem and its relationship to the n and n+2 ladder problem. This will certainly be an important chapter to the eventual book of hex :)