Difference between revisions of "Talk:Wheel"

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(Further simplifcation to 8x10.)
(Added a 7x7 example)
 
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By filling in the [[captured cell|captured]] areas, Wurfmaul's position can be further simplified to  [https://hexworld.org/board/#9x10,g1a1i2a2e4a3i4a4e5a5i5a6d6a7e6a8i6a9i7b1:pb2:pb3:pb4:pb5:pb6:pb7:p:p:pb8:pc1:pc2:pc3:pc4:pc5:pc6:pd1:pd2:pd3:pd4:pd5:pe1:pe2:pe3:pf1:pf5:pg5:ph5:pg6:pd7:pe7:pf8 this]. [[User:Selinger|Selinger]] ([[User talk:Selinger|talk]]) 17:51, 18 April 2023 (UTC)
 
By filling in the [[captured cell|captured]] areas, Wurfmaul's position can be further simplified to  [https://hexworld.org/board/#9x10,g1a1i2a2e4a3i4a4e5a5i5a6d6a7e6a8i6a9i7b1:pb2:pb3:pb4:pb5:pb6:pb7:p:p:pb8:pc1:pc2:pc3:pc4:pc5:pc6:pd1:pd2:pd3:pd4:pd5:pe1:pe2:pe3:pf1:pf5:pg5:ph5:pg6:pd7:pe7:pf8 this]. [[User:Selinger|Selinger]] ([[User talk:Selinger|talk]]) 17:51, 18 April 2023 (UTC)
Using [[Theorems about templates|corner clipping]], we can reduce the number of empty hexes by 4 more, and reduce the size to 8x10: [https://hexworld.org/board/#8x10,f1a1h2a2d4a3h4a4d5a5h5a6c6a7d6a8h6b1h7b2a9b3h9b4:pb5:pb6:pc1:pc2:pc3:p:p:pc4:pc5:pc7:pd1:pd2:pd3:pd7:pe5:pe8:pf5:pf6:pg5:ph10 here]. [[User:Selinger|Selinger]] ([[User talk:Selinger|talk]]) 21:35, 18 April 2023 (UTC)
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Using [[Theorems about templates|corner clipping]], we can reduce the number of empty hexes by 4 more, and reduce the size to 8x10: [https://hexworld.org/board/#8x10,f1a1h2a2d4a3h4a4d5a5h5a6c6a7d6a8h6b1h7b2a9b3h9b4:pb5:pb6:pc1:pc2:pc3:p:p:pc4:pc5:pc7:pd1:pd2:pd3:pd7:pe5:pe8:pf5:pf6:pg5:ph10:pe1 here]. [[User:Selinger|Selinger]] ([[User talk:Selinger|talk]]) 21:35, 18 April 2023 (UTC)
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In fact, the following is a 7x7 example, which also works for the [[Interior_template#The_hammock|hammock]] and not just the wheel, and uses only 5 empty cells outside of the hammock:
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<hexboard size="7x7"
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  contents="B a1--e1 g1 a2 b2 a3 f3 e4 b5 c5 c6 d6 g6 g7--e7 f5 R f1 d2 g2--g5 e3 a4--a7--d7 b6 c4 d5
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            E A:(c3 e2 d4) B:d3"
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  />
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[[User:Selinger|Selinger]] ([[User talk:Selinger|talk]]) and [[User:Demer|Demer]] ([[User talk:Demer|talk]]) 00:26, 19 April 2023 (UTC)

Latest revision as of 00:26, 19 April 2023

This pattern is not what I would call a Wheel. It is at most a Broken wheel. I think somebody called it a U-turn. — taral

Yes, granted. My initial feeling was that it was important to point out that this related position is weak, but maybe it's not important enough to merit mention on this page. We could always create a "U-turn" page, or just refer to the example in the tutorial. turing 08:18, 10 Feb 2005 (CET)


Blue should never attempt to intrude by playing at B. I believe this is false. Here's a simple example (Blue to move) where Blue's only winning moves are at one of (*). In particular, d2 is winning for Blue, while all of the hexes marked A are losing.

abcde12345AAA

Now, this might not be very interesting because playing at d2 just delays the "real" winning d4 move. A more interesting question: is there a position where playing at B is the unique winning move? This seems trickier than it looks, in many winning positions there's a useless bridge intrusion that also preserves the win for instance, so the move at B isn't unique. Intuitively, I think that in order for the intrusion at B to be better than all intrusions at A, the intrusion at B needs to serve the function of asking a question.

"lazyplayer" from LittleGolem showed me a position loosely based from this game. I was surprised when I put this in KataHex, because it was the first position I saw where an intrusion at B (specifically, f10) was winning but intrusions at A were losing. But, f10 might not be unique; KataHex thinks c11 or i5 could also win. I played around a bit to see if I could make f10 unique — it was tricky to get KataHex to not think c11 was also winning or close to 50%. Eventually, I found a position where KataHex thinks, after about 50k visits, that f10 is winning (90.3% win rate), and the second best move has only an 8.6% win rate. So it appears that this last position has a unique winning move at B, but I only have KataHex evaluations and not a solid proof. (Note for reproducibility: I used the katahex_model_20220618.bin.gz net with a high value of analysisWideRootNoise, 0.5, to reduce blind spots.)

It would be nice if someone could come up with an example small enough that one can prove the intrusion at B is the unique winning move. I suspect the smallest example might not even fit on a 5x5 or 6x6 board, since it's a fickle matter getting the condition to hold. Hexanna (talk) 01:09, 17 April 2023 (UTC)

Here is a simplification for which the dfpn solver can prove that B is the only winning move. Surely it can be simplified further. --Wurfmaul (talk) 22:07, 17 April 2023 (UTC) Actually the top-most row and right-most column of this position can simply be discarded to reach a 10x10 position. --Wurfmaul (talk) 00:56, 18 April 2023 (UTC)

By filling in the captured areas, Wurfmaul's position can be further simplified to this. Selinger (talk) 17:51, 18 April 2023 (UTC) Using corner clipping, we can reduce the number of empty hexes by 4 more, and reduce the size to 8x10: here. Selinger (talk) 21:35, 18 April 2023 (UTC)

In fact, the following is a 7x7 example, which also works for the hammock and not just the wheel, and uses only 5 empty cells outside of the hammock:

abcdefg1234567AABA

Selinger (talk) and Demer (talk) 00:26, 19 April 2023 (UTC)