Difference between revisions of "User:Bobson8"
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(1) "Red always loses regardless of who goes first" if and only if b >= 2a+1 | (1) "Red always loses regardless of who goes first" if and only if b >= 2a+1 | ||
| − | (In other words, Red's average edge length | + | (In other words, Red's average edge length <= Blue's+1). |
(2) "Red always wins regardless of who goes first" if and only if b <= 2a-3 | (2) "Red always wins regardless of who goes first" if and only if b <= 2a-3 | ||
| − | (In other words, Red's average edge length | + | (In other words, Red's average edge length >= Blue's-1). |
Indeed, this conjecture is an analog of the results on parallelogram boards. I have verified this conjecture for a=1,2,3, and KataHex suggested that a=4,5,6 are possibly true. | Indeed, this conjecture is an analog of the results on parallelogram boards. I have verified this conjecture for a=1,2,3, and KataHex suggested that a=4,5,6 are possibly true. | ||
Revision as of 18:40, 21 April 2024
7th row single-stone pre-template
The following pattern guarantees a stone on 7th row to connect to the edge. It was verified by me using HexProver developped by mirefek. Whether this connection is minimal is not known. This partially answers the single-stone 7th row edge template problem (The template exists but the shape is unknown).
To start with, one notice that Red can connect to the edge with a variant of edge template V1b from both sides.
Thus Blue has these possible blocks (a~j, mirrored positions are removed).
For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.
A conjecture on Trapezoid boards
Conjecture. For an isosceles trapezoid board of edge lengths a (Red), b (Blue), a+b-1 (Red), b (Blue),
(1) "Red always loses regardless of who goes first" if and only if b >= 2a+1 (In other words, Red's average edge length <= Blue's+1).
(2) "Red always wins regardless of who goes first" if and only if b <= 2a-3 (In other words, Red's average edge length >= Blue's-1).
Indeed, this conjecture is an analog of the results on parallelogram boards. I have verified this conjecture for a=1,2,3, and KataHex suggested that a=4,5,6 are possibly true.
To prove this conjecture one need to verify
(1) When b=2a-3, Red wins even if Blue goes first.
(2) When b=2a-2, Red loses if Blue goes first.
(3) When b=2a, Red wins if Red goes first.
(4) When b=2a+1, Red loses even if Red goes first.
We demo some small cases (a=1,2,3).
(a,b)=(1,2): Red wins with a bridge.
(a,b)=(1,3): Blue edges form a span template.
(a,b)=(2,1): Red edges form a bridge template.
(a,b)=(2,2): Blue wins with a chain of two bridges.
(a,b)=(2,4): Red wins with a chain of a bridge and a ziggurat.
(a,b)=(2,5): Blue edges form a template. One possible defence for Red:
(a,b)=(3,3): Red edges are connected by the edge template IV2a.
(a,b)=(3,4): Blue wins with a chain of two ziggurats.
(a,b)=(3,6): Red can win with this template:
(a,b)=(3,7): Blue edges form a pre-template. There are two template within it, removing cells either with * or with + (due to Quasar). One possible defence for Red:
