# This includes a series of questions.

One of the most typical examples is shown as the picture below.
An half-infinite Hex board with an one row of black stones. Black stones are separated by one white stone. Can black connect the two parts of black stones, which means this white stone is trapped in a finite area, or the white stone can escape infinitely far?

## One similar but weaker problem:

Can black connect the two parts? Until now, my special-trained KataHex(a Hex AI based on alphazero-like algorithm) think black can not connect with a board width of 51.

# More questions:

## 2. The angle of the above examples are 180°, if the answers are white wins, can black win if the angle is smaller?

One known result(verified using Mohex): minimum area to trap a single stone on 120° angle:

#### One conjecture:

No matter how many white stones initially, black wins if angle<180°, white wins if angle>=180°.

## 3. For arbitrarily large N, does N-th edge template exists?

N-th edge template is the minimum area to connect one stone to the edge at a distance of N-1.
Example: 6-th row edge template

If exists, what is the minimum width f(N) of the template? How fast does f(N) growth?
It has been proved that the minimum width 2~7th edge template are f(2)=2 f(3)=4 f(4)=7 f(5)=10 f(6)=14 f(7)=17, which can be found at HexWiki: https://www.hexwiki.net/index.php/Edge_templates_with_one_stone
I trained a special KataHex for this problem, and got some results not strict but very sure: f(8)=26, f(9)=35, f(10)=44 (f(10) is not quite sure)

The two pictures below are analyzing 8th edge template using KataHex, indicating f(8)=26

## 4.Unstoppable patterns

For the following 3 pictures, can black reach the red line? And can white reach the blue line?
If both of black and white cannot, then they are called "unstoppable patterns". Are there any other unstoppable patterns other than these three?

## Switchback strategy of "ladder escape problem"

There is a strategy to force the ladder turn to left, then the upper-right part become unreachable (red line).
If there is an extra stone very far from the edge. I guess this strategy can be used for multiple times and finally reaches the extra stone, then the two part can be connected. But it's not obvious because when using the switchback the second time, it is a terraced ladder at 2nd/4th rows now. So black has more choices (picture).

• See also the open questions at the end of "Infinite Hex is a Draw" JD Hamkins and D Leonessi, math.colgate.edu/~integers/xg6/xg6.pdf. Your question is similar to our trapezoid questions. Commented May 1 at 16:51
• In particular, do you know the answer when the angle is 90 degrees? This amounts essentially to an open question in our paper. Commented May 1 at 16:53
• It is also open whether Red wins on the full infinite board with the advantage of a single extra stone, or even finitely many stones, or indeed much more. If even weak versions of your conjecture are true, you will settle several open questions. Commented May 1 at 16:56
• @JoelDavidHamkins I have read your paper and found your trapezoid questions. However I have no idea how to solve these questions. The only thing I can do is training KataHex for some specific openings. It seems to show that the critical point is 180°. With one extra black stone even on 10th row will help black win the game
– hzy
Commented May 1 at 17:46
• @JoelDavidHamkins White can escape to infinity if there are no extra black stones. No matter where black plays, white can just play symmetrically.(proved by Bobson in discord server). But this proof can't work if there are any extra black stones.
– hzy
Commented May 1 at 19:08