User talk:Fjan2ej57w

Hello :D

Here is the way to make a hex board:

 [<][hexboard size=]["][*^v][x][*<>] 
 ["]//set size
 [contents=]["][*colour] [*position] 
 ["]//put stones
 [/>]

then I will get this:

 [<][hexboard size=]["][5][x][8]["]
 [contents=]["][R] [c4]["][/>]

then I will get this:

 [<][hexboard size=]["3x4"]
 [visible=]["][area(][a3,d3,d1,c1] 
 [)"]//set bondary by vertex
 [edges=]["][bottom]["]
 [coords=]["][none]["]//not coloring
 [contents=]["][R c1]["]
 [/>]

then I will get a ziggurat:

 [<][hexboard size=]["6x7"]
 [coords="none"]
 [egdes="bottom"]  
 [visible=]["] 
 [area(a6,g6,g4,f5,e4,d5,c4,b5,a4)]["]
 [contents="R f5 d5 b5"]
 [/>]

a longer rampart :p

some questions about hex

1.Is there a 2nd row ladder escape template that contains only empty cells? (It's equivalent to ask that given enough space, is a 2nd row ladder able to connect itself to the edge. 2.find templates that escape ladders on both sides,e.g.,Edge template II escapes 2nd and 3rd row ladders on both size at the same time. 3.is it possible to get a switchback from a 3rd row ladder? 4.Does single stone 8th row edge template exist? Is there a better way to analyze such templates? 5.is it possible to totally surround a single stone on an infinite board? I.e., to make the group coming from that stone remain bounded. 6.Find the ways to prolong a losing game as much as possible.(or shorten a winning game as much as possible) 7.Which leads to the question that how much space of an empty board would be filled if both sides play optimally.(I guess if n(approaching infinity) is the length of the board, then the game(with or without swap)would end in n^(3/2)•k moves, where k is a constant depending on whether the swap rule is used or not) 8.Bridge ladders are common in actual games, so there must be some conclusions about it, comparing such "interior" bridges to an interval of edge. 9.there is the concept of capture. And it can be used to determine the shape of an edge template. https://www.hexwiki.net/index.php/Theorems_about_templates more complicated and frequently occurring conditions of capturing and their corresponding strategy may be very helpful.

independent subgames

In some cases, a game is seperated into two or more independent subgames, just like the example above. In order to win, the only way for red is to connect his group to both sides. On the other hand, if red fails to connect his group to either side, blue is able to connect within that region instead. The board is actually seperated into two independent parts by red stones parallel to his own edge. No matter what happens in one part, the other part won't be affected at all.(except that the question of who would play first in a certain region) The upper part:

And the lower part:

Sometimes, the board would be seperated into more than two independent parts. The position below is one example:

Now the board is seperated into at least 3 independent parts: 1.

2.

3.

Blue will achieve the final victory if and only if she wins 1 or both 2 and 3. On the other hand, red will win if and only if he wins 1 and either 2 or 3.

It actually doesn't matter what's the relative position of these subgames or how they are interconnected in the original board. What We really care is their logical relationships, that is, the winning condition for both players.

These examples looks more or less artificial, but the phenomen of independent subgames actually occurs quite often in actual game play. By merely identifying such patterns, a game would be much easier to analyze in some cases. The following are some examples with subgames having more vague boundaries.

some thoughts about hex

Beginners are too eager to push ladders. For the attacker, this seems to oversimplify the game(since a ladder could be used in many different ways). For the defender, it's generally better to play elsewhere until the ladder becomes threatening enough.

It's not good in general to handle the ladder in such early stage of the game. By defending this ladder, red has actually assumed that d8 is already connected to the top edge and by pushing the ladder, blue has also assumed this, and that he gets a 2nd row ladder escape on the right, which is currently not the case.

Finding all 2nd row ladder escapes may be at least as hard as finding all 3rd row ladder escapes. Consider the following position, blue to move.

Assume Red's ladder stone is already connected to the top edge. The connection is not shown, but it does not use any space on the 2nd and 3rd row, nor any space within the carrier of the potential ladder escape template. Notice that red has a 2nd row ladder at first, but would only get a 3rd row ladder to the right of that blue stone in the middle:

That means we could at least get one 2nd row ladder escape template from any 3rd row ladder escape template by simply adding the following pattern to the left of it:

In accordence with Theory of ladder escapes, the cell marked with "+"are not part of the carrier, they just indicate how such templates are glued together.

There is a reason why I say this is just a thought. I actually assumed in this argument that red could only get a 3rd row ladder from a 2nd row ladder after the pattern above, but it's cureently unkown if red could pivot instead of yielding at 10, and somehow find a way to connect the gap marked with "*":