Chess IS solvable.I 'll present an imaginary proof: Suppose that: a)We have a book that has as many blank pages as we want b)Each blank page has enough space to hold the moves of an entire game
The algorithm is this: First we fill the book writing on each page every possible game(Very large book).As each game was a winner or is stalemate we sort the pages in this way: 1,4,7,... are the pages containing games that white wins 2,5,8,... are the pages containing games that black wins 3,6,9,... are the pages containing stalemates. Now let the game begin.Suppose we play the black pieces and we have(and so does the opponent)a copy of the above book. White makes a move.We open our book and rip off the pages that contain a game that does not start with that move.Then from the remaining pages we rip off the pages that are 1 mod 3 and 0 mod 3 equivalent(sorry for my bad english)that is we rip the pages containing games that white wins and the stalemates. From the remaining pages we choose one and play the next move. White does the some.Then we do the same for white's second move and white does the same and so on. Now I have a question : In our initial book are the number of pages that white wins equal to the number of pages that black wins? If this is true then theoretically white will always win because: 1)at each turn players rip equal number of pages 2)black rips first 3)The game is finite 4)At each turn at least one page is ripped So black will run out of pages first so black will have no moves.And white will have only one page remaining (the page containing the game that just has been played with number equivalent to 1 mod 3(white winning page))
If my question takes 'no' for an answer then,sorry, I cant think anymore at this moment...
Slashdot Mirror
Chess IS solvable.I 'll present an imaginary proof:
.We open our book and rip off the pages that contain a game that does not start with that move.Then from the remaining pages we rip off the pages that are 1 mod 3 and 0 mod 3 equivalent(sorry for my bad english)that is we rip the pages containing games that white wins and the stalemates.
,sorry, I cant think anymore at this moment...
Suppose that:
a)We have a book that has as many blank pages as we want
b)Each blank page has enough space to hold the moves of an entire game
The algorithm is this:
First we fill the book writing on each page every possible game(Very large book).As each game was a winner or is stalemate we sort the pages in this way:
1,4,7,... are the pages containing games that white wins
2,5,8,... are the pages containing games that black wins
3,6,9,... are the pages containing stalemates.
Now let the game begin.Suppose we play the black pieces and we have(and so does the opponent)a copy
of the above book.
White makes a move
From the remaining pages we choose one and play the next move.
White does the some.Then we do the same for white's second move and white does the same and so on.
Now I have a question :
In our initial book are the number of pages that white wins equal to the number of pages that black wins?
If this is true then theoretically white will always win because:
1)at each turn players rip equal number of pages
2)black rips first
3)The game is finite
4)At each turn at least one page is ripped
So black will run out of pages first so black will have no moves.And white will have only one page remaining (the page containing the game that just has been played with number equivalent to 1 mod 3(white winning page))
If my question takes 'no' for an answer then
Kokoland- koko@otenet.gr