| | Brickgame [connect 4] algorithm David Eaton
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| | Ok, I'm a geek. I admit that. But now that I've got that out of the way, lemme say I've always loved connect 4, just because it's such a more strategic game (like checkers, rather) than one typically accounts for-- rather than say, tic-tac-toe, (...) (24 years ago, 30-Jan-01, to lugnet.off-topic.geek)
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| | | | Re: Brickgame [connect 4] algorithm Dan Boger
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| | | | (...) heh, I liked it. Never thought of trying a connect 4 game... I did write a tic-tac-toe that learned from experiance... hmmm... I'm trying to think of a way to quantify the killing play (one that leaves you two ways to win next turn) without (...) (24 years ago, 30-Jan-01, to lugnet.off-topic.geek)
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| | | | | | Re: Brickgame [connect 4] algorithm David Eaton
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| | | | | (...) I think I'd quantify that to something along the lines of (tell me if this sounds wrong): A. If there exists a line of 5 or more available spaces wherein my opponent has not taken any of them: B - If I have 0 of these spaces (excluding the (...) (24 years ago, 31-Jan-01, to lugnet.off-topic.geek)
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| | | | Re: Brickgame [connect 4] algorithm Matthew Miller
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| | | | (...) I remember reading some years ago that this game has actually been solved; that is, there is a way for whoever goes first to win every time. *does google search* Yeah. Victor Allis of Vrije Universiteit in Amsterdam. (URL) (...) (24 years ago, 31-Jan-01, to lugnet.off-topic.geek)
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| | | | | | Re: Brickgame [connect 4] algorithm David Eaton
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| | | | (...) Cool! I rather imagined that such might be possible-- And actually I suppose it wouldn't be *all* that difficult to write an algorithm to prove so, seeing that a game lasts no longer than 49 moves, and at each step, there are a max of 8 moves. (...) (24 years ago, 31-Jan-01, to lugnet.off-topic.geek)
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