Author: Ricardo Gibert
Date: 19:20:16 01/13/00
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On January 13, 2000 at 21:39:49, Ricardo Gibert wrote: >On January 12, 2000 at 18:34:49, Dan Ellwein wrote: > >>Hi >> >>Just wanted to bounce this off of the group and see if this is an accurate >>representation of how many (non redundant) pawn positions there are in chess... >> >> >>(0,8)(1,7)(2,6)(3,5)(4,4) 8P 48x47x46x45x44x43x42x41 x5 = _______ >>(0,7)(1,6)(2,5)(3,4) 7P 48X47X46X45X44X43X42 x4 = _______ >>(0,6)(1,5)(2,4)(3,3) 6p 48X47X46X45X44X43 x4 = _______ >>(0,5)(1,4)(2,3) 5P 48x47x46x45x44 x3 = _______ >>(0,4)(1,3)(2,2) 4P 48x47x46x45 x3 = _______ >>(0,3)(1,2) 3P 48x47x46 x2 = _______ >>(0,2)(1,1) 2P 48x47 x2 = _______ >>(0,1) 1P 48 x1 = _______ >> >> >>Number of non redundant pawn postions in chess -- TOTAL: _______ >> >>haven't done the math on it yet, but it looks like about 75 trillion... >> >>thanks... >> >>PilgrimDan > >I wrote a short little program that calculates the number of pawn positions to >be 8e10. The vast majority of positions represented by this number are illegal. >so this is an upper bound. However, I'm a C newbie so I probably made an error. > >In any case, there is a quite simple argument that shows the upper bound to be >less than 1.8e19: Consider a 48 bit bitmap giving the locations of all the pawns >on the board. Since at most 16 of these bits can be set, we can use a 16 bit >bitmap to give color. That's 64 bits total or 1.8e19 pawn positions. > >Of course, we are overcounting like crazy. The 48 bit location bitmap includes >positions with 17 to 48 pawns! Also, the 16 bit color bitmap includes those >positions with more than 8 white or 8 black pawns. > >Taking all this into account I calculated the 8e10 result. With a little >friendly cajoling, I may be persuaded to explain how I calculated this result. >In the meantime, I will double check my result. 8e10 is an error. I found and corrected the overflow problem I feared and came up with 6.1e12. I'll triple check now...
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