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Title: Equilateral Triangle Post by Michael_Dagg on Dec 4th, 2005, 11:59am In how many ways can three vertices of an n-dimensional cube be chosen so that the chosen vertices form an equilateral triangle? |
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Title: Re: Equilateral Triangle Post by Barukh on Dec 5th, 2005, 11:19am IMHO this is related to Hamming distances. 3 vertices on n-dimensional cube form an equilateral triangle if their corresponding 0-1 encodings have the same mutual Hamming distance. For instance, the following 3 vertices in 4-dimensional space: 0000, 0011, 0101. I wonder why this question was placed in General section? :-/ |
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Title: Re: Equilateral Triangle Post by Eigenray on Aug 17th, 2006, 12:28pm Nearly forgot about this one! If x,y,z are in {0,1}n, let A be the set of bits where x,y differ, and let B the set where x,z differ. Then y,z differ in the symmetric difference of A and B, which has cardinality |A\B|+|B\A| = |A| + |B| - 2|A n B|. The points x,y,z form an equilateral triangle iff |A| = |B| = |A|+|B| - 2|A n B|, so that |A| = |B| = 2|A n B| = 2k, say, is even. The first point, x, can be picked in 2n ways. Then we pick subset A in (nC2k) ways, and this determines y. Finally, we pick B by specifying AnB, in (2kCk) ways, and B\A, in (n-2kCk) ways, and this determines z. Now we've counted each triple 3! times, so divide by 6. Thus the number of equilateral triangles of side length sqrt(2k) is 2n (nC2k) (2kCk) (n-2kCk)/6 = 2nn!/[6(n-3k)!k!3]. The total number of triangles is the sum over all k > 0 (note the above is 0 if 3k > n). The sequence starts 0, 0, 8, 64, 320, 2240, 17920, 121856, 831488, 6215680, and doesn't seem to be in [link=http://www.research.att.com/~njas/sequences/?q=8%2C64%2C320]Sloane[/link]. |
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