############################################################################### This Text File is Public Domain. This is available in INTERNET. http://www.cli.di.unipi.it/~velucchi/diff.txt This is for review/abstract in journals and magazines in all the World. I am available for more technical details. IF YOU PRINT THIS FILE OR REFERENCE ON, WRITE ME PLEASE. ############################################################################### "Different Dispositions in the ChessBoard" by Timothy CHOW, Mario VELUCCHI (16 Nov. 1995) PROBLEM: How many different dispositions I have in an order-N chessboard, if I place K Queens? ########## The answer is in the "Polya's Theorem" The wanted number is 1/8th of the coefficient of a^K * b^(N^2-K) in these polynomials: if N is even p(a,b,N) := (a+b)^(N^2)+2*(a+b)^N*(a^2+b^2)^((N^2-N)/2)+3*(a^2+b^2)^(N^2/2)+2*(a^4+b^4)^(N^2/4) if N is odd p(a,b,N) := (a+b)^(N^2)+2*(a+b)*(a^4+b^4)^((N^2-1)/4)+(a+b)*(a^2+b^2)^((N^2-1)/2)+4*(a+b)^N*(a^2+b^2)^((N^2-N)/2) Note: the total number of dispositions is equal to: Binomial coefficient(N^2, K) and the number of different dispositions is near 1/8th. Samples (K = N): for N = 1 different dispositions = 1 for N = 2 different dispositions = 2 for N = 3 different dispositions = 16 for N = 4 different dispositions = 252 for N = 5 different dispositions = 6814 for N = 6 different dispositions = 244344 for N = 7 different dispositions = 10746377 for N = 8 different dispositions = 553319048 for N = 9 different dispositions = 32611596056 for N = 10 different dispositions = 2163792255680 ... ... ... ... . ... ... ... ... ... . ... for N = 14 different dispositions = 110066314584030859544 for N = 15 different dispositions = 11375695977099383509351 ... ... ... ... . ... ... ... ... ... . ... Mario VELUCCHI Via Emilia, 106 I-56121 Pisa - ITALY e-mail: velucchi@cli.di.unipi.it