Dark Photon Posted May 13, 2006 Posted May 13, 2006 on dotted paper a 3 x 3 square . . . . . . . . . you can get 8 different ways of linking 2 dots diagonally at a 45 degree angle. and 2 ways of linking 3 dots diagonally on a 3 by 4 square . . . . . . . . . . . . you can link 2 in 12 different ways and link 3 in 4 different ways on a 4 by 4 you can link 2 in 18 ways , 3 in 8 and 4 in 2. assuming the verticle hieght of the square is x and the width is y. and x is the number of dots we need to link. so can anyone formulate a formula using x y and w to give me the number of diagonal combinations on N by N sqare. the formulae for verticle and horizonal linkages are as follows respectvly: w(y-(x-1)), y(w-(x+1))
Dak Posted May 13, 2006 Posted May 13, 2006 well, in each length you can make one-less diagonal than there are dots: horizontal, 4 dots, 3 diags: . . . . .\.\.\. . . . . . . . . vertical, same: . . . . .\. . . .\. . . .\. . . and, of course, there are two types of diagonal: \ and /... So I guess it's 2((w-1)(h-1)) {w = width in dots, h = height in dots}. diagonals spanning more than two dots should be dedusable in the same way.
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