Dark Photon Posted May 13, 2006 Share 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)) Link to comment Share on other sites More sharing options...
Dark Photon Posted May 13, 2006 Author Share Posted May 13, 2006 little help? Link to comment Share on other sites More sharing options...
Dak Posted May 13, 2006 Share 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. Link to comment Share on other sites More sharing options...
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