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))

little help?

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.