On Fri, May 24, 2013 at 12:28 PM, Robert Herman <[email protected]> wrote:
> I was helping my son with his maths homework, and I thought it would be a
> great chance to utilize J. Here is the problem: Given three sides, x, y and
> z, where x<y<z and a perimeter of 57, with z < 1/2 of the perimeter, how
> many triangles can you make that fit that criteria. Somebody created a perl
> script with nested loops:
>
> for $x (1..57/2) {
> for $y ($x+1..57/2) {
> for $z ($y+1..57/2) {
> if ($x + $y + $z eq 57) {
> ++$count;
> print "$count: $x $y $z\n";
> }}}}
> print "Total: $count";
>
>
> I thought I could start to do it this way in J:
>
> x =. i..26
> y =. >:x
> z =. >:y
>
> NB. Here's where I get stuck in trying to do a nested loop in J
> +/" x y z NB. just to see if I could get it to sum each column
You are good up to this point.
As Elton Wang points out, you can use x+/y+/z to generate all possible
permutations for x y and z. But note that this does not distinguish
between different versions of the same triangle. In other words,
these are probably all the same triangle:
18+(i.!3) A. i.3
18 19 20
18 20 19
19 18 20
19 20 18
20 18 19
20 19 18
The perl program avoids this issue by enforcing that $x < $y and $y < $z.
In J you could generate all the possibilities, sort them (achieving
the perl constraint) and then discard duplicates. Here's some hints
about that approach:
x=:1+i.26
y=:1+x
z=:y
$,{x;y;z
17576
$~./:~&.>,{x;y;z
3626
$>{x;y;z
26 26 26 3
$>,{x;y;z
17576 3
$>~./:~&.>,{x;y;z
3626 3
> I am stuck in the world of imperative programming and loops. I'm not
> looking for the solution, but just a hint in which direction to take it.
> I'm still fascinated with J, and hope to grasp array programming. Thank you.
There are other ways of approaching this problem, also. Conceptually
speaking, you do not really have three independent values. Once you
have two values, you know the third (57-x+y), and you might need to
discard some cases (where z is negative, and where y is not greater
than x).
I hope this helps.
--
Raul
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