While loop runs forever in JavaScript

Question

I've writen a function in JavaScript that checks through all the combinations of three digits between 1 and 9 and gives me the number of combinations that follow this pattern √(x^2 + y^2 + z^2) = a natural number (a full number like 24 or 34 but not 2.54)

√ = square root , ^2 = to the power of 2,

My problem is that whenever I run the function the computer gets stuck and the function never ends so it doesn't return an answer. I would very much appreciate if someone could tell me whats wrong with it (I'm running my programs on the chrome browser console)

function mmd() {
    var chk = false;
    var a = 1;
    var b = 1;
    var c = 1; 
    var d = 1;
    var e = 0;
    while(chk != true) {
        d = Math.sqrt(Math.pow(a, 2)+Math.pow(b, 2)+Math.pow(c, 2));
        if( d == d.toFixed(0)) {
            e++;
        }
        else {
            if((b == 9) && (a == 9) && (c == 9)) {chk = true;}
            else if((a == 9) && (b == 9)) {c++;}
            else if(b == 9) {b = 1; a++;}
            else if(c == 9) {c = 1; b++;}
            else if(c < 9) {c++;}
        }
    }
    return e
}

you only change the check if all the numbers equal 9 but you keep changing b and c back to 1 over and over — omarjmh
– omarjmh, Commented Apr 28, 2016 at 20:31
if (d == d.toFixed(0)), you do e++; and nothing else. If this condition is met even once, it will always stay true, and your loop will keep doing e++ for ever (well, until it crashes). — blex
– blex, Commented Apr 28, 2016 at 20:34
Isn't this off topic? I'd suggest using the code review stack exchange website. — evolutionxbox
– evolutionxbox, Commented Apr 28, 2016 at 20:36
@evolutionxbox If it's broken, it's for Stack Overflow. If it's not, but needs quality check, performance etc., then it's for Code Review. — Joseph
– Joseph, Commented Apr 28, 2016 at 20:41
Side note: There's no reason to use a runnable snippet for code that doesn't run. (You should include your code but it may confuse people that you're hiding the actual code and that the snippet doesn't produce results.) — BSMP
– BSMP, Commented Apr 28, 2016 at 21:06

blex · Accepted Answer · 2016-04-28 21:24:02Z

3

This part of the code is causing it to never end:

if (d == d.toFixed(0)){} else {}

If the result of the formula is an integer, you add 1 to e, but you don't increment the other variables, because of the else. It keeps doing e++ for ever. So you need to remove that else.

I also took the liberty or removing that chk variable, and instead used while(true), which will be ended by a return of the final result:

function mmd() {
    var a = 1, b = 1, c = 1, d, e = 0;

    while(true) {
        d = Math.sqrt(Math.pow(a, 2)+Math.pow(b, 2)+Math.pow(c, 2));
        if( d == parseInt(d, 10)) {
            e++;
        }

        if((b == 9) && (a == 9) && (c == 9)) {return e;}
        else if((a == 9) && (b == 9)) {c++;}
        else if(b == 9) {b = 1; a++;}
        else if(c == 9) {c = 1; b++;}
        else {c++;}
    }
}

alert(mmd());

edited Apr 28, 2016 at 21:24

answered Apr 28, 2016 at 20:54

blex

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2 Comments

Yogi Over a year ago

up voted. Yet, might there be a more elegant test than: d == d.toFixed(0) ?

blex Over a year ago

@Roberto Thank you, I've now replaced it with parseInt, which does seem more natural to me, but I don't have any argument for it being better than toFixed().

IMTheNachoMan · Accepted Answer · 2016-04-28 21:08:46Z

It gets stuck once it hits the e++ block and never increases a, b, or c.

function mmd()
{
    var keepGoing = true;
    var a = 1, b = 1, c = 1, d, e = 0;
    while(keepGoing)
    {
        // calculate d
        d = Math.sqrt(Math.pow(a, 2) + Math.pow(b, 2) + Math.pow(c, 2));

        // check if it is a whole number
        if(d == d.toFixed(0)) e++;

        // if we're done then stop
        if(a == 9 && b == 9 && c == 9){ keepGoing = false; }

        // if c is less than 9 then increase it
        else if(c < 9){ c++; }

        // if c is 9 and b is less than 9 then set c back to 1 and increase b
        else if(b < 9){ c = 1; b++; }

        // if c is 9 and b is 9 then set both back to 1 and increase a
        else if(a < 9){ c = b = 1; a++; }
    }
    return e;
}

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