javascript中最快的斜边？

I've seen a number of questions about simulations and animations in javascript, which often involve calculating the hypotenuse:

hypot = Math.sqrt(x*x + y*y);

Since cartesian coordinates are the weapon of choice in most of these engines, these calculations are needed to find the distance between pairs of points, etc. So any speedup in calculating the hypotenuse could be a great help to many projects.

To that end, can you see a faster method than the simple implementation above? I found an approximation which was marginally faster in Chrome, but turned out to be much slower in Firefox, based on this approximation function in SuperCollider.

Edit 2015-08-15: I've switched the accepted answer to being the Math.hypot one; I suspect the pragmatic approach at present would be to use Math.hypot or a synthesized hypot function if not available, and to compare against the square (per sch's answer) if that is sufficient and Math.hypot is not available.

In ECMAScript ES6 you can use Math.hypot:

// ES5 support

Math.hypot = Math.hypot || function(x, y){ return Math.sqrt(x*x + y*y) }

var x = 3, y = 4;

document.write(Math.hypot(x, y))

Edit: You can run this test on a blank tab, are 2 million operations with both methods, the results are very good, it is 24% faster.

var i, tmp, x = 55, y = 66, end, ini = performance.now();

// Math.sqrt operation
i = 0;
ini = performance.now();
tmp = 0;
while(i++ < 2000000){
    tmp += Math.sqrt(x*x + y*y)
}
end = performance.now();
console.log(tmp, "Math.sqrt operation: " + (end - ini) + " ms");

// Math.hypot

i = 0;
ini = performance.now();
tmp = 0;
while(i++ < 2000000){
    tmp += Math.hypot(x, y)
}
end = performance.now();

console.log(tmp, "Math.hypot: " + (end - ini) + " ms");

Note: In this test, it's used ES6's Math.hypot.

</div>

Often, you don't need to compute the square root and hypot^2 = x*x + y*y is enough. This is the case for example if you want to compare the distances and don't need the actual values.

You can look to equality of x and y. If are equals you can calculate hypotenuse as (x + y)/sqrt(2) where sqrt(2)is a constant.

So this method can be used for case where x = y. For other cases it can be used with maximum imprecision of ~41%. This is a big error. But when you specify allowable error limits you can use this method. For example, if define allowable error to 5% you can get that b must be between 0.515*a and 1.942*a.

So if you don't need perfect imprecision of your calculations, you can improve performance of calculations with range of values.

By analogy you can look to equality of x or y to zero. And with some accuracy calculate hypotenuse more faster for this cases.

P.S. I've read about this in the one russian article.

An important point that many do not know:

hypot = Math.sqrt(x*x + y*y);

That works in theory, but in practice it may fail. If x is so large that x*x overflows, the code will produce an infinite result.

Here’s how to compute sqrt(xx + yy) without risking overflow.

max = maximum(|x|, |y|)
min = minimum(|x|, |y|)
r = min / max
return max*sqrt(1 + r*r)

Reference and complete text: John D. Cook - http://www.johndcook.com/blog/2010/06/02/whats-so-hard-about-finding-a-hypotenuse/