Math.hypot()
The Math.hypot()
function returns the square root of the
sum of squares of its arguments, that is:
1, v_2, \dots, v_n)} = \sqrt{\sum{i=1}^n v_i^2} = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}
Syntax
Math.hypot()
Math.hypot(value0)
Math.hypot(value0, value1)
Math.hypot(value0, value1, ... , valueN)
Parameters
value1, value2, ...
-
Numbers.
Return value
The square root of the sum of squares of the given arguments. If at least one of the
arguments cannot be converted to a number, NaN
is returned.
Description
Calculating the hypotenuse of a right triangle, or the magnitude of a complex number,
uses the formula Math.sqrt(v1*v1 + v2*v2)
, where v1 and v2 are the lengths
of the triangle's legs, or the complex number's real and complex components. The
corresponding distance in 2 or more dimensions can be calculated by adding more squares
under the square root: Math.sqrt(v1*v1 + v2*v2 + v3*v3 + v4*v4)
.
This function makes this calculation easier and faster; you call
Math.hypot(v1, v2)
, or Math.hypot(v1, v2, v3, v4, ...)
.
Math.hypot
also avoids overflow/underflow problems if the magnitude of
your numbers is very large. The largest number you can represent in JS is
Number.MAX_VALUE
, which is around 10^308. If your numbers are
larger than about 10^154, taking the square of them will result in Infinity.
For example, Math.sqrt(1e200*1e200 + 1e200*1e200) = Infinity
. If you use
hypot()
instead, you get better answer:
Math.hypot(1e200, 1e200) = 1.4142...e+200
. This is also true with very
small numbers. Math.sqrt(1e-200*1e-200 + 1e-200*1e-200) = 0
, but
Math.hypot(1e-200, 1e-200) = 1.4142...e-200
.
Because hypot()
is a static method of Math
, you always use it
as Math.hypot()
, rather than as a method of a Math
object you
created (Math
is not a constructor).
If no arguments are given, the result is +0. If any of the arguments is ±Infinity, the
result is Infinity. If any of the arguments is NaN (unless another argument is
±Infinity), the result is NaN. If at least one of the arguments cannot be converted to a
number, the result is NaN
.
With one argument, Math.hypot()
is equivalent to Math.abs()
.
Examples
Using Math.hypot()
Math.hypot(3, 4); // 5
Math.hypot(3, 4, 5); // 7.0710678118654755
Math.hypot(); // 0
Math.hypot(NaN); // NaN
Math.hypot(NaN, Infinity); // Infinity
Math.hypot(3, 4, 'foo'); // NaN, since +'foo' => NaN
Math.hypot(3, 4, '5'); // 7.0710678118654755, +'5' => 5
Math.hypot(-3); // 3, the same as Math.abs(-3)
Polyfill
A naive approach that does not handle overflow/underflow issues:
if (!Math.hypot) Math.hypot = function() {
var y = 0, i = arguments.length, containsInfinity = false;
while (i--) {
var arg = arguments[i];
if (arg === Infinity || arg === -Infinity)
containsInfinity = true
y += arg * arg
}
return containsInfinity ? Infinity : Math.sqrt(y)
}
A polyfill that avoids underflows and overflows:
if (!Math.hypot) Math.hypot = function () {
var max = 0;
var s = 0;
var containsInfinity = false;
for (var i = 0; i < arguments.length; ++i) {
var arg = Math.abs(Number(arguments[i]));
if (arg === Infinity)
containsInfinity = true
if (arg > max) {
s *= (max / arg) * (max / arg);
max = arg;
}
s += arg === 0 && max === 0 ? 0 : (arg / max) * (arg / max);
}
return containsInfinity ? Infinity : (max === 1 / 0 ? 1 / 0 : max * Math.sqrt(s));
};
Specifications
Specification |
---|
ECMAScript Language Specification # sec-math.hypot |
Browser compatibility
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