# Why Does MATLAB Have the Function hypot?7

Posted by Loren Shure,

Sometimes I am asked questions like "why did you introduce function xyz into MATLAB? And sometimes I have a compelling answer, even for something that looks simple on the surface. Consider the function hypot.

### What Does hypot Compute?

hypot essentially computes the square root of the sum of the squares of 2 inputs. So what's the big deal, right? For "reasonable" input values, there is no big issue.

fhypot = @(a,b) sqrt(abs(a).^2+abs(b).^2);

### Other Ways to Compute hypot

Let's create a set of values and compute hypot 3 ways for these: with hypot, fhypot, and sqrt(2)*x. These should all give the same answers, provide a is real and positive.

a = logspace(0,5,6)
format long
aHypot = hypot(a,a)
aFhypot = fhypot(a,a)
aSqrt2 = sqrt(2)*a
a =
1          10         100        1000       10000      100000
aHypot =
1.0e+005 *
Columns 1 through 3
0.000014142135624   0.000141421356237   0.001414213562373
Columns 4 through 6
0.014142135623731   0.141421356237310   1.414213562373095
aFhypot =
1.0e+005 *
Columns 1 through 3
0.000014142135624   0.000141421356237   0.001414213562373
Columns 4 through 6
0.014142135623731   0.141421356237310   1.414213562373095
aSqrt2 =
1.0e+005 *
Columns 1 through 3
0.000014142135624   0.000141421356237   0.001414213562373
Columns 4 through 6
0.014142135623731   0.141421356237310   1.414213562373095

### Results

The results are the same, to within round-off, for the 3 methods here. But what happens if the magnitude of a is larger, near realmax perhaps?

realmax
ans =
1.797693134862316e+308

For my computer, a 32-bit Windows machine, realmax for doubles is on the order of 10^308. Let's see what happens if a value nearly that large is used with the different versions of hypot.

a = 1e308
aHypot = hypot(a,a)
aFhypot = fhypot(a,a)
aSqrt2 = sqrt(2)*a
a =
1.000000000000000e+308
aHypot =
1.414213562373095e+308
aFhypot =
Inf
aSqrt2 =
1.414213562373095e+308

What you can see is that the straight-forward method returns Inf instead of a finite answer. And that's why hypot was added to MATLAB, to compute the hypotenuse robustly, avoiding both underflow and overflow.

Get the MATLAB code

Published with MATLAB® 7.5

Quan replied on : 1 of 7

That was a pretty interesting post Loren. I’m not sure if it will affect the way I do things around here, but it’s something that could come in useful in the future. Thanks for the tip!

Duane Hanselman replied on : 2 of 7

Before hypot existed, I used a=abs(complex(a,b)); In doing so, I was able to tap the robustness of the abs() function.

Loren replied on : 3 of 7

Duane-

Thanks for pointing out that abs is also implemented robustly.

–Loren

Hal K replied on : 4 of 7

Loren,
Does hypot(a,a) any different from norm([a a])?

Hal

Loren replied on : 5 of 7

Hal-

norm is also careful to scale results, like abs and hypot.

–Loren

Shahn replied on : 6 of 7

Hi Loren

I have a binary image and I want to find the distance between every object. I have already computed the centroid for every object. Can Hypot be used for many objects ???

Loren replied on : 7 of 7

Shahn-

You can use hypot if that is what makes sense. Here’s the help page. And here’s a few details from it:

c = hypot(a,b) returns the element-wise result of the following equation, computed to avoid underflow and overflow:

c = sqrt(abs(a).^2 + abs(b).^2)

Inputs a and b must follow these rules:

* Both a and b must be single- or double-precision, floating-point arrays.

* The sizes of the a and b arrays must either be equal, or one a scalar and the other nonscalar. In the latter case, hypot expands the scalar input to match the size of the nonscalar input.

Best wishes,
Loren

These postings are the author's and don't necessarily represent the opinions of MathWorks.