Comments on: The science of suckiness

Question: The science of suckiness

handybitesize — Mon, 08 Jun 2009 10:07:15 -0800

I want to calculate the mean of a matrix, but I don't want a value, I want co-ordinates. The following math question has lots of terminology that I'm not qualified to use i.e. most of it is wrong. I hope you can see through this and help me. I will be using the words matrix, mean and transformation without fully understanding them. For this I apologize.

Imagine a 10x10 matrix of integers, I wish to find the 'point of balance'. That is, if the integers were weights and the matrix a sheet of plywood, it is where I would place a pivot for the sheet to remain in balance (assume the plywood has no weight).

I don't want to work it our mechanically (using trig and moments and things of that ilk, besides the weight thing is an analogy, I'm looking at luminosity), I want to apply a matrix transformation if possible.

However my knowledge doesn't even let me begin to search on the type of transformation I need.

I have played with the mean of the rows and cols. In a 2x2 matrix for example

10,20
4,200

x = ((1 * 10) + (2 * 20) + (1 * 4) + (2 * 200)) / (10 + 20 + 4 + 200) = 1.94017094
y = ((1 * 10) + (2 * 4) + (1 * 20) + (2 * 200)) / (10 + 20 + 4 + 200) = 1.87179487

which looks right, but doesn't feel right. I'm sure we are missing something.

Any pointers/equations/corrections are very appreciated.

(The 'suckiness' in the title refers to the larger integers "sucking" the balance point towards them, not that I suck at science, which I do)

By: 7segment

7segment — Mon, 08 Jun 2009 10:21:26 -0800

What you're talking about is a weighted mean, and, in your example, you're doing it correctly. I'm not sure why your example doesn't feel right to you, but the 200 "sucks" a lot (it's around a factor of 10 larger than the other entries, and the coordinates you realize are only about 10% away from its location). Note that this answer is preserved by rotating the matrix or reflecting it over its axes (x' = 1.05982906, y' = 1.12820513). Also note some extreme cases... for any x, the "center" of the matrix

x x
x x

is [1*x + 2*x]/2x for both coordinates, or (1.5,1.5) as you'd expect. Also, the "center" of

x 0
0 0

or any reflection/rotation thereof will be the location of x.

By: milqman

milqman — Mon, 08 Jun 2009 10:28:35 -0800

The center of mass formula should work here (which is what you are implementing). I think you have it right. There might be something more sophisticated you can do but I don't know what that would be without knowing the type of pattern you are looking for.

So, without knowing more, I'd say you are right. Watch your coordinates though, you want to make sure that you keep careful track of your row and column indexes!

By: Lemurrhea

Lemurrhea — Mon, 08 Jun 2009 10:35:16 -0800

Yeah you've got it right. You'll find it a bit easier in a 10x10 case to use the sum of the columns when finding the x-coord and use the sum of the rows when finding the y-coord.

So if you had:

x₀₀ x₀₁ ... x₀₉
x₁₀ x₁₁ ... x₁₉
.
.
x₉₀x₉₁ ... x₉₉¹

You'd want x_middle = [1*(x₀₀ + x₁₀ + ... + x₉₀) + 2*(x₀₁...) + ...]/SUM.
It'll be a lot easier for you to work with. But you've definitely got it.

1: starting at 0 for consistent formatting.

By: mbrubeck

mbrubeck — Mon, 08 Jun 2009 10:54:27 -0800

See also Center of mass on Wikipedia, which agrees with everything above.

By: handybitesize

handybitesize — Mon, 08 Jun 2009 10:58:02 -0800

Oh wow - multiple "ur doing it right" replies before I got home. Thanks all. I'll run it against some edge cases now, because maybe my its my eyes that's misinterpreting the expected result

By: Netzapper

Netzapper — Mon, 08 Jun 2009 11:24:24 -0800

Having done a lot of image processing and computer vision...

Since your math is right, but you say "maybe my its my eyes that's misinterpreting the expected result", it's possible that you're just barking up the wrong tree. I've definitely spent the better part of a couple days working out a particularly tricky convolution, only to find that the visual result is nothing like what I was imagining.