Find the minimum sum of errors for every permutation

I have an (m x n) matrix that is filled with error percentages.

Example:

error = [ 2.1 2.3 1.9 2.0;

1.8 1.9 2.3 2.1;

2.0 1.9 1.8 2.3];

Each row/column can only contain 1 final error percentage.

I need the best permutation that minimizes the sum of all of the chosen errors.

The actual matrix is on the scale of about 100x100 so writing nested for loops and if statements just doesn't seem viable for each row/column.

Is there a more efficient way of calculating the sums of every permutation?

I don't need the whole code written out, just a general approach.

Thanks!

[James Tursa]

Please provide more detail. What is a "final error percentage"? What exactly are the "chosen errors"? How are you allowed to permute the elements? What would be the sum for the given matrix above? Etc.

A complete example, rather than a wordy description, would help us greatly understand.

[the cyclist]

Here is a guess at what you mean:

Choose the set of values that has the smallest sum, where you cannot choose values that share a row or column. So, in your example,

error = [ 2.1 2.3 1.9 2.0;

1.8 1.9 2.3 2.1;

2.0 1.9 1.8 2.3];

the solution (I think) would be

error = [ 2.1 2.3 1.9 2.0;

1.8 1.9 2.3 2.1;

2.0 1.9 1.8 2.3];

which gives a total error of 5.6.

Is that right?

[em7]

@James Tursa, The example was dumbed down. The gyst is that each element in the matrix will consist of an error (in decimal form). Each row represents some different objects and each column represents some objects that the row can possibly "map" to, hence an error that is calculated using mean squared.

I've been using nested for loops like this:

for i = 1:size(error, 1)

for j = 1:size(error, 2)

end

end

But I need to create several more loops to account for the permutations; far more than I care to type out if I don't have to to save space. I feel like the math is on the tip of my tongue but I can't remember any from school for the life of me.

@the cyclist, your solution would be the correect solution, but I was curious if there's a function or mathematical approach to solving that rather than brute force since I would be operating on several 10's of rows/columns.

[the cyclist]

Understood that you want an efficient algorithm. I was just trying to clarify the problem you are trying to solve.

Answer 1

[Steven Lord]

Take a look at the matchpairs function.

[em7]

oh this is EXACTLY what i'm looking for, but it's for a R2019 and I have R2018.

Can't find anything similar in file exchange either.

[Torsten]

Can't find anything similar in file exchange either.

Strange. My google search gives many, many matches:

linear assignment problem & matlab

[em7]

oh! you're absolutely right. thanks!