Sorting a binary 2D matrix?

Question

I'm looking for some pointers here as I don't quite know where to start researching this one.

I have a 2D matrix with 0 or 1 in each cell, such as:

  1 2 3 4
A 0 1 1 0
B 1 1 1 0
C 0 1 0 0
D 1 1 0 0

And I'd like to sort it so it is as "upper triangular" as possible, like so:

  4 3 1 2
B 0 1 1 1
A 0 1 0 1
D 0 0 1 1
C 0 0 0 1

The rows and columns must remain intact, i.e. elements can't be moved individually and can only be swapped "whole".

I understand that there'll probably be pathological cases where a matrix has multiple possible sorted results (i.e. same shape, but differ in the identity of the "original" rows/columns.)

So, can anyone suggest where I might find some starting points for this? An existing library/algorithm would be great, but I'll settle for knowing the name of the problem I'm trying to solve!

I doubt it's a linear algebra problem as such, and maybe there's some kind of image processing technique that's applicable.

Any other ideas aside, my initial guess is just to write a simple insertion sort on the rows, then the columns and iterate that until it stabilises (and hope that detecting the pathological cases isn't too hard.)

More details: Some more information on what I'm trying to do may help clarify. Each row represents a competitor, each column represents a challenge. Each 1 or 0 represents "success" for the competitor on a particular challenge.

By sorting the matrix so all 1s are in the top-right, I hope to then provide a ranking of the intrinsic difficulty of each challenge and a ranking of the competitors (which will take into account the difficulty of the challenges they succeeded at, not just the number of successes.)

Note on accepted answer: I've accepted Simulated Annealing as "the answer" with the caveat that this question doesn't have a right answer. It seems like a good approach, though I haven't actually managed to come up with a scoring function that works for my problem.

Answer 1

Here's a starting point:

Convert each row from binary bits into a number

Sort the numbers in descending order.

Then convert each row back to binary.

Answer 2

Basic algorithm:

1. Determine the row sums and store values. Determine the column sums and store values.
2. Sort the row sums in ascending order. Sort the column sums in ascending order.

Hopefully, you should have a matrix with as close to an upper-right triangular region as possible.

Answer 3

An Algorithm based upon simulated annealing can handle this sort of thing without too much trouble. Not great if you have small matrices which most likely hae a fixed solution, but great if your matrices get to be larger and the problem becomes more difficult.

(However, it also fails your desire that insertions can be done incrementally.)

Preliminaries

1. Devise a performance function that "scores" a matrix - matrices that are closer to your triangleness should get a better score than those that are less triangle-y.

2. Devise a set of operations that are allowed on the matrix. Your description was a little ambiguous, but if you can swap rows then one op would be SwapRows(a, b). Another could be SwapCols(a, b).

The Annealing loop

I won't give a full exposition here, but the idea is simple. You perform random transformations on the matrix using your operations. You measure how much "better" the matrix is after the operation (using the performance function before and after the operation). Then you decide whether to commit that transformation. You repeat this process a lot.

Deciding whether to commit the transform is the fun part: you need to decide whether to perform that operation or not. Toward the end of the annealing process, you only accept transformations that improved the score of the matrix. But earlier on, in a more chaotic time, you allow transformations that don't improve the score. In the beginning, the algorithm is "hot" and anything goes. Eventually, the algorithm cools and only good transforms are allowed. If you linearly cool the algorithm, then the choice of whether to accept a transformation is:

public bool ShouldAccept(double cost, double temperature, Random random) {
    return Math.Exp(-cost / temperature) > random.NextDouble();
}

You should read the excellent information contained in Numerical Recipes for more information on this algorithm.

Long story short, you should learn some of these general purpose algorithms. Doing so will allow you to solve large classes of problems that are hard to solve analytically.

Scoring algorithm

This is probably the trickiest part. You will want to devise a scorer that guides the annealing process toward your goal. The scorer should be a continuous function that results in larger numbers as the matrix approaches the ideal solution.

How do you measure the "ideal solution" - triangleness? Here is a naive and easy scorer: For every point, you know whether it should be 1 or 0. Add +1 to the score if the matrix is right, -1 if it's wrong. Here's some code so I can be explicit (not tested! please review!)

int Score(Matrix m) {
    var score = 0;
    for (var r = 0; r < m.NumRows; r++) {
        for (var c = 0; c < m.NumCols; c++) {
            var val = m.At(r, c);
            var shouldBe = (c >= r) ? 1 : 0;
            if (val == shouldBe) {
                score++;
            }
            else {
                score--;
            }
        }
    }
    return score;
}

With this scoring algorithm, a random field of 1s and 0s will give a score of 0. An "opposite" triangle will give the most negative score, and the correct solution will give the most positive score. Diffing two scores will give you the cost.

If this scorer doesn't work for you, then you will need to "tune" it until it produces the matrices you want.

This algorithm is based on the premise that tuning this scorer is much simpler than devising the optimal algorithm for sorting the matrix.

Answer 4

I came up with the below algorithm, and it seems to work correctly.

Phase 1: move rows with most 1s up and columns with most 1s right.

1. First the rows. Sort the rows by counting their 1s. We don't care if 2 rows have the same number of 1s.
2. Now the columns. Sort the cols by counting their 1s. We don't care if 2 cols have the same number of 1s.

Phase 2: repeat phase 1 but with extra criterions, so that we satisfy the triangular matrix morph.
Criterion for rows: if 2 rows have the same number of 1s, we move up the row that begin with fewer 0s.

Criterion for cols: if 2 cols have the same number of 1s, we move right the col that has fewer 0s at the bottom.

---

Example:

Phase 1

  1 2 3 4                     1 2 3 4                   4 1 3 2
A 0 1 1 0                   B 1 1 1 0                 B 0 1 1 1
B 1 1 1 0  - sort rows->    A 0 1 1 0  - sort cols->  A 0 0 1 1
C 0 1 0 0                   D 1 1 0 0                 D 0 1 0 1
D 1 1 0 0                   C 0 1 0 0                 C 0 0 0 1

Phase 2

  4 1 3 2                     4 1 3 2
B 0 1 1 1                   B 0 1 1 1
A 0 0 1 1  - sort rows->    D 0 1 0 1  - sort cols-> "completed"
D 0 1 0 1                   A 0 0 1 1
C 0 0 0 1                   C 0 0 0 1

Edit: it turns out that my algorithm doesn't give proper triangular matrices always.
For example:

Phase 1

   1 2 3 4                    1 2 3 4                
A  1 0 0 0                  B 0 1 1 1                
B  0 1 1 1 - sort rows->    C 0 0 1 1  - sort cols-> "completed"
C  0 0 1 1                  A 1 0 0 0                
D  0 0 0 1                  D 0 0 0 1

Phase 2

   1 2 3 4                    1 2 3 4                   2 1 3 4
B  0 1 1 1                  B 0 1 1 1                 B 1 0 1 1
C  0 0 1 1 - sort rows->    C 0 0 1 1  - sort cols->  C 0 0 1 1
A  1 0 0 0                  A 1 0 0 0                 A 0 1 0 0
D  0 0 0 1                  D 0 0 0 1                 D 0 0 0 1
                           (no change)

(*) Perhaps a phase 3 will increase the good results. In that phase we place the rows that start with fewer 0s in the top.

Answer 5

Treat rows as binary numbers, with the leftmost column as the most significant bit, and sort them in descending order, top to bottom

Treat the columns as binary numbers with the bottommost row as the most significant bit and sort them in ascending order, left to right.

Repeat until you reach a fixed point. Proof that the algorithm terminates left as an excercise for the reader.