Abbi,
You have a couple of options:
If you know for sure that a certain solution should be favored, you can deduct the cost of one of those elements by an eps. The total cost will be 10-eps and that would cause Munkres to choose that one.
c1 = % your matrix
c1(1) = c1(1)-1e-15;
assignmunkres(c1,10)
ans =
4×2 uint32 matrix
1 1
3 2
2 3
4 4
The Munkres algorithm is described by the paper that Munkres wrote, please see the help/doc page.
You can use the other assignment functions we provide: matchpairs, assignjv, assignauction. These may return a different result when the cost is exactly the same. For example:
c1 = % your matrix
assignjv(c1,10)
ans =
4×2 uint32 matrix
1 1
2 3
3 2
4 4
However, the most robust is to simply look at more than optimal solution. You can use assignkbest. This algorithm uses Murti's algorithm to return the top k optimal solutions. For example:
[assignments, ~, ~, cost] = assignkbest(c, 10, 5)
Returns the following result:
assignments =
5×1 cell array
{4×2 uint32}
{4×2 uint32}
{4×2 uint32}
{4×2 uint32}
{4×2 uint32}
cost =
10
10
10
10
12
This indicates that there are 4 solutions with the exact same cost (10).
