Covering the zeros with minimum lines in the Hungarian Method - algorithm

Question

I am trying to follow the steps of covering the zeros with the minimum number of lines in the Hungarian Method as follows:
Tick all unassigned rows.
If the ticked row has zeros, then tick the correspondent column.
Within the ticked column, if there is an assignment, then tick the correspondent row.
Draw a line above each un-ticked row and ticked column.
Repeat for each unassigned row.
Then find Theta (which is the smallest uncovered value)
The problem is when I do that, I still have zeros uncovered! causing Theta to be zero and go to an infinite loop!
For example, If we take the following matrix 25 by 25):
1 5 5 2 3 1 2 3 2 4 5 2 3 1 5 5 2 3 1 5 1 4 3 2 5
5 5 3 2 3 2 5 1 4 3 2 5 3 2 4 5 2 5 2 1 1 4 1 2 5
5 1 4 3 2 5 1 1 4 1 2 5 2 2 3 4 1 4 5 3 2 4 5 2 5
1 1 4 1 2 5 3 2 4 5 2 5 5 5 1 5 1 5 5 2 2 3 4 1 4
3 2 4 5 2 5 2 2 3 4 1 4 5 4 2 1 3 2 5 5 5 1 5 1 5
2 2 3 4 1 4 5 5 1 5 1 5 5 5 2 5 5 1 4 5 4 2 1 3 2
5 5 1 5 1 5 5 5 3 2 3 2 1 5 5 1 5 1 5 5 5 2 5 5 1
5 4 2 1 3 2 5 1 4 3 2 5 5 5 4 2 1 3 2 5 1 4 3 2 5
5 5 2 5 5 1 1 1 4 1 2 5 1 5 5 2 5 5 1 1 1 4 1 2 5
2 4 5 3 4 2 3 2 4 5 2 5 2 2 4 5 3 4 2 3 2 4 5 2 5
2 2 5 5 1 3 2 2 3 4 1 4 2 2 2 5 5 1 3 2 2 3 4 1 4
4 1 5 4 5 3 5 5 1 5 1 5 5 4 1 5 4 5 3 5 5 1 5 1 5
5 1 4 3 2 5 3 2 4 5 2 5 5 5 1 4 3 2 5 3 2 4 5 2 5
1 1 4 1 2 5 2 2 3 4 1 4 1 1 1 4 1 2 5 2 2 3 4 1 4
3 2 4 5 2 5 5 5 1 5 1 5 4 3 2 4 5 2 5 5 5 1 5 1 5
2 2 3 4 1 4 5 4 2 1 3 2 1 2 2 3 4 1 4 5 4 2 1 3 2
5 5 1 5 1 5 5 5 2 5 5 1 2 5 5 1 5 1 5 5 5 2 5 5 1
5 1 4 3 2 5 3 5 1 4 3 2 5 3 5 2 2 3 5 2 2 3 2 5 3
3 4 1 4 1 1 1 1 1 4 1 2 5 5 1 4 3 2 5 1 4 1 2 5 2
1 5 5 2 3 1 5 3 2 4 5 2 5 1 1 4 1 2 5 2 4 5 2 5 5
5 5 3 2 3 2 2 2 2 3 4 1 4 3 2 4 5 2 5 2 3 4 1 4 3
5 1 4 3 2 5 2 5 5 1 5 1 5 2 2 3 4 1 4 5 1 5 1 5 5
1 1 4 1 2 5 2 5 4 2 1 3 2 5 5 1 5 1 5 4 2 1 3 2 1
3 2 4 5 2 5 1 5 5 2 5 5 1 5 4 2 1 3 2 5 2 5 5 1 3
2 2 3 4 1 4 1 2 4 5 3 4 2 5 5 2 5 5 1 4 5 3 4 2 2
After subtracting minimum row and column values as steps 1 and 2 from the Hungarian method, I get:
0 4 4 1 2 0 1 2 1 3 4 1 2 0 4 4 1 2 0 4 0 3 2 1 4
4 4 2 1 2 1 4 0 3 2 1 4 2 1 3 4 1 4 1 0 0 3 0 1 4
4 0 3 2 1 4 0 0 3 0 1 4 1 1 2 3 0 3 4 2 1 3 4 1 4
0 0 3 0 1 4 2 1 3 4 1 4 4 4 0 4 0 4 4 1 1 2 3 0 3
2 1 3 4 1 4 1 1 2 3 0 3 4 3 1 0 2 1 4 4 4 0 4 0 4
1 1 2 3 0 3 4 4 0 4 0 4 4 4 1 4 4 0 3 4 3 1 0 2 1
4 4 0 4 0 4 4 4 2 1 2 1 0 4 4 0 4 0 4 4 4 1 4 4 0
4 3 1 0 2 1 4 0 3 2 1 4 4 4 3 1 0 2 1 4 0 3 2 1 4
4 4 1 4 4 0 0 0 3 0 1 4 0 4 4 1 4 4 0 0 0 3 0 1 4
0 2 3 1 2 0 1 0 2 3 0 3 0 0 2 3 1 2 0 1 0 2 3 0 3
1 1 4 4 0 2 1 1 2 3 0 3 1 1 1 4 4 0 2 1 1 2 3 0 3
3 0 4 3 4 2 4 4 0 4 0 4 4 3 0 4 3 4 2 4 4 0 4 0 4
4 0 3 2 1 4 2 1 3 4 1 4 4 4 0 3 2 1 4 2 1 3 4 1 4
0 0 3 0 1 4 1 1 2 3 0 3 0 0 0 3 0 1 4 1 1 2 3 0 3
2 1 3 4 1 4 4 4 0 4 0 4 3 2 1 3 4 1 4 4 4 0 4 0 4
1 1 2 3 0 3 4 3 1 0 2 1 0 1 1 2 3 0 3 4 3 1 0 2 1
4 4 0 4 0 4 4 4 1 4 4 0 1 4 4 0 4 0 4 4 4 1 4 4 0
4 0 3 2 1 4 2 4 0 3 2 1 4 2 4 1 1 2 4 1 1 2 1 4 2
2 3 0 3 0 0 0 0 0 3 0 1 4 4 0 3 2 1 4 0 3 0 1 4 1
0 4 4 1 2 0 4 2 1 3 4 1 4 0 0 3 0 1 4 1 3 4 1 4 4
4 4 2 1 2 1 1 1 1 2 3 0 3 2 1 3 4 1 4 1 2 3 0 3 2
4 0 3 2 1 4 1 4 4 0 4 0 4 1 1 2 3 0 3 4 0 4 0 4 4
0 0 3 0 1 4 1 4 3 1 0 2 1 4 4 0 4 0 4 3 1 0 2 1 0
2 1 3 4 1 4 0 4 4 1 4 4 0 4 3 1 0 2 1 4 1 4 4 0 2
1 1 2 3 0 3 0 1 3 4 2 3 1 4 4 1 4 4 0 3 4 2 3 1 1
Then when we do the assignment, we will have 23 assignments instead of 25, so we do the mentioned earlier covering zeros based on the above steps, I would get the following:
The bold cells are the ones covered according to the above steps.
Notice that there are still zeros uncovered causing the infinite loop as it will be selected next.
Please help me.
Thank you in advance

Answer 1

You can use min-cost maximum flow algorithm for solving the problem when each worker may accomplish two tasks.
At first, let's see how to solve standard assignment problem using min-cost max flow. Create a bipartite graph where workers are in one part and tasks are in another. Put an edge with the capacity 1 and cost cost_ij between worker i and task j for all i, j. Then add a source S and edges from source to every worker of capacity 1 and cost 0. Similarly, add a sink T and edges from every task to sink of same cost and capacity. Then, if you find a min-cost max flow from S to T, then its value will be the total assignment cost.
So, if you allow each worker to select two tasks then edges from source to workers should be with capacity 2. This addition to the algorithm will solve your problem in the optimal way regardless to the given constraint on maximum difference.
However, at the moment I do not know the solution for the task with given restriction on every possible input. If your input values are something special, you may say it in the response and we'll think about special cases of the problem.