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

i trying follow steps of covering zeros minimum number of lines in hungarian method follows:

1. tick unassigned rows.
2. if ticked row has zeros, tick correspondent column.
3. within ticked column, if there assignment, tick correspondent row.
4. draw line above each un-ticked row , ticked column.
5. repeat each unassigned row.
6. then find theta (which smallest uncovered value)

the problem when that, still have zeros uncovered! causing theta 0 , go infinite loop!

for example, if take following matrix 25 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 , column values steps 1 , 2 hungarian method, 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 assignment, have 23 assignments instead of 25, mentioned earlier covering zeros based on above steps, following:

the bold cells ones covered according above steps.

notice there still zeros uncovered causing infinite loop selected next.

please me.

thank in advance

you can use min-cost maximum flow algorithm solving problem when each worker may accomplish 2 tasks.

at first, let's see how solve standard assignment problem using min-cost max flow. create bipartite graph workers in 1 part , tasks in another. put edge capacity 1 , cost cost_ij between worker , task j i, j. add source s , edges source every worker of capacity 1 , cost 0. similarly, add sink t , edges every task sink of same cost , capacity. then, if find min-cost max flow s t, value total assignment cost.

so, if allow each worker select 2 tasks edges source workers should capacity 2. addition algorithm solve problem in optimal way regardless given constraint on maximum difference.

however, @ moment not know solution task given restriction on every possible input. if input values special, may in response , we'll think special cases of problem.