Assignment Algorithm

Assignment Algorithm-24
Rotate the matrix so that there are at least as many columns as rows and let k=min(n,m). If there is no starred zero in the row containing this primed zero, Go to Step 5.Otherwise, cover this row and uncover the column containing the starred zero.Unstar each starred zero of the series, star each primed zero of the series, erase all primes and uncover every line in the matrix. Step 6: Add the value found in Step 4 to every element of each covered row, and subtract it from every element of each uncovered column.

Rotate the matrix so that there are at least as many columns as rows and let k=min(n,m). If there is no starred zero in the row containing this primed zero, Go to Step 5.Otherwise, cover this row and uncover the column containing the starred zero.Unstar each starred zero of the series, star each primed zero of the series, erase all primes and uncover every line in the matrix. Step 6: Add the value found in Step 4 to every element of each covered row, and subtract it from every element of each uncovered column.

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Some of these descriptions require careful interpretation.

In Step 4, for example, the possible situations are, that there is a noncovered zero which get primed and if there is no starred zero in its row the program goes onto Step 5.

By applying Rule 4 to the step-algorithm we decide to make each step its own procedure.

Now we can apply Rule 8 by using a case statement in a loop to control the ordering of step execution.

We will assume that the cost matrix C(i,j) has already been loaded with the first index referring to the row number and the second index referring to the column number.

For each row of the matrix, find the smallest element and subtract it from every element in its row. We can define a local variable called minval that is used to hold the smallest value in a row.Since each worker can perform only one job and each job can be assigned to only one worker the assignments constitute an independent set of the matrix C. A brute-force algorithm for solving the assignment problem involves generating all independent sets of the matrix C, computing the total costs of each assignment and a search of all assignment to find a minimal-sum independent set.An arbitrary assignment is shown above in which worker a is assigned job q, worker b is assigned job s and so on. The complexity of this method is driven by the number of independent assignments possible in an nxn matrix.We also define two vectors R_cov and C_cov that are used to "cover" the rows and columns of the cost matrix C. If we have found (and starred) K independent zeros then we are done. Otherwise, cover this row and uncover the column containing the starred zero.In the nested loop (over indices i and j) we check to see if C(i,j) is a zero value and if its column or row is not already covered. Continue in this manner until there are no uncovered zeros left.DONE: Assignment pairs are indicated by the positions of the starred zeros in the cost matrix.If C(i,j) is a starred zero, then the element associated with row i is assigned to the element associated with column j.There are n choices for the first assignment, n-1 choices for the second assignment and so on, giving n! Therefore, this approach has, at least, an exponential runtime complexity.As each assignment is chosen that row and column are eliminated from consideration.The following 6-step algorithm is a modified form of the original Munkres' Assignment Algorithm (sometimes referred to as the Hungarian Algorithm).This algorithm describes to the manual manipulation of a two-dimensional matrix by starring and priming zeros and by covering and uncovering rows and columns.

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