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Assignment Problem - Let C be an nxn matrix representing the costs of each of n workers to perform any of n jobs. The assignment problem is to assign jobs to workers so as to minimize the total cost. 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. The complexity of this method is driven by the number of independent assignments possible in an nxn matrix. There are n choices for the first assignment, n-1 choices for the second assignment and so on, giving n! possible assignment sets. Therefore, this approach has, at least, an exponential runtime complexity.

As each assignment is chosen that row and column are eliminated from consideration. The question is raised as to whether there is a better algorithm. In fact there exists a polynomial runtime complexity algorithm for solving the assignment problem developed by James Munkre's in the late 1950's despite the fact that some references still describe this as a problem of exponential complexity.
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.
This
is because, at the time of publication (1957), few people had access to
a computer and the algorithm was exercised by hand.
Step 0: Create an nxm matrix called the cost matrix in which each element represents the cost of assigning one of n workers to one of m jobs. Rotate the matrix so that there are at least as many columns as rows and let k=min(n,m).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. The other possible way out of Step 4 is that there are no noncovered zeros at all, in which case the program goes to Step 6.Step 1: For each row of the matrix, find the smallest element and subtract it from every element in its row. Go to Step 2.
Step 2: Find a zero (Z) in the resulting matrix. If there is no starred zero in its row or column, star Z. Repeat for each element in the matrix. Go to Step 3.
Step 3: Cover each column containing a starred zero. If K columns are covered, the starred zeros describe a complete set of unique assignments. In this case, Go to DONE, otherwise, Go to Step 4.
Step 4: Find a noncovered zero and prime it. 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. Continue in this manner until there are no uncovered zeros left. Save the smallest uncovered value and Go to Step 6.
Step 5: Construct a series of alternating primed and starred zeros as follows. Let Z0 represent the uncovered primed zero found in Step 4. Let Z1 denote the starred zero in the column of Z0 (if any). Let Z2 denote the primed zero in the row of Z1 (there will always be one). Continue until the series terminates at a primed zero that has no starred zero in its column. Unstar each starred zero of the series, star each primed zero of the series, erase all primes and uncover every line in the matrix. Return to Step 3.
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. Return to Step 4 without altering any stars, primes, or covered lines.
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.
At first it may seem
that
the erratic nature of this algorithm would make its implementation
difficult.
However, we can apply a few general rules of programming style to
simplify
this problem. The same rules can be applied to any step-algorithm.
Good Programming Style and Design Practices
1. Strive to create readable source code through the use of blank lines, comments and spacing.2. Use consistent naming conventions, for variable and constant identifiers and subprograms.
3. Use consistent indentation and always indent the bodies of conditionals and looping constructs.
4. Place logically distinct computations in their own execution blocks or in separate subprograms.
5. Don't use global variables inside subprograms except where such use is clear and improves readability and efficiency.
6. Use local variables where appropriate and try to limit the creation of unnecessary identifiers in the main program.
7. Open I/O files only when needed and close them as soon as they are no longer required.
8. Work to keep the level of nesting of conditionals and loops at a minimum.
9. Use constant identifiers instead of hardwiring for-loop and array ranges in the body of the code with literal values.
10. When you feel that things are getting out of control, start over. Re-coding is good coding.

Step 1
For each row of the
matrix,
find the smallest element and subtract it from every element in its
row.
Go to Step 2. We can define a local variable called minval
that
is used to hold the smallest value in a row. Notice that there
are
two loops over the index j appearing inside an outer loop over
the
index i. The first inner loop over the index j searches
a
row for the minval. Once minval has been found
this
value is subtracted from each element of that row in the second inner
loop
over j. The value of step is set to 2 just before
stepone
ends.

Step 2
Find a zero (Z) in
the
resulting matrix. If there is no starred zero in its row or
column,
star Z. Repeat for each element in the matrix. Go to Step 3. In
this
step,
we
introduce the mask matrix M, which in the same dimensions
as the cost matrix and is used to star and prime zeros of the cost
matrix.
If M(i,j)=1 then C(i,j) is a starred zero, If M(i,j)=2 then
C(i,j)
is a primed zero. We also define two vectors R_cov and C_cov that
are used to "cover" the rows and columns of the cost matrix C. 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. If not
then
we star this zero (i.e. set M(i,j)=1) and cover its row and column
(i.e.
set R_cov(i)=1 and C_cov(j)=1). Before we go on to Step 3, we
uncover
all rows and columns so that we can use the cover vectors to help us
count
the number of starred zeros.

Step 3
Cover each column containing a starred zero. If K columns are covered, the starred zeros describe a complete set of unique assignments. In this case, Go to DONE, otherwise, Go to Step 4. Once we have searched the entire cost matrix, we count the number of independent zeros found. If we have found (and starred) K independent zeros then we are done. If not we procede to Step 4.

Find a noncovered zero and prime it. 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. Continue in this manner until there are no uncovered zeros left. Save the smallest uncovered value and Go to Step 6.


Step 5
Construct a series of alternating primed and starred zeros as follows. Let Z0 represent the uncovered primed zero found in Step 4. Let Z1 denote the starred zero in the column of Z0 (if any). Let Z2 denote the primed zero in the row of Z1 (there will always be one). Continue until the series terminates at a primed zero that has no starred zero in its column. Unstar each starred zero of the series, star each primed zero of the series, erase all primes and uncover every line in the matrix. Return to Step 3. You may notice that Step 5 seems vaguely familiar. It is a verbal description of the augmenting path algorithm (for solving the maximal matching problem) which we discussed in Lecture 3. We decompose the operations of this step into a main procedure and five relatively simple subprograms.


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. Return to Step 4 without
altering
any stars, primes, or covered lines. Notice that this step uses the
smallest uncovered value in the cost matrix to modify the matrix.
Even though this step refers to the value being found in Step 4 it is
more
convenient to wait until you reach Step 6 before searching for this
value.
It may seem that since the values in the cost matrix are being altered,
we would lose sight of the original problem. However, we are only
changing certain values that have already been tested and found not to
be elements of the minimal assignment. Also we are only changing
the values by an amount equal to the smallest value in the cost matrix,
so we will not jump over the optimal (i.e. minimal assignment) with
this
change.

For Step 6, there is one supporting method to find the smallest uncovered value (described in Step 4).

The
code used above is available
in a complete Microsoft Visual Studio .NET 2010 C# project - munkres.zip. This is a
console application that can read a
text-file containing the cost matrix values arranged in a
two-dimensional array or generate test matrices. The input file
should be placed in the same folder as the calling executable.
This project application can generate a random value cost matrix or a
worst-case test matrix C(i,j) = i*j. Setting the value of each
element C(i,j) of the cost matrix to the value i*j ensures that the
maximum number of operations will be required to find the minimal
assignment (which is the back diagonal). The locations of the
ones (1's) in the associated mask matrix M correspond to the assignment
pairs selected by Munkres' Algorithm. The following is an example
run of the algorithm on a 3x3 worst-case test matrix.
An Example Execution of Munkres' Algorithm
|
Jobs = {p, q, r} Cost of assigning job j to work i
is |
![]() 1. Step 0 |
![]() 2. Step 1 |
![]() 3. Step 2 |
![]() 4. Step 3 |
![]() 5. Step 4 |
![]() 6. Step 6 |
![]() 7. Step 4 |
![]() 8. Step 5 |
![]() 9. Step 3 |
![]() 10. Step 4 |
![]() 11. Step 6 |
![]() 12. Step 4 |
![]() 13. Step 6 |
![]() 14. Step 4 |
![]() 15. Step 5 |
![]() 16. Step 3 |
![]() 17. Done |
This example illustrates the method of implementing a step-algorithm. It also serves to demonstrate why we do not attempt to implement every algorithm discussed in this course. :-)
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