*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.

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 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).

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 Z_{0}represent the uncovered primed zero found in Step 4. Let Z_{1}denote the starred zero in the column of Z_{0}(if any). Let Z_{2}denote the primed zero in the row of Z_{1}(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*

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.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.

A old implementation of Munkres - This implementation is

In each pass of the loop the step procedure called sets the value of stepnum for the next pass. When the algorithm is finished the value of stepnum is set to some value outside the range 1..6 so that done will be set to true and the program will end. In the completed program the tagged (starred) zeros flag the row/column pairs that have been assigned to each other. We will discuss the implementation of a procedure for each step of Munkres' Algorithm below. We will assume that the cost matrixproceduremunkresis

n : constant integer := 20;-- num rows/columns (this version is hard-wired for square matrices)

C : is array(1..n,1..n) of float;-- cost matrix

M : is array(1..n,1..n) of integer;-- a mask matrix to indicate primed (= 2) and starred (=1) zeros in C

Row,Col : is array(1..n) of integer;-- maintains record of which row/columns are covered.

stepnum : integer;-- covered = 1, non-covered = 0

done : boolean;

functionstep1(stepnum:in outinteger)is

:

functionstep2(stepnum:in outinteger)is

:

functionstep3(stepnum:in outinteger)is

:

begin

done:=false;

stepnum:=1;

whilenot(done)loop

casestepnumis

when1 => step1(stepnum);

when2 => step2(stepnum);

when3 => step3(stepnum);

when4 => step4(stepnum);

when5 => step5(stepnum);

when6 => step6(stepnum);

whenothers=> done:=true;

end case;

end loop;

endmunkres;

*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.

**procedure**
stepone(step : **in out** integer) **is**

` minval : integer;`

` begin`

` minval:=C(i,1);`

` for j in 2..n loop`

` for j in 1..n loop`

` end loop;`

*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.

` procedure steptwo(step: in out integer)
is`

` for i in 1..n loop`

` for i in 1..n loop`

` end steptwo;`

*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.

` procedure stepthree(step : in out
integer)
is`

*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.*
In
this step, statements such as "find a noncovered zero" are clearly
distinct
operations that deserve their own functional blocks. We have
decomposed
this step into a main procedure and three subprograms (2 procedures and
a boolean function).

**procedure**
stepfour(step : **in out** integer) **is**

` row,col : integer;`

` done : boolean;`

` procedure find_a_zero(row,col : out
integer) is`

` function star_in_row(row : integer) return
boolean is`

` procedure find_star_in_row(row, col :
in
out integer) is`

` begin`

*Step 5*

*Construct a series
of
alternating primed and starred zeros as follows. Let Z _{0}
represent the uncovered primed zero found in Step 4. Let Z_{1}
denote the starred zero in the column of Z_{0} (if any). Let Z_{2}
denote the primed zero in the row of Z_{1} (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.

**procedure**
stepfive(step : **in out
**integer)
**is**

` count : integer;`

` done : boolean;`

` r,c : integer;`

` procedure find_star_in_col(c : in
integer;
r : in out integer) is`

` procedure find_prime_in_row(r : in
integer;
c : in out integer) is`

` procedure convert_path is`

` procedure clear_covers is`

` procedure erase_primes is`

` begin`

*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.

**procedure**
stepsix(step : **in out** integer) **is**

` minval : integer;`

` procedure find_smallest(minval : out
integer) is`

` begin`

*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. :-)

Answers
to Frequently Asked Questions

- The algorthm will work even when the minimum values in two or more rows are in the same column.
- The algorithm will work even when two or more of the rows contain the same values in the the same order.
- The algorithm will work even when all the values are the same
(although the result is not very interesting).

- Munkres Assignment Algorithm is not exponential run time or intractable; it is of a low order polynomial run time.
- Optimality is guaranteed in Munkres Assignment Algorithm.

- Setting the cost matrix to C(i,j) = i*j makes a good testing matrix for this problem.
- In this algorithm the range of indices is[1..n] rather than [0..n-1]. Using the latter range makes the sample code less than helpful in your algorithm implementation.
- Step 3 is an example of the greedy method. If the minimum values are all in different rows then their positions represent the minimal pairwise assignments.
- Step 5 is an example of the Augmenting Path Algorithm (Stable Marriage Problem).
- Step 6 is an example of contraint relaxation. It is "giving up" on a particular cost and raising the constraint by the least amount possible.
- If your implementation is jumping between Step 4 and Step 6 without entering Step 5, you probably have not properly dealt with recognizing that there are no uncovered zeros in Step 4.
- In the matrix M 1=starred zero and 2=primed zero. So, if C[i,j] is a starred zero we would set M[i,j]=1. All other elements in M are set to zero
- The Munkres assignment algorithm can be implemented as a sparse matrix, but you will need to ensure that the correct (optimal) assignment pairs are active in the sparse cost matrix C
- Munkres Assignment can be applied to TSP, pattern matching, track initiation, data correlation, and (of course) any pairwise assignment application.
- Munkres can be extended to rectangular arrays (i.e. more jobs
than workers, or more workers than jobs) .

- The best way to find a maximal assignment is to replace the
values c
_{i,j}in the cost matrix with C[i,j] = bigval - c_{i,j}. - Original Reference: Algorithms for Assignment and Transportation Problems, James Munkres, Journal of the Society for Industrial and Applied Mathematics Volume 5, Number 1, March, 1957
- Extension to Rectangular Arrays Ref: F. Burgeois and J.-C.
Lasalle. An extension of the Munkres algorithm for the assignment
problem to rectangular matrices. Communications of the ACM, 142302-806,
1971.