Hungarian algorithm

Let us call a function ${\displaystyle y:(S\cup T)\to \mathbb {R} }$ a potential if ${\displaystyle y(i)+y(j)\leq c(i,j)}$ for each ${\displaystyle i\in S,j\in T}$. The value of potential ${\displaystyle y}$ is the sum of the potential over all vertices: ${\displaystyle \sum _{v\in S\cup T}y(v)}$.

The cost of each perfect matching is at least the value of each potential: the total cost of the matching is the sum of costs of all edges; the cost of each edge is at least the sum of potentials of its endpoints; since the matching is perfect, each vertex is an endpoint of exactly one edge; hence the total cost is at least the total potential.

The Hungarian method finds a perfect matching and a potential such that the matching cost equals the potential value. This proves that both of them are optimal. In fact, the Hungarian method finds a perfect matching of tight edges: an edge ${\displaystyle ij}$ is called tight for a potential ${\displaystyle y}$ if ${\displaystyle y(i)+y(j)=c(i,j)}$. Let us denote the subgraph of tight edges by ${\displaystyle G_{y}}$. The cost of a perfect matching in ${\displaystyle G_{y}}$ (if there is one) equals the value of ${\displaystyle y}$.

During the algorithm we maintain a potential ${\displaystyle y}$ and an orientation of ${\displaystyle G_{y}}$ (denoted by ${\displaystyle {\overrightarrow {G_{y}}}}$) which has the property that the edges oriented from T to S form a matching M. Initially, y is 0 everywhere, and all edges are oriented from S to T (so M is empty). In each step, either we modify y so that its value increases, or modify the orientation to obtain a matching with more edges. We maintain the invariant that all the edges of M are tight. We are done if M is a perfect matching.

In a general step, let ${\displaystyle R_{S}\subseteq S}$ and ${\displaystyle R_{T}\subseteq T}$ be the vertices not covered by M (so ${\displaystyle R_{S}}$ consists of the vertices in S with no incoming edge and ${\displaystyle R_{T}}$ consists of the vertices in T with no outgoing edge). Let ${\displaystyle Z}$ be the set of vertices reachable in ${\displaystyle {\overrightarrow {G_{y}}}}$ from ${\displaystyle R_{S}}$ by a directed path only following edges that are tight. This can be computed by breadth-first search.

If ${\displaystyle R_{T}\cap Z}$ is nonempty, then reverse the orientation of a directed path in ${\displaystyle {\overrightarrow {G_{y}}}}$ from ${\displaystyle R_{S}}$ to ${\displaystyle R_{T}}$. Thus the size of the corresponding matching increases by 1.

If ${\displaystyle R_{T}\cap Z}$ is empty, then let

${\displaystyle \Delta$ :=\min\{c(i,j)-y(i)-y(j):i\in Z\cap S,j\in T\setminus Z\}.}

${\displaystyle \Delta }$ is positive because there are no tight edges between ${\displaystyle Z\cap S}$ and ${\displaystyle T\setminus Z}$. Increase y by ${\displaystyle \Delta }$ on the vertices of ${\displaystyle Z\cap S}$ and decrease y by ${\displaystyle \Delta }$ on the vertices of ${\displaystyle Z\cap T}$. The resulting y is still a potential, and although the graph ${\displaystyle G_{y}}$ changes, it still contains M (see the next subsections). We orient the new edges from S to T. By the definition of ${\displaystyle \Delta }$ the set Z of vertices reachable from ${\displaystyle R_{S}}$ increases (note that the number of tight edges does not necessarily increase).

We repeat these steps until M is a perfect matching, in which case it gives a minimum cost assignment. The running time of this version of the method is O(n⁴): M is augmented n times, and in a phase where M is unchanged, there are at most n potential changes (since Z increases every time). The time sufficient for a potential change is ${\displaystyle O(n^{2})}$.

Given ${\displaystyle n}$ workers and tasks, and an n×n matrix containing the cost of assigning each worker to a task, find the cost minimizing assignment.

First the problem is written in the form of a matrix as given below

a1	a2	a3	a4
b1	b2	b3	b4
c1	c2	c3	c4
d1	d2	d3	d4

where a, b, c and d are the workers who have to perform tasks 1, 2, 3 and 4. a1, a2, a3, a4 denote the penalties incurred when worker "a" does task 1, 2, 3, 4 respectively. The same holds true for the other symbols as well. The matrix is square, so each worker can perform only one task.

Step 1

Then we perform row operations on the matrix. To do this, the lowest of all a_i (i belonging to 1-4) is taken and is subtracted from each element in that row. This will lead to at least one zero in that row (We get multiple zeros when there are two equal elements which also happen to be the lowest in that row). This procedure is repeated for all rows. We now have a matrix with at least one zero per row. Now we try to assign tasks to agents such that each agent is doing only one task and the penalty incurred in each case is zero. This is illustrated below.

0	a2'	a3'	a4'
b1'	b2'	b3'	0
c1'	0	c3'	c4'
d1'	d2'	0	d4'

The zeros that are indicated as 0 are the assigned tasks.

Step 2

Sometimes it may turn out that the matrix at this stage cannot be used for assigning, as is the case for the matrix below.

0	a2'	a3'	a4'
b1'	b2'	b3'	0
0	c2'	c3'	c4'
d1'	0	d3'	d4'

In the above case, no assignment can be made. Note that task 1 is done efficiently by both agent a and c. Both can't be assigned the same task. Also note that no one does task 3 efficiently. To overcome this, we repeat the above procedure for all columns (i.e. the minimum element in each column is subtracted from all the elements in that column) and then check if an assignment is possible.

In most situations this will give the result, but if it is still not possible then we need to keep going.

Step 3

All zeros in the matrix must be covered by marking as few rows and/or columns as possible. The following procedure is one way to accomplish this:

First, assign as many tasks as possible.

• Row 1 has one zero, so it is assigned. The 0 in row 3 is crossed out because it is in the same column.
• Row 2 has one zero, so it is assigned.
• Row 3's only zero has been crossed out, so nothing is assigned.
• Row 4 has two uncrossed zeros. Either one can be assigned, and the other zero is crossed out.

Alternatively, the 0 in row 3 may be assigned, causing the 0 in row 1 to be crossed instead.

0'	a2'	a3'	a4'
b1'	b2'	b3'	0'
0	c2'	c3'	c4'
d1'	0'	0	d4'

Now to the drawing part.

• Mark all rows having no assignments (row 3).
• Mark all columns having zeros in newly marked row(s) (column 1).
• Mark all rows having assignments in newly marked columns (row 1).
• Repeat the steps outlined in the previous 2 bullets until there are no new rows or columns being marked.

×
0'	a2'	a3'	a4'	×
b1'	b2'	b3'	0'
0	c2'	c3'	c4'	×
d1'	0'	0	d4'

Now draw lines through all marked columns and unmarked rows.

×
0'	a2'	a3'	a4'	×
b1'	b2'	b3'	0'
0	c2'	c3'	c4'	×
d1'	0'	0	d4'

The aforementioned detailed description is just one way to draw the minimum number of lines to cover all the 0s. Other methods work as well.

Step 4

From the elements that are left, find the lowest value. Subtract this from every unmarked element and add it to every element covered by two lines.

Repeat steps 3–4 until an assignment is possible; this is when the minimum number of lines used to cover all the 0s is equal to max(number of people, number of assignments), assuming dummy variables (usually the max cost) are used to fill in when the number of people is greater than the number of assignments.

Basically you find the second minimum cost among the remaining choices. The procedure is repeated until you are able to distinguish among the workers in terms of least cost.

• R.E. Burkard, M. Dell'Amico, S. Martello: Assignment Problems (Revised reprint). SIAM, Philadelphia (PA.) 2012. ISBN 978-1-61197-222-1
• M. Fischetti, "Lezioni di Ricerca Operativa", Edizioni Libreria Progetto Padova, Italia, 1995.
• R. Ahuja, T. Magnanti, J. Orlin, "Network Flows", Prentice Hall, 1993.
• S. Martello, "Jeno Egerváry: from the origins of the Hungarian algorithm to satellite communication". Central European Journal of Operational Research 18, 47–58, 2010

• Bruff, Derek, The Assignment Problem and the Hungarian Method (matrix formalism).
• Mordecai J. Golin, Bipartite Matching and the Hungarian Method (bigraph formalism), Course Notes, Hong Kong University of Science and Technology.
• Hungarian maximum matching algorithm (both formalisms), in Brilliant website.
• R. A. Pilgrim, Munkres' Assignment Algorithm. Modified for Rectangular Matrices, Course notes, Murray State University.
• Mike Dawes, The Optimal Assignment Problem, Course notes, University of Western Ontario.
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• Lecture: Fundamentals of Operations Research - Assignment Problem - Hungarian Algorithm, Prof. G. Srinivasan, Department of Management Studies, IIT Madras.
• Extension: Assignment sensitivity analysis (with O(n^4) time complexity), Liu, Shell.
• Solve any Assignment Problem online, provides a step by step explanation of the Hungarian Algorithm.