Hungarian Method
The Hungarian method is a computational optimization technique that addresses the assignment problem in polynomial time and foreshadows following primal-dual alternatives. In 1955, Harold Kuhn used the term “Hungarian method” to honour two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry. Let’s go through the steps of the Hungarian method with the help of a solved example.
Hungarian Method to Solve Assignment Problems
The Hungarian method is a simple way to solve assignment problems. Let us first discuss the assignment problems before moving on to learning the Hungarian method.
What is an Assignment Problem?
A transportation problem is a type of assignment problem. The goal is to allocate an equal amount of resources to the same number of activities. As a result, the overall cost of allocation is minimised or the total profit is maximised.
Because available resources such as workers, machines, and other resources have varying degrees of efficiency for executing different activities, and hence the cost, profit, or loss of conducting such activities varies.
Assume we have ‘n’ jobs to do on ‘m’ machines (i.e., one job to one machine). Our goal is to assign jobs to machines for the least amount of money possible (or maximum profit). Based on the notion that each machine can accomplish each task, but at variable levels of efficiency.
Hungarian Method Steps
Check to see if the number of rows and columns are equal; if they are, the assignment problem is considered to be balanced. Then go to step 1. If it is not balanced, it should be balanced before the algorithm is applied.
Step 1 – In the given cost matrix, subtract the least cost element of each row from all the entries in that row. Make sure that each row has at least one zero.
Step 2 – In the resultant cost matrix produced in step 1, subtract the least cost element in each column from all the components in that column, ensuring that each column contains at least one zero.
Step 3 – Assign zeros
- Analyse the rows one by one until you find a row with precisely one unmarked zero. Encircle this lonely unmarked zero and assign it a task. All other zeros in the column of this circular zero should be crossed out because they will not be used in any future assignments. Continue in this manner until you’ve gone through all of the rows.
- Examine the columns one by one until you find one with precisely one unmarked zero. Encircle this single unmarked zero and cross any other zero in its row to make an assignment to it. Continue until you’ve gone through all of the columns.
Step 4 – Perform the Optimal Test
- The present assignment is optimal if each row and column has exactly one encircled zero.
- The present assignment is not optimal if at least one row or column is missing an assignment (i.e., if at least one row or column is missing one encircled zero). Continue to step 5. Subtract the least cost element from all the entries in each column of the final cost matrix created in step 1 and ensure that each column has at least one zero.
Step 5 – Draw the least number of straight lines to cover all of the zeros as follows:
(a) Highlight the rows that aren’t assigned.
(b) Label the columns with zeros in marked rows (if they haven’t already been marked).
(c) Highlight the rows that have assignments in indicated columns (if they haven’t previously been marked).
(d) Continue with (b) and (c) until no further marking is needed.
(f) Simply draw the lines through all rows and columns that are not marked. If the number of these lines equals the order of the matrix, then the solution is optimal; otherwise, it is not.
Step 6 – Find the lowest cost factor that is not covered by the straight lines. Subtract this least-cost component from all the uncovered elements and add it to all the elements that are at the intersection of these straight lines, but leave the rest of the elements alone.
Step 7 – Continue with steps 1 – 6 until you’ve found the highest suitable assignment.
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Hungarian Method Example
Use the Hungarian method to solve the given assignment problem stated in the table. The entries in the matrix represent each man’s processing time in hours.
\(\begin{array}{l}\begin{bmatrix} & I & II & III & IV & V \\1 & 20 & 15 & 18 & 20 & 25 \\2 & 18 & 20 & 12 & 14 & 15 \\3 & 21 & 23 & 25 & 27 & 25 \\4 & 17 & 18 & 21 & 23 & 20 \\5 & 18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} \)
With 5 jobs and 5 men, the stated problem is balanced.
\(\begin{array}{l}A = \begin{bmatrix}20 & 15 & 18 & 20 & 25 \\18 & 20 & 12 & 14 & 15 \\21 & 23 & 25 & 27 & 25 \\17 & 18 & 21 & 23 & 20 \\18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} \)
Subtract the lowest cost element in each row from all of the elements in the given cost matrix’s row. Make sure that each row has at least one zero.
\(\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 5 & 10 \\6 & 8 & 0 & 2 & 3 \\0 & 2 & 4 & 6 & 4 \\0 & 1 & 4 & 6 & 3 \\2 & 2 & 0 & 3 & 4 \\\end{bmatrix}\end{array} \)
Subtract the least cost element in each Column from all of the components in the given cost matrix’s Column. Check to see if each column has at least one zero.
\(\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 3 & 7 \\6 & 8 & 0 & 0 & 0 \\0 & 2 & 4 & 4 & 1 \\0 & 1 & 4 & 4 & 0 \\2 & 2 & 0 & 1 & 1 \\\end{bmatrix}\end{array} \)
When the zeros are assigned, we get the following:
The present assignment is optimal because each row and column contain precisely one encircled zero.
Where 1 to II, 2 to IV, 3 to I, 4 to V, and 5 to III are the best assignments.
Hence, z = 15 + 14 + 21 + 20 + 16 = 86 hours is the optimal time.
Practice Question on Hungarian Method
Use the Hungarian method to solve the following assignment problem shown in table. The matrix entries represent the time it takes for each job to be processed by each machine in hours.
\(\begin{array}{l}\begin{bmatrix}J/M & I & II & III & IV & V \\1 & 9 & 22 & 58 & 11 & 19 \\2 & 43 & 78 & 72 & 50 & 63 \\3 & 41 & 28 & 91 & 37 & 45 \\4 & 74 & 42 & 27 & 49 & 39 \\5 & 36 & 11 & 57 & 22 & 25 \\\end{bmatrix}\end{array} \)
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Frequently Asked Questions on Hungarian Method
What is hungarian method.
The Hungarian method is defined as a combinatorial optimization technique that solves the assignment problems in polynomial time and foreshadowed subsequent primal–dual approaches.
What are the steps involved in Hungarian method?
The following is a quick overview of the Hungarian method: Step 1: Subtract the row minima. Step 2: Subtract the column minimums. Step 3: Use a limited number of lines to cover all zeros. Step 4: Add some more zeros to the equation.
What is the purpose of the Hungarian method?
When workers are assigned to certain activities based on cost, the Hungarian method is beneficial for identifying minimum costs.
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Hungarian Algorithm for Assignment Problem | Set 1 (Introduction)
- For each row of the matrix, find the smallest element and subtract it from every element in its row.
- Do the same (as step 1) for all columns.
- Cover all zeros in the matrix using minimum number of horizontal and vertical lines.
- Test for Optimality: If the minimum number of covering lines is n, an optimal assignment is possible and we are finished. Else if lines are lesser than n, we haven’t found the optimal assignment, and must proceed to step 5.
- Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.
Try it before moving to see the solution
Explanation for above simple example:
An example that doesn’t lead to optimal value in first attempt: In the above example, the first check for optimality did give us solution. What if we the number covering lines is less than n.
Time complexity : O(n^3), where n is the number of workers and jobs. This is because the algorithm implements the Hungarian algorithm, which is known to have a time complexity of O(n^3).
Space complexity : O(n^2), where n is the number of workers and jobs. This is because the algorithm uses a 2D cost matrix of size n x n to store the costs of assigning each worker to a job, and additional arrays of size n to store the labels, matches, and auxiliary information needed for the algorithm.
In the next post, we will be discussing implementation of the above algorithm. The implementation requires more steps as we need to find minimum number of lines to cover all 0’s using a program. References: http://www.math.harvard.edu/archive/20_spring_05/handouts/assignment_overheads.pdf https://www.youtube.com/watch?v=dQDZNHwuuOY
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Hungarian algorithm for solving the assignment problem ¶
Statement of the assignment problem ¶.
There are several standard formulations of the assignment problem (all of which are essentially equivalent). Here are some of them:
There are $n$ jobs and $n$ workers. Each worker specifies the amount of money they expect for a particular job. Each worker can be assigned to only one job. The objective is to assign jobs to workers in a way that minimizes the total cost.
Given an $n \times n$ matrix $A$ , the task is to select one number from each row such that exactly one number is chosen from each column, and the sum of the selected numbers is minimized.
Given an $n \times n$ matrix $A$ , the task is to find a permutation $p$ of length $n$ such that the value $\sum A[i]\left[p[i]\right]$ is minimized.
Consider a complete bipartite graph with $n$ vertices per part, where each edge is assigned a weight. The objective is to find a perfect matching with the minimum total weight.
It is important to note that all the above scenarios are " square " problems, meaning both dimensions are always equal to $n$ . In practice, similar " rectangular " formulations are often encountered, where $n$ is not equal to $m$ , and the task is to select $\min(n,m)$ elements. However, it can be observed that a "rectangular" problem can always be transformed into a "square" problem by adding rows or columns with zero or infinite values, respectively.
We also note that by analogy with the search for a minimum solution, one can also pose the problem of finding a maximum solution. However, these two problems are equivalent to each other: it is enough to multiply all the weights by $-1$ .
Hungarian algorithm ¶
Historical reference ¶.
The algorithm was developed and published by Harold Kuhn in 1955. Kuhn himself gave it the name "Hungarian" because it was based on the earlier work by Hungarian mathematicians Dénes Kőnig and Jenő Egerváry. In 1957, James Munkres showed that this algorithm runs in (strictly) polynomial time, independently from the cost. Therefore, in literature, this algorithm is known not only as the "Hungarian", but also as the "Kuhn-Mankres algorithm" or "Mankres algorithm". However, it was recently discovered in 2006 that the same algorithm was invented a century before Kuhn by the German mathematician Carl Gustav Jacobi . His work, About the research of the order of a system of arbitrary ordinary differential equations , which was published posthumously in 1890, contained, among other findings, a polynomial algorithm for solving the assignment problem. Unfortunately, since the publication was in Latin, it went unnoticed among mathematicians.
It is also worth noting that Kuhn's original algorithm had an asymptotic complexity of $\mathcal{O}(n^4)$ , and only later Jack Edmonds and Richard Karp (and independently Tomizawa ) showed how to improve it to an asymptotic complexity of $\mathcal{O}(n^3)$ .
The $\mathcal{O}(n^4)$ algorithm ¶
To avoid ambiguity, we note right away that we are mainly concerned with the assignment problem in a matrix formulation (i.e., given a matrix $A$ , you need to select $n$ cells from it that are in different rows and columns). We index arrays starting with $1$ , i.e., for example, a matrix $A$ has indices $A[1 \dots n][1 \dots n]$ .
We will also assume that all numbers in matrix A are non-negative (if this is not the case, you can always make the matrix non-negative by adding some constant to all numbers).
Let's call a potential two arbitrary arrays of numbers $u[1 \ldots n]$ and $v[1 \ldots n]$ , such that the following condition is satisfied:
(As you can see, $u[i]$ corresponds to the $i$ -th row, and $v[j]$ corresponds to the $j$ -th column of the matrix).
Let's call the value $f$ of the potential the sum of its elements:
On one hand, it is easy to see that the cost of the desired solution $sol$ is not less than the value of any potential.
Lemma. $sol\geq f.$
The desired solution of the problem consists of $n$ cells of the matrix $A$ , so $u[i]+v[j]\leq A[i][j]$ for each of them. Since all the elements in $sol$ are in different rows and columns, summing these inequalities over all the selected $A[i][j]$ , you get $f$ on the left side of the inequality, and $sol$ on the right side.
On the other hand, it turns out that there is always a solution and a potential that turns this inequality into equality . The Hungarian algorithm described below will be a constructive proof of this fact. For now, let's just pay attention to the fact that if any solution has a cost equal to any potential, then this solution is optimal .
Let's fix some potential. Let's call an edge $(i,j)$ rigid if $u[i]+v[j]=A[i][j].$
Recall an alternative formulation of the assignment problem, using a bipartite graph. Denote with $H$ a bipartite graph composed only of rigid edges. The Hungarian algorithm will maintain, for the current potential, the maximum-number-of-edges matching $M$ of the graph $H$ . As soon as $M$ contains $n$ edges, then the solution to the problem will be just $M$ (after all, it will be a solution whose cost coincides with the value of a potential).
Let's proceed directly to the description of the algorithm .
Step 1. At the beginning, the potential is assumed to be zero ( $u[i]=v[i]=0$ for all $i$ ), and the matching $M$ is assumed to be empty.
Step 2. Further, at each step of the algorithm, we try, without changing the potential, to increase the cardinality of the current matching $M$ by one (recall that the matching is searched in the graph of rigid edges $H$ ). To do this, the usual Kuhn Algorithm for finding the maximum matching in bipartite graphs is used. Let us recall the algorithm here. All edges of the matching $M$ are oriented in the direction from the right part to the left one, and all other edges of the graph $H$ are oriented in the opposite direction.
Recall (from the terminology of searching for matchings) that a vertex is called saturated if an edge of the current matching is adjacent to it. A vertex that is not adjacent to any edge of the current matching is called unsaturated. A path of odd length, in which the first edge does not belong to the matching, and for all subsequent edges there is an alternating belonging to the matching (belongs/does not belong) - is called an augmenting path. From all unsaturated vertices in the left part, a depth-first or breadth-first traversal is started. If, as a result of the search, it was possible to reach an unsaturated vertex of the right part, we have found an augmenting path from the left part to the right one. If we include odd edges of the path and remove the even ones in the matching (i.e. include the first edge in the matching, exclude the second, include the third, etc.), then we will increase the matching cardinality by one.
If there was no augmenting path, then the current matching $M$ is maximal in the graph $H$ .
Step 3. If at the current step, it is not possible to increase the cardinality of the current matching, then a recalculation of the potential is performed in such a way that, at the next steps, there will be more opportunities to increase the matching.
Denote by $Z_1$ the set of vertices of the left part that were visited during the last traversal of Kuhn's algorithm, and through $Z_2$ the set of visited vertices of the right part.
Let's calculate the value $\Delta$ :
Lemma. $\Delta > 0.$
Suppose $\Delta=0$ . Then there exists a rigid edge $(i,j)$ with $i\in Z_1$ and $j\notin Z_2$ . It follows that the edge $(i,j)$ must be oriented from the right part to the left one, i.e. $(i,j)$ must be included in the matching $M$ . However, this is impossible, because we could not get to the saturated vertex $i$ except by going along the edge from j to i. So $\Delta > 0$ .
Now let's recalculate the potential in this way:
for all vertices $i\in Z_1$ , do $u[i] \gets u[i]+\Delta$ ,
for all vertices $j\in Z_2$ , do $v[j] \gets v[j]-\Delta$ .
Lemma. The resulting potential is still a correct potential.
We will show that, after recalculation, $u[i]+v[j]\leq A[i][j]$ for all $i,j$ . For all the elements of $A$ with $i\in Z_1$ and $j\in Z_2$ , the sum $u[i]+v[j]$ does not change, so the inequality remains true. For all the elements with $i\notin Z_1$ and $j\in Z_2$ , the sum $u[i]+v[j]$ decreases by $\Delta$ , so the inequality is still true. For the other elements whose $i\in Z_1$ and $j\notin Z_2$ , the sum increases, but the inequality is still preserved, since the value $\Delta$ is, by definition, the maximum increase that does not change the inequality.
Lemma. The old matching $M$ of rigid edges is valid, i.e. all edges of the matching will remain rigid.
For some rigid edge $(i,j)$ to stop being rigid as a result of a change in potential, it is necessary that equality $u[i] + v[j] = A[i][j]$ turns into inequality $u[i] + v[j] < A[i][j]$ . However, this can happen only when $i \notin Z_1$ and $j \in Z_2$ . But $i \notin Z_1$ implies that the edge $(i,j)$ could not be a matching edge.
Lemma. After each recalculation of the potential, the number of vertices reachable by the traversal, i.e. $|Z_1|+|Z_2|$ , strictly increases.
First, note that any vertex that was reachable before recalculation, is still reachable. Indeed, if some vertex is reachable, then there is some path from reachable vertices to it, starting from the unsaturated vertex of the left part; since for edges of the form $(i,j),\ i\in Z_1,\ j\in Z_2$ the sum $u[i]+v[j]$ does not change, this entire path will be preserved after changing the potential. Secondly, we show that after a recalculation, at least one new vertex will be reachable. This follows from the definition of $\Delta$ : the edge $(i,j)$ which $\Delta$ refers to will become rigid, so vertex $j$ will be reachable from vertex $i$ .
Due to the last lemma, no more than $n$ potential recalculations can occur before an augmenting path is found and the matching cardinality of $M$ is increased. Thus, sooner or later, a potential that corresponds to a perfect matching $M^*$ will be found, and $M^*$ will be the answer to the problem. If we talk about the complexity of the algorithm, then it is $\mathcal{O}(n^4)$ : in total there should be at most $n$ increases in matching, before each of which there are no more than $n$ potential recalculations, each of which is performed in time $\mathcal{O}(n^2)$ .
We will not give the implementation for the $\mathcal{O}(n^4)$ algorithm here, since it will turn out to be no shorter than the implementation for the $\mathcal{O}(n^3)$ one, described below.
The $\mathcal{O}(n^3)$ algorithm ¶
Now let's learn how to implement the same algorithm in $\mathcal{O}(n^3)$ (for rectangular problems $n \times m$ , $\mathcal{O}(n^2m)$ ).
The key idea is to consider matrix rows one by one , and not all at once. Thus, the algorithm described above will take the following form:
Consider the next row of the matrix $A$ .
While there is no increasing path starting in this row, recalculate the potential.
As soon as an augmenting path is found, propagate the matching along it (thus including the last edge in the matching), and restart from step 1 (to consider the next line).
To achieve the required complexity, it is necessary to implement steps 2-3, which are performed for each row of the matrix, in time $\mathcal{O}(n^2)$ (for rectangular problems in $\mathcal{O}(nm)$ ).
To do this, recall two facts proved above:
With a change in the potential, the vertices that were reachable by Kuhn's traversal will remain reachable.
In total, only $\mathcal{O}(n)$ recalculations of the potential could occur before an augmenting path was found.
From this follow these key ideas that allow us to achieve the required complexity:
To check for the presence of an augmenting path, there is no need to start the Kuhn traversal again after each potential recalculation. Instead, you can make the Kuhn traversal in an iterative form : after each recalculation of the potential, look at the added rigid edges and, if their left ends were reachable, mark their right ends reachable as well and continue the traversal from them.
Developing this idea further, we can present the algorithm as follows: at each step of the loop, the potential is recalculated. Subsequently, a column that has become reachable is identified (which will always exist as new reachable vertices emerge after every potential recalculation). If the column is unsaturated, an augmenting chain is discovered. Conversely, if the column is saturated, the matching row also becomes reachable.
To quickly recalculate the potential (faster than the $\mathcal{O}(n^2)$ naive version), you need to maintain auxiliary minima for each of the columns:
$minv[j]=\min_{i\in Z_1} A[i][j]-u[i]-v[j].$
It's easy to see that the desired value $\Delta$ is expressed in terms of them as follows:
$\Delta=\min_{j\notin Z_2} minv[j].$
Thus, finding $\Delta$ can now be done in $\mathcal{O}(n)$ .
It is necessary to update the array $minv$ when new visited rows appear. This can be done in $\mathcal{O}(n)$ for the added row (which adds up over all rows to $\mathcal{O}(n^2)$ ). It is also necessary to update the array $minv$ when recalculating the potential, which is also done in time $\mathcal{O}(n)$ ( $minv$ changes only for columns that have not yet been reached: namely, it decreases by $\Delta$ ).
Thus, the algorithm takes the following form: in the outer loop, we consider matrix rows one by one. Each row is processed in time $\mathcal{O}(n^2)$ , since only $\mathcal{O}(n)$ potential recalculations could occur (each in time $\mathcal{O}(n)$ ), and the array $minv$ is maintained in time $\mathcal{O}(n^2)$ ; Kuhn's algorithm will work in time $\mathcal{O}(n^2)$ (since it is presented in the form of $\mathcal{O}(n)$ iterations, each of which visits a new column).
The resulting complexity is $\mathcal{O}(n^3)$ or, if the problem is rectangular, $\mathcal{O}(n^2m)$ .
Implementation of the Hungarian algorithm ¶
The implementation below was developed by Andrey Lopatin several years ago. It is distinguished by amazing conciseness: the entire algorithm consists of 30 lines of code .
The implementation finds a solution for the rectangular matrix $A[1\dots n][1\dots m]$ , where $n\leq m$ . The matrix is 1-based for convenience and code brevity: this implementation introduces a dummy zero row and zero column, which allows us to write many cycles in a general form, without additional checks.
Arrays $u[0 \ldots n]$ and $v[0 \ldots m]$ store potential. Initially, they are set to zero, which is consistent with a matrix of zero rows (Note that it is unimportant for this implementation whether or not the matrix $A$ contains negative numbers).
The array $p[0 \ldots m]$ contains a matching: for each column $j = 1 \ldots m$ , it stores the number $p[j]$ of the selected row (or $0$ if nothing has been selected yet). For the convenience of implementation, $p[0]$ is assumed to be equal to the number of the current row.
The array $minv[1 \ldots m]$ contains, for each column $j$ , the auxiliary minima necessary for a quick recalculation of the potential, as described above.
The array $way[1 \ldots m]$ contains information about where these minimums are reached so that we can later reconstruct the augmenting path. Note that, to reconstruct the path, it is sufficient to store only column values, since the row numbers can be taken from the matching (i.e., from the array $p$ ). Thus, $way[j]$ , for each column $j$ , contains the number of the previous column in the path (or $0$ if there is none).
The algorithm itself is an outer loop through the rows of the matrix , inside which the $i$ -th row of the matrix is considered. The first do-while loop runs until a free column $j0$ is found. Each iteration of the loop marks visited a new column with the number $j0$ (calculated at the last iteration; and initially equal to zero - i.e. we start from a dummy column), as well as a new row $i0$ - adjacent to it in the matching (i.e. $p[j0]$ ; and initially when $j0=0$ the $i$ -th row is taken). Due to the appearance of a new visited row $i0$ , you need to recalculate the array $minv$ and $\Delta$ accordingly. If $\Delta$ is updated, then the column $j1$ becomes the minimum that has been reached (note that with such an implementation $\Delta$ could turn out to be equal to zero, which means that the potential cannot be changed at the current step: there is already a new reachable column). After that, the potential and the $minv$ array are recalculated. At the end of the "do-while" loop, we found an augmenting path ending in a column $j0$ that can be "unrolled" using the ancestor array $way$ .
The constant INF is "infinity", i.e. some number, obviously greater than all possible numbers in the input matrix $A$ .
To restore the answer in a more familiar form, i.e. finding for each row $i = 1 \ldots n$ the number $ans[i]$ of the column selected in it, can be done as follows:
The cost of the matching can simply be taken as the potential of the zero column (taken with the opposite sign). Indeed, as you can see from the code, $-v[0]$ contains the sum of all the values of $\Delta$ , i.e. total change in potential. Although several values of $u[i]$ and $v[j]$ could change at once, the total change in the potential is exactly equal to $\Delta$ , since until there is an augmenting path, the number of reachable rows is exactly one more than the number of the reachable columns (only the current row $i$ does not have a "pair" in the form of a visited column):
Connection to the Successive Shortest Path Algorithm ¶
The Hungarian algorithm can be seen as the Successive Shortest Path Algorithm , adapted for the assignment problem. Without going into the details, let's provide an intuition regarding the connection between them.
The Successive Path algorithm uses a modified version of Johnson's algorithm as reweighting technique. This one is divided into four steps:
- Use the Bellman-Ford algorithm, starting from the sink $s$ and, for each node, find the minimum weight $h(v)$ of a path from $s$ to $v$ .
For every step of the main algorithm:
- Reweight the edges of the original graph in this way: $w(u,v) \gets w(u,v)+h(u)-h(v)$ .
- Use Dijkstra 's algorithm to find the shortest-paths subgraph of the original network.
- Update potentials for the next iteration.
Given this description, we can observe that there is a strong analogy between $h(v)$ and potentials: it can be checked that they are equal up to a constant offset. In addition, it can be shown that, after reweighting, the set of all zero-weight edges represents the shortest-path subgraph where the main algorithm tries to increase the flow. This also happens in the Hungarian algorithm: we create a subgraph made of rigid edges (the ones for which the quantity $A[i][j]-u[i]-v[j]$ is zero), and we try to increase the size of the matching.
In step 4, all the $h(v)$ are updated: every time we modify the flow network, we should guarantee that the distances from the source are correct (otherwise, in the next iteration, Dijkstra's algorithm might fail). This sounds like the update performed on the potentials, but in this case, they are not equally incremented.
To deepen the understanding of potentials, refer to this article .
Task examples ¶
Here are a few examples related to the assignment problem, from very trivial to less obvious tasks:
Given a bipartite graph, it is required to find in it the maximum matching with the minimum weight (i.e., first of all, the size of the matching is maximized, and secondly, its cost is minimized). To solve it, we simply build an assignment problem, putting the number "infinity" in place of the missing edges. After that, we solve the problem with the Hungarian algorithm, and remove edges of infinite weight from the answer (they could enter the answer if the problem does not have a solution in the form of a perfect matching).
Given a bipartite graph, it is required to find in it the maximum matching with the maximum weight . The solution is again obvious, all weights must be multiplied by minus one.
The task of detecting moving objects in images : two images were taken, as a result of which two sets of coordinates were obtained. It is required to correlate the objects in the first and second images, i.e. determine for each point of the second image, which point of the first image it corresponded to. In this case, it is required to minimize the sum of distances between the compared points (i.e., we are looking for a solution in which the objects have taken the shortest path in total). To solve, we simply build and solve an assignment problem, where the weights of the edges are the Euclidean distances between points.
The task of detecting moving objects by locators : there are two locators that can't determine the position of an object in space, but only its direction. Both locators (located at different points) received information in the form of $n$ such directions. It is required to determine the position of objects, i.e. determine the expected positions of objects and their corresponding pairs of directions in such a way that the sum of distances from objects to direction rays is minimized. Solution: again, we simply build and solve the assignment problem, where the vertices of the left part are the $n$ directions from the first locator, the vertices of the right part are the $n$ directions from the second locator, and the weights of the edges are the distances between the corresponding rays.
Covering a directed acyclic graph with paths : given a directed acyclic graph, it is required to find the smallest number of paths (if equal, with the smallest total weight) so that each vertex of the graph lies in exactly one path. The solution is to build the corresponding bipartite graph from the given graph and find the maximum matching of the minimum weight in it. See separate article for more details.
Tree coloring book . Given a tree in which each vertex, except for leaves, has exactly $k-1$ children. It is required to choose for each vertex one of the $k$ colors available so that no two adjacent vertices have the same color. In addition, for each vertex and each color, the cost of painting this vertex with this color is known, and it is required to minimize the total cost. To solve this problem, we use dynamic programming. Namely, let's learn how to calculate the value $d[v][c]$ , where $v$ is the vertex number, $c$ is the color number, and the value $d[v][c]$ itself is the minimum cost needed to color all the vertices in the subtree rooted at $v$ , and the vertex $v$ itself with color $c$ . To calculate such a value $d[v][c]$ , it is necessary to distribute the remaining $k-1$ colors among the children of the vertex $v$ , and for this, it is necessary to build and solve the assignment problem (in which the vertices of the left part are colors, the vertices of the right part are children, and the weights of the edges are the corresponding values of $d$ ). Thus, each value $d[v][c]$ is calculated using the solution of the assignment problem, which ultimately gives the asymptotic $\mathcal{O}(nk^4)$ .
If, in the assignment problem, the weights are not on the edges, but on the vertices, and only on the vertices of the same part , then it's not necessary to use the Hungarian algorithm: just sort the vertices by weight and run the usual Kuhn algorithm (for more details, see a separate article ).
Consider the following special case . Let each vertex of the left part be assigned some number $\alpha[i]$ , and each vertex of the right part $\beta[j]$ . Let the weight of any edge $(i,j)$ be equal to $\alpha[i]\cdot \beta[j]$ (the numbers $\alpha[i]$ and $\beta[j]$ are known). Solve the assignment problem. To solve it without the Hungarian algorithm, we first consider the case when both parts have two vertices. In this case, as you can easily see, it is better to connect the vertices in the reverse order: connect the vertex with the smaller $\alpha[i]$ to the vertex with the larger $\beta[j]$ . This rule can be easily generalized to an arbitrary number of vertices: you need to sort the vertices of the first part in increasing order of $\alpha[i]$ values, the second part in decreasing order of $\beta[j]$ values, and connect the vertices in pairs in that order. Thus, we obtain a solution with complexity of $\mathcal{O}(n\log n)$ .
The Problem of Potentials . Given a matrix $A[1 \ldots n][1 \ldots m]$ , it is required to find two arrays $u[1 \ldots n]$ and $v[1 \ldots m]$ such that, for any $i$ and $j$ , $u[i] + v[j] \leq a[i][j]$ and the sum of elements of arrays $u$ and $v$ is maximum. Knowing the Hungarian algorithm, the solution to this problem will not be difficult: the Hungarian algorithm just finds such a potential $u, v$ that satisfies the condition of the problem. On the other hand, without knowledge of the Hungarian algorithm, it seems almost impossible to solve such a problem.
This task is also called the dual problem of the assignment problem: minimizing the total cost of the assignment is equivalent to maximizing the sum of the potentials.
Literature ¶
Ravindra Ahuja, Thomas Magnanti, James Orlin. Network Flows [1993]
Harold Kuhn. The Hungarian Method for the Assignment Problem [1955]
James Munkres. Algorithms for Assignment and Transportation Problems [1957]
Practice Problems ¶
UVA - Crime Wave - The Sequel
UVA - Warehouse
SGU - Beloved Sons
UVA - The Great Wall Game
UVA - Jogging Trails
- Alessandro Minisini (92.97%)
- Oleksandr Kulkov (7.03%)
Hungarian Method: Assignment Problem
Hungarian Method is an efficient method for solving assignment problems .
This method is based on the following principle:
- If a constant is added to, or subtracted from, every element of a row and/or a column of the given cost matrix of an assignment problem, the resulting assignment problem has the same optimal solution as the original problem.
Hungarian Algorithm
The objective of this section is to examine a computational method - an algorithm - for deriving solutions to the assignment problems. The following steps summarize the approach:
Steps in Hungarian Method
1. Identify the minimum element in each row and subtract it from every element of that row.
2. Identify the minimum element in each column and subtract it from every element of that column.
3. Make the assignments for the reduced matrix obtained from steps 1 and 2 in the following way:
- For every zero that becomes assigned, cross out (X) all other zeros in the same row and the same column.
- If for a row and a column, there are two or more zeros and one cannot be chosen by inspection, then you are at liberty to choose the cell arbitrarily for assignment.
4. An optimal assignment is found, if the number of assigned cells equals the number of rows (and columns). In case you have chosen a zero cell arbitrarily, there may be alternate optimal solutions. If no optimal solution is found, go to step 5.
5. Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix obtained from step 3 by adopting the following procedure:
- Mark all the rows that do not have assignments.
- Mark all the columns (not already marked) which have zeros in the marked rows.
- Mark all the rows (not already marked) that have assignments in marked columns.
- Repeat steps 5 (i) to (iii) until no more rows or columns can be marked.
- Draw straight lines through all unmarked rows and marked columns.
You can also draw the minimum number of lines by inspection.
6. Select the smallest element from all the uncovered elements. Subtract this smallest element from all the uncovered elements and add it to the elements, which lie at the intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment.
7. Go to step 3 and repeat the procedure until you arrive at an optimal assignment.
For the time being we assume that number of jobs is equal to number of machines or persons. Later in the chapter, we will remove this restrictive assumption and consider a special case where no. of facilities and tasks are not equal.
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The Hungarian method is a computational optimization technique that addresses the assignment problem in polynomial time and foreshadows following primal-dual alternatives. In 1955, Harold Kuhn used the term "Hungarian method" to honour two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry. Let's go through the steps of the Hungarian method with the help of a solved example.
The Hungarian algorithm, aka Munkres assignment algorithm, utilizes the following theorem for polynomial runtime complexity (worst case O(n 3)) and guaranteed optimality: If a number is added to or subtracted from all of the entries of any one row or column of a cost matrix, then an optimal assignment for the resulting cost matrix is also an ...
The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods.It was developed and published in 1955 by Harold Kuhn, who gave it the name "Hungarian method" because the algorithm was largely based on the earlier works of two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry.
Hungarian method for assignment problem Step 1. Subtract the entries of each row by the row minimum. Step 2. Subtract the entries of each column by the column minimum. Step 3. Make an assignment to the zero entries in the resulting matrix. A = M 17 10 15 17 18 M 6 10 20 12 5 M 14 19 12 11 15 M 7 16 21 18 6 M −10
The matrix below shows the cost of assigning a certain worker to a certain job. The objective is to minimize the total cost of the assignment. Below we will explain the Hungarian algorithm using this example. Note that a general description of the algorithm can be found here. Step 1: Subtract row minima.
General description of the algorithm. This problem is known as the assignment problem. The assignment problem is a special case of the transportation problem, which in turn is a special case of the min-cost flow problem, so it can be solved using algorithms that solve the more general cases. Also, our problem is a special case of binary integer ...
In this lesson we learn what is an assignment problem and how we can solve it using the Hungarian method.
The assignment problem deals with assigning machines to tasks, workers to jobs, soccer players to positions, and so on. The goal is to determine the optimum assignment that, for example, minimizes the total cost or maximizes the team effectiveness. The assignment problem is a fundamental problem in the area of combinatorial optimization.
The Hungarian Method: The following algorithm applies the above theorem to a given n × n cost matrix to find an optimal assignment. Step 1. Subtract the smallest entry in each row from all the entries of its row. Step 2. Subtract the smallest entry in each column from all the entries of its column. Step 3.
The Hungarian algorithm can be seen as the Successive Shortest Path Algorithm, adapted for the assignment problem. Without going into the details, let's provide an intuition regarding the connection between them. The Successive Path algorithm uses a modified version of Johnson's algorithm as reweighting technique.
Example 1: Hungarian Method. The Funny Toys Company has four men available for work on four separate jobs. Only one man can work on any one job. The cost of assigning each man to each job is given in the following table. The objective is to assign men to jobs in such a way that the total cost of assignment is minimum. Job.
The Hungarian method is a combinatorial optimization algorithm which solves the assignment problem in polynomial time . Later it was discovered that it was a primal-dual Simplex method.. It was developed and published by Harold Kuhn in 1955, who gave the name "Hungarian method" because the algorithm was largely based on the earlier works of two Hungarian mathematicians: Denes Konig and Jeno ...
The Hungarian Method for Solving the Assignment Problem We're ready to state the Hungarian method now that we've seen a couple of examples. Initialize the algorithm: { Subtract the lowest row value from each row. { For each column, subtract the lowest value. Steps 1 and 2 create zeros to start the algorithm o .
THEOREM 4. There is an adequate budget and an assignment such that the total allotment of the budget equals the number of jobs assigned to qualified individuals. Since Theorem 3 implies that the assignment of Theorem 4 is optimal, we have provided the following answer to the Simple Assignment Problem: H. W. KUHN.
1. INTRODUCTION The Hungarian Method [ 11 is an algorithm for solving assignment problems that is based on the work of D. Konig and J. Egervgry. In one possible interpretation, an assignment problem asks for the best assignment of a set of persons to a set of jobs, where the feasible assignments are ranked by the total scores or ratings of the ...
Hungarian Algorithm Steps. To use the Hungarian Algorithm, we first arrange the activities and people in a matrix with rows being people, columns being activity, and entries being the costs. Once ...
THE HUNGARIAN METHOD FOR THE ASSIGNMENT. PROBLEM'. H. W. Kuhn. Bryn Y a w College. Assuming that numerical scores are available for the perform- ance of each of n persons on each of n jobs, the "assignment problem" is the quest for an assignment of persons to jobs so that the sum of the. n scores so obtained is as large as possible.
Hungarian Method is an efficient method for solving assignment problems. This method is based on the following principle: If a constant is added to, or subtracted from, every element of a row and/or a column of the given cost matrix of an assignment problem, the resulting assignment problem has the same optimal solution as the original problem.
The Hungarian method provides an efficient solution to assignment problems. But first, let's understand what an assignment problem is. Defining an Assignment Problem. An assignment problem is a type of transportation problem where the goal is to assign resources to tasks in such a way that the total cost of assignment is minimized or the total ...
Here is the video about assignment problem - Hungarian method with algorithm.NOTE: After row and column scanning, If you stuck with more than one zero in th...
Textbooks: https://amzn.to/2VgimyJhttps://amzn.to/2CHalvxhttps://amzn.to/2Svk11kIn this video, we'll talk about how to solve a special case of the transporta...
Solve an assignment problem online. Fill in the cost matrix of an assignment problem and click on 'Solve'. The optimal assignment will be determined and a step by step explanation of the hungarian algorithm will be given. Fill in the cost matrix ():
Algorithm & Example-1. Algorithm. Hungarian Method Steps (Rule) Step-1: If number of rows is not equal to number of columns, then add dummy rows or columns with cost 0, to make it a square matrix. Step-2: a. Identify the minimum element in each row and subtract it from each element of that row. b. Identify the minimum element in each column and ...