munkres assignment algorithm python

Introduction

Assignment problem.

Let C be an n by n matrix representing the costs of each of n workers to perform any of n jobs. The assignment problem is to assign jobs to workers in a way that minimizes the total cost. Since each worker can perform only one job and each job can be assigned to only one worker the assignments represent an independent set of the matrix C .

One way to generate the optimal set is to create all permutations of the indexes necessary to traverse the matrix so that no row and column are used more than once. For instance, given this matrix (expressed in Python):

You could use this code to generate the traversal indexes:

After the call to permute(), the results matrix would look like this:

You could then use that index matrix to loop over the original cost matrix and calculate the smallest cost of the combinations:

While this approach works fine for small matrices, it does not scale. It executes in O( n !) time: Calculating the permutations for an n x n matrix requires n ! operations. For a 12x12 matrix, that’s 479,001,600 traversals. Even if you could manage to perform each traversal in just one millisecond, it would still take more than 133 hours to perform the entire traversal. A 20x20 matrix would take 2,432,902,008,176,640,000 operations. At an optimistic millisecond per operation, that’s more than 77 million years.

The Munkres algorithm runs in O( n ^3) time, rather than O( n !). This package provides an implementation of that algorithm.

This version is based on http://csclab.murraystate.edu/~bob.pilgrim/445/munkres.html

This version was written for Python by Brian Clapper from the algorithm at the above web site. (The Algorithm:Munkres Perl version, in CPAN, was clearly adapted from the same web site.)

Construct a Munkres object:

Then use it to compute the lowest cost assignment from a cost matrix. Here’s a sample program:

Running that program produces:

The instantiated Munkres object can be used multiple times on different matrices.

Non-square Cost Matrices

The Munkres algorithm assumes that the cost matrix is square. However, it’s possible to use a rectangular matrix if you first pad it with 0 values to make it square. This module automatically pads rectangular cost matrices to make them square.

• The module operates on a copy of the caller’s matrix, so any padding will not be seen by the caller.
• The cost matrix must be rectangular or square. An irregular matrix will not work.

Calculating Profit, Rather than Cost

The cost matrix is just that: A cost matrix. The Munkres algorithm finds the combination of elements (one from each row and column) that results in the smallest cost. It’s also possible to use the algorithm to maximize profit. To do that, however, you have to convert your profit matrix to a cost matrix. The simplest way to do that is to subtract all elements from a large value. For example:

The munkres module provides a convenience method for creating a cost matrix from a profit matrix. By default, it calculates the maximum profit and subtracts every profit from it to obtain a cost. If, however, you need a more general function, you can provide the conversion function; but the convenience method takes care of the actual creation of the matrix:

So, the above profit-calculation program can be recast as:

Disallowed Assignments

You can also mark assignments in your cost or profit matrix as disallowed. Simply use the munkres.DISALLOWED constant.

Running this program produces:

http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html

• Harold W. Kuhn. The Hungarian Method for the assignment problem. Naval Research Logistics Quarterly , 2:83-97, 1955.
• Harold W. Kuhn. Variants of the Hungarian method for assignment problems. Naval Research Logistics Quarterly , 3: 253-258, 1956.
• Munkres, J. Algorithms for the Assignment and Transportation Problems. Journal of the Society of Industrial and Applied Mathematics , 5(1):32-38, March, 1957.
• http://en.wikipedia.org/wiki/Hungarian_algorithm

Getting and installing munkres

Because munkres is available via PyPI , if you have pip installed on your system, installing munkres is as easy as running this command:

WARNING: As of version 1.1.0, munkres no longer supports Python 2. If you need to use it with Python 2, install an earlier version (e.g., 1.0.12):

Installing from source

You can also install munkres from source. Either download the source (as a zip or tarball) from http://github.com/bmc/munkres/downloads , or make a local read-only clone of the Git repository using one of the following commands:

Once you have a local munkres source directory, change your working directory to the source directory, and type:

To install it somewhere other than the default location (such as in your home directory) type:

Documentation

Consult the API documentation for details. The API documentation is generated from the source code, so you can also just browse the source .

• http://csclab.murraystate.edu/bob.pilgrim/445/munkres.html

This module is released under the Apache Software License, version 2. See the license file for details.

• Data Structures
• Linked List
• Binary Tree
• Binary Search Tree
• Segment Tree
• Disjoint Set Union
• Fenwick Tree
• Red-Black Tree
• Advanced Data Structures
• Hungarian Algorithm for Assignment Problem | Set 2 (Implementation)
• Clustering Coefficient in Graph Theory
• Maximum number of edges in Bipartite graph
• Types of Graphs with Examples
• Count of nodes with maximum connection in an undirected graph
• Erdos Renyl Model (for generating Random Graphs)
• Program to find the number of region in Planar Graph
• Maximize count of nodes disconnected from all other nodes in a Graph
• Find node having maximum number of common nodes with a given node K
• Convert the undirected graph into directed graph such that there is no path of length greater than 1
• Ways to Remove Edges from a Complete Graph to make Odd Edges
• Cost of painting n * m grid
• Number of Simple Graph with N Vertices and M Edges
• Count of Disjoint Groups by grouping points that are at most K distance apart
• Program to check if N is a Concentric Hexagonal Number
• Program to check if N is a Octagonal Number
• Graph Data Structure And Algorithms
• Program to check if N is a Icosidigonal Number
• Program to check if N is a Octadecagon number

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 Maximum Matching Algorithm

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The Hungarian matching algorithm , also called the Kuhn-Munkres algorithm, is a \(O\big(|V|^3\big)\) algorithm that can be used to find maximum-weight matchings in bipartite graphs , which is sometimes called the assignment problem . A bipartite graph can easily be represented by an adjacency matrix , where the weights of edges are the entries. Thinking about the graph in terms of an adjacency matrix is useful for the Hungarian algorithm.

A matching corresponds to a choice of 1s in the adjacency matrix, with at most one 1 in each row and in each column.

The Hungarian algorithm solves the following problem:

In a complete bipartite graph \(G\), find the maximum-weight matching. (Recall that a maximum-weight matching is also a perfect matching.)

This can also be adapted to find the minimum-weight matching.

Say you are having a party and you want a musician to perform, a chef to prepare food, and a cleaning service to help clean up after the party. There are three companies that provide each of these three services, but one company can only provide one service at a time (i.e. Company B cannot provide both the cleaners and the chef). You are deciding which company you should purchase each service from in order to minimize the cost of the party. You realize that is an example of the assignment problem, and set out to make a graph out of the following information: \(\quad\) Company\(\quad\) \(\quad\) Cost for Musician\(\quad\) \(\quad\) Cost for Chef\(\quad\) \(\quad\) Cost for Cleaners\(\quad\) \(\quad\) Company A\(\quad\) \(\quad\) $108\(\quad\) \(\quad\) $125\(\quad\) \(\quad\) $150\(\quad\) \(\quad\) Company B\(\quad\) \(\quad\) $150\(\quad\) \(\quad\) $135\(\quad\) \(\quad\) $175\(\quad\) \(\quad\) Company C\(\quad\) \(\quad\) $122\(\quad\) \(\quad\) $148\(\quad\) \(\quad\) $250\(\quad\) Can you model this table as a graph? What are the nodes? What are the edges? Show Answer The nodes are the companies and the services. The edges are weighted by the price.

What are some ways to solve the problem above? Since the table above can be thought of as a \(3 \times 3\) matrix, one could certainly solve this problem using brute force, checking every combination and seeing what yields the lowest price. However, there are \(n!\) combinations to check, and for large \(n\), this method becomes very inefficient very quickly.

The Hungarian Algorithm Using an Adjacency Matrix

The hungarian algorithm using a graph.

With the cost matrix from the example above in mind, the Hungarian algorithm operates on this key idea: 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 optimal assignment for the original cost matrix.

The Hungarian Method [1] Subtract the smallest entry in each row from all the other entries in the row. This will make the smallest entry in the row now equal to 0. Subtract the smallest entry in each column from all the other entries in the column. This will make the smallest entry in the column now equal to 0. Draw lines through the row and columns that have the 0 entries such that the fewest lines possible are drawn. If there are \(n\) lines drawn, an optimal assignment of zeros is possible and the algorithm is finished. If the number of lines is less than \(n\), then the optimal number of zeroes is not yet reached. Go to the next step. Find the smallest entry not covered by any line. Subtract this entry from each row that isn’t crossed out, and then add it to each column that is crossed out. Then, go back to Step 3.

Solve for the optimal solution for the example in the introduction using the Hungarian algorithm described above. Here is the initial adjacency matrix: Subtract the smallest value in each row from the other values in the row: Now, subtract the smallest value in each column from all other values in the column: Draw lines through the row and columns that have the 0 entries such that the fewest possible lines are drawn: There are 2 lines drawn, and 2 is less than 3, so there is not yet the optimal number of zeroes. Find the smallest entry not covered by any line. Subtract this entry from each row that isn’t crossed out, and then add it to each column that is crossed out. Then, go back to Step 3. 2 is the smallest entry. First, subtract from the uncovered rows: Now add to the covered columns: Now go back to step 3, drawing lines through the rows and columns that have 0 entries: There are 3 lines (which is \(n\)), so we are done. The assignment will be where the 0's are in the matrix such that only one 0 per row and column is part of the assignment. Replace the original values: The Hungarian algorithm tells us that it is cheapest to go with the musician from company C, the chef from company B, and the cleaners from company A. We can verify this by brute force. 108 + 135 + 250 = 493 108 + 148 + 175 = 431 150 + 125 + 250 = 525 150 + 148 + 150 = 448 122 + 125 + 175 = 422 122 + 135 + 150 = 407. We can see that 407 is the lowest price and matches the assignment the Hungarian algorithm determined. \(_\square\)

The Hungarian algorithm can also be executed by manipulating the weights of the bipartite graph in order to find a stable, maximum (or minimum) weight matching. This can be done by finding a feasible labeling of a graph that is perfectly matched, where a perfect matching is denoted as every vertex having exactly one edge of the matching.

How do we know that this creates a maximum-weight matching?

A feasible labeling on a perfect match returns a maximum-weighted matching. Suppose each edge \(e\) in the graph \(G\) connects two vertices, and every vertex \(v\) is covered exactly once. With this, we have the following inequality: \[w(M’) = \sum_{e\ \epsilon\ E} w(e) \leq \sum_{e\ \epsilon\ E } \big(l(e_x) + l(e_y)\big) = \sum_{v\ \epsilon\ V} l(v),\] where \(M’\) is any perfect matching in \(G\) created by a random assignment of vertices, and \(l(x)\) is a numeric label to node \(x\). This means that \(\sum_{v\ \epsilon\ V}\ l(v)\) is an upper bound on the cost of any perfect matching. Now let \(M\) be a perfect match in \(G\), then \[w(M) = \sum_{e\ \epsilon\ E} w(e) = \sum_{v\ \epsilon\ V}\ l(v).\] So \(w(M’) \leq w(M)\) and \(M\) is optimal. \(_\square\)

Start the algorithm by assigning any weight to each individual node in order to form a feasible labeling of the graph \(G\). This labeling will be improved upon by finding augmenting paths for the assignment until the optimal one is found.

A feasible labeling is a labeling such that

\(l(x) + l(y) \geq w(x,y)\ \forall x \in X, y \in Y\), where \(X\) is the set of nodes on one side of the bipartite graph, \(Y\) is the other set of nodes, \(l(x)\) is the label of \(x\), etc., and \(w(x,y)\) is the weight of the edge between \(x\) and \(y\).

A simple feasible labeling is just to label a node with the number of the largest weight from an edge going into the node. This is certain to be a feasible labeling because if \(A\) is a node connected to \(B\), the label of \(A\) plus the label of \(B\) is greater than or equal to the weight \(w(x,y)\) for all \(y\) and \(x\).

A feasible labeling of nodes, where labels are in red [2] .

Imagine there are four soccer players and each can play a few positions in the field. The team manager has quantified their skill level playing each position to make assignments easier.

How can players be assigned to positions in order to maximize the amount of skill points they provide?

The algorithm starts by labeling all nodes on one side of the graph with the maximum weight. This can be done by finding the maximum-weighted edge and labeling the adjacent node with it. Additionally, match the graph with those edges. If a node has two maximum edges, don’t connect them.

Although Eva is the best suited to play defense, she can't play defense and mid at the same time!

If the matching is perfect, the algorithm is done as there is a perfect matching of maximum weights. Otherwise, there will be two nodes that are not connected to any other node, like Tom and Defense. If this is the case, begin iterating.

Improve the labeling by finding the non-zero label vertex without a match, and try to find the best assignment for it. Formally, the Hungarian matching algorithm can be executed as defined below:

The Hungarian Algorithm for Graphs [3] Given: the labeling \(l\), an equality graph \(G_l = (V, E_l)\), an initial matching \(M\) in \(G_l\), and an unmatched vertex \(u \in V\) and \(u \notin M\) Augmenting the matching A path is augmenting for \(M\) in \(G_l\) if it alternates between edges in the matching and edges not in the matching, and the first and last vertices are free vertices , or unmatched, in \(M\). We will keep track of a candidate augmenting path starting at the vertex \(u\). If the algorithm finds an unmatched vertex \(v\), add on to the existing augmenting path \(p\) by adding the \(u\) to \(v\) segment. Flip the matching by replacing the edges in \(M\) with the edges in the augmenting path that are not in \(M\) \((\)in other words, the edges in \(E_l - M).\) Improving the labeling \(S \subseteq X\) and \(T \subseteq Y,\) where \(S\) and \(T\) represent the candidate augmenting alternating path between the matching and the edges not in the matching. Let \(N_l(S)\) be the neighbors to each node that is in \(S\) along edges in \(E_l\) such that \(N_l(S) = \{v|\forall u \in S: (u,v) \in E_l\}\). If \(N_l(S) = T\), then we cannot increase the size of the alternating path (and therefore can't further augment), so we need to improve the labeling. Let \(\delta_l\) be the minimum of \(l(u) + l(v) - w(u,v)\) over all of the \(u \in S\) and \(v \notin T\). Improve the labeling \(l\) to \(l'\): If \(r \in S,\) then \(l'(r) = l(r) - \delta_l,\) If \(r \in T,\) then \(l'(r) = l(r) + \delta_l.\) If \(r \notin S\) and \(r \notin T,\) then \(l'(r) = l(r).\) \(l'\) is a valid labeling and \(E_l \subset E_{l'}.\) Putting it all together: The Hungarian Algorithm Start with some matching \(M\), a valid labeling \(l\), where \(l\) is defined as the labelling \(\forall x \in X, y \in Y| l(y) = 0, l(x) = \text{ max}_{y \in Y}(w\big(x, y)\big)\). Do these steps until a perfect matching is found \((\)when \(M\) is perfect\():\) (a) Look for an augmenting path in \(M.\) (b) If an augmenting path does not exist, improve the labeling and then go back to step (a).

Each step will increase the size of the matching \(M\) or it will increase the size of the set of labeled edges, \(E_l\). This means that the process will eventually terminate since there are only so many edges in the graph \(G\). [4]

When the process terminates, \(M\) will be a perfect matching. By the Kuhn-Munkres theorem , this means that the matching is a maximum-weight matching.

The algorithm defined above can be implemented in the soccer scenario. First, the conflicting node is identified, implying that there is an alternating tree that must be reconfigured.

There is an alternating path between defense, Eva, mid, and Tom.

To find the best appropriate node, find the minimum \(\delta_l\), as defined in step 4 above, where \(l_u\) is the label for player \(u,\) \(l_v\) is the label for position \(v,\) and \(w_{u, v}\) is the weight on that edge.

The \(\delta_l\) of each unmatched node is computed, where the minimum is found to be a value of 2, between Tom playing mid \((8 + 0 – 6 = 2).\)

The labels are then augmented and the new edges are graphed in the example. Notice that defense and mid went down by 2 points, whereas Eva’s skillset got back two points. However, this is expected as Eva can't play in both positions at once.

Augmenting path leads to relabeling of nodes, which gives rise to the maximum-weighted path.

These new edges complete the perfect matching of the graph, which implies that a maximum-weighted graph has been found and the algorithm can terminate.

The complexity of the algorithm will be analyzed using the graph-based technique as a reference, yet the result is the same as for the matrix-based one.

Algorithm analysis [3] At each \(a\) or \(b\) step, the algorithm adds one edge to the matching and this happens \(O\big(|V|\big)\) times. It takes \(O\big(|V|\big)\) time to find the right vertex for the augmenting (if there is one at all), and it is \(O\big(|V|\big)\) time to flip the matching. Improving the labeling takes \(O\big(|V|\big)\) time to find \(\delta_l\) and to update the labelling accordingly. We might have to improve the labeling up to \(O\big(|V|\big)\) times if there is no augmenting path. This makes for a total of \(O\big(|V|^2\big)\) time. In all, there are \(O\big(|V|\big)\) iterations each taking \(O\big(|V|\big)\) work, leading to a total running time of \(O\big(|V|^3\big)\).

• Matching Algorithms
• Bruff, D. The Assignment Problem and the Hungarian Method . Retrieved June 26, 2016, from http://www.math.harvard.edu/archive/20_spring_05/handouts/assignment_overheads.pdf
• Golin, M. Bipartite Matching & the Hungarian Method . Retrieved Retrieved June 26th, 2016, from http://www.cse.ust.hk/~golin/COMP572/Notes/Matching.pdf
• Grinman, A. The Hungarian Algorithm for Weighted Bipartite Graphs . Retrieved June 26, 2016, from http://math.mit.edu/~rpeng/18434/hungarianAlgorithm.pdf
• Golin, M. Bipartite Matching & the Hungarian Method . Retrieved June 26, 2016, from http://www.cse.ust.hk/~golin/COMP572/Notes/Matching.pdf

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Hungarian Algorithm Introduction & Python Implementation

How to use hungarian method to resolve the linear assignment problem..

By Eason on 2021-08-02

In this article, I will introduce how to use Hungarian Method to resolve the linear assignment problem and provide my personal Python code solution.

So… What is the linear assignment problem?

The linear assignment problem represents the need to maximize the available resources (or minimize the expenditure) with limited resources. For instance, below is a 2D matrix, where each row represents a different supplier, and each column represents the cost of employing them to produce a particular product. Each supplier can only specialize in the production of one of these products. In other words, only one element can be selected for each column and row in the matrix, and the sum of the selected elements must be minimized (minimized cost expense).

The cost of producing different goods by different producers:

Indeed, this is a simple example. By trying out the possible combinations, we can see that the smallest sum is 13, so supplier A supplies Bubble Tea , supplier B supplies milk tea, and supplier C supplies Fruit Tea . However, such attempts do not follow a clear rule and become inefficient when applied to large tasks. Therefore, the next section will introduce step by step the Hungarian algorithm, which can be applied to the linear assignment problem.

Hungarian Algorithm & Python Code Step by Step

In this section, we will show how to use the Hungarian algorithm to solve linear assignment problems and find the minimum combinations in the matrix. Of course, the Hungarian algorithm can also be used to find the maximum combination.

Step 0. Prepare Operations

First, an N by N matrix is generated to be used for the Hungarian algorithm (Here, we use a 5 by 5 square matrix as an example).

The above code randomly generates a 5x5 cost matrix of integers between 0 and 10.

If we want to find the maximum sum, we could do the opposite. The matrix to be solved is regarded as the profit matrix, and the maximum value in the matrix is set as the common price of all goods. The cost matrix is obtained by subtracting the profit matrix from the maximum value. Finally, the cost matrix is substituted into the Hungarian algorithm to obtain the minimized combination and then remapped back to the profit matrix to obtain the maximized sum value and composition result.

The above code randomly generates a 5x5 profit matrix of integers between 0 and 10 and generate a corresponding cost matrix

By following the steps above, you can randomly generate either the cost matrix or the profit matrix. Next, we will move into the introduction of the Hungarian algorithm, and for the sake of illustration, the following sections will be illustrated using the cost matrix shown below. We will use the Hungarian algorithm to solve the linear assignment problem of the cost matrix and find the corresponding minimum sum.

Example cost matrix:

Step 1. Every column and every row subtract its internal minimum

First, every column and every row must subtract its internal minimum. After subtracting the minimum, the cost matrix will look like this.

Cost matrix after step 1:

And the current code is like this:

Step 2.1. Min_zero_row Function Implementation

At first, we need to find the row with the fewest zero elements. So, we can convert the previous matrix to the boolean matrix(0 → True, Others → False).

Transform matrix to boolean matrix:

Corresponding Boolean matrix:

Therefore, we can use the “min_zero_row” function to find the corresponding row.

The row which contains the least 0:

Third, mark any 0 elements on the corresponding row and clean up its row and column (converts elements on the Boolean matrix to False). The coordinates of the element are stored in mark_zero.

Hence, the boolean matrix will look like this:

The boolean matrix after the first process. The fourth row has been changed to all False.

The process is repeated several times until the elements in the boolean matrix are all False. The below picture shows the order in which they are marked.

The possible answer composition:

Step 2.2. Mark_matrix Function Implementation

After getting Zero_mat from the step 2–1, we can check it and mark the matrix according to certain rules. The whole rule can be broken down into several steps:

• Mark rows that do not contain marked 0 elements and store row indexes in the non_marked_row
• Search non_marked_row element, and find out if there are any unmarked 0 elements in the corresponding column
• Store the column indexes in the marked_cols
• Compare the column indexes stored in marked_zero and marked_cols
• If a matching column index exists, the corresponding row_index is saved to non_marked_rows
• Next, the row indexes that are not in non_marked_row are stored in marked_rows

Finally, the whole mark_matrx function is finished and then returns marked_zero , marked_rows , marked_cols. At this point, we will be able to decide the result based on the return information.

If we use the example cost matrix, the corresponding marked_zero , marked_rows, and marked_cols are as follows:

• marked_zero : [(3, 2), (0, 4), (1, 1), (2, 0), (4, 3)]
• marked_rows : [0, 1, 2, 3, 4]
• marked_cols : []

Step 3. Identify the Result

At this step, if the sum of the lengths of marked_rows and marked_cols is equal to the length of the cost matrix, it means that the solution of the linear assignment problem has been found successfully, and marked_zero stores the solution coordinates. Fortunately, in the example matrix, we find the answer on the first try. Therefore, we can skip to step 5 and calculate the solution.

However, everything is hardly plain sailing. Most of the time, we will not find the solution on the first try, such as the following matrix:

After Step 1 & 2 , the corresponding matrix, marked_rows, and marked_cols are as follows:

The sum of the lengths of Marked_Rows and Marked_Cols is 4 (less than 5).

Apparently, the sum of the lengths is less than the length of the matrix. At this time, we need to go into Step 4 to adjust the matrix.

Step 4. Adjust Matrix

In Step 4, we're going to put the matrix after Step 1 into the Adjust_Matrix function . Taking the latter matrix in Step 3 as an example, the matrix to be modified in Adjust_Matrix is:

The whole function can be separated into three steps:

• Find the minimum value for an element that is not in marked_rows and not in marked_cols . Hence, we can find the minimum value is 1.

• Subtract the elements which not in marked_rows nor marked_cols from the minimum values obtained in the previous step.

• Add the element in marked_rows , which is also in marked_cols , to the minimum value obtained by Step 4–1.

Return the adjusted matrix and repeat Step 2 and Step 3 until the conditions satisfy the requirement of entering Step 5.

Step 5. Calculate the Answer

Using the element composition stored in marked_zero , the minimum and maximum values of the linear assignment problem can be calculated.

The minimum composition of the assigned matrix and the minimum sum is 18.

The maximum composition of the assigned matrix and the maximum sum is 43.

The code of the Answer_Calculator function is as follows:

The complete code is as follows:

Hungarian algorithm - Wikipedia

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scipy.optimize.linear_sum_assignment #

Solve the linear sum assignment problem.

The cost matrix of the bipartite graph.

Calculates a maximum weight matching if true.

An array of row indices and one of corresponding column indices giving the optimal assignment. The cost of the assignment can be computed as cost_matrix[row_ind, col_ind].sum() . The row indices will be sorted; in the case of a square cost matrix they will be equal to numpy.arange(cost_matrix.shape[0]) .

for sparse inputs

The linear sum assignment problem [1] is also known as minimum weight matching in bipartite graphs. A problem instance is described by a matrix C, where each C[i,j] is the cost of matching vertex i of the first partite set (a ‘worker’) and vertex j of the second set (a ‘job’). The goal is to find a complete assignment of workers to jobs of minimal cost.

Formally, let X be a boolean matrix where \(X[i,j] = 1\) iff row i is assigned to column j. Then the optimal assignment has cost

where, in the case where the matrix X is square, each row is assigned to exactly one column, and each column to exactly one row.

This function can also solve a generalization of the classic assignment problem where the cost matrix is rectangular. If it has more rows than columns, then not every row needs to be assigned to a column, and vice versa.

This implementation is a modified Jonker-Volgenant algorithm with no initialization, described in ref. [2] .

New in version 0.17.0.

https://en.wikipedia.org/wiki/Assignment_problem

DF Crouse. On implementing 2D rectangular assignment algorithms. IEEE Transactions on Aerospace and Electronic Systems , 52(4):1679-1696, August 2016, DOI:10.1109/TAES.2016.140952

jk-munkres 1.2.0

pip install jk-munkres Copy PIP instructions

Released: Feb 20, 2023

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License: Apache Software License

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Project description

Munkres implementation for python.

Introduction

The Munkres module provides an O(n^3) implementation of the Munkres algorithm (also called the Hungarian algorithm or the Kuhn-Munkres algorithm). The algorithm models an assignment problem as an NxM cost matrix, where each element represents the cost of assigning the ith worker to the jth job, and it figures out the least-cost solution, choosing a single item from each row and column in the matrix, such that no row and no column are used more than once.

This particular implementation is based on https://csclab.murraystate.edu/~bob.pilgrim/445/munkres.html .

See the docs on the project page for more details.

WARNING : As of version 1.1.0, munkres no longer supports Python 2. If you need to use this package with Python 2, install an earlier version. See the installation instructions for details.

© 2008-2019 Brian M. Clapper

Licensed under the Apache License, Version 2.0. See LICENSE for details.

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Table of Contents

Introduction.

The Munkres algorithm described here aims to find an optimal solution which minimizes the total cost of the assignments .

For example, it can be used to match tracked (red) and detected (green) image points:

The following example also available in tutorial-munkres-assignment.cpp shows how to use Munkres algorithm:

The tutorial starts with 10 random image points (red) which represent our tracked points:

Then, by clicking on the image, the user is able to simulate the detected points (green):

Once the "fake" detection are selected, a cost matrix is built by defining the cost of assigning a track to a detection point as the Euclidean distance between them:

Finally, Munkres is ran to assign tracked points with the detected points (blue lines):

• Munkres Assignment Algorithm

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Python wrapper of a C++ implementation of the Munkres assignment algorithm.

jfrelinger/cython-munkres-wrapper

Folders and files, repository files navigation, munkres readme.

Munkres calculates the minimum cost assignment of the assignment using the Hungarian/Munkres algorithm.

Can handle non-square cost matricies, using algorithm provded by Bougeois and Lassalle in An extension of the Munkres Algorithm for the Assignment Problem to Rectangular Matrices .

which should print out :

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assignmunkres

Munkres global nearest neighbor assignment algorithm

Description

[ assignments , unassignedrows , unassignedcolumns ] = assignmunkres( costmatrix , costofnonassignment ) returns a table of assignments of detections to tracks using the Munkres algorithm. The Munkres algorithm obtains an optimal solution to the global nearest neighbor (GNN) assignment problem. An optimal solution minimizes the total cost of the assignments.

The cost of each potential assignment is contained in the cost matrix, costmatrix . Each matrix entry represents the cost of a possible assignments. Matrix rows represent tracks and columns represent detections. All possible assignments are represented in the cost matrix. The lower the cost, the more likely the assignment is to be made. Each track can be assigned to at most one detection and each detection can be assigned to at most one track. If the number of rows is greater than the number of columns, some tracks are unassigned. If the number of columns is greater than the number of rows, some detections are unassigned. You can set an entry of costmatrix to Inf to prohibit an assignment.

costofnonassignment represents the cost of leaving tracks or detections unassigned. Higher values increase the likelihood that every existing object is assigned.

The function returns a list of unassigned tracks, unassignedrows , and a list of unassigned detections, unassignedcolumns

collapse all

Assign Detections to Tracks Using Munkres Algorithm

Use assignMunkres to assign three detections to two tracks.

Start with two predicted track locations in x-y coordinates.

Assume three detections are received. At least one detection will not be assigned.

Construct a cost matrix by defining the cost of assigning a detection to a track as the Euclidean distance between them. Set the cost of non-assignment to 0.2.

Use the Auction algorithm to assign detections to tracks.

Display the assignments.

Show that there are no unassigned tracks.

Display the unassigned detections.

Plot detection to track assignments.

The track to detection assignments are:

Detection 1 is assigned to track 1.

Detection 2 is assigned to track 2.

Detection 3 is not assigned.

Input Arguments

Costmatrix — cost matrix real-valued m -by- n.

Cost matrix, specified as an M -by- N matrix. M is the number of tracks to be assigned and N is the number of detections to be assigned. Each entry in the cost matrix contains the cost of a track and detection assignment. The matrix may contain Inf entries to indicate that an assignment is prohibited. The cost matrix cannot be a sparse matrix.

Data Types: single | double

costofnonassignment — cost of non-assignment of tracks and detections scalar

Cost of non-assignment, specified as a scalar. The cost of non-assignment represents the cost of leaving tracks or detections unassigned. Higher values increase the likelihood that every object is assigned. The value cannot be set to Inf .

The costofnonassignment is half of the maximum cost that a successful assignment can have.

Output Arguments

Assignments — assignment of tracks to detections integer-valued l -by-2 matrix.

Assignment of detections to track, returned as an integer-valued L -by-2 matrix where L is the number of assignments. The first column of the matrix contains the assigned track indices and the second column contains the assigned detection indices.

Data Types: uint32

unassignedrows — Indices of unassigned tracks integer-valued P -by-1 column vector

Indices of unassigned tracks, returned as an integer-valued P -by-1 column vector.

unassignedcolumns — Indices of unassigned detections integer-valued Q -by-1 column vector

Indices of unassigned detections, returned as an integer-valued Q -by-1 column vector.

[1] Samuel S. Blackman and Popoli, R. Design and Analysis of Modern Tracking Systems . Artech House: Norwood, MA. 1999.

Extended Capabilities

C/c++ code generation generate c and c++ code using matlab® coder™., version history.

Introduced in R2018b

• assignauction | assignjv | assignkbest | assignkbestsd | assignsd | assignTOMHT | trackerTOMHT | trackerGNN

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