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Hungarian Algorithm for Linear Assignment Problems (V2.3)

version 1.4 (3.27 KB) by

An extremely fast implementation of the Hungarian algorithm on a native Matlab code.

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This is an extremely fast implementation of the famous Hungarian algorithm (aslo known as Munkres' algorithm). It can solve a 1000 x 1000 problem in about 20 seconds in a Core Duo (T2500 @ 2.00GHz) XP laptop with Matlab 2008a, which is about 2.5 times faster than the mex code "assignmentoptimal" in FEX ID 6543, about 6 times faster than the author's first version in FEX ID 20328, and at least 30 times faster than other Matlab implementations in the FEX.

The code can also handle rectangular prolems and problems with forbiden allocations.

The new version (V2.3)is able to conduct a partial assignment if a full assignment is not feasible.

For more details of the Hungarian algorithm, visit

Comments and Ratings (46)


echo (view profile)

love it .

David Franco

It worked very well.


Val (view profile)

Warning, this is highly buggy when working with zero-weight edges. Example:
x = [1 1; 2 4; 2 5; 2 6; 7 7];
costMat = inf(max(x(:,1)), max(x(:,2)));
for itEdge = 1:size(x, 1)
costMat(x(itEdge, 1), x(itEdge, 2)) = 0;
[assignment,cost] = munkres(costMat);
matchedEdges = sum(assignment>0)
Here, we should see three matched edges instead of two. When changing the edge costs to 1 instead of 0, it works
PS: Your old version works fine.


Val (view profile)

Warning, this is highly buggy when working with zero-weight edges. Example:
x = [1 1; 2 4; 2 5; 2 6; 7 7];
costMat = inf(max(x(:,1)), max(x(:,2)));
for itEdge = 1:size(x, 1)
costMat(x(itEdge, 1), x(itEdge, 2)) = 0;
[assignment,cost] = munkres(costMat);
matchedEdges = sum(assignment>0)

Here, we should see three matched edges instead of two. When changing the edge costs to 1 instead of 0, it works.

Rui Wu

Rui Wu (view profile)

Rupert Thomas


David (view profile)

Hi people, who know the maximal number this file can deal with?

Roy Sung

Prakhar Sinha

be careful it gives completely false results for 40*40 matrix !


Matteo (view profile)

Hi people,
I a question about the hungarian algorithm. this algorithm is optimal algorithm for the assignment problem, and the time complexity is O(n^3), right?
But , if the input is the multidimensional matrix, it's possible to use the hungarian algorithm? how does it change the algorithm and the time complexity ?


Kui (view profile)

Note that only after I posted did I see the discussion about JIT and profiler results... I got my numbers by looking at the profiler and not with tic/toc or external timer. Perhaps the profiler is misleading?


Kui (view profile)

I have found a code optimization for the outerplus function using bsxfun. The following code is equivalent but about an order of magnitude faster:

function [minval,rIdx,cIdx]=outerplus(M,x,y)

xPlusYc = bsxfun(@plus, x, y);
M = M - xPlusYc;
minval = min(M(:));
[rIdx, cIdx] = find(M == minval);

For my 60x60 test cases, this reduces the runtime of the overall algorithm by about 30%.

Alexander Farley

kindly help me by writing short and generalized code for assignment method of scheduling...

Shengjie Guo

I was wrong about the above comment. It completes, but takes a very long time. For instance, it took 287s for a 100 x 983 matrix. This matrix was from my specific problem. All the weights were real and there were no zero weights.

On the other hand, on the same system it completes a 400 x 400 random in 4 seconds.

Could the specific values in the input matrix cause such a drastic difference in performance?

I checked

>> munkres([1 0])

ans =

1 0

Could you fix this problem?


Syed (view profile)

I am thankful to you for pointing this out as I was comparing your method with another algorithm and it was giving output columnwise.

Yi Cao

Yi Cao (view profile)

Well, I can see what you try to do is to increase the cost of selected assignment then to find next best assignment. However, you made a wrong change. The assignment results in dicated row 1 assigned with colume 3, but you miss understood as column 1 assigned with row 3. Wish this helps.


Syed (view profile)

Thanks for a nice implementation
I used the code for following matrix
I have the following matrix
67 98 65 95 79 82
50 40 91 66 57 35
61 74 85 112 39 79
41 63 72 97 39 56
56 55 83 91 59 50
66 34 98 70 69 54

gives me:
Result= 3 4 5 1 6 2

I change the Matrix to
67 98 65 205 79 82
50 40 91 66 57 145
171 74 85 112 39 79
41 173 72 97 39 56
56 55 193 91 59 50
66 34 98 70 179 54

gives me:
Result= 3 4 5 1 6 2
cost is also same in both cases.
could u please tell me the reason.

Ricky Wong

Hi, Yi. Thanks for your effort.
However, the function as below is missing
taht prevents me from using your code on Matlab...

bsxfun(@minus, dMat, minR)



DQ LIU (view profile)

Great work, Thanks a lot~~
BTW, I compared your implementation with Niclas Borlin's implementation ( The results are not the same. Yours seems better. Given his is a classic implementation, do you know why? or did you did some improvements over the classic ones?


Yi Cao

Yi Cao (view profile)


Thanks for pointing out the JV algorithm. You may wish to know that I have implemented a Matlab version of the JV algorithm in
Certainly, it will not be as fast as the mex code, but it has no limitation on the cost matrix to be integer. If you test it, please let me know how it works with your application.


Thanks... You might be right about the memory part. Although I was cleaning the workspace, I was using a MEX code which might have a memory leak (dont know if this is possible).

I ended up using the Jonker-Volgenant shortest augmenting path algorithm from ''. This solves the same problem and appeared way faster.

Yi Cao

Yi Cao (view profile)

Hi Anoop Balan,

Thanks for commenting on the code. The algorithm is polynomial. This means the computation time will growth about n^p with n the size of the problem and p some constant depending on CPU speed, memory and software implementation. On my PC, Intel Core2 Quad CPU (Q9300) 2.5 GHz with 4 GB RAM, with XP and Matlab 2009b, I got the following results:

n = 1000, cpu time = 18 sec
n = 2000, cpu time = 150 sec
n = 4000, cpu time = 1216 sec

Another machin

n = 1000, cpu time = 128 sec
n = 2000, cpu time = 1024 sec

So roughly, p = 3 (2^3 = 8). I do not know why your machin has so large p (>5). When you solve a large size problem, make sure you start with a clean workspace to avoid any unnecessary overhead for swaping memory with hard disk.


Thanks a lot for for sharing this code!

This works very well for sizes around 1000x1000 (around 35 secs). However, for a 1500x1500 problem it takes around 260 secs and for a 2000x2000 problem it takes around 1670 seconds.

My problem sets are of size 3000x3000 - 4000x4000. Would you know of any approximation algorithms that work well with such sizes?


James (view profile)

V. Poor

Oliver Woodford


John D'Errico

John D'Errico (view profile)


Yi Cao

Yi Cao (view profile)

Oh, yes. It was my mistake. A=-PROFIT gives the maximum of sum(PROFIT), but A=1./PROFIT results in the maximum of 1/sum(1./PROFIT), which is different from sum(PROFIT). Sorry for this.

ek de

ek de (view profile)

Hi Yi,

Using COST = 1./PROFIT doesn't seem to give the same results as with COST = -PROFIT. Try out the code below. In some cases the solutions result with different profits.
for i = 1:100
m = ceil(rand*20)+1;
n = ceil(rand*20)+1;
a = rand(m,n)+eps;
[assign1 cost1] = munkres(-a);
[assign2 cost2] = munkres(1./a);
if ~all(assign1 == assign2)
disp('Different assignments');
if ~all(sort(assign1) == sort(assign2))
disp('Assignment vectors do not agree on permutations');
assign1=assign1(assign1~=0);m1 = length(assign1);
assign2=assign2(assign2~=0);m2 = length(assign2);
if m1 ~= m2
disp('Assignment Vecs not compatible');
cost1 = sum(sum((a.*accumarray([(1:m1)' assign1'],ones(m1,1),[m n]))));
cost2 = sum(sum((a.*accumarray([(1:m1)' assign2'],ones(m1,1),[m n]))));
disp(sprintf('Cost difference = %f',abs(cost1 - cost2)/cost1));

Yi Cao

Yi Cao (view profile)

Yes, the code works with negative cost. You can either use negative cost or use reciprocal, i.e. COST = 1./PROFIT if you do not have zero profit elements.

ek de

ek de (view profile)

Hi Yi,

Will this work if I want to solve the maximum weight matching? That is, if we have a profit matrix rather than a cost, and we want to maximize the profit rather than minimize the cost. I assume that negating the cost matrix should work but was wondering if you could confirm this.
BTW, this is really fast!

Hey Yi
that was a fast bugfix for your fast algorithm.
switched back to it.
Thank you!

Yi Cao

Yi Cao (view profile)

Thanks Immanuel. The bug has been fixed.

Hi Yi, I used your algorithm and it did good work until I encountered a possible bug.
As long as the input matrix ist big enough it works fine, but when the matrix consists of only 2 entries the algorithm creates a wrong match. For example the matrix [1 0], the matching should be 1->2, a return of [2], but the algorithm returns 1->1 ([1]).
Caused by lack of time I switched for the moment to another algorithm, but it would be nice if you could post a bugfix.
In addition, I use it on R2006b, I can't test it on newer versions, so maybe this is related to the old one.

search search

cool, really fast. Thank you.
I test it using Matlab 2007a, will 2008a faster?

Yi Cao

Hi, Laszlo:

Thanks for comments.
The outerplus function uses JIT acceleration. If you use an old version Matlab, which does not support JIT, then you have to convert it to a vectorized version. Note, if you use profile view, this function will always be slower than vectorized. One way to do vectorization is as you described. Alternatively, you can use bsxfun, which should be faster than repmat.

Laszlo Sragner

Hi, good stuff, but it is much slower on my machine. It seems outerplus() is the culprit.
Is it possible that this implementation has some 64bit issues?

This helped me a bit:

function [minval,rIdx,cIdx]=outerplus3(M,x,y)



a bug fixed


The new version implements particial assignment if a full assignment is not feasible.


Update to improve efficiency further.


Bug fix

MATLAB Release
MATLAB 7.12 (R2011a)

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