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## permn(V, N, K)

version 6.1 (5.94 KB) by

Permutations with repetition, all or a subset

4.59459
35 Ratings

Updated

permn - permutations with repetition
Using two input variables V and N, M = permn(V,N) returns all
permutations of N elements taken from the vector V, with repetitions.
V can be any type of array (numbers, cells etc.) and M will be of the
same type as V. If V is empty or N is 0, M will be empty. M has the
size numel(V).^N-by-N.

When only a subset of these permutations is needed, you can call permn
with 3 input variables: M = permn(V,N,K) returns only the K-ths
permutations. The output is the same as M = permn(V,N) ; M = M(K,:),
but it avoids memory issues that may occur when there are too many
combinations. This is particulary useful when you only need a few
permutations at a given time. If V or K is empty, or N is zero, M will
be empty. M has the size numel(K)-by-N.

[M, I] = permn(...) also returns an index matrix I so that M = V(I).

Examples:
M = permn([1 2 3],2) % returns the 9-by-2 matrix:
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3

M = permn([99 7],4) % returns the 16-by-4 matrix:
99 99 99 99
99 99 99 7
99 99 7 99
99 99 7 7
...
7 7 7 99
7 7 7 7

M = permn({'hello!' 1:3},2) % returns the 4-by-2 cell array
'hello!' 'hello!'
'hello!' [1x3 double]
[1x3 double] 'hello!'
[1x3 double] [1x3 double]

V = 11:15, N = 3, K = [2 124 21 99]
M = permn(V, N, K) % returns the 4-by-3 matrix:
% 11 11 12
% 15 15 14
% 11 15 11
% 14 15 14
% which are the 2nd, 124th, 21st and 99th permutations
% Check with permn using two inputs
M2 = permn(V,N) ; isequal(M2(K,:),M)
% Note that M2 is a 125-by-3 matrix

% permn can be used generate a binary table, as in
B = permn([0 1],5)

NB Matrix sizes increases exponentially at rate (n^N)*N.

allcomb, permpos on the File Exchange

mahmoud afifi

Farinaz

Matthew Daunt

Farrukh Pervez

### Farrukh Pervez (view profile)

A very quick algorithm to find the permutations. A great work

Ceyhun KIRIMLI

### Ceyhun KIRIMLI (view profile)

or as a full function as follows;

function result=seqrep(V,N)
l=length(V);
a=zeros(l^N,N);
for i=1:N
a(:,i)=reshape(reshape(repmat((1:l),1,l^(N-1)),l^i,l^(N-i))',l^N,1);
end
result=V(a);

Ceyhun KIRIMLI

### Ceyhun KIRIMLI (view profile)

i guess this can be done using only 1 for loop as shown below;

for i=1:n
a(:,i)=reshape(reshape(repmat((1:l),1,l^(n-1)),l^i,l^(n-i))',l^n,1);
end

where n= number of elements (as N here) and
l= length of V

Mohsen Mansouryar

GALO ALBUJA

### GALO ALBUJA (view profile)

guglielmo fortuni

Triveni

### Triveni (view profile)

A=[30 30 30 30 30 30 60 60 60 60 60 60 60 60 30 30 30 30 30 30];

[c,b] = hist(A,unique(A));
N= length(A);
x = permn(b,N); %function permn

I have matrix A. i want to sort x. i need c(1,1) no of angle of 30. and c(1,2) no of angle of 60. means i need 12 nos of 30 and 8 nos of 60 in every permutations. How can i do it...please help me. Just like perm command works. but it's limited to 10. i need 20.

I also have another program but it's only works with 0 and 1.

A=[1 1 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1];

n2 = numel(A);

b2 = nnz(A);

M2 = 0:ones(1,n2)*pow2(n2-1:-1:0)';

z2 = rem(floor(M2(:)*pow2(1-n2:0)),2);

out = z2(sum(z2,2) == b2,:);

Jos (10584)

### Jos (10584) (view profile)

On second thoughts, I change the name as well. Thanks!

Jos (10584)

### Jos (10584) (view profile)

I know, Stephen. I decided to keep the original name for backward compatibility. Note that this function is more than 10 years old and still regularly downloaded.
I will update the description though!

Stephen Cobeldick

### Stephen Cobeldick (view profile)

This submission is misnamed:

* If the order doesn't matter, it is a Combination.

* If the order does matter it is a Permutation.

It is clear from the example that this submission generates what is sometimes called "permutations with repetitions":

http://en.wikipedia.org/wiki/Permutation

If this submission truly calculated the combinations then the sets {0,0,1}, {0,1,0}, and {1,0,0} are all equivalent, and would not all appear in the output shown in the example.

http://en.wikipedia.org/wiki/Combination

http://mathworld.wolfram.com/Combination.html

or for a simpler introduction:

https://www.mathsisfun.com/combinatorics/combinations-permutations.html

Qi An

### Qi An (view profile)

saved me sometime!

ted p teng

### ted p teng (view profile)

Patrizio Manganiello

### Patrizio Manganiello (view profile)

Thank you very much!

mya

### mya (view profile)

verry useful, thank you

Han

### Han (view profile)

Repetitive combination, great!

Jos (10584)

### Jos (10584) (view profile)

@Marc, thanks for your interest and suggestions for improvement. Yet, for code compatibility (and fairness) you need to implement to inverse order step as well :-)

Marc Lalancette

### Marc Lalancette (view profile)

Nice code, thanks! Just for fun, slight to medium speed improvement (sorry my k = your N):

%[Y{k:-1:1}] = ndgrid(X) ;
%Y = reshape(cat(k+1,Y{:}),[],k) ;
%fill Y directly:
nX = numel(X);
Y(nX^k, k) = X(1); % Initialize to give shape and data type.
s(1:k-1) = {ones(1, nX)};
for i = 1:k
Y( ((i-1)*nX^k + 1) : (i*nX^k) ) = X(s{1:i-1}, :, s{i:k-1});
X = permute(X, [1:i-1, i+1, i]);
end
% This has inverse order for the combinations as you had, easy to change.

Jos (10584)

### Jos (10584) (view profile)

@ Talaria, this is impossible as it would require a memory of (at least) 150*(2^150) bits which is about 2700000000000000000000000000000000 Terabyte ...

However, it is unlikely that you need all those combinations at once; perhaps you can think of another approach to your problem, for instance, by drawing a few random combinations at a time?

Talaria

### Talaria (view profile)

works great, but is there a way to beat the limitation? for example i want to be able to return the following:
M = COMBN([0 1],150)
is this possible?

Owen Brimijoin

### Owen Brimijoin (view profile)

The core of this function is a lovely little computational gem. Thank you.

Manu

### Manu (view profile)

Excellent and fast!

Excellent, just what I needed. Thanks!

Fabio

### Fabio (view profile)

Does anyone have an idea on how to create a function like "combn" but where the generated matrix contains only monotone rows? I could do it manually by removing non-monotone rows after having built the matrix, but when the size of the matrix becomes large i got out of memory error. Any idea?

John D'Errico

Oleg Komarov

### Oleg Komarov (view profile)

Can't find a better suited H1-line. Neat help, simple example, history (maybe more details on the dates...) and past algorithms too.
I don't use combinatorics very much but I always wonder why matlab doesn't cover it esplicitly. A must have.

Oleg

Jan Simon

### Jan Simon (view profile)

Nice and comapct program! H1-line, help, history, comments in the code, efficient. Thanks!

But can be made still a little bit faster:
X = repmat({X},1,N) ;
[X{1:N}] = ndgrid(X{:}) ;
X = fliplr(reshape(cat(N+1,X{:}),[],N)) ;
1. If X is filled in reverse order with [X{N:-1:1}], FLIPLR can be omitted => 25% faster for COMBN(1:10, 5) (see my comment for ALLCOMB)
2. Inside NDGRID the same operations are performed for each input, but all inputs are equal. Inlining and simplyfying NDGRID leads to (REPMAT can be omitted then):
C = cell(1, N);
X = X(:);
L = length(X);
s(1:N - 1) = L;
X = reshape(X(:, ones(1, prod(s))), [L, s]);
C{N} = X;
for iC = 2:N
C{N - iC + 1} = permute(X, [2:iC, 1, iC + 1:N]);
end
X = reshape(cat(N+1, C{:}), [], N);
=> 40-50% faster.
To my surprise it is *slower* to create the output array at once and insert the permuted subarrays directly.

Philip L

Carlos Baiz

### Carlos Baiz (view profile)

Thanks!

jose caceres

any idea of how to get the combinations without repetition of previous combinations, and without re-using the elements in a a selection?

Riccardo Bevilacqua

Exactly what I was looking for!

Antonio Silva

Thanks a million!

Does the job...

Jos the author

Roger Stafford pointed out that, due to the IEEE 754 standard, the floor of a floating point (as used in COMBN's algorithm) may lead to faulty results for very specific inputs.
The update incorporates his excellent solution to this potential problem. Thanks Roger!
The update (COMBN version 3.2) should be up soon ...

Kaijun Wang

It is very useful, thanks !

Wolfgang Zieroth

Fast and (almost) exactly what I needed!

... I only need those combinations which sum to (say) one. I can, of course, first create all combinations with this program and then find(m*ones(N,1)==1). Yet I bounce very quickly against maximum variable size for Matlab. Any Suggestions?

ShenKae Wu

Thanks a lot

Vangelis R

I am very surprised this is not a build in function. You saved my day thanks a lot!

B E

Thanks! Exactly what I needed.

Peter Navé

What I needed!

Jürgen Womser-Schütz

Very fast! Very helpful to solve my actual problem.

David Torres

Great tool. Easy to use.

Richard Vance

Easy to use, fast, and suitable for incorporating into other programs. Good documentation well organized. It really helped me solve some problems.

yakir gagnon

Jos vdG

Jamie, use COMBN(UNIQUE(V),N)

Jamie Baier

Does what it says, but there really should be an option to treat elements as non-unique.