File Exchange

image thumbnail

permn(V, N, K)

version 6.1 (5.94 KB) by

Permutations with repetition, all or a subset

4.58333
34 Ratings

136 Downloads

Updated

View License

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.
 
    See also perms, nchoosek
             allcomb, permpos on the File Exchange

Comments and Ratings (47)

Farinaz

Matthew Daunt

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

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);

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

GALO ALBUJA

Triveni

Please help me to solve it

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

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)

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 :-)

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

%instead of:
%[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

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

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

Manu

Manu (view profile)

Excellent and fast!

Brad Ridder

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

John D'Errico (view profile)

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

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!

Ohad N

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.

Updates

6.1

spelling corrections

6.1

Incorporated the functionality of permnsub, allowing for returning a subset rather than all permutations as well.

5.1

Renamed file into PERMN, fixed small bug, extended help section

1.3

Reference to COMBNSUB for large combinatorial problems.

1.2

corrected to give column vector output for N=1. (error pointed out by Wilson via email).

1.1

modified slightly based on suggestions by Jan Simon (thanks!)

arggg updated with older version ...

new (very fast) algorithm

new faster algorithm

MATLAB Release
MATLAB 7 (R14)

Download apps, toolboxes, and other File Exchange content using Add-On Explorer in MATLAB.

» Watch video