PERMPOS

all possible permutations of M values in N positions

Description

PERMPOS - all possible permutations of M values in N positions
    B = PERMPOS(M,N) where M and N are non-negative integers produces a
    logical (N!/M!(N-M)!)-by-N matrix in which each row contains a unique
    permation of M trues and (N-M) falses. Note that each column will hence
    contain (N-1) trues.
    
    B = PERMPOS(V,N) where V is a vector of length M produces a matrix
    where each row a contains the values of V in preserved order, but
    uniquely permuted at the N columns. The remaining positions are zero.
 
    Examples:
      permpos(2,4) % ->
       % 1 1 0 0
       % 1 0 1 0
       % 1 0 0 1
       % 0 1 1 0
       % 0 1 0 1
       % 0 0 1 1
 
      permpos([-4 9 3],4) % >
       % -4 9 3 0
       % -4 9 0 3
       % -4 0 9 3
       % 0 -4 9 3
 
    See also NCHOOSEK, PERMS, RANDPERM, TRUE, FALSE
    and COMBN, ALLCOMB, BALLATSQ, NONES (Matlab File Exchange)

Version 2.1 (march 2007)

• MATLAB release: MATLAB 6.5 (R13)

[Dimitri Shvorob]

Trying to permute a (20x1) vector:
??? Error using ==> permpos
This function is not usefull for more than 14 entries
Why impose the limit? And why not disclose it upfront?

[Jos (the author)]

Dimitri, you're right. This is a somewhat harsh, undocumented and unnecessary restriction. I will update the file shortly.

[brad mcevoy]

does exactly what i'm looking for. good job and thanks.

[Berta Neto]

It does exactly what I was needing.
Thanks

[ahmad ababnah]

This code is so good.

[Amanda]

I've been struggling with a matlab problem of combinations and permutations for a while and came across your post here. I thought you might be able to help me out.

Let's say I have a vector say A =[1 2 3 4] and want to find every way I can choose all the elements of A to fit in some number of bins (let's say 3). Bin 1 could have (1, 2, 3), Bin 2 could have (4) or (0) and Bin 3 could have (4) or (0) as well. The problem isn't quite solved by "nchoosek" or "permpos" because each bin could have more or less elements than the previous bin and the element choice for each bin depends on the elements not already chosen. I know that the number of possible combinations is # of elements in A^#of bins (in this case 81), because I solved it out by hand. I am looking for an easier way to do this! I'm sorry if this sounds confusing!

If you can point me in the right direction or offer me any help I would greatly appreciate it!

[Jos (10584)]

Amanda, could you contact me directly? See any of my recent submissions for an address. Jos

[Anders]

Exactly what is needed when finding syndromes for the corresponding bit error patterns when making a decoder for block codes

[Richard Crozier]

Very useful, thanks.

[Nima Moshtagh]