Comments on: All combinations from vectors

By: Zoran Zegarac

Zoran Zegarac — Mon, 29 Nov 2010 07:50:01 +0000


Yes, Yuri what is pp?
I do not know if this would work with my impdance matrix problem. 
There are 3 varibles, and each can have a value from 1 to 3,
and all possible sets of 2 of this variables are:
11
22
33
12
13
21
31
23
32
All in all,  3*3= 9, which is easy to do manually.

Now, doing the same as above, but now with  6 variables, each one  can have value form 1 to 9, making a total  of
9*9*9*9*9*9=531441.
Has anybody got and idea, please?

Zoran

By: eliza

eliza — Fri, 23 Apr 2010 03:09:25 +0000

Yury, What is PP?

By: Yury

Yury — Thu, 25 Mar 2010 11:42:14 +0000

Let NN=[2 1 2] (each element is number of combinations) and we need matrix of all combinations such as:
1 1 1
1 1 2
2 1 1
2 1 2
Here is solution:

for i=1:prod(NN)
  t=i-1;
  for j=length(PP):-1:1
    M(i,j)=mod(t,NN(j))+1;
    t = fix(t/NN(j));
  end
end

By: Doug

Doug — Thu, 25 Feb 2010 20:57:36 +0000

@Marina,

What you are asking for is not permutation exactly. I am not sure that what you are asking for has a specific name in statistics, so you will have to write your own MATLAB code to solve this.

It should not be too hard. I am assuming none of the N groups that hold M objects is to be empty, and that every object is in a group. If so, you can calculate how many different ways to group things without perms.m

Doug

By: Marina Zahari

Marina Zahari — Tue, 23 Feb 2010 23:27:02 +0000

I have tried to use perms.m for my problem but it didn’t work. Suppose that we have 4 objects, A, B, C, and D. And I would like to assign them to 3 unique labels/ groupings – 1,2 or 3. So all possible unrepetitve combinations will be:

1 1 2 3 (meaning A and B in group 1, C in group 2 and D in group 3)
1 2 1 3
1 2 3 1
2 1 1 3
2 1 3 1
2 3 1 1.

perms.m gave me the extras,
1 1 3 2, 1 3 1 2, 1 3 2 1, 3 1 1 2, 3 1 2 1 and 3 2 1 1 which I do not need. For example 1 1 3 2 still assign A ann B in group 1.

If you could point me in the right direction, I would greatly appreciate it.

Thank you
Marina

By: Mithat Perköz

Mithat Perköz — Tue, 02 Jun 2009 10:27:54 +0000

Thank you very much, I was just looking for an algorithm like this!

By: Doug

Doug — Tue, 28 Oct 2008 17:15:41 +0000

I think you need to look at the problem as such. Each element of the vector can reside in exactly one of three bins.

3 * 3 * 3 * 3 or numBins ^ numElements

-Doug

By: Amanda

Amanda — Mon, 27 Oct 2008 18:40:40 +0000

Doug,

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” 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! 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!

Thank you,

Amanda Palazzo