Efficiently generating unique pairs of integers

Easy, peasy, when viewed in the proper way.

You wish to generate n pairs of integers, [p,q], such that p and q lie in the interval [1,m], and p

How many possible pairs are there? The total number of pairs is just m*(m-1)/2. (I.e., the sum of the numbers from 1 to m-1.)

So we could generate n random integers in the range [1,m*(m-1)/2]. Randperm does this nicely. (Older matlab releases do not allow the second argument to randperm.)

k = randperm(m/2*(m-1),n);

(Note that I've written this expression with m in a funny way, dividing by 2 in perhaps a strange place. This avoids precision problems for some values of m near the upper limits.)

Now, if we associate each possible pair [p,q] with one of the integers in k, we can work backwards, from the integers generated in k, to a pair [p,q]. Thus the first few pairs in that list are:

{[1,2], [1,3], [2,3], [1,4], [2,4], [3,4], ..., [m-1,m]}

We can think of them as the elements in a strictly upper triangular array of size m by m, thus those elements above the main diagonal.

q = floor(sqrt(8*(k-1) + 1)/2 + 1/2);
p = k - q.*(q-1)/2;

See that these formulas recover p and q from the unrolled elements in k. We can convince ourselves that this does indeed work, but perhaps a simple way here is just this test:

k = 1:21;
q = floor(sqrt(8*(k-1) + 1)/2 + 3/2);
p = k - (q-1).*(q-2)/2;
[k;p;q]'

ans =
     1     1     2
     2     1     3
     3     2     3
     4     1     4
     5     2     4
     6     3     4
     7     1     5
     8     2     5
     9     3     5
    10     4     5
    11     1     6
    12     2     6
    13     3     6
    14     4     6
    15     5     6
    16     1     7
    17     2     7
    18     3     7
    19     4     7
    20     5     7
    21     6     7

Another way of testing it is to show that all pairs get generated for a small case.

m = 5;
n = 10;
k = randperm(m/2*(m-1),n);
q = floor(sqrt(8*(k-1) + 1)/2 + 3/2);
p = k - (q-1).*(q-2)/2;

sortrows([p;q]',[2 1])
ans =
     1     2
     1     3
     2     3
     1     4
     2     4
     3     4
     1     5
     2     5
     3     5
     4     5

Yup, it looks like everything works perfectly. Now try it for some large numbers for m and n to test the time used.

tic
m = 1e6;
n = 100000;
k = randperm(m/2*(m-1),n);
q = floor(sqrt(8*(k-1) + 1)/2 + 3/2);
p = k - (q-1).*(q-2)/2;
toc

Elapsed time is 0.014689 seconds.

This scheme will work for m as large as roughly 1e8, before it fails due to precision errors in double precision. The exact limit should be m no larger than 134217728 before m/2*(m-1) exceeds 2^53. A nice feature is that no rejection for repeat pairs need be done.

The following code does what you need:

n = 10000;
m = 500;
my_list = unique(sort(round(rand(n,2)*m),2),'rows');
my_list = my_list(find((my_list(:,1)==my_list(:,2))==0),:);
%temp = my_list;    %In case you want to check what you initially generated.
while(size(my_list,1)~=n)
    %my_list = unique([my_list;sort(round(rand(1,2)*m),2)],'rows');
    %Changed as per @jucestain's suggestion.
    my_list = unique([my_list;sort(round(rand((n-size(my_list,1)),2)*m),2)],'rows');
    my_list = my_list(find((my_list(:,1)==my_list(:,2))==0),:);
end

This is more of a general approach rather then a matlab solution.

How about you do the following first you fill a vector like the following.

x[n] = rand()
x[n + 1] = x[n] + rand() %% where rand can be equal to 0.

Then you do the following again

x[n][y] = x[n][y] + rand() + 1

And if

x[n] == x[n+1]

You would make sure that the same pair is not already selected.

After you are done you can run a permutation algorithm on the matrix if you want them to be randomly spaced.

This approach will give you all the possibility or 2 integer pairs, and it runs in O(n) where n is the height of the matrix.