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random symmetric logical matrix

Asked by Raviteja

Raviteja (view profile)

on 12 Sep 2011

How to generate a random logical symmetric matrix?

(binary symmetric matrix)



Raviteja (view profile)



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1 Answer

Answer by Andrei Bobrov

Andrei Bobrov (view profile)

on 12 Sep 2011
Accepted answer
a = triu(rand(5));
out = a + tril(a',-1)>.5

variant 2

a = rand(5)
b = a+a'
out = b >= mean(b(:))


Andrei Bobrov

Andrei Bobrov (view profile)

on 12 Sep 2011

Hi Derek! Thanks.
Another variant:
a = rand(5)
out = (a+a')/2>.5

Derek O'Connor

Derek O'Connor (view profile)

on 12 Sep 2011

I like variant 3 even better. Concise but not obscure. Excellent!

Walter Roberson

Walter Roberson (view profile)

on 12 Sep 2011

I don't see why the mean() version would work without bias. Suppose that all of the random numbers generated were less than 1/4, so even after adding transpose each total was less than 1/2. Clearly the output for such a matrix should be complete logical 0s. But unless all of the values generated were identical, at least one of the sums is going to be greater than the mean, so if one compares to mean(b(:)) one would get logical 1 in that position. (And if all of the values _are_ identical then they would all be equal to the mean to within roundoff and so the returned result would either be all 0 or all 1 depending which way the mean() rounded.)

It seems to me, then, that for variant 2, instead of comparing to mean(b(:)), one should be comparing to 1

a = rand(5);
b = a + a';
out = b >= 1;

Which can then be made more concise as

a = rand(5);
out = (a+a') >= 1;

Note: the uniform distributions all generate (2^b - 2) different possible numbers, with b=53 for all the generators except one legacy generator that has b=24 . The possible numbers are spaced 2^(-b) apart. The generators cannot generate exactly 0 or exactly 1. If you do a quick counting test, you will see that 0.5 exactly is the first number of the second half of the count, so instead of comparing for > 0.5, you should be comparing for >= 0.5 .

Andrei Bobrov

Andrei Bobrov (view profile)

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