# random symmetric logical matrix

조회 수: 2 (최근 30일)
Raviteja 2011년 9월 12일
댓글: John D'Errico 2018년 5월 21일
How to generate a random logical symmetric matrix?
(binary symmetric matrix)

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### 채택된 답변

Andrei Bobrov 2011년 9월 12일
a = triu(rand(5));
out = a + tril(a',-1)>.5
variant 2
a = rand(5)
b = a+a'
out = b >= mean(b(:))
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Derek O'Connor 2011년 9월 12일
I like variant 3 even better. Concise but not obscure. Excellent!
Walter Roberson 2011년 9월 12일
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 .

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### 추가 답변 (1개)

kika198513 2018년 5월 21일
Hi all! Do you have an idea about how to generate a random symmetric logical matrix having a fixed number of 1s over each row and each column? The final matrix should be huge (60000x60000) with very few 1s (8 within each row and each column). I cannot apply the idea above for two reasons: the huge size of the matrix and the fact that in "out = a + tril(a',-1)>.5" I cannot control the final number of 1s at the levels of rows and columns. Many thanks!
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John D'Errico 2018년 5월 21일

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