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Bruno Luong
on 8 Sep 2019

I correct the flaw of dpb method

U=randn(10,1);

V=randn(10,1);

ps=U.'*V;

a=sqrt(abs(ps));

U=(sign(ps)/a)*U;

V=(1/a)*V;

dpb
on 8 Sep 2019

>> uv=randn(1000,2);

>> uv=uv/sqrt(sum(uv(:,1).*uv(:,2)));

>> uv(:,1).'*uv(:,2)

ans =

1.0000

>> mean(uv)

ans =

0.0025 0.0091

>>

dpb
on 8 Sep 2019

Bruno Luong
on 8 Sep 2019

the sqrt() might be imaginary (1/2 chance)

>> uv=randn(10,2); uv=uv/sqrt(sum(uv(:,1).*uv(:,2)))

uv =

0.0000 + 0.3856i 0.0000 + 0.0703i

0.0000 + 0.9468i 0.0000 - 1.0239i

0.0000 + 0.0416i 0.0000 - 0.9650i

0.0000 + 0.1552i 0.0000 + 0.0931i

0.0000 - 0.1422i 0.0000 + 0.4418i

0.0000 - 0.7345i 0.0000 + 1.1256i

0.0000 - 0.4265i 0.0000 - 0.4973i

0.0000 - 0.9666i 0.0000 - 0.0169i

0.0000 + 0.5398i 0.0000 + 0.8718i

0.0000 + 0.6400i 0.0000 + 0.2477i

John D'Errico
on 8 Sep 2019

Normally distributed, with zero mean. You don't say if the sample mean should be zero, or the population mean zero. Normal samples from randn have a zero popiulation mean. But the sample mean just has an expected value of zero. A subtle difference that few people seem to appreciate when they lack the proper background.

Anyway, it is easy enough to scale them so the dot product is also 1. I could keep them as separate vectors, since that seems to have confused you, but you need to LEARN TO USE MATRICES.

uv = randn(1000,2);

dp = uv(:,1)' * uv(:,2);

uv = uv/sqrt(abs(dp));

if dp < 0

uv(:,1) = -uv(:,1);

end

As you can see, the dot product is 1. Note that the code I gave you will not end up 50% of the time with an imaginary square root.

uv(:,1)'*uv(:,2)

ans =

1

mean(uv)

ans =

-0.00248985528111534 -0.00172138844051432

The sample means are not identically zero, but you were not specific there, and that is simply repaired.

Do you really think you need disrinctly named vectors? LEARN TO USE MATLAB! The mat in MATLAB stand for matrix.

u = uv(:,1);

v = uv(:,2);

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