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From: Felipe Uribe on 12 Apr 2010 23:43
I'm programming an optimization algorithm, for this I generate a series of binary numbers by vectors with the following command:
x=round(rand(1,10))
x=
0 1 1 0 1 0 0 1 1 1

Then convert it to decimal using the command:
y=binvec2dec(x)
y=
918

By this command I always get positive integers, but I need to evaluate all kinds of numbers in my cost function... My question is, how do I make when converting the binary vector, get numbers with decimals and negatives numbers?
From: Walter Roberson on 13 Apr 2010 00:22
Felipe Uribe wrote:
> I'm programming an optimization algorithm, for this I generate a series
> of binary numbers by vectors with the following command:
> x=round(rand(1,10))
> x=
> 0 1 1 0 1 0 0 1 1 1
>
> Then convert it to decimal using the command:
> y=binvec2dec(x)
> y=
> 918
>
> By this command I always get positive integers, but I need to evaluate
> all kinds of numbers in my cost function... My question is, how do I
> make when converting the binary vector, get numbers with decimals and
> negatives numbers?

WordSize = 32; %or 64
x = rand(1,WordSize) >= 0.5; %binary vector
xd = sum(x .^ (WordSize-1:-1:0));
if WordSize == 32
y = typecast(uint32(xd),'single');
else
y = typecast(uint64(xd),'double');
end

y will end up as a signed single precision floating point number if
WordSize is 32, and as a signed double precision floating point number
otherwise.

Be warned that this procedure can potentially generate a representation
for +infinity, for -infinity, or for one of the NaN (Not A Number).

Also be warned that not very many of the numbers so generated will be
close to 1 -- e.g., only about 1 in 2^12 will be between 1 and 2.
From: James Tursa on 13 Apr 2010 02:42
Walter Roberson <roberson(a)hushmail.com> wrote in message <hq0rhm$nds$1(a)canopus.cc.umanitoba.ca>...
> Felipe Uribe wrote:
> > I'm programming an optimization algorithm, for this I generate a series
> > of binary numbers by vectors with the following command:
> > x=round(rand(1,10))
> > x=
> > 0 1 1 0 1 0 0 1 1 1
> >
> > Then convert it to decimal using the command:
> > y=binvec2dec(x)
> > y=
> > 918
> >
> > By this command I always get positive integers, but I need to evaluate
> > all kinds of numbers in my cost function... My question is, how do I
> > make when converting the binary vector, get numbers with decimals and
> > negatives numbers?
>
> WordSize = 32; %or 64
> x = rand(1,WordSize) >= 0.5; %binary vector

At this point x is a bunch of 0's and 1's

> xd = sum(x .^ (WordSize-1:-1:0));

So what is the point of raising the 0's and 1's to a power? You're going to get the same 0's and 1's you started with. Looks to me like the max value you are going to get for xd is 32 (or 64 respectively). I think you meant something like sum(x.*(2.^(etc))), at least for the 32 case. But this scheme breaks down for 64-bit floating point case because you will lose precision in the sum (low order bits get lost).

> if WordSize == 32
> y = typecast(uint32(xd),'single');
> else
> y = typecast(uint64(xd),'double');
> end

So I don't think your typecast is going to do what you expect, particularly for the 64-double case.

I think we need to get more clarification from OP as to what kind of numbers he wants as a result. What range? Uniformly distributed? Random integers or floating point numbers? etc.

James Tursa
From: Walter Roberson on 13 Apr 2010 10:37
James Tursa wrote:
> I think
> you meant something like sum(x.*(2.^(etc))), at least for the 32 case.
> But this scheme breaks down for 64-bit floating point case because you
> will lose precision in the sum (low order bits get lost).

You are right about both issues. I was posting after my bed-time again :-(

Both issues can be fixed by using bitset to create the uint* number
instead of using sum() and casting it to uint* afterwards.
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