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Iterate through binary vector

Posted on 2008-10-19
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Last Modified: 2013-11-25
Hi,

I've got a binary vector N long and I want to be able to loop through all possible values of the vector and do some processing on it.
For example if N = 3;

001
Process
010
Process
100
Process
101
Process
011
Process
110
Process
111
Process

But would like to be able to do this in some sort of loop construction with N. So can then scale up to bigger sized vectors!

Any suggestions?
James
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Question by:James_h1023
8 Comments
 
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Expert Comment

by:Frosty555
ID: 22753387
I feel that a good direction to go with this is to instead use an N-dimensional matrix, instead of a single column vector filled with values. Fill each dimension of the matrix with alternating zeros and ones and then loop through each entry in the matrix in the correct way to iterate every possibility.

It would be easy to fill every alternative entry. To do it with a simple vector you can do something like:

x = zeros(50,1);
x(1:2:length(x)) = x(1:2:length(x)) + 1;

But I'm shaky on my matlab and I don't know if this is really the right way to do it. Some food for thought though.
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Author Comment

by:James_h1023
ID: 22756146
Hi, thanks for a quick response.
Not quite sure I understand what your suggesting.
Can you clarify?

James
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Expert Comment

by:Frosty555
ID: 22771904
Ah nevermind that's a stupid way of doing it. You only need a two dimensional matrix. What what I babbling about before?

You'll need to define a n by 2^n matrix using the zeros operator:

x = zeros(2^n, n);

Now, each row is a potential vector for you to analyze.

You need to fill in the columns following this pattern (for a 4-bit example)

0000
1000
0100
1100
0010
1010
0110
1110
0001
1001
0101
1101
0011
1011
0111
1111

Notice how the first column is just 0, 1, 0, 1. The second column is 0, 0, 1, 1, 0, 0, 1, 1. The third column is 0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1 etc. The number of consecutive numbers multiplies by two every time. What does this look like? It looks to me just like a square wave, or a clock signal.

The first column is just a clock signal with a period of 2
The second column is a clock signal with a period of 4
The third column is a clock signal with a period of 8
... etc.

You can also, conveniently, use matrix operations in MatLab to set the value of an entire column in one fell swoop, e.g.

x(:, 1) = [a clock signal with period 2]
x(:, 2) = [a clock signal with period 4]
x(:, 3) = [a clock signal with period 8]

etc.

All we need to do now is figure out how to synthesize that "clock signal" in matlab, which SURELY must be an easy thing to do. In fact, there's even a function called "square()" that is ALMOST what you want:

http://www.mathworks.com/access/helpdesk/help/toolbox/signal/index.html?/access/helpdesk/help/toolbox/signal/square.html

But it needs some tweaking to get anything other than a period of 2 to work. I'll keep hunting but this should put you on the right track. If you implement it this way, your function will not only be effective, and short, but also blazingly fast since you used almost entirely matrix operations to do it.
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Author Comment

by:James_h1023
ID: 22771918
Absolutely wonderful, I shall implement this when I have some time and have a look at the link :)

Thanks
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Expert Comment

by:Frosty555
ID: 22771936
Now, the square function isn't quite right. It just generates 0,1,0,1,0,1. The period is 2. Always.

You'll run into trouble generating clock cycles with periods other than 2, but I'm sure there's some trickery you can do to puff out a matrix that looks like [0,1,0,1] into something like [0,0,0,1,1,1,0,0,0,1,1,1].
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Accepted Solution

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Frosty555 earned 200 total points
ID: 22771998
Our only problem now is generating "clock" functions with period N. You might want to create a whole other matlab function to do just this, call it myclock() or something.

It might be possible to do it by creating an N by N matrix, the first row filled with all zeros, then the second filled with all ones, then all zeros etc. Then call "reshape" to turn that N x N matrix into an N² x 1 matrix.

E.g. create a matrix that looks like this:

0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1

Using subsequent calls to the zeros() and ones() function:

x = zeros(n,n)
x(1,:) = zeros(1,n)
x(2,:) = ones(2,n)
x(3,:) = zeros(3,n)
x(4,:) = ones(4,n)
.... etc up until column "n". Use a FOR loop to accomplish this.

And then call reshape() on that matrix:

e.g.
   reshape(x,[],1)

will result in a column vector:

0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1

wrap all that stuff into a function and use it to create the "clock" vectors we need for the first part of this problem.
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Expert Comment

by:yuk99
ID: 23601292
I propose a solution using binary representation of integers. It works up to N=54, what will be 2^54-1 ~1.8e16 possible vectors.
You can get all the vectors with dec2bin(1:2^N-1)-48 (easy to get out of memory). 48 is ascii code for 0.

N=3;
for i=1:2^N-1
    dec2bin(i,N)-48
end

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Expert Comment

by:WillDC
ID: 38967888
To create a matrix M of size R by C, with R indexed from zero to exp(2,C)-1, and which includes all possible binary vectors of size C, you can set M[r,c] equal to zero if ((r mod period(c)) < period(c)/2) evaluates to TRUE, and 1 otherwise. Period(c) is just 2 for the first column, 4 for the second, 8 for the third, 16, etc.
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