All possible combination based on 2^n but with 1 and -1

Question

BeeTiaw 2021-5-14

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评论： Dyuman Joshi 2021-5-14

Hi all, thank you for those of you who have answered my question below.

https://uk.mathworks.com/matlabcentral/answers/829758-all-possible-combination-based-on-2-n

I have a slightly different question but still does not know how to achieve this.

Again, I want to create a matrix containing all possible combination. Example of this is shown below. The size of the matrix depends on the number of variable n and the total of combination should follow the . The example below is valid for

and, hence, the total number of rows is 8. The value of each element is 1 and -1.

How to create this matrix automatically depending on the number of variable n?

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Jan 2021-5-14

Your example contain [1,1,1] twice.

BeeTiaw 2021-5-14

编辑：BeeTiaw 2021-5-14

Yes. It is correct. That is the desired output.

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Answer 1

Dyuman Joshi 2021-5-14

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y=dec2bin([7 4 2 1])-'0';
y(y==0)=-1;
z =[y; flipud(y)]

This only works for this particular example. If you want a generalised answer, give more examples.

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BeeTiaw 2021-5-14

I am still trying to understand why we put [7 4 2 1].

Dyuman Joshi 2021-5-14

Because only this combination corresponds to the desired result.

That's why I mentioned - "This only works for this particular example. If you want a generalised answer, give more examples"

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Answer 2

Daniel Pollard 2021-5-14

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You could take the answer from your previous question, subtract 0.5 and multiply by 2. Your accepted answer was

n = 3;
m = dec2bin(0:pow2(n)-1)-'0' % limited precision
m = 8×3
   0     0
   0     1
   1     0
   1     1
   0     0
   0     1
   1     0
   1     1

which then becomes

n = 3;
m = dec2bin(0:pow2(n)-1)-'0'; % limited precision
m = 2*(m-0.5)
m = 8×3
    -1    -1    -1
    -1    -1     1
    -1     1    -1
    -1     1     1
     1    -1    -1
     1    -1     1
     1     1    -1
     1     1     1

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Answer 3

Jan 2021-5-14

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编辑：Jan 2021-5-14

在 MATLAB Online 中打开

dec2bin creates a CHAR vector, while -'0' converts it to a double again. This indirection costs some time. The direct approach:

n = 3;
m = rem(floor((0:2^n-1).' ./ 2 .^ (0:n-1)), 2)
m = 8×3
     0     0     0
     1     0     0
     0     1     0
     1     1     0
     0     0     1
     1     0     1
     0     1     1
     1     1     1
pool   = [1, -1];      % Arbitrary values
result = pool(m + 1)   % Add 1 to use m as index
result = 8×3
     1     1     1
    -1     1     1
     1    -1     1
    -1    -1     1
     1     1    -1
    -1     1    -1
     1    -1    -1
    -1    -1    -1