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

3 views (last 30 days)
Hi all, thank you for those of you who have answered my question below.
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?
  2 Comments
BeeTiaw
BeeTiaw on 14 May 2021
Edited: BeeTiaw on 14 May 2021
Yes. It is correct. That is the desired output.

Sign in to comment.

Answers (3)

Dyuman Joshi
Dyuman Joshi on 14 May 2021
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.
  2 Comments
BeeTiaw
BeeTiaw on 14 May 2021
I am still trying to understand why we put [7 4 2 1].
Dyuman Joshi
Dyuman Joshi on 14 May 2021
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"

Sign in to comment.


Daniel Pollard
Daniel Pollard on 14 May 2021
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 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 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

Jan
Jan on 14 May 2021
Edited: Jan on 14 May 2021
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
  1 Comment
BeeTiaw
BeeTiaw on 14 May 2021
This is not creating the desired output as per the figure. The combination is different.

Sign in to comment.

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!