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

3 次查看(过去 30 天)
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?

回答(3 个)

Dyuman Joshi
Dyuman Joshi 2021-5-14
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 个评论
Dyuman Joshi
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"

请先登录,再进行评论。


Daniel Pollard
Daniel Pollard 2021-5-14
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 2021-5-14
编辑:Jan 2021-5-14
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

类别

Help CenterFile Exchange 中查找有关 Resizing and Reshaping Matrices 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by