Need to test all permutations

CD
2019 4 6
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更新时间：2019 4 6
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Say I have the following:
A = [1 2 3], B = [1 2], C = [4],
D = C + B*A
I need D to contain all permutations of C+B*A:
D1 = 4+1*1
D2 = 4+1*2
D3 = 4+1*3
D4 = 4+2*1
D5 = 4+2*2
D6 = 4+2*3
What is this called and what tools are available to perform this on bigger equations?

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Star Strider
Star Strider 2019-4-6

0 个投票

One approach is to use the ndgrid (link) function:
A = [1 2 3];
B = [1 2];
C = [4];
[p1,p2,p3] = ndgrid(A,B,C);
P = [p3(:),p2(:),p1(:)]; % Components
D = p3(:) + p2(:).*p1(:)
The ‘P’ assignment simply displays the components used in the ‘D’ calculation. It is not otherwise necessary for the code.
I cannot guarantee that this will easily scale to other problems. It works here.

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CD
CD 2019-4-6
Star Strider,
Very nice example.
How would one expand/tweak your method to work with a numerator and denominator:
A = [1 2 3];
B = [1 2];
C = [4];
Num = C+A*B;
Den = C+A*B;
D = Num/Den;
D should be a column of all '1's.
Star Strider
Star Strider 2019-4-6
And it is:
Num = p3(:) + p1(:).*p2(:);
Den = p3(:) + p1(:).*p2(:);
D = Num./Den
D =
1
1
1
1
1
1
CD
CD 2019-4-6
Star Strider,
Ok, the equation that I need to solve is pretty complicated.
Can your method be applied or do you recommend something else?
fs = 100000;
Ts = 1/fs;
Vref = 5;
Vo = Vref;
Vg = [22 32]
D = [Vo/Vg(1) Vo/Vg(2)]
Dp = 1-D
LL = 100e-6;
L = [(LL-LL*.1) LL (LL*.1+LL)]
Io = [2 4];
R = [(Vo/Io(1)) (Vo/Io(2))]
M1 = [(Vg(1)-Vo)/L(1) (Vg(1)-Vo)/L(2) (Vg(1)-Vo)/L(3) (Vg(2)-Vo)/L(1) (Vg(2)-Vo)/L(2) (Vg(2)-Vo)/L(3)]
M2 = [Vo/L(1) Vo/L(2) Vo/L(3)]
Ma = M2(2)
% Not real equations, numbers within "[]" represent row or element length
% Example M1 contains 6 elements, M2 contains 2 elements, together Fm should be a 6*3 or 18 element deep column
Fm = [1/((Ma+((M1[6]-M2[3])/2))*Ts);
% Ggo should be a 2*3*6*3*18*2*2 element deep column, I'm looking for the max value in this column
Gg0 = D[2]*(((2*Ma-M2[3])/(2*Ma+M1[6]-M2[3]))/(1+(Fm[18]*Vo)/(D[2]*R[2])));
Star Strider
Star Strider 2019-4-6
I can’t understand your notation. My method might work, although I can’t guarantee it. I’m not certain how to construct the matrices for that calculation, since I don’t know how many would be required.
It might be best to use nested loops (or repeated bsxfun calls) for that, especially since the dimensions of the variables in the calculation are significantly different.
Star Strider
Star Strider 2019-4-6
编辑：Star Strider 2019-4-6
Craig Dekker’s Answer moved to this Comment:
Thank you.
I'll try breaking the equation into several nested loops.
Star Strider
Star Strider 2019-4-6
As always, my pleasure.
That may not be absolutely necessary.
I simply don’t understand your notation well enough to figure out how to use the ndgrid approach with it.
Experiment with something like this:
Dv1 = [1 2];
M2v1 = [3 4 5];
M1v1 = [6 7 8 9 10 11];
Fmv1 = (12:12+17);
Rv1 = [30 31];
[p1,p2,p3,p4,p5] = ndgrid(Dv1,M2v1,M1v1,Fmv1,Rv1);
Rv = p5(:);
Fmv = p4(:);
M1v = p3(:);
M2v = p2(:);
Dv = p1(:);
It seems that it might work, although I can’t guarantee it. I may also have missed something, since this code produces a series of (1296x1) vectors, while:
veclen = 2*3*6*3*18*2*2;
would have 7776 elements.
If the ndgrid approach works, it would definitely be faster than nested for loops. The challenge then comes in reshaping the output vector into a matrix that is meaningful in terms of the argument vectors. Appropriately indexing the result is the principal reason I would err on the side of the loops.

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