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Find all combinaisons of sum of two vector that corresponds to the value of a third vector

Asked by anthony dall'agnol on 18 Jun 2019
Latest activity Edited by Stephen Cobeldick on 18 Jun 2019
Hi community !
I have three vectors A, B and C.
A = [1561.28, 1561.64, 1562.00, 1562.36, 1562.72]; % signal
B = [1064.00, 1064.36, 1064.72, 1065.08, 1065.44]; % pump
C = [632.89, 633.02, 633.14, 633.27, 633.40]; % converted
I want to check all the possible combinaisons of A and B (see next function) that correspond to a value of C
func = 1./(1./A + 1./B) >> C
For example a correct anwser in the result vector would be [3 1 1] for :
1/( 1/A(3) + 1/B(1) ) = C(1)
I've already check some functions like nchoosecrit() but it only take one scalar as an result argument so I don't know.

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3 Answers

Answer by Matt J
on 18 Jun 2019
Edited by Matt J
on 18 Jun 2019
 Accepted Answer

tolerance=1e-6;
[A,B,C]=deal(A(:),B(:),C(:)); %Make everything a column vector
[Ar,Br,Cr]=deal(A,B.' , reshape(C,1,1,[]) );
lidx = abs( 1./Ar + 1./Br.' - 1./Cr ) <= tolerance ;
[i,j,k]=ind2sub(size(lidx), find(lidx));
locations=[i,j,k],
combinations=[A(i),B(j),C(k)]

  3 Comments

This is nicer indeed thank you, is there any chances to have the C value corresponding to each combinations ? Something like :
[i,j,k] = ..
combinations=[A(i),B(j),C(k)];

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Answer by Adam
on 18 Jun 2019

Well, what you wrote:
1./(1./A + 1./B)
will give you answers you can compare to C as e.g.
abs( 1./(1./A + 1./B) - C )
to give an absolute difference of each from C which you can then test against some threshold to see of they are close enough to equal.

  1 Comment

I don't see how doing this
1./(1./A + 1./B)
is going to give me all combinaisons of what I want. Because this operation will just do
for ii = 1:end
someVector = 1/(1/A(ii) + 1/B(ii));
end
And I want something more like
for ii = 1:end
for jj = 1:end
someVector = 1/(1/A(ii) + 1/B(jj));
end
end
and at the end be able to have the index of the value that respect the function. Ideally I would like something like :
[value, index] = nchoosecrit(someVector, @(x,y) function(x,y) == someResult);

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Answer by Stephen Cobeldick on 18 Jun 2019
Edited by Stephen Cobeldick on 18 Jun 2019

Simpler to use ndgrid:
>> A = [1561.28, 1561.64, 1562.00, 1562.36, 1562.72]; % signal
>> B = [1064.00, 1064.36, 1064.72, 1065.08, 1065.44]; % pump
>> C = [632.89, 633.02, 633.14, 633.27, 633.40]; % converted
>> [Aa,Ba,Ca] = ndgrid(A,B,C);
>> idx = abs(1./(1./Aa + 1./Ba) - Ca) < 0.01; % adjust tolerance to suit data.
>> [X,Y,Z] = ind2sub(size(idx),find(idx));
>> [X,Y,Z] % indices corresponding to A,B,C:
ans =
3 1 1
1 2 1
3 2 2
1 3 2
5 2 3
3 3 3
5 3 4
3 4 4
3 5 5
And checking (e.g. the last row):
>> 1/( 1/A(3) + 1/B(5) )
ans = 633.40
>> C(5)
ans = 633.40

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