How to match matrix elements for a condition?

I am working on App Designer. I have a matrix and want to match the elements 3 by 3 with a condition. The condition is that addition of 3 elements must be near at a value.

I already can match them by making the addition from minimum to maximum, but it does not seem to be optimal. So, I want them to be around a value.

[Dyuman Joshi]

Use tolerance to compare -

in = [3 3.3 3.6 3.9];
check = 3.5;
tol = 0.3;

out = abs(in-check)<tol

out = 1x4 logical array
   0   1   1   0

[piston_pim_offset]

I'll try right now

Answer 1

[Torsten]

If A becomes larger, this brute-force way of solving will become intractable.

A = [15 25 36 17 48 59 31 64 18 21 97 84 31 64 15];
target = 100;
C = nchoosek(1:numel(A),3);
P = arrayfun(@(i)sum(A(C(i,1:3))),1:size(C,1));
[~,I] = sort(abs(P-target));
sums = arrayfun(@(i)sum(A(C(I(i),:))),1:numel(I));
sums = sums(sums==sums(1));
n = numel(sums);
result = unique(sort(A(C(I(1:n),:)),2),'rows')

result = 2x3

15 21 64 21 31 48

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[piston_pim_offset]

That's it. Thank you all in advance!

Answer 2

[Hassaan]

Assuming a 1-dimensional array. Will check for sums of three consecutive elements that are close to a given value (target_sum) within a tolerance (tol).

% Example matrix (1D array)
A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

% Target sum to match
target_sum = 15;

% Tolerance for matching sum
tol = 0.5;

% Length of the array
n = length(A);

% Pre-allocate the logical output array for matched cases
matches = false(1, n-2);  % (n-2) because the last two elements can't start a triplet

% Iterate over the array to find matching triplets
for i = 1:n-2
    % Calculate the sum of the current and next two elements
    current_sum = sum(A(i:i+2));
    
    % Check if the current sum is within the tolerance of the target sum
    if abs(current_sum - target_sum) <= tol
        matches(i) = true;
    end
end

% Print matched triplets and their indices
matched_indices = find(matches);
for idx = matched_indices
    fprintf('Match found at indices [%d, %d, %d]: [%d, %d, %d]\n', idx, idx+1, idx+2, A(idx), A(idx+1), A(idx+2));
end

Match found at indices [4, 5, 6]: [4, 5, 6]

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[piston_pim_offset]

Your code does not include all possible combinations. Better to use nchoosek(A(:),3) .

[Dyuman Joshi]

How is the addition of elements to be done?

say the matrix is -

[1 2 3;
 4 5 6;
 7 8 9]

Which combinations are to be considered here?

[piston_pim_offset]

@Dyuman Joshi matrix is 1 by N (1xN), and all possible additions are considered. Then the results are sorted with the condition given as being nearest to 100.

For example,

A is 1x15 matrix as : A = [15 25 36 17 48 59 31 64 18 21 97 84 31 64 15]

I need all possible triple additions that meet the condition, and l need them sorted from the nearest to farther to 100.

[Torsten]

A = [15 25 36 17 48 59 31 64 18 21 97 84 31 64 15];
C = nchoosek(1:numel(A),3);
P = arrayfun(@(i)sum(A(C(i,1:3))),1:size(C,1));
[B,I] = sort(abs(P-100));
A(C(I(1),:))  % One of the six best combinations

ans = 1x3

15 64 21

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arrayfun(@(i)sum(A(C(I(i),:))),1:numel(I))

ans = 1x455

100 100 100 100 100 100 99 99 101 99 101 99 99 99 102 102 98 102 102 98 97 97 97 97 97 103 97 103 97 104

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[piston_pim_offset]

@Torsten that's almost what I was looking for. Is there a way to eliminate the emenents used for the next iteration?