How to match matrix elements for a condition?

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piston_pim_offset on 22 Apr 2024
Commented: piston_pim_offset on 24 Apr 2024
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 on 22 Apr 2024
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 on 22 Apr 2024
I'll try right now

Torsten on 23 Apr 2024
Edited: Torsten on 23 Apr 2024
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 on 24 Apr 2024
That's it. Thank you all in advance!

Hassaan on 22 Apr 2024
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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Torsten on 23 Apr 2024
Edited: Torsten on 23 Apr 2024
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 on 23 Apr 2024
@Torsten that's almost what I was looking for. Is there a way to eliminate the emenents used for the next iteration?