Dear MathWorks Community,
I have written a short script that generates a matrix containing the lengths of the sides of right triangles, where all lengths are integers (e.g. the 3-4-5-triangle). After generation, the matrix is sorted by rows, so that any triangles with the same legs are removed (e.g. 3-4-5 is considered equal to 4-3-5), sides of the triangle are ordered by descending length (e.g. 3-4-5 becomes 5-4-3), and the rows are ordered in ascending order, by first column. See code-snippet # 2 below. Code-snippet # 2 is executed before code-snippet # 1.
My question is: How can I remove any triangles that are congruent from the resulting list? Preferably by removing the larger congruent triangles and keeping the smaller (e.g. remove 6-8-10 and keep 3-4-5).
I have tried doing the following, with limited succes: (code-snippet # 1)
for l = 1 : 1 : size(K1,1) % Go through l from 1 to the number of rows in matrix K1, in steps of 1.
for m = 1 : 1 : size(K1,1)% Go through m from 1 to the number of rows in matrix K1, in steps of 1.
leg1L = K1(l,2);% Variable leg1L is the 1st leg of the triangle described in K1 on row l.
leg2L = K1(l,3);% Variable leg2L is the 2nd leg of the triangle described in K1 on row l.
leg1M = K1(m,2);% Variable leg1M is the 1st leg of the triangle described in K1 on row m.
leg2M = K1(m,3);% Variable leg2M is the 2nd leg of the triangle described in K1 on row m.
frac1 = leg1L/leg1M;% Variable frac1 is:
% the 1st leg of the triangle described in K1 on row l,
% divided by
% 1st leg of the triangle described in K1 on row m.
frac2 = leg2L/leg2M;% Variable frac2 is:
% the 2nd leg of the triangle described in K1 on row l,
% divided by % the 2nd leg of the triangle described in K1 on row m.
if ( (l ~= m) && (frac1 == int64(frac1)) && (frac2 == int64(frac2)) )% IF l is not equal to m AND frac1 is an integer AND frac2 is an integer.
K2(l,:) = [0 0 0];% Row l of K2 equals zeroes.
K2(m,:) = [0 0 0];% and row m of K2 equals zeroes.
elseif ( (l ~= m) && (frac1 ~= int64(frac1)) && (frac2 ~= int64(frac2)) )% IF l is not equal to m AND frac1 is not an integer AND frac2 is not an integer.
K2(m,:) = K1(m,:);% Row m of K2 equals row m of K1.
end endendIt removes 3-4-5, 5-12-13 and 75-100-125, which is correct, but not entirely the desired result, as the first two of these sets are smaller than e.g. 6-8-10 and 10-24-26. It also fails to remove many congruent triangles, e.g. 12-16-20, which is simply 2*(6-8-10).
The code for generating, sorting and ordering the matrix is as follows, code-snippet # 2:
K = zeros(1e4,3);% Matrix for combination is allocated, to speed up for-loops.
k = 0;% Loop-counter is allocated.
for b = 1:1:100% Go through b from 1 to 100 in steps of 1.
for c = 1:1:100% Go through c from 1 to 100 in steps of 1.
a = sqrt(b^2 + c^2);% a is defined as the squareroot of the sum of the squares of b and c.
if ( a == int64(a) )% If a equals the integer of a
k = k+1;% Add 1 to the loop-counter k
K(k,:)=[a, b, c];% And write a, b and c to the k'th line of matrix K.
end endend K = K(any(K,2),:);% Remove any zero-rows of matrix K.
K = sortrows(K,1);% Sort rows of matrix K by descending order after the values in the first
% column (the hypotenuse).
K1 = (zeros(size(K)));% Allocate a zero-matrix the size of K called K1, for manipulating data
% from K.
% The purpose is to remove any copy-rows, where c and by have simple
% swapped places, so that 5 = sqrt(3^2 + 4^2) will be seen as equal to
% 5 = sqrt(4^2 + 3^2).
for i = 1 : 1 : size(K,1)% Go through i from 1 to the number of rows in matrix K, in steps of 1.
vec = K(i,:);% The vector vec equals the i'th row of matrix K.
vec = sort(vec,'descend');% The vector vec is sorted in descending order.
K1(i,:) = vec;% The i'th row of K1 equals vec. The loop continues to the next number,
% until it has gone through all of the rows of matrix K.
end K1 = unique(K1,'rows');% Any rows in matrix K1 that are identical, are now removed.
The entire script can be found in the attachment. Note that it is slightly different; in this post I have removed some non-essential code for clarity.
Thank you for your help, and may you have a Happy New Year!
Best Answer