Write a function that finds all the Three Pythagoras less than n?

function result = checkPyth(n)
y=n;
for i =1:1:length(y)
    for j=1:1:length(y)
        for k=1:1:length(y)
        while (y(i)).^2+(y(j)).^2==(y(k)).^2
            fprintf('true\n')
            return
        end
        while (y(i)).^2+(y(j)).^2~=(y(k)).^2
            fprintf('false\n')
            return
        end
    end
    end
end

This is the function I have so far, but it keeps returning false.

This function will first check if there exists numbers less than n such that i^2+j^2=k^2 with i,j,k<=n

Answer 1

[Steven Lord]

Hint: how many times do these two loops run their loop bodies, and for which values?

y = 5;
disp('k1')
for k1 = 1:1:length(y)
    disp(k1)
end

disp('k2')
for k2 = 1:1:y
    disp(k2)
end

Do you want your loops to work like the k1 loop or the k2 loop?

[Mairav Simon]

function output= findPyth(n)

for a=1:n;

for b=1:n;

for c=1:n;

a<=b<=c;

if c==sqrt(a.^2 +b.^2);

disp([a b c])

end

end

end

end

end

so now i have this, which works, but it gives me a lot of doubles. How do I tell the code I want [3 4 5] and [4 3 5] to be equal

[Steven Lord]

Ideally you'd detect [3 4 5] when a = 3 and b = 4. So when a = 4, do you even want to check the case b = 3?

The "limits" of an inner loop variable can use the current value of an outer loop variable(s).

Z = zeros(5);
for row = 1:5
    for col = row:5
        Z(row, col) = 1;
    end
end
disp(Z)

[John D'Errico]

You can even go a little further, aince there are NO Pythagorean triples where a == b, since then a^2 + b^2 = 2*a^2, and we know that 2*a^2 can never be a perfect square itself. Essentially that just means there are no isosceles Pythagorean triangles. So the inner loop can start at row+1.

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