hard one

tricky

very challenging

function flag = isTherePythagoreanTriple(a, b, c, d)
flag = false;
com_bin = nchoosek([a b c d],3);
com_bin = com_bin(:,1) .^ 2 + com_bin(:,2) .^ 2 == com_bin(:,3) .^ 2;
flag = logical(sum(com_bin));
end

M=[a,b,c,d]
c=nchoosek(M,3);
c=c(sum(c(:,1:2).^2,2)-c(:,3).^2==0,:);
if isempty(c)
flag = false;
else
flag = true;
end

function flag = isTherePythagoreanTriple(a, b, c, d)
flag = false;
com_bin = nchoosek([a b c d],3);
com_bin = com_bin(:,1) .^ 2 + com_bin(:,2) .^ 2 == com_bin(:,3) .^ 2;
flag = logical(sum(com_bin));
end

This is not a valid solution to this puzzle!!

This is a smart solution that is not cheating the system.

Logical enough!

smart!

Great work!

Lol saying great work on your own solution :P

Great work :)

Haha that's how we roll!

why the leading solution size is always 10, how could that possible?

Yeah, I know this is cheating, taking advantage of the limited test suite. I just wanted to see what the other small scores were about.

i need to change this

hardcore brute force

Should work for any case and doesn't contain str2num or regexp :)

Fifth test cannot be true because 3.5^2 = 12.25
and no integer after squaring and adding will give such result

how can this be a leading solution?? Are all the cody problems like this?

I don't understand this solution. Can you explain it ?

Still without str2num or regexp

this would fail on assert(isequal(isTherePythagoreanTriple(1,4,7,9),false))

The benefit of a limited test suite.

Without using str2num or regexp