MATLAB code for Hamacher sum - matlab

Question

The hamacher sum is :
I wrote the following MATLAB code for the above function
function f=hamachersum(x,y)
f = zeros(numel(x),1);
for j=1:numel(x)
if x(j)==1 && y(j)==1
f(j,1)=1;
else
f(j,1)=(x(j)+y(j)-2*(x(j)*y(j)))/(1-(x(j)*y(j)));
end
end
end
Then I want to test t3=hamachersum(t1,t2)
My input values t1, t2 are
t1
t1 =
1.0000
0
1.0000
1.0000
1.0000
1.0000
NaN
0.8167
1.0000
1.0000
1.0000
0.4667
NaN
1.0000
1.0000
1.0000
NaN
NaN
1.0000
1.0000
1.0000
NaN
0.0250
1.0000
t2
t2 =
1.0000
0.5524
1.0000
1.0000
1.0000
1.0000
NaN
0
1.0000
1.0000
1.0000
1.0000
NaN
1.0000
1.0000
1.0000
NaN
NaN
0.6032
1.0000
1.0000
NaN
0.9973
0.7260
The result is
t3 =
1.2000
0.5524
1.0000
1.0000
1.0000
1.0000
NaN
0.8167
2.0000
2.0000
1.0000
1.0000
NaN
0.6667
1.0000
1.0769
NaN
NaN
1.0000
1.0000
1.0000
NaN
0.9973
1.0000
Why do I get values above 1. As this is a fuzzy operator it can't have values above 1.
Is there something wrong in my code?

Answer 1

I am probably answering this against my better judgement.
Caveat: I am not familiar with the Hamacher sum so my approach to answering this is strictly based on the equation in your question.
Is there something wrong in my code?
Your code produces 1.0 when I run it on MATLAB R2014a for inputs of 1.0 and 1.0, I'm presuming that is correct as you have an explicit condition for it. I cant produce the results you are are seeing in your question.
However, I felt compelled to provide a more efficient implementation of the equation
function h = hamachersum(mu_a, mu_b)
h = (mu_a + mu_b - (2 .* mu_a .* mu_b)) ./ (1 - mu_a .* mu_b);
% h(isnan(h)) = 1.0; % Included this line to show you how to remove NaN
end
Note: I've included % h(isnan(h)) = 1.0; to show you how to handle cases when mu_a and mu_b are both 1.0 as you have explicitly handled this in your question (rather poorly might I add).
Comparing floating point numbers is not reliable even in MATLAB and could be part of the reason why you are receiving the results you are. A better way to check what the value of a floating point number is would be to use
if (x - 1.0 > 1e-15)
fprintf(1, "x == 1.0");
else
fprintf(1, "x ~= 1.0");
end
If x is equal to 1.0 down to machine precision this expression will be true otherwise it will be false.