Why NaN even though all elements considered

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Abhishek H P
Abhishek H P on 25 May 2016
Edited: Stephen23 on 25 May 2016
I am comparing every column with other columns but getting the answer in R(4,2) and R(3,1)
for i=1:nrow-1
a=data(1,i);
c=data(2,i);
e=data(3,i);
g=data(4,i);
for j=[1:i-1,i+1:nrow]
b=a-data(1,j);
d=c-data(2,j);
f=e-data(3,j);
h=g-data(4,j);
kU=b'*b;
kL=d'*d;
pU=f'*f;
pL=h'*h;
S=(kU'*kU+kL'*kL)/(kU+kL)+(pU'*pU+pL'*pL)/(pU+pL);
R(i,j) = S;
R(j,i) = S;
end
end
end
trial(A)
ans =
0 1.8299 NaN 1.8299
1.8299 0 1.8299 NaN
NaN 1.8299 0 1.8299
1.8299 NaN 1.8299 0
when the input
A = 2.0000 2.2000 2.0000 2.2000
3.0000 3.1000 3.0000 3.1000
2.0000 2.2000 2.0000 2.2000
3.0000 3.1000 3.0000 3.1000
  2 Comments
Stephen23
Stephen23 on 25 May 2016
Edited: Stephen23 on 25 May 2016
Your algorithm divides zero by zero. This happens for these pairs of (i,j): (1,3), (2,4), (3,1). What do you expect division of zero by zero to do?
Roger Stafford
Roger Stafford on 25 May 2016
Actually it is zero divided by zero that is producing the NaNs here. A non-zero divided by zero will produce inf or -inf,

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