min function on two arrays

I've read the documentation on the min function, but still don't understand how it works on two arrays. I have the following:

dS=zeros(1,N)

dS=min(cA(i+1,:),cB(i+1,:))

where cA and cB are equally sized arrays. Doesn't the min function just take the value at equivalent locations in cA or cB that is the lowest. So if position 10,50 in cA is 5 and the same position in cB is 3, min returns 3 in the resulting row vector, dS, at column 50. Is that correct?

[James Tursa]

Yes. Are you getting a result you don't expect?

[Robert Demyanovich]

No, I'm just a MatLab beginner trying to understand some complex (for me, anyway) code.

What if dS had been defined as an array? I guess the result is the same because in both cA and cB, the i+1 row has been specified.

[James Tursa]

This line creates dS as a vector:

dS=zeros(1,N)

Then this line completely overwrites the dS you just created and instead assigns dS the result of the min( ) function call:

dS=min(cA(i+1,:),cB(i+1,:))

I.e., the first line is completely useless and accomplishes nothing because it gets overwritten in the second line.

And yes, cA(i:1,:) and cB(i+1,:) will be a row vectors if i is a scalar.

Answer 1

[SALAH ALRABEEI]

    % 
  dS=min([cA(i+1,:),cB(i+1,:)])

[Walter Roberson]

The above would take the single overall minimum of the two rows i+1 of cA and i+1 of cB . cA(i+1,:) is a row and so is cB(i+1,:), and [] of two rows creates a combined row, so the min() is going to be applied to a vector; it would take the minimum of that resulting vector.

[SALAH ALRABEEI]

Oops, yes it must be semicol; Thanks.

    % correction
  dS=min([cA(i+1,:);cB(i+1,:)])

[Walter Roberson]

That would work in the case that cA and cB have the same number of columns.

The original code

dS=min(cA(i+1,:),cB(i+1,:))

would also work if one of cA or cB had a single column and the other one did not.