How to do this while loop

[Walter Roberson]

What should the loop do if no whatevers have absolute value <= 1 ?

I cannot tell what it is that the absolute value is being taken for.

Is it correct that you expect your while loop to take only one trip and the whole algorithm would stop immediately because it would detect that abs(A(1)-B(1)) is <= 1 and that single entry is all you need?

[Robin Li]

False if no abs value <=1

Yea. I want the while loop to stop and spit out TRUE as soon as it finds out abs(A(1)-B(1)) is <= 1

[Robin Li]

And also, not only abs(A(1)-B(1)) is <= 1; it can also abs(A(2)-B(1)) is <= 1. Are there easy ways to do so?

[Walter Roberson]

Are you to produce a vector the same size as A? A vector the same size as B? Are you permitted to use vectorized operations at all?

Answer 1

[Amit]

We can do it in this way :

A = [1 2 3]; B = [2 4 5];
LenA = 1;
while LenA <= length(A)
    if(A(LenA)-B)<=1
        out = true;
        break
    else
        out = false;
    end
    LenA =LenA+1;
end
disp(out)

Hope it helps.

[Walter Roberson]

We discourage people from prividing complete solutions to homework problems.

Also, this code does not impliment the original requirement to use abs().

if(A(LenA)-B)<=1

A(LenA) would be a scalar, but B is a vector, so A(LenA)-B would be a vector, and the result of comparing to 1 would be a vector. if statements are only considered true if all elements of the array of values being tested are non-zero. The above code is therefore looking to see whether there is some element of A that is within 1 of every element of B. It is not at all clear to me that is what the homework is intended to test.