How to do this while loop
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4 个评论
Walter Roberson
2020-3-5
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
2020-3-5
Robin Li
2020-3-5
Walter Roberson
2020-3-5
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?
采纳的回答
Amit
2020-3-5
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.
1 个评论
Walter Roberson
2020-3-5
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.
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