How to solve the following exercise?

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I have to solve the following problem:
Consider a matrix M with only the numbers 1 to 9 as elements.
M =
2 9 3 2 4
8 6 4 8 5
5 7 1 6 4
9 8 9 5 1
Consider one of the elements M(i,j) that's not on the edge of the matrix. Such element always has 8 neighbours. If M(i,j) > 1 and each number from 1 to M(i,j)-1 is one of the 8 neighbours, we say that element is neighboring. If M(i,j) = 1, the element is automatically neighboring.
For example, M(2,2) is neighboring because 1,2,3,4 and 5 are one of the element's neighbours. M(3,4) on the other hand isn't neighboring because 2 and 3 don't occur around the element.
Now, I have to write a function that has 3 inputs: a matrix M and a row- and column index. The function has to control whether de element is neighboring or not and has a logical 0 or 1 as output.
Guillaume on 14 Jan 2020
@Ellen, you've got the correct algorithm. As you suspec and Bob pointed out, the construction of n can be done in just one line with simple indexing.
@Bob, no the code also works for 1. m would be empty, so ismember will return empty. sum(empty) is 0 which is also the length of empty.
For the record, the one-liner I was talking about is
res = all(ismember(1:matrix(row, col)-1, matrix(row-1:row+1, col-1:col+1)))
Ellen De Jonghe
Ellen De Jonghe on 14 Jan 2020
I indeed forgot to check the case where M(i,j) = 1, I will try to figure that out.

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Accepted Answer

Andrei Bobrov
Andrei Bobrov on 14 Jan 2020
Edited: Andrei Bobrov on 14 Jan 2020
function out = find_neighbor(M,i,j)
out = all(ismember(1:M(i,j)-1,M(i-1:i+1,j-1:j+1)));
Guillaume on 14 Jan 2020
The if is not needed, out is true anyway, if M(i,j) is 1, since all([]) is true.
Andrei Bobrov
Andrei Bobrov on 14 Jan 2020
Thanks Guillaume! I'm fixed.

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More Answers (1)

Ellen De Jonghe
Ellen De Jonghe on 14 Jan 2020
That's a very easy one! Thanks.
I seem to always make my scripts much longer than it has to.


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