How to solve the following exercise?

5 vues (au cours des 30 derniers jours)
Ellen De Jonghe
Ellen De Jonghe le 14 Jan 2020
Modifié(e) : Andrei Bobrov le 14 Jan 2020
Hye!
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.
  9 commentaires
Guillaume
Guillaume le 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 le 14 Jan 2020
Thanks!
I indeed forgot to check the case where M(i,j) = 1, I will try to figure that out.

Connectez-vous pour commenter.

Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 14 Jan 2020
Modifié(e) : Andrei Bobrov le 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)));
end
  2 commentaires
Guillaume
Guillaume le 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 le 14 Jan 2020
Thanks Guillaume! I'm fixed.

Connectez-vous pour commenter.

Plus de réponses (1)

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

Catégories

En savoir plus sur Matrix Indexing dans Help Center et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by