How to solve the following exercise?

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.

[darova]

Please show your attempts

[Ellen De Jonghe]

That's the problem, I ain't got one. I just don't see it :).

[darova]

I don't want to do all your work for you. No one is gonna pay me for this

[Bob Thompson]

What are the steps that you think you need to accomplish conceptually?

How familiar are you with MATLAB?

[Guillaume]

That's the problem, I ain't got one. I just don't see it :).

First step: write down the algorithm?

• How would you list the 8 neighbours
• Once you've listed the neighbours, how would you check that they make up the set from 1 to M(i, j)?

Depending on which functions you're allowed to use, the whole thing can be done in just one line, so it's not a very hard problem.

[Ellen De Jonghe]

Thanks for your logical insight. I'm always surprised how simple things are. But I always seem to make problems more complicated, especially when using MATLAB (btw: I'm only using it for a couple of weeks right now).

I made this script. It works, but I'm sure it can be done easier, especially the part where I create the array with the neighbours (n).

function [res] = isNeighboring(matrix,row,column)
n = zeros(1,8);
n(1,1) = matrix(row+1,column+1);
n(1,2) = matrix(row+1,column);
n(1,3) = matrix(row+1, column-1); 
n(1,4) = matrix(row, column-1);
n(1,5) = matrix(row, column+1);
n(1,6) = matrix(row-1, column+1);
n(1,7) = matrix(row-1, column);
n(1,8) = matrix(row-1, column-1);
m = 1:matrix(row,column)-1;
p = ismember(m,n);
res = sum(p) == length(m);
end 

[Bob Thompson]

That's definitely a really good start. I have a couple of thoughts to add.

1) How does your code respond to the selected element being 1? I think you may have a problem there.

2) You can always just take the entire 3x3 around the selected element. Even if it contains the element that shouldn't be a problem with your logic check.

n = matrix(row-1:row+1,column-1:column+1);

[Guillaume]

@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]

Thanks!

I indeed forgot to check the case where M(i,j) = 1, I will try to figure that out.

Answer 1

[Andrei Bobrov]

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

[Guillaume]

The if is not needed, out is true anyway, if M(i,j) is 1, since all([]) is true.

[Andrei Bobrov]

Thanks Guillaume! I'm fixed.

Answer 2

[Ellen De Jonghe]

That's a very easy one! Thanks.

I seem to always make my scripts much longer than it has to.