Circularly exclude elements.

Hello all,

I'm trying to compactly write a line of code for a interesting program I'm working on. Fixing , suppose we have the following code:

x = 1:n;
non_adj = zeros(n,n-2);
for k=1:n
    non_adj(k,:) = []; %in here comes the confusion
end

In non_adj(k,:) I would like to fill in x without the elements k-1 and k+1. However, these elements are regarded circularily, i.e. if I have , then I want (order of elements irrelevant). I know I can do this with if statements handling the cases, but I was wondering if you all had a nicer way to one liner the construction.

I attempted the following construction

non_adj(k,:) = [k, k+2:mod(k-3,n)+1, mod(k+1,n)+1:k-2];

However this fails, for example, where . The above would construct the array , where in fact I wanted .

Do you have any ideas? If it helps, the context is that I'm trying to construct a vertex set minus the neighbors of vertex k, which happen to be (circularily shifted for ).

EDIT: I've realized I could use set differencing, i.e.: (however, you still might come up with better ideas).

non_adj(k,:) = setdiff([1:n], 1 + [mod(k-2,n) mod(k,n)]);

[madhan ravi]

explicitly state your desired output

[Abhijit Chowdhary]

If , then I want . If , then I want . Otherwise I want , and I'm hoping for a one-line to cover these scenarios.

Answer 1

[David Goodmanson]

Hi Abhijit,

try

x  = 1:n;
x(mod([k-1 k+1]-1,n)+1) = []

[Abhijit Chowdhary]

Hello David,

This indeed works for the case, however, it seems that it doesn't for other k. Thank you, though!

[David Goodmanson]

Hi Abhijit,

I thought that for a given k, you wanted to keep k, exclude k-1 and k+1 and keep the rest of them, doing so in a circular fashion. Could you give me a quick example of where the code above doesn't work? I am having trouble finding one (n.b. both lines of code have to be run for each new value of k).

[Abhijit Chowdhary]

Ah, no it's my fault, you're right. I did not realize that x was being changed when you do this, and blindly continued to test on it. Yes, this indeed works, thanks!