# Find distance between to elements of a "circular" vector

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Jérôme on 16 Jul 2013
Commented: Andrea Ramazzina on 13 Oct 2017
Hi folks;
I'm being stuck on something that I feel is not so difficult, but I can't find a proper way to do it.
Let's say I have the following vector : v = [6 7 1 3 9 6 8 10]
And let's say I want to know the distance, i.e. number of cells, between the position of 7 and the position of 10. At first glance, it is 6, since starting from 7, you have to move 6 times to the right to get to 10. However, I want to do as if the vector was "circular", like if both ends were connected. So by going two steps to the left, we could reach 10 from 7. The algorithm I want to write should give me the shortest distance, whether it is going to the left or the right.
Do you people se what I mean? I'm not sure it is clear. Hope you can help me. :)

Muthu Annamalai on 16 Jul 2013
Assuming your list is unique you can get the linear positions of two numbers as,
p1 = find( v == n1 )
p2 = find( v == n2 )
% ensure p2 >= p1 always
if ( p2(1) < p1(1) )
t = p1(1); p1 = p2(1); p2 = p1;
end
% forward dist
dist = p2 - p1 + 1
% rev dist
dist2 = p1 - 1 + length(v) - p2
dist = min(dist,dist2)
Andrea Ramazzina on 13 Oct 2017
t = p1(1); p1 = p2(1); p2 = t;
I think you had a mistake, this should be the right one

Azzi Abdelmalek on 16 Jul 2013
Edited: Azzi Abdelmalek on 16 Jul 2013
v = [6 7 1 3 9 6 8 10]
a1=7;
id1=2
a2=10 % number to find
vi1=[v v]
ii1=find(vi1==a2)-id1
idx1=min(ii1(find(ii1>0)))
vi2=fliplr([v v])
id2=numel(v)-id1+1
ii2=find(vi2==a2)-id2
idx2=min(ii2(find(ii2>0)))
idx=min(idx1,idx2)