# finding neighbor of a position

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Mohammad Golam Kibria on 26 Jun 2011
Hi, I have a matrix.
I =
1 0 0
2 5 0
0 0 3
0 0 0
I know the position of 5 in I is 6 linear index.
is there any easy function to have the 8 other neighbors of 5. Thanks

Oleg Komarov on 26 Jun 2011
EDITED: should be fine now
I =[ 1 0 0
2 5 0
0 0 3
0 0 0];
l = 8;
sz = size(I);
% row, col subs of center
[r,c] = ind2sub(sz,l); % c = ceil(l/4); r = mod(l,4)+ c*sz(1);
% Calculate 8 neighbors
neigh(1:8,1:2) = [r+[-1;0;1;-1;1;-1;0;1] c+[-1;-1;-1;0;0;1;1;1] ];
% Only those in the range
neigh = neigh(all(neigh,2) & neigh(:,1) <= sz(1) & neigh(:,2) <= sz(2),:);
% Convert to position
idx = (neigh(:,2)-1)*sz(1) + neigh(:,1);
Mohammad Golam Kibria on 28 Jun 2011
Thanks

Sean de Wolski on 27 Jun 2011
idx = find(conv2(double(I==5),ones(3),'same'))
%This includes the 6, but that could easily be taken care of with setdiff.
Mohammad Golam Kibria on 28 Jun 2011
Thanks

Wolfgang Schwanghart on 26 Jun 2011
Mohammad Golam Kibria on 28 Jun 2011
thanks,this also works fine for me

Andrei Bobrov on 26 Jun 2011
idl = 6;
idxs = ...
nonzeros(bsxfun(@plus,idl - [1 0 -1]',size(I,1)*[-1 0 1]).*[1 1 1;1 0 1;1 1 1])
CORRECTED 06/27/2011 10:05 MSK
idl = 6;
s = size(I);
I0 = zeros(s+2);
I0(2:end-1,2:end-1) = reshape(1:numel(I),s);
idxs = nonzeros(I0(bsxfun(@plus,find(I0==idl) - [1 0 -1]',(s(1)+2)*[-1 0 1])).*[1 1 1;1 0 1;1 1 1])
MORE variant (06/27/2011 11:12 MSK)
s = size(I);
[ii jj] = ind2sub(s,idl);
v = [-1 -1 -1;0 0 0;1 1 1];
R=ii+v;
C=jj+v';
loc = (R<=s(1) & R>=1&C<=s(2) & C>=1&[1 1 1;1 0 1;1 1 1])>0;
idxl = sub2ind(s,R(loc),C(loc));
MORE variant 2 (06/27/2011 11:35 MSK) with idea of Oleg
s = size(I);
[ii jj] = ind2sub(s,idl);
R = ii + [-1 0 1 -1 1 -1 0 1];
C = jj + [-1 -1 -1 0 0 1 1 1];
loc = (R<=s(1) & R>=1&C<=s(2) & C >= 1 )>0;
idxl = sub2ind(size(I),R(loc),C(loc));
LAST variant (06/27/2011 13:43 MSK)
I1 = zeros(size(I));
I1(idl)=1;
idx = find(bwdist(I1,'chessboard')==1)
or
idx = find(bwdist(I==5,'chessboard')==1)
Mohammad Golam Kibria on 28 Jun 2011
Thanks this also works fine for me