arrange rows according to common values present between the rows in an array in matlab - arrays

Question

i have an array of 800 rows with 3 columns. I want to arrange it in such a way that the row coming after the first row contains at least 2 common values as the first and the next row after that one also contains at least 2 common values as the second.
Example; row 1 = 2 4 5 common row= row 30 = 2 5 13
so first arrangement;
2 4 5
2 5 13
the next row will be one common to the new row 2
example row 4= 13 45 5
therefore the arrangement will now be;
2 4 5
2 5 13
13 45 5
etc
currently, i have this code that groups those common together in cells and then displays them one after another. the problem with this is that the array contains more than one commonality.. for example, row 1 could have 2 rows that have a common value as it, this code brings all those rows together into one array and does the same for the second row.. how can i make the code i make the code do what i explained in the first paragraph;
here is the code;
% Data
A = connections;
% Engine
[m, n] = size(A);
groups =[];
ng = 0;
for k=1:m-1
u = unique(A(k,:)); % representation of kth row
[in, J] = ismember(A(k:end,:),u);
l = m-k+1;
r = repmat((1:l).', n, 1);
c = accumarray([r(in) J(in)],1,[l n]); % count
c = bsxfun(#min,c,c(1,:)); % clip
rows = sum(c,2)>=2; % check if at least 2 elements are common
rows(1) = false;
if any(rows)
ng = ng+1;
rows(1) = true;
groups = (k-1) + find(rows);
end
end
please note that the rest of the code below could be added only if groups was a cell array, bt it has been changed to be a normal array as written above.. so to test the code, there is no need to add the code below.
% Remove the tail
groups(ng+1:end) = [];
% Display
for k=1:1:length(groups)
gk = groups{k};
for r = gk'
fprintf('%d %d %d %d\n', r, A(r,:));
end
end

Answer 1

This is the Hamiltonian path problem. A Hamiltonian path or traceable path is a path that visits each vertex exactly once. Here you have 800 vertices (rows) each have one or more neighbors. The condition for neighborhood is that a vertex having 2 or more values common with other vertex.
to solve the problem You should create an adjacency matrix corresponding to the graph structure then finding an Hamiltonian path on the graph.
Here is a way to create the adjacency matrix:
[ m n] = size(A);
%generate all 2 element subsets of the set [1 2 3 ] to check if a row
% has two elemnts common with other row
x = nchoosek(1:n,2);
%generate indices for generating two element rows
col = repmat(x,m,1);
M = (1:m).';
row = kron(M,ones(size(x)));
IDX = sub2ind([m n] ,row, col);
%convert A with [m , 3] to a [(m * 3), 2)]matrix
% then sort each row to be comparable with other rows
B=sort(A(IDX),2);
%indices of rows that have two common elements is found
[~,I,II] = unique(B,'rows', 'first');
convert index of the [(m * 3), 2)]matrix to [m , 3] matrix
I = uint32(I-1) / n + 1;
%create adjacency matrix
adjacency= sparse(I(II(:)), M(II(:)),1,m,m);
%make the matrix symmetric
adjacency = adjacency | adjacency.';
% a vertex can not be neighbor of itself
adjacency(speye(m))=false;
You should implement your own version of the Hamiltonian path algorithm or find in the web.
However the Hamiltonian path problem can be converted to a special form of the travelling salesman problem by adding a vertex to the graph and making all other vertices neighbor to it.
So some changes should be applied to the adjacency matrix:
adjacency = [adjacency; ones(1,m)];
adjacency = [adjacency, ones(n+1,1)];
adjacency (end) = 0;
In Mathworks you can find an example that shows how to solve the travelling salesman problem using binary integer programming. However the method provided doesn't seems straightforward. So you may found other one or implement your own.