How can I get permutations of a vector satisfying the given condition?

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I need to get all permutations of an array (A=[1:24] ) satisf[y]ing given conditions.The very next number after the given number can be only as per the condition mentioned. For example next number after '1' can be the non zero elements in the first row of the given matrix B. So each row represents the possible 'next numbers' of the number representing the row.
So after '1' 2 & 13 are only possible
after '2' (second row) 1,3,14&13 are possible.
after 23 (23rd row) 22,24,11&12 are possible .
B= [0 2 0 13 0 0;
1 3 0 14 0 13;
2 4 0 15 0 14;
3 5 0 16 0 15;
4 6 0 17 0 16;
5 7 0 18 0 17;
6 8 0 19 0 18;
7 9 0 20 0 19;
8 10 0 21 0 20;
9 11 0 22 0 21;
10 12 0 23 0 22;
11 0 0 24 0 23;
0 14 1 0 2 0;
13 15 2 0 3 0;
14 16 3 0 4 0;
15 17 4 0 5 0;
16 18 5 0 6 0;
17 19 6 0 7 0;
18 20 7 0 8 0;
19 21 8 0 9 0;
20 22 9 0 10 0;
21 23 10 0 11 0;
22 24 11 0 12 0;
23 0 12 0 0 0];
I am not able to apply this condtion after finding all permutation of vector A as 'perms' function can not generate all possible permutation due to storge and time limitations.
  4 Commenti
Torsten
Torsten il 1 Ago 2022
Should all paths start with "1" or with one of the nodes "2" or "13" ?
If all paths start with "1", why is "1" still listed in the matrix as "next node" option ?
Bruno Luong
Bruno Luong il 1 Ago 2022
Modificato: Bruno Luong il 1 Ago 2022
@Torsten I see [2, 13] as next of 1 and nowhere 1.
For your question, see the discussion comments under my answer.

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Risposta accettata

Bruno Luong
Bruno Luong il 1 Ago 2022
Modificato: Bruno Luong il 1 Ago 2022
There are 9918 permutations according to this code
B= [0 2 0 13 0 0;
1 3 0 14 0 13;
2 4 0 15 0 14;
3 5 0 16 0 15;
4 6 0 17 0 16;
5 7 0 18 0 17;
6 8 0 19 0 18;
7 9 0 20 0 19;
8 10 0 21 0 20;
9 11 0 22 0 21;
10 12 0 23 0 22;
11 0 0 24 0 23;
0 14 1 0 2 0;
13 15 2 0 3 0;
14 16 3 0 4 0;
15 17 4 0 5 0;
16 18 5 0 6 0;
17 19 6 0 7 0;
18 20 7 0 8 0;
19 21 8 0 9 0;
20 22 9 0 10 0;
21 23 10 0 11 0;
22 24 11 0 12 0;
23 0 12 0 0 0];
P = AllHpath(B);
fprintf('there are %d permutations\n', size(P,1))
there are 9918 permutations
% Display 10 solutions
P([1:5 end-4:end],:)
ans = 10×24
1 2 3 4 5 6 7 8 9 10 11 12 24 23 22 21 20 19 18 17 16 15 14 13 1 2 13 14 15 16 17 18 19 20 21 22 23 24 12 11 10 9 8 7 6 5 4 3 1 2 13 14 15 3 4 5 6 7 8 9 10 11 12 24 23 22 21 20 19 18 17 16 1 2 13 14 15 3 4 5 16 17 18 19 20 21 22 23 24 12 11 10 9 8 7 6 1 2 13 14 15 3 4 5 16 17 18 6 7 8 9 10 11 12 24 23 22 21 20 19 1 13 2 14 3 15 4 16 5 17 6 18 7 19 8 20 9 21 10 22 23 11 12 24 1 13 2 14 3 15 4 16 5 17 6 18 7 19 8 20 9 21 10 22 11 12 24 23 1 13 2 14 3 15 4 16 5 17 6 18 7 19 8 20 9 21 10 22 11 12 23 24 1 13 2 14 3 15 4 16 5 17 6 18 7 19 8 20 9 21 10 22 11 23 24 12 1 13 2 14 3 15 4 16 5 17 6 18 7 19 8 20 9 21 10 22 11 23 12 24
function P = AllHpath(B)
P = Hpath_helper(B, 1, size(B,1));
end
function P = Hpath_helper(B, i, n)
if n == 1
P = i;
else
Bi = B(i,:);
next = Bi(Bi>0);
P = zeros(0,n);
if ~isempty(next)
B(B==i) = 0;
P = [];
for k = next
Pk = Hpath_helper(B, k, n-1);
P = [P; [repmat(i,size(Pk,1),1), Pk]]; %#ok
end
end
end
end
  3 Commenti
Bruno Luong
Bruno Luong il 1 Ago 2022
Modificato: Bruno Luong il 1 Ago 2022
Then change (loop on) the second argument "1" in this command, which is the starting number
P = Hpath_helper(B, 1, size(B,1))
the function becomes
function P = AllHpath(B)
n = size(B,1);
CP = arrayfun(@(start) Hpath_helper(B, start, n), 1:n, 'unif', 0);
P = cat(1,CP{:});
end
there are 48860 permutations

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Più risposte (1)

Steven Lord
Steven Lord il 31 Lug 2022
My first thought would be to construct a digraph from your data then try to find Hamiltonian paths in that digraph. Or there may be a way to use the digraph to solve your underlying problem without having to determine all permutations of its nodes. What is that underlying problem you're trying to solve?
  8 Commenti
dpb
dpb il 1 Ago 2022
Modificato: dpb il 3 Ago 2022
Given the problem and the geometry, you can simply solve for one set of connected tubes and by symmetry know the answer for the rest...
.....
13
2
14
3
.....
You could, if wanted, also do the ends as second geometry.

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