Problem Detail: I need to find a Hamiltonian cycle in a directed graph using propositional logic, and to solve it by sat solver. So after I couldn’t find a working solution, I found a paper that describes how to construct a CNF formula to find an Hamiltonian path: Xi,j – node j is in position i in the path List of constraints:
- Each node j must appear in the path x1j ∨ x2j ∨ · · · ∨ xnj – for every node j
- No node j appears twice in the path ¬xij ∨ ¬xkj for all i, j, k with i 6= k.
- Every position i on the path must be occupied – xi1 ∨ xi2 ∨ · · · ∨ xin for each i
- No two nodes j and k occupy the same position in the path – ¬xij ∨ ¬xik for all i, j, k with j != k
- Nonadjacent nodes i and j cannot be adjacent in the path – ¬xki ∨ ¬xk+1,j for all (i, j) !∈ E and k = 1, 2, . . . , n − 1.
My question is, how can I find Hamiltonian cycle using these constraints? I understand that I need to check if there’s a cycle (v1==vn), that’s one thing (But I got constraint 2). Second, it’s a directed graph and I don’t know how can I assure that the vertices would be in the right order of the edges, I thought about this: Every two nodes must have edges – Xki ^ Xk+1j for each (i,j)∈ E and k = 1, 2, . . . , n − 1. But it doesn’t seem to work, any help would be appreciated. EDIT What I did was to add another constraint –
- No edges from the set: (i,j)!∈ E that are the last and the first
Asked By : Dor Cohen
Answered By : D.W.
Just add the constraint that $x_{1j}=x_{nj}$ for all $j$, and make a special exception to constraint 2 so that you omit constraint 2 in the case where $i=1$ and $k=n$. Everything works fine for a directed graph. You just need to interpret constraint 5 appropriately.
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