OKlibrary  0.2.1.6
ExactTransversals.hpp File Reference

Plans regarding exact transversals. More...

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## Detailed Description

Plans regarding exact transversals.

A transversal of a hypergraph is called exact if it intersects every hyperedge exactly once. Denote it by ETr(G).

Todo:
The basic algorithm
• For every hypergraph G the set ETr(G) can be enumerated in the following obvious way:
```template <class Hypergraph>
void enumerate_exact_transversals(const Hypergraph G, std::ostream& out, const Hypergraph::vertex_set U) {
// Assumption: V(G) intersect U is empty.
// Returns the minimal exact transversals of G, with U added, on out (without repetition).
if (G contains empty hyperedge) return;
if (G has no hyperedge) {
out << U;
return;
}
choose v in V(G); // (*)
{ // the minimal exact transversals containing v
let S be the star of v in G;
let H be the set of vertices in S;
let G' be G with the hyperedges of S removed and the vertices from H crossed out;
enumerate_exact_transversals(G', out, U union {v});
}
{ // the minimal exact transversals not containing v
let G' be G with the vertex v crossed out;
enumerate_exact_transversals(G', out, U);
}
return;
}
```
• Remarks:
1. If a linear order on V(G) is given and at (*) the smallest element of V(G) is taken, then the transversals are output in lexicographical order.
2. If the output would be given as a complete list, then parameter U wouldn't be needed, but it could be added to the outputs obtained from recursive calls. However here we need U, since we output the transversals one at a time.
• Has this algorithm only polynomial delay for inputs G which are exact transversal hypergraphs ?
• Is deciding whether a hypergraph has an exact transversal NP-complete?
• By dualisation this decision problem is equivalent to decide, whether a hypergraph has a matching (a set of disjoint hyperedges) which is also an edge cover (a set of hyperedges covering all edges).
• By complementation it is equivalent to the problem whether an independent set I in a hypergraph (a set of vertices not containing any hyperedge) exists, such that for every hyperedge H all but one vertices of H are in I.
• Apparently the proof of NP-completeness is in [Karp 1972]; one should investigate the proof (also implementing the reduction).
Todo:
Exact transversal hypergraphs (that is, Tr(G) = ETr(G))
• Implement the decision algorithm for the special class (? open) given in [Galesi, Kullmann 2004]; this exist already as mupad-function "ist_speziell_exakt_transversal" in Mupad/Orthogonal.mup.
• One would guess that the above general algorithm is actually more efficient than the algorithm from [Eiter 1994]? (Exists as mupad-function "Transversals_exakt" in Mupad/Orthogonal.mup.)
Todo:
Generalisation to boolean CNF
• As Tr(G) was generalised to Tr(F) (see HypergraphTransversals/plans/DirectTransversalEnumeration.hpp), we can also generalise ETr(G) to ETr(F), that is the set of all minimal partial assignments (w.r.t. the standard partial ordering) which satisfy in every clause of F exactly one literal.
• Is Tr(F) = ETr(F) (F is exact transversal) also decidable in polynomial time?
• It seems the above algorithm for computing ETr(G) (for arbitrary G) cannot be (easily) generalised to (boolean) clause-sets F? (The problem is the minimality condition.)
• For F which are exact transversal, can we compute Tr(F) in polynomial time?
• As remarked above, to decide whether for a positive clause-set an exactly satisfying partial assignment exists is NP-complete; what about a 1-regular hitting clause-set --- is it still NP-complete?!?
• The point here is, that finding an exactly satisfying partial assignment for a clause-set F just means to find a clause which clashes with every clause from F in exactly one literal (and so, if we can find exactly satisfying partial assignment for 1-regular hitting clause-sets efficiently, then we can "grow" 1-regular hitting clause-sets; see HittingClauseSets/plans/ExtendingHittingClauseSets.hpp).
• It must be emphasised, that an "exactly satisfying assignment" is a partial assignment; the corresponding satisfiability problem, finding an exactly satisfying (partial) assignment, should be called PXSAT (where XSAT is the standard abbreviation for "exact SAT", which uses total assignments).
• For example ((a or b) and (a or not b)) is PXSAT-satisfiable, but not XSAT-satisfiable.
• Of course, from XSAT follows PXSAT, but the reverse direction does not hold in general, as the example shows:
1. It holds for positive clause-sets (here we can extend a partial assignment to a total one by setting all other variables to 0).
2. PXSAT seems to be a harder than SAT (it is not a "SAT-problem" in the sense of "systems with partial instantiation" (see [Kullmann, Annals of Mathematics and Artificial Intelligence 40: 303-352, 2004]), since an extension of an exactly satisfying assignment need not to be exactly satisfying again).
• Reduction to standard SAT
1. For positive clause-sets we can just use the XSAT -> SAT translation, which adds all binary clauses excluding the possibility of two satisfied literals in a clause.
2. This, of course, is the same as translating exact transversal for hypergraphs into SAT (the satisfying assignments of the translation, after removal of variables set to 0, correspond 1-1 to the exact transversals).
3. For general (boolean) CNF F a natural translation into a monosigned CNF SAT problem over variables with domain {0,1,*}, where * means "unassigned", translates every clause C to the corresponding clause C', where literal x becomes x' stating that the underlying variable v must be equal to 0 (if x is negative) or 1 (if x is positive), and adds clauses (v <> e1) of (w <> e2) for clauses C' (as just constructed) containing literals (x = e1) and (w = e2) (for different variables v, w).
4. The translated F' has only total satisfying assignments, corresponding exactly to the exactly satisfying (partial) assignments of F.
Todo:
Exact covers

Definition in file ExactTransversals.hpp.