Dominik Scheder
ETH Zurich
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Featured researches published by Dominik Scheder.
symposium on the theory of computing | 2011
Robin A. Moser; Dominik Scheder
Schoening in 1999 presented a simple randomized algorithm for k-SAT with running time a<sup>n</sup> * poly(n) for a = 2(k-1)/k. We give a deterministic version of this algorithm running in time a<sup>n+o(n)</sup>.
latin american symposium on theoretical informatics | 2008
Dominik Scheder
Most deterministic algorithms for NP-hard problems are splitting algorithms: They split a problem instance into several smaller ones, which they solve recursively. Often, the algorithm has a choice between several splittings. For 3-SAT, we show that choosing wisely which splitting to apply, one can avoid encountering too many worst-case instances. This improves the currently best known deterministic worst case running time for 3-SAT from O(1.473n) to O(1.465n), n being the number of variables in the input formula.
symposium on theoretical aspects of computer science | 2011
Timon Hertli; Robin A. Moser; Dominik Scheder
A critical variable of a satisfiable CNF formula is a variable that has the same value in all satisfying assignments. Using a simple case distinction on the fraction of critical variables of a CNF formula, we improve the running time for 3-SAT fromO(1.32216 n ) by Rolf [10] toO(1.32153 n ). Using a dierent
Efficient Algorithms | 2009
Heidi Gebauer; Robin A. Moser; Dominik Scheder; Emo Welzl
We consider boolean formulas in conjunctive normal form (CNF). If all clauses are large, it needs many clauses to obtain an unsatisfiable formula; moreover, these clauses have to interleave. We review quantitative results for the amount of interleaving required, many of which rely on the Lovasz Local Lemma, a probabilistic lemma with many applications in combinatorics. In positive terms, we are interested in simple combinatorial conditions which guarantee for a CNF formula to be satisfiable. The criteria obtained are nontrivial in the sense that even though they are easy to check, it is by far not obvious how to compute a satisfying assignment efficiently in case the conditions are fulfilled; until recently, it was not known how to do so. It is also remarkable that while deciding satisfiability is trivial for formulas that satisfy the conditions, a slightest relaxation of the conditions leads us into the territory of NP-completeness. Several open problems remain, some of which we mention in the concluding section.
symposium on theoretical aspects of computer science | 2010
Dominik Scheder
We call a CNF formula linear if any two clauses have at most one variable in common. We show that there exist unsatisfiable linear k-CNF formulas with at most 4k^2 4^k clauses, and on the other hand, any linear k-CNF formula with at most 4^k/(8e^2k^2) clauses is satisfiable. The upper bound uses probabilistic means, and we have no explicit construction coming even close to it. One reason for this is that unsatisfiable linear formulas exhibit a more complex structure than general (non-linear) formulas: First, any treelike resolution refutation of any unsatisfiable linear k-CNF formula has size at least 2^(2^(k/2-1))
theory and applications of satisfiability testing | 2008
Dominik Scheder; Philipp Zumstein
. This implies that small unsatisfiable linear k-CNF formulas are hard instances for Davis-Putnam style splitting algorithms. Second, if we require that the formula F have a strict resolution tree, i.e. every clause of F is used only once in the resolution tree, then we need at least a^a^...^a clauses, where a is approximately 2 and the height of this tower is roughly k.
Electronic Notes in Discrete Mathematics | 2007
Claudia Käppeli; Dominik Scheder
A pair of clauses in a CNF formula constitutes a conflict if there is a variable that occurs positively in one clause and negatively in the other. Clearly, a CNF formula has to have conflicts in order to be unsatisfiable--in fact, there have to be many conflicts, and it is the goal of this paper to quantify how many. An unsatisfiable k-CNF has at least 2k clauses; a lower bound of 2k for the number of conflicts follows easily. We improve on this trivial bound by showing that an unsatisfiable k-CNF formula requires Ω(2.32k) conflicts. On the other hand there exist unsatisfiable k-CNF formulas with O(4k log3 k/k) conflicts. This improves the simple bound O(4k) arising from the unsatisfiable k-CNF formula with the minimum number of clauses.
international workshop and international workshop on approximation, randomization, and combinatorial optimization. algorithms and techniques | 2013
Dominik Scheder; Li-Yang Tan
Abstract A CNF formula is called k-satisfiable if every subformula containing at most k clauses is satisfiable. Let s k be the largest ratio s such that for any k-satisfiable formula F, there is an assignment satisfying at least s | F | clauses. We denote the analogous ratio for formulas with weighted clauses by r k . The numbers r k have already been studied, but little has been known about s k . We show that s k and r k differ for k = 2 , 3 . For k = 2 , we show that s 2 = 2 3 , which is larger than r 2 = ( 5 − 1 ) / 2 ≈ 0.619 , the inverse of the golden ratio. Further, we show that r 3 = 2 3 21 / 31 ⩽ s 3 ⩽ 7 / 10 .
theory and applications of satisfiability testing | 2007
Dominik Scheder; Philipp Zumstein
We study the relationship between the average sensitivity and density of k-CNF formulas via the isoperimetric function \(\varphi:[0,1]\to{I\!\!R}\),
international symposium on algorithms and computation | 2013
Dominik Scheder
Fix a set S ⊆ {0, 1}* of exponential size, e.g. |S ∩ {0, 1}n| ∈ Ω(αn), α > 1. The S-SAT problem asks whether a propositional formula F over variables v1, . . . , vn has a satisfying assignment (v1, . . . , vn) ∈ {0, 1}n ∩ S. Our interest is in determining the complexity of S-SAT. We prove that S-SAT is NP-complete for all context-free sets S. Furthermore, we show that if S-SAT is in P for some exponential S, then SAT and all problems in NP have polynomial circuits. This strongly indicates that satisfiability with exponential families is a hard problem. However, we also give an example of an exponential set S for which the S-SAT problem is not NP-hard, provided P ≠ NP.