Directed Hamiltonicity parameterized by the largest independent set - PowerPoint PPT Presentation

Directed Hamiltonicity parameterized by the largest independent set Andreas Björklund, Petteri Kaski, and Yiannis Koutis

A directed graph

A Hamiltonian cycle

Known Hamiltonicity detection results NP-hard! Undirected Hamiltonicity can be decided in O(1.657 n ) time, and O(1.415 n ) in bipartite graphs, both in polynomial space [B14]. Directed Hamiltonicity in bipartite graphs can be decided in O(1.888 n ) time using exponential space [CKN13].

Open problem Can directed Hamiltonicity be decided in O(c n ) time for some c<2? 
 (For comparison, the parity of the number of Hamiltonian cycles can be decided in O(1.619 n ) time [BH13].)

Our result Directed Hamiltonicity can be decided in O(3 n-mis(G) ) time and polynomial space, where mis(G) is the size of the largest independent set in G. In particular, for bipartite directed graphs we get O(1.733 n ) time. We will only show the bipartite case in this presentation.

Why care about an incremental result? More than just a shaving from O(1.888 n ) to O(1.733 n ): The first O(c n ) time algorithm for c<2 using only polynomial space. Allows efficient parallel computation. Smooth parameterization “immediately above” the bipartite case.

Examples of speedups based on tabulation Space matters Max Cut in O(2 ⍵ n/3 ) time [W04]. Uses O(2 2n/3 ) space. n/3 ) [BK16]. No known Can be reduced to O(2 n ). polynomial space faster than O(2 n ) time [BHK08]. Uses O(2 n ) Chromatic number in O(2 n ) [BHKK11 space. Can be reduced to O(1.3 ]. No known n ). polynomial space faster than O(2.2 n/2 ) time [HS74]. Uses O(2 n/2 ) space. Subset sum in O(2 n/ 4 ) space [SS79]. Was no known Can be reduced to O(2 n ), now there is: 
 polynomial space faster than O(2 Recall Jesper’s talk!

Key ideas in the new algorithm Familiar Algebraic fingerprinting: Construct a polynomial that is not identically zero whenever there is a Hamiltonian cycle. k ) so that an even Characteristic two: Use the field GF(2 counting of any arc subsets not representing Hamiltonian cycles will cancel. Determinants: Use the facts that a determinant counts matchings over a characteristic two field and that determinants are polynomial time computable. Inclusion—exclusion: Sum over all subsets to sieve for set covers.

Key ideas in the new algorithm Novel? Label arcs by any of their endpoints: Every arc will be thought of as either belonging to its origin or its target. Determinant matrix embedding: Use a new interpretation of the rows and column of the matrix. Each row represents a vertex *and* an orientation, either incoming or outgoing arcs. Columns represent labels.

The basic case: a bipartite graph

A special vertex s

A selected arc is labeled by one of its two endpoint Notation: either Arc is associated with its origin or Arc is associated with its target

Counting arc labeled cycle cover We will count arc labeled cycle covers, i.e. subsets of arcs labeled by vertices such that: Every vertex has precisely one incoming and one outgoing arc. Every vertex is the label of exactly one arc. The special vertex is always a label of an outgoing arc.

An arc labeled cycle cover s

Hamiltonian cycles can be labeled only in one way s s forces a label orientation!

Non-Hamiltonian cycle covers can be labeled in an even number of ways s s Origin labeled Target labeled Any cycle not through s has two label association orientations.

The determinant summation formula Call the blue vertices B. For every arc (i,j) we introduce a variable x i,j . The set O is the allowed origins in B, and the set T is the allowed targets in B. X Y X x i,j = det( M O,T ) . O,T ⊆ B H ∈ HamCycles(G) ( i,j ) ∈ H Inclusion-exclusion over two sets, but in fact only terms where the union of O and T covers B will contribute, hence 3 |B| terms.

The structure of the matrix M O,T Labels e w x s u v a b c d t f s . 0 0 0 0 0 0 0 0 0 0 0 s . . 0 0 0 0 . 0 0 0 0 0 t 0 . 0 . 0 0 0 . 0 0 0 0 t 0 . 0 . 0 0 0 . 0 0 0 0 u 0 0 . 0 . 0 0 0 . 0 0 0 u Arcs . . 0 0 . 0 0 0 . 0 0 0 v 0 . 0 . . 0 0 0 0 . 0 0 v 0 0 0 . 0 . 0 0 0 . 0 0 w 0 0 0 0 . 0 0 0 0 0 . 0 w 0 0 . 0 0 0 0 0 0 0 . 0 x 0 0 0 0 0 . 0 0 0 0 0 . x 0 0 0 0 . . 0 0 0 0 0 .

The left side of the matrix Labels e w x s u v a b c d t f s . 0 0 0 0 0 0 0 0 0 0 0 s . . 0 0 0 0 . 0 0 0 0 0 ⇢ : b ∈ O x b,t t 0 . 0 . 0 0 0 . 0 0 0 0 0 : otherwise . t 0 . 0 . 0 0 0 . 0 0 0 0 u 0 0 . 0 . 0 0 0 . 0 0 0 u Arcs . . 0 0 . 0 0 0 . 0 0 0 v 0 . 0 . . 0 0 0 0 . 0 0 v 0 0 0 . 0 . 0 0 0 . 0 0 w 0 0 0 0 . 0 0 0 0 0 . 0 ⇢ : c ∈ T w x w,c 0 0 . 0 0 0 0 0 0 0 . 0 0 : otherwise. x 0 0 0 0 0 . 0 0 0 0 0 . x 0 0 0 0 . . 0 0 0 0 0 .

The right side of the matrix Labels e w x s u v a b c d t f s . 0 0 0 0 0 0 0 0 0 0 0 s . . 0 0 0 0 . 0 0 0 0 0 t 0 . 0 . 0 0 0 . 0 0 0 0 X x t,i t 0 . 0 . 0 0 0 . 0 0 0 0 u 0 0 . 0 i ∈ T . 0 0 0 . 0 0 0 u Arcs . . 0 0 . 0 0 0 . 0 0 0 v 0 . 0 . . 0 0 0 0 . 0 0 v 0 0 0 . 0 . 0 0 0 . 0 0 X x i,w w 0 0 0 0 . 0 0 0 0 0 . 0 i ∈ O w 0 0 . 0 0 0 0 0 0 0 . 0 x 0 0 0 0 0 . 0 0 0 0 0 . x 0 0 0 0 . . 0 0 0 0 0 .

A monomial in the determinant expansion Labels e w x s u v a b c d t f s . 0 0 0 0 0 0 0 0 0 0 0 s . . 0 0 0 0 a 0 0 0 0 0 t 0 . 0 . 0 0 0 d 0 0 0 0 t 0 . 0 . 0 0 0 . 0 0 0 0 u 0 0 . 0 . 0 0 0 . 0 0 0 u Arcs . . 0 0 . 0 0 0 a 0 0 0 v 0 . 0 . . 0 0 0 0 . 0 0 v 0 0 0 . 0 . 0 0 0 f 0 0 w 0 0 0 0 . 0 0 0 0 0 . 0 w 0 0 . 0 0 0 0 0 0 0 c 0 x 0 0 0 0 0 . 0 0 0 0 0 . x 0 0 0 0 . . 0 0 0 0 0 f

A monomial in the determinant expansion s a b t u c v e d w f x Not necessarily a cycle cover!

Non-cycle covers will be counted an even number of times s a b t u c There are no targets in b and e: Counted for T’s with and without v e b and e! d w f x

Non-cycle covers will be counted an even number of times s a b There is no target in t u c e and no origin in d: Counted for T’s with and without e, and v e O’s with and d w without d! f x

Bipartite algorithm Polynomial Identity Testing Randomly assign x uv from GF(2 k ). X Compute det( M O,T ) . O,T ⊆ B O ∪ T = B If non-zero conclude Hamiltonian.

Mis parameterised algorithm Assume n even Compute matching. If no perfect matching output non-Hamiltonian. Find mis by any sufficiently fast algorithm. Trying all subsets of endpoints of the perfect matching will do (O(3 n/2 ) time). Do P .I.T. on a determinant formula very similar to the bipartite case.

Thank you for listening!

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