Directed Hamiltonicity parameterized by the largest independent set - - PowerPoint PPT Presentation
Directed Hamiltonicity parameterized by the largest independent set - - PowerPoint PPT Presentation
Directed Hamiltonicity parameterized by the largest independent set Andreas Bjrklund, Petteri Kaski, and Yiannis Koutis A directed graph A Hamiltonian cycle Known Hamiltonicity detection results NP-hard! Undirected Hamiltonicity can be
A directed graph
A Hamiltonian cycle
Undirected Hamiltonicity can be decided in O(1.657n) time, and O(1.415n) in bipartite graphs, both in polynomial space [B14]. Directed Hamiltonicity in bipartite graphs can be decided in O(1.888n) time using exponential space [CKN13].
Known Hamiltonicity detection results
NP-hard!
(For comparison, the parity of the number of Hamiltonian cycles can be decided in O(1.619n) time [BH13].)
Open problem
Can directed Hamiltonicity be decided in O(cn) time for some c<2?
Directed Hamiltonicity can be decided in O(3n-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(1.733n) time.
Our result
We will only show the bipartite case in this presentation.
More than just a shaving from O(1.888n) to O(1.733n): The first O(cn) time algorithm for c<2 using only polynomial space. Allows efficient parallel computation. Smooth parameterization “immediately above” the bipartite case.
Why care about an incremental result?
Max Cut in O(2⍵n/3) time [W04]. Uses O(2
2n/3) space.
Can be reduced to O(2
n/3) [BK16]. No known
polynomial space faster than O(2
n).
Chromatic number in O(2
n) time [BHK08]. Uses O(2 n)
- space. Can be reduced to O(1.3
n) [BHKK11
]. No known polynomial space faster than O(2.2
n).
Subset sum in O(2
n/2) time [HS74]. Uses O(2 n/2) space.
Can be reduced to O(2
n/ 4) space [SS79]. Was no known
polynomial space faster than O(2
n), now there is:
Recall Jesper’s talk!
Examples of speedups based on tabulation
Space matters
Algebraic fingerprinting: Construct a polynomial that is not identically zero whenever there is a Hamiltonian cycle. Characteristic two: Use the field GF(2
k) so that an even
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
Familiar
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
- utgoing arcs. Columns represent labels.
Key ideas in the new algorithm
Novel?
The basic case: a bipartite graph
A special vertex
s
A selected arc is labeled by one of its two endpoint
either
- r
Arc is associated with its origin Arc is associated with its target
Notation:
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
- utgoing arc.
Counting arc labeled cycle cover
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
Any cycle not through s has two label association orientations.
Origin labeled Target labeled
Call the blue vertices B. For every arc (i,j) we introduce a variable xi,j. The set O is the allowed origins in B, and the set T is the allowed targets in B.
The determinant summation formula
X
H∈HamCycles(G)
Y
(i,j)∈H
xi,j = X
O,T ⊆B
det(MO,T ).
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 MO,T
s t u v w x s s t t u u v v w w x x a b c d e f
Labels Arcs
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The left side of the matrix
s s t t u u v v w w x x
⇢ xb,t : b ∈ O : otherwise.
s t u v w x a b c d e f
Labels Arcs
⇢ xw,c : c ∈ T : otherwise.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The right side of the matrix
s s t t u u v v w w x x
X
i∈T
xt,i
X
i∈O
xi,w
s t u v w x a b c d e f
Labels Arcs
. . . a . . d . . . . . . . . . a . . . . . . f . . . c . . . . f A monomial in the determinant expansion
s s t t u u v v w w x x s t u v w x a b c d e f
Labels Arcs
A monomial in the determinant expansion
s t u v x w a b c d e f
Not necessarily a cycle cover!
Non-cycle covers will be counted an even number of times
s t u v x w a b c d e f
There are no targets in b and e: Counted for T’s with and without b and e!
Non-cycle covers will be counted an even number of times
s t u v x w a b c d e f
There is no target in e and no origin in d: Counted for T’s with and without e, and O’s with and without d!
Randomly assign xuv from GF(2k). Compute If non-zero conclude Hamiltonian.
Bipartite algorithm
X
O,T ⊆B O∪T =B
det(MO,T ).
Polynomial Identity Testing
Compute matching. If no perfect matching
- utput non-Hamiltonian.
Find mis by any sufficiently fast algorithm. Trying all subsets of endpoints of the perfect matching will do (O(3n/2) time). Do P .I.T. on a determinant formula very similar to the bipartite case.
Mis parameterised algorithm
Assume n even
Thank you for listening!
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- exclusion. SICOMP 2008.
[BHKK11 ] A.Björklund, T.Husfeldt, P .Kaski, and M.Koivisto. Covering and packing in linear
- space. IPL 2011.
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[HS74] E.Horowitz and S.Sahni. Computing partitions with applications to the knapsack
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[SS79] R.Schroeppel and A.Shamir. A T*S2=0 (2n) time/space tradeoff for certain NP- complete problems. FOCS 1979. [W04] R.Williams. A new algorithm for optimal 2-constraint satisfaction and its
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