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Constrained Bipartite Vertex Cover: The Easy Kernel is Essentially Tight
Bart M. P. Jansen February 18th, STACS 2016, OrlΓ©ans, France
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The Constrained Bipartite Vertex Cover problem
Input: Bipartite graph πΊ=(π΄βͺπ΅, πΈ) and π π΄ , π π΅ ββ Question: Does πΊ have a vertex cover π such that πβ©π΄ β€ π π΄ , πβ©π΅ β€ π π΅ ? NP-complete, applications in reconfigurable VLSI Differs from Constrained Minimum Bipartite Vertex Cover: Is there a minimum vertex cover π in πΊ for which πβ©π΄ β€ π π΄ and πβ©π΅ β€ π π΅ ? π π΄ =3 π π΅ =4
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The easy kernel π π΄ =2 π π΅ =4 π π΄ =3 π π΅ =4
If there is a vertex πβπ΄ with deg π > π π΅ : If π is not in the cover, then all πβs neighbors must be We cannot afford that, since there are more than π π΅ Any solution contains π: delete π and decrease π π΄ by 1 Similarly, if there is a vertex πβπ΅ with deg π > π π΄ : Delete π, decrease π π΅ by one π π΄ =2 π π΅ =4 π π΄ =3 π π΅ =4
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Analysis of the easy kernel
If (πΊ= π΄,π΅,πΈ , π π΄ , π π΅ ) is exhaustively reduced: The π π΄ vertices from π΄ cover at most π π΅ edges each The π π΅ vertices from π΅ cover at most π π΄ edges each A yes-instance has at most 2 π π΄ β
π π΅ edges Therefore at most 4( π π΄ β
π π΅ ) vertices This easy kernel was first given by Evans (1981) Also helps to get fast FPT algorithms π( π π΄ + π π΅ + π π΄ + π π΅ β
π)) by Fernau and Niedermeier Implemented and re-engineered by Bai and Fernau Can the kernel size be improved?
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Our results (I) Both the number of vertices and edges of the easy kernel is essentially tight in terms of the product π π΄ β
π π΅ If ππ is not in ππππ/ππππ¦, there is no polynomial-time algorithm that reduces an instance (πΊ= π΄,π΅,πΈ , π π΄ , π π΅ ) of Con. Bip. VC to an instance ( πΊ β² = π΄ β² , π΅ β² , πΈ β² , π π΄ β² , π π΅ β² ) such that the instances are equivalent, π π΄ β² β€ π π΄ π 1 , π π΅ β² β€ π π΅ π 1 , and π πΊ β² βπ( π π΄ β
π π΅ 1βπ ) for some π>0
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Our results (II) Both the number of vertices and edges of the easy kernel is essentially tight in terms of the product π π΄ β
π π΅ If ππ is not in ππππ/ππππ¦, there is no polynomial-time algorithm that reduces an instance (πΊ= π΄,π΅,πΈ , π π΄ , π π΅ ) of Con. Bip. VC to an instance π₯ of an arbitrary problem πΏ such that the instances are equivalent, π₯ βπ( π 2βπ ) for some π>0, where πβ|π πΊ | Implies that the number of edges in an instance of Con. Bip. VC cannot be reduced to π π π΄ β
π π΅ 1βπ , unless ππβ ππππ/ππππ¦
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Proof: Vertex lower bound
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NP-Completeness proof
Reduction from a Clique instance (πΊ,π) [Kuo and Fuchsβ87] Build πΊ β² by subdividing each edge by a new vertex Let π΄β² be the original vertices of πΊ and π΅β the subdividers Put πβ² π΄ βπ and πβ² π΅ β πΈ πΊ β π 2 π=4 πβ² π΄ =4 π π΅ β² =7
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Key construction There is an algorithm with the following specifications Input: A list of π‘ graphs πΊ 1 , β¦, πΊ π‘ with exactly π vertices each, where π is even and π‘ is a power of two Output: A bipartite graph πΊ β² = π΄ β² βͺ π΅ β² , πΈ β² along with integers π π΄ β² , π π΅ β² such that: βπβ π‘ such that πΊ π contains a clique of size π/2 β βvertex cover π of πΊβ with πβ© π΄ β² β€ π π΄ β² , πβ© π΅ β² β€ π π΅ β² π π΄ β² βπ( π 2 log π‘) π π΅ β² βπ(πβ
π‘) The running time is polynomial in π‘ and π
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Sketch of key construction
Input: A list of π‘ graphs πΊ 1 , β¦, πΊ π‘ with exactly π vertices each, where π is even and π‘ is a power of two. Output: Blocks in π΄ represent 0/1 bit values for logβ‘π‘ bit positions Blocks in π΅ represent input graphs Connect π΅ π to bit values based on binary expansion of π Connect π΅ π to subdividers for edges that do not exist in πΊ π Weight π 2 per block, 2log π‘ blocks Weight π per block, π‘ blocks
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Sketch of key construction
Input: A list of π‘ graphs πΊ 1 , β¦, πΊ π‘ with exactly π vertices each, where π is even and π‘ is a power of two. Output: Bit selectors force one block π΅ π to be chosen In the canonical part, this disables the subdividers for non-existing edges in πΊ π This makes the left part act as the result of the NP-completeness reduction for Clique-instance πΊ π Weight π 2 per block, 2log π‘ blocks Weight π per block, π‘ blocks Summary. We embed π‘ instances of Clique into one instance of Constrained Bipartite Vertex Cover with π π΄ βπ( π 2 log π‘) and π π΅ βπ(π‘β
π) that is yes if and only if a Clique-input is yes
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Complementary witness lemma
Lemma saying: [Dell & van Melkebeek, 2010] Sufficiently good compression for hard problems implies an unlikely complexity-theoretic collapse Lemma (simplified version). Let πΏββ Ξ£ β be a language. If there is a polynomial-time algorithm as follows: Input: list of π‘= π 100 graphs πΊ 1 ,β¦, πΊ π‘ with π vertices each, Output: string π₯ β such that π₯ β β πΏ β² if and only if some πΊ π has a clique of size π/2, and π₯ β βπ π 100 , then ππβππππ/ππππ¦. By the pigeon-hole principle, there is an input from which only π(1) bits remain
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Simplified vertex lower bound
If ππ is not in ππππ/ππππ¦, there is no polynomial-time algorithm that reduces an instance (πΊ= π΄,π΅,πΈ , π π΄ , π π΅ ) of Con. Bip. VC to an instance ( πΊ β² = π΄ β² , π΅ β² , πΈ β² , π π΄ β² , π π΅ β² ) such that the instances are equivalent, π π΄ β² β€ π π΄ , π π΅ β² β€ π π΅ , and π πΊ β² βπ( π π΄ β
π π΅ ) Proof. Assume such a kernelization algorithm π¦ exists Using π¦, the key construction, and the easy kernel, we build a compression algorithm for Clique instances ππβππππ/ππππ¦ by complementary witness lemma
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Compression algorithm for Clique instances
πΊ 1 πΊ 2 πΊ π‘= π 100 π π΄ βπ( π 2 log π‘) π π΅ βπ πβ
π‘ =π( π ) In: OR β¦ Reduce |V| to π π π΄ β
π π΅ β€ π (π 103 log π ) π¦ deg πβ π΅ β β€ π π΄ β² βπ( π 2 log π) π΄β π΅β² β€|πβ²|βπ( π 93 ) Out: Evans πΈ β β€ π΅ β β
deg πβ π΅ β πΈ β β€ π΅ β² β
π π΄ πΈ β βπ( π 93 β
π 2 log π) βπ π 96 π π΄ β² β€ π π΄ βπ( π 2 log π)
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Discussion The given proof contains all the ideas of the general proof
More careful math makes it work for all π>0 We rely crucially on the fact that the budgets do not increase The lower bound construction produces a lopsided graph One side is much larger than the other Result II (sparsification bound) uses a balanced construction Both sides of the bipartite graph have the same size
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Conclusion The easy kernel for Con. Bip. VC is essentially tight in terms of π π΄ β
π π΅ Both the number of vertices and edges Compare to the classic Vertex Cover case: The easy kernel (Bussβ rule) gives tight bounds on the number of edges The number of vertices in the easy kernel can be improved to 2π Open problems: Does Constrained Bipartite Vertex Cover have a kernel with π( π π΄ + π π΅ ) vertices? Does Feedback Vertex Set admit a kernel with π( π 2βπ ) vertices? THANK YOU!
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