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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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Presentation on theme: "Constrained Bipartite Vertex Cover: The Easy Kernel is Essentially Tight Bart M. P. Jansen February 18th, STACS 2016, OrlΓ©ans, France."β€” Presentation transcript:

1 Constrained Bipartite Vertex Cover: The Easy Kernel is Essentially Tight
Bart M. P. Jansen February 18th, STACS 2016, OrlΓ©ans, France

2 The Constrained Bipartite Vertex Cover problem
Input: Bipartite graph 𝐺=(𝐴βˆͺ𝐡, 𝐸) and π‘˜ 𝐴 , π‘˜ 𝐡 βˆˆβ„• Question: Does 𝐺 have a vertex cover 𝑆 such that π‘†βˆ©π΄ ≀ π‘˜ 𝐴 , π‘†βˆ©π΅ ≀ π‘˜ 𝐡 ? NP-complete, applications in reconfigurable VLSI Differs from Constrained Minimum Bipartite Vertex Cover: Is there a minimum vertex cover 𝑆 in 𝐺 for which π‘†βˆ©π΄ ≀ π‘˜ 𝐴 and π‘†βˆ©π΅ ≀ π‘˜ 𝐡 ? π‘˜ 𝐴 =3 π‘˜ 𝐡 =4

3 The easy kernel π‘˜ 𝐴 =2 π‘˜ 𝐡 =4 π‘˜ 𝐴 =3 π‘˜ 𝐡 =4
If there is a vertex π‘Žβˆˆπ΄ with deg π‘Ž > π‘˜ 𝐡 : If π‘Ž is not in the cover, then all π‘Žβ€™s neighbors must be We cannot afford that, since there are more than π‘˜ 𝐡 Any solution contains π‘Ž: delete π‘Ž and decrease π‘˜ 𝐴 by 1 Similarly, if there is a vertex π‘βˆˆπ΅ with deg 𝑏 > π‘˜ 𝐴 : Delete 𝑏, decrease π‘˜ 𝐡 by one π‘˜ 𝐴 =2 π‘˜ 𝐡 =4 π‘˜ 𝐴 =3 π‘˜ 𝐡 =4

4 Analysis of the easy kernel
If (𝐺= 𝐴,𝐡,𝐸 , π‘˜ 𝐴 , π‘˜ 𝐡 ) is exhaustively reduced: The π‘˜ 𝐴 vertices from 𝐴 cover at most π‘˜ 𝐡 edges each The π‘˜ 𝐡 vertices from 𝐡 cover at most π‘˜ 𝐴 edges each A yes-instance has at most 2 π‘˜ 𝐴 β‹… π‘˜ 𝐡 edges Therefore at most 4( π‘˜ 𝐴 β‹… π‘˜ 𝐡 ) vertices This easy kernel was first given by Evans (1981) Also helps to get fast FPT algorithms 𝑂( π‘˜ 𝐴 + π‘˜ 𝐡 + π‘˜ 𝐴 + π‘˜ 𝐡 ⋅𝑛)) by Fernau and Niedermeier Implemented and re-engineered by Bai and Fernau Can the kernel size be improved?

5 Our results (I) Both the number of vertices and edges of the easy kernel is essentially tight in terms of the product π‘˜ 𝐴 β‹… π‘˜ 𝐡 If 𝑁𝑃 is not in π‘π‘œπ‘π‘ƒ/π‘π‘œπ‘™π‘¦, there is no polynomial-time algorithm that reduces an instance (𝐺= 𝐴,𝐡,𝐸 , π‘˜ 𝐴 , π‘˜ 𝐡 ) of Con. Bip. VC to an instance ( 𝐺 β€² = 𝐴 β€² , 𝐡 β€² , 𝐸 β€² , π‘˜ 𝐴 β€² , π‘˜ 𝐡 β€² ) such that the instances are equivalent, π‘˜ 𝐴 β€² ≀ π‘˜ 𝐴 𝑂 1 , π‘˜ 𝐡 β€² ≀ π‘˜ 𝐡 𝑂 1 , and 𝑉 𝐺 β€² βˆˆπ‘‚( π‘˜ 𝐴 β‹… π‘˜ 𝐡 1βˆ’πœ– ) for some πœ–>0

6 Our results (II) Both the number of vertices and edges of the easy kernel is essentially tight in terms of the product π‘˜ 𝐴 β‹… π‘˜ 𝐡 If 𝑁𝑃 is not in π‘π‘œπ‘π‘ƒ/π‘π‘œπ‘™π‘¦, there is no polynomial-time algorithm that reduces an instance (𝐺= 𝐴,𝐡,𝐸 , π‘˜ 𝐴 , π‘˜ 𝐡 ) of Con. Bip. VC to an instance π‘₯ of an arbitrary problem 𝐿 such that the instances are equivalent, π‘₯ βˆˆπ‘‚( 𝑛 2βˆ’πœ– ) for some πœ–>0, where 𝑛≔|𝑉 𝐺 | Implies that the number of edges in an instance of Con. Bip. VC cannot be reduced to 𝑂 π‘˜ 𝐴 β‹… π‘˜ 𝐡 1βˆ’πœ– , unless π‘π‘ƒβŠ† π‘π‘œπ‘π‘ƒ/π‘π‘œπ‘™π‘¦

7 Proof: Vertex lower bound

8 NP-Completeness proof
Reduction from a Clique instance (𝐺,π‘˜) [Kuo and Fuchs’87] Build 𝐺 β€² by subdividing each edge by a new vertex Let 𝐴′ be the original vertices of 𝐺 and 𝐡’ the subdividers Put π‘˜β€² 𝐴 β‰”π‘˜ and π‘˜β€² 𝐡 ≔ 𝐸 𝐺 βˆ’ π‘˜ 2 π‘˜=4 π‘˜β€² 𝐴 =4 π‘˜ 𝐡 β€² =7

9 Key construction There is an algorithm with the following specifications Input: A list of 𝑑 graphs 𝐺 1 , …, 𝐺 𝑑 with exactly 𝑛 vertices each, where 𝑛 is even and 𝑑 is a power of two Output: A bipartite graph 𝐺 β€² = 𝐴 β€² βˆͺ 𝐡 β€² , 𝐸 β€² along with integers π‘˜ 𝐴 β€² , π‘˜ 𝐡 β€² such that: βˆƒπ‘–βˆˆ 𝑑 such that 𝐺 𝑖 contains a clique of size 𝑛/2 ⇔ βˆƒvertex cover 𝑆 of 𝐺’ with π‘†βˆ© 𝐴 β€² ≀ π‘˜ 𝐴 β€² , π‘†βˆ© 𝐡 β€² ≀ π‘˜ 𝐡 β€² π‘˜ 𝐴 β€² βˆˆπ‘‚( 𝑛 2 log 𝑑) π‘˜ 𝐡 β€² βˆˆπ‘‚(𝑛⋅𝑑) The running time is polynomial in 𝑑 and 𝑛

10 Sketch of key construction
Input: A list of 𝑑 graphs 𝐺 1 , …, 𝐺 𝑑 with exactly 𝑛 vertices each, where 𝑛 is even and 𝑑 is a power of two. Output: Blocks in 𝐴 represent 0/1 bit values for log⁑𝑑 bit positions Blocks in 𝐡 represent input graphs Connect 𝐡 𝑖 to bit values based on binary expansion of 𝑖 Connect 𝐡 𝑖 to subdividers for edges that do not exist in 𝐺 𝑖 Weight 𝑛 2 per block, 2log 𝑑 blocks Weight 𝑛 per block, 𝑑 blocks

11 Sketch of key construction
Input: A list of 𝑑 graphs 𝐺 1 , …, 𝐺 𝑑 with exactly 𝑛 vertices each, where 𝑛 is even and 𝑑 is a power of two. Output: Bit selectors force one block 𝐡 𝑖 to be chosen In the canonical part, this disables the subdividers for non-existing edges in 𝐺 𝑖 This makes the left part act as the result of the NP-completeness reduction for Clique-instance 𝐺 𝑖 Weight 𝑛 2 per block, 2log 𝑑 blocks Weight 𝑛 per block, 𝑑 blocks Summary. We embed 𝑑 instances of Clique into one instance of Constrained Bipartite Vertex Cover with π‘˜ 𝐴 βˆˆπ‘‚( 𝑛 2 log 𝑑) and π‘˜ 𝐡 βˆˆπ‘‚(𝑑⋅𝑛) that is yes if and only if a Clique-input is yes

12 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 πΏβ€™βŠ† Ξ£ βˆ— be a language. If there is a polynomial-time algorithm as follows: Input: list of 𝑑= 𝑛 100 graphs 𝐺 1 ,…, 𝐺 𝑑 with 𝑛 vertices each, Output: string π‘₯ βˆ— such that π‘₯ βˆ— ∈ 𝐿 β€² if and only if some 𝐺 𝑖 has a clique of size 𝑛/2, and π‘₯ βˆ— βˆˆπ‘‚ 𝑛 100 , then π‘π‘ƒβŠ†π‘π‘œπ‘π‘ƒ/π‘π‘œπ‘™π‘¦. By the pigeon-hole principle, there is an input from which only 𝑂(1) bits remain

13 Simplified vertex lower bound
If 𝑁𝑃 is not in π‘π‘œπ‘π‘ƒ/π‘π‘œπ‘™π‘¦, there is no polynomial-time algorithm that reduces an instance (𝐺= 𝐴,𝐡,𝐸 , π‘˜ 𝐴 , π‘˜ 𝐡 ) of Con. Bip. VC to an instance ( 𝐺 β€² = 𝐴 β€² , 𝐡 β€² , 𝐸 β€² , π‘˜ 𝐴 β€² , π‘˜ 𝐡 β€² ) such that the instances are equivalent, π‘˜ 𝐴 β€² ≀ π‘˜ 𝐴 , π‘˜ 𝐡 β€² ≀ π‘˜ 𝐡 , and 𝑉 𝐺 β€² βˆˆπ‘‚( π‘˜ 𝐴 β‹… π‘˜ 𝐡 ) Proof. Assume such a kernelization algorithm 𝒦 exists Using 𝒦, the key construction, and the easy kernel, we build a compression algorithm for Clique instances π‘π‘ƒβˆˆπ‘π‘œπ‘π‘ƒ/π‘π‘œπ‘™π‘¦ by complementary witness lemma

14 Compression algorithm for Clique instances
𝐺 1 𝐺 2 𝐺 𝑑= 𝑛 100 π‘˜ 𝐴 βˆˆπ‘‚( 𝑛 2 log 𝑑) π‘˜ 𝐡 βˆˆπ‘‚ 𝑛⋅𝑑 =𝑂( 𝑛 ) In: OR … Reduce |V| to 𝑂 π‘˜ 𝐴 β‹… π‘˜ 𝐡 ≀ 𝑂 (𝑛 103 log 𝑛 ) 𝒦 deg π‘βˆˆ 𝐡 βˆ— ≀ π‘˜ 𝐴 β€² βˆˆπ‘‚( 𝑛 2 log 𝑛) 𝐴’ 𝐡′ ≀|𝑉′|βˆˆπ‘‚( 𝑛 93 ) Out: Evans 𝐸 βˆ— ≀ 𝐡 βˆ— β‹… deg π‘βˆˆ 𝐡 βˆ— 𝐸 βˆ— ≀ 𝐡 β€² β‹… π‘˜ 𝐴 𝐸 βˆ— βˆˆπ‘‚( 𝑛 93 β‹… 𝑛 2 log 𝑛) βˆˆπ‘‚ 𝑛 96 π‘˜ 𝐴 β€² ≀ π‘˜ 𝐴 βˆˆπ‘‚( 𝑛 2 log 𝑛)

15 Discussion The given proof contains all the ideas of the general proof
More careful math makes it work for all πœ–>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

16 Conclusion The easy kernel for Con. Bip. VC is essentially tight in terms of π‘˜ 𝐴 β‹… π‘˜ 𝐡 Both the number of vertices and edges Compare to the classic Vertex Cover case: The easy kernel (Buss’ 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 𝑂( π‘˜ 𝐴 + π‘˜ 𝐡 ) vertices? Does Feedback Vertex Set admit a kernel with 𝑂( π‘˜ 2βˆ’πœ– ) vertices? THANK YOU!


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