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Parameterized Complexity of Conflict-free Graph Coloring
WADS 2019 Hans L. Bodlaender1, Sudeshna Kolay2, and Astrid Pieterse3 1 Utrecht University, The Netherlands 2 Ben Gurion University of Negev, Israel 3 Eindhoven University of Technology, The Netherlands
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Conflict-free graph coloring
π-CNCF-Coloring Input: A graph πΊ Question: Is it possible to assign every vertex in πΊ a color from {1,β¦,π}, such that for all π£, there is a color occurring exactly once in π[π£]? Related: π-ONCF-Coloring Considers open neighborhoods instead π-Closed Neighborhood Conflict-Free Coloring z z y y u u v v w w x x
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CNCF-Coloring How many colors do we need to CNCF-color
A clique on π vertices? 2 u u v v w w y y x x
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CNCF-Coloring How many colors do we need to CNCF-color
A clique on π vertices? A bipartite graph? 2 2 w w u u x x v v y y
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CNCF-Coloring How many colors do we need to CNCF-color
A clique on π vertices? A bipartite graph? An odd cycle? 2 2 u u v v w w y y x x
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CNCF-Coloring How many colors do we need to CNCF-color
A clique on π vertices? A bipartite graph? An odd cycle? Petersen graph 2 2 2
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CNCF-Coloring How many colors do we need to CNCF-color
A clique on π vertices? A bipartite graph? An odd cycle? Petersen graph A π-colorable graph? 2 2 k 2 β€ k β€ 2 β€ 3 β€ 4 Observation Any proper coloring is a CNCF-Coloring Each vertex is the uniquely colored vertex in its closed neighborhood
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Background Special case of conflict-free coloring set systems
Sets given by closed (or open) neighborhoods Frequency assignment problems Studied from a combinatorial perspective CNCF-coloring π-vertex graph takes π( log 2 π ) colors [Pach, Tardos 2009] Tight [Glebov, SzabΓ³, Tardos, 2014]
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Combinatorial results
As seen earlier π πΆπ (πΊ)β€π(πΊ) Immediately π πΆπ πΊ β€π(πΊ)β€π‘π€ πΊ +1β€ππ£π πΊ +1β€π£π πΊ +1 The same does not hold for π ππ . We obtain the following π ππ πΊ β€2π‘π€ πΊ +1 π ππ πΊ β€ππ£π πΊ +3 π ππ πΊ β€π£π πΊ +1 π ππ πΊ β€π£π(πΊ), unless πΊ is a star or edge-star β¦ β¦
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Background NP-hard for πβ₯2 [Gargano and Rescigno, TCS, 2015]
Study parameterized complexity and kernelizability FPT parameterized by [Gargano and Rescigno, TCS, 2015] Vertex Cover Neighborhood diversity Treewidth
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Algorithmic results 2 π 2 π‘π€ π π 1 algorithm
for both π-ONCF and π-CNCF coloring Lower bounds for πβ₯3 No πβπ π‘π€ π π 1 algorithm under SETH For both problems For π=2 No 2 π π‘π€ π π 1 algorithm under ETH
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Kernelization results
Kernelization parameterized by Vertex Cover 2-CNCF-Coloring has a polynomial kernel (up next) π-CNCF-Coloring has no polynomial kernel1 for πβ₯3 π-ONCF-Coloring has no polynomial kernel1 for πβ₯2 Both results proven by cross-composition See 1Unless ππβππππ/poly Polynomial kernelization There is a polynomial-time algorithm that does the following Input Instance (π₯,π) Output Instance ( π₯ β² , π β² ) such that: π₯ β² β€ππππ¦ π πβ²β€ππππ¦(π) ( π₯ β² , π β² ) is a yes-instance if and only if (π₯,π) is Efficient Small bound Correct
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A general reduction rule
For π-CNCF-Coloring [Based on Gargano and Rescigno, TCS 2015, Lemma 6] Start at 17 min
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Reducing number of twins
Let π β² βπ. Suppose there are >π+1 vertices π£βπ with π π£ =πβ². Mark π+1, remove the others. π=3 s ο ο ο ο ο πβ
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Correctness Let πΊβ² be the resulting graph
Suppose πΊ is π-CNCF-Colorable Color πΊβ² similarly, ensuring that each vertex in π keeps its conflict-free neighbor (if any) πΊ πΊβ² π s π s πβ πβ
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Correctness Suppose πΊβ² is π-CNCF-Colorable Use the same coloring on πΊ
Let π£ be a removed vertex, color π£ using a color already used twice on marked vertices Observe that such a color exists! πΊ πΊβ² π s π s πβ πβ
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Effectiveness Suppose we apply this rule exhaustively, for all π β² βπ
Number of vertices in π π by definition Number of vertices not in π Number of degree-π vertices not in π At most π+1 π π =π( π π ) Total π( 2 π ) Exponential, unless we can somehow bound the number of high-degree vertices
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Polynomial kernel for 2-CNCF-Coloring
Parameterized by Vertex Cover size Start at 17 min
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A generalized kernel We give a generalized kernel to π-Polynomial root CSP Input: A set πΏ of equalities over variables π, where each equality is of the form π π₯ 1 ,β¦, π₯ π =0, where π is a polynomial of degree at most π Parameter: The number of variables π Question: Does there exist an assignment π:πβ{0,1} satisfying all equalities in πΏ? Example of 2-Poly root CSP: { π₯ 1 + π₯ 2 β1=0, π₯ 1 β π₯ 2 +π₯ 2 β π₯ 3 =0} Satisfied by π π₯ 1 =π π₯ 3 =0, π π₯ 2 =1 Theorem [Jansen and Pieterse, MFCS 2016] π-Polynomial root CSP has a kernel with π( π π ) equalities, where π is the number of variables That is a subset of the original set of equalities!
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Kernelization: General idea
Three steps Reduce the number of low-degree vertices Rewrite the problem to an instance of π-Poly root CSP For some constant π Using not too many variables Tricky! Apply (known) kernelization result for π-Poly root CSP π-Polynomial root CSP has a kernel with π( #π£πππ π ) equalities
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Removing low-degree vertices
Apply known reduction rule: For every π β² βπ of size at most 2 Mark 3 vertices π’βπ such that π π’ =πβ² If there are less than three, mark all Delete all unmarked vertices of degree 1 and 2 Number of vertices in π π (by definition) Number of degree-β€2 vertices not in π At most 3k+3 π 2 = π(π 2 ) Number of degree-β₯3 vertices not in π Possibly many ο
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Removing low-degree vertices
Add these to π Technically, this increases |π| to π( π 2 ) We ignore this for simplicity
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Rewriting: Basics Creating an instance of π-Poly root CSP (for some π)
For each vertex π£, create variables π π£ and π π£ π π£ =1 means π£ is red, π π£ =1 means it is blue Add the constraint that π π£ + π π£ =1 A constraint on the coloring of π[π£] for all π£ Exactly one blue, or exactly one red vertex Thus, π’βπ[π£] π π’ =1 , or π’βπ[π£] π π’ =1 For all π£ add the constraint 1β π’βπ π£ π π’ 1β π’βπ π£ π π’ =0
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Rewriting: Continued So far, variables π π£ , π π£ π£βπ πΊ , constraints
For all π£: π π£ + π π£ =1 For all π£: 1β π’βπ π£ π π’ 1β π’βπ π£ π π’ =0 Hereby The two problem instances are equivalent We use low-degree polynomials (degree-2) As many variables as vertices Using the known kernel for π-Poly root CSP gives a kernel of size π( π 2 ) (useless) Plan: reduce the number of variables to O(π)?
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Reducing the number of variables
Recall, each vertex π£βπ has degree at least 3 In a valid 2-CNCF coloring, there are only the following options for the coloring of π(π£) The coloring of π£ is determined by the coloring of π(π£)! But, the coloring of π(π£) is not known S S S S π£ π£ π£ π£
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Reducing the number of variables
Write π π£ as π π π’ 1 ,β¦, π π’ π , π π’ 1 ,β¦, π π’ π if π π£ ={ π’ 1 ,β¦, π’ π } For low-degree polynomial π Then π π£ =1 β π π£ =1βπ(β¦) Substituting π π£ by π(β¦), reduces the number of variables to 2 π =2π
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Finding π First, ensure that the neighborhood of π£βπ is βokβ
All blue, all red, one red, or one blue Done by an additional equality of degree 4 for π£βπ ( π’βπ π£ π π’ )( π’βπ π£ π π’ )(1β π’βπ π£ π π’ )(1β π’βπ π£ π π’ ) =0 Defining π π π€ + π π₯ + π π¦ + π π§ =4 implies π π£ =0 π π€ + π π₯ + π π¦ + π π§ =3 implies π π£ =1 π π€ + π π₯ + π π¦ + π π§ =1 implies π π£ =0 π π€ + π π₯ + π π¦ + π π§ =0 implies π π£ =1 Let π such that π π₯ =0 if π₯β{1,4} and π π₯ =1 if π₯β{0,3} Substitute π π£ =π π π€ ,π π₯ , π π¦ , π π§ =π( π π€ + π π₯ + π π¦ + π π§ ) π€ π₯ π£ π¦ π§
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Finding π In general, for π£βπ, find polynomial π s.t.
π π₯ =0 if π₯β{1, |π π£ |} π π₯ =1 if π₯β{0, |π π£ |β1} Use interpolating polynomial for 1,0 , π π£ ,0 , 0,1 ,{ π π£ β1,1} Has degree 3 Let π π π£ ={ π’ 1 ,β¦, π’ π } Substitute π π£ by π π π’ 1 + π π’ 2 +β¦+ π π’ π in all equalities. Note: the colors in the formulas in the first two lines are wrong. You can change them, but powerpoint refuses to save these changes. How nice.
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Resulting Poly root CSP instance
Variables π π£ , π π£ π£βπ , constraints For all π£: π π£ + π π£ =1 For all π£: 1β π’βπ π£ π π’ 1β π’βπ π£ π π’ =0 With for π£βπ π π£ substituted by π(β¦), π π£ by 1βπ(β¦) For all π£βπ: ( π’βπ π£ π π’ )( π’βπ π£ π π’ )(1β π’βπ π£ π π’ )(1β π’βπ π£ π π’ ) =0
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Kernelization We obtained a π-Poly root CSP instance π=6
On 2π variables (actually, π variables suffices) That is equivalent to the original instance Apply kernel for 6-Poly root CSP Instance with π #π£πππ π =π( π 6 ) equalities Can be encoded in π( π 10 ) bits Theorem 2-CNCF-Coloring parameterized by Vertex Cover has a generalized kernel of size π( π 10 ) Can be turned into normal kernel of polynomial size
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Conclusion 2-CNCF-Coloring parameterized by vertex cover has a polynomial kernel π-CNCF-Coloring for πβ₯3 and π-ONCF-Coloring do not1 Not even for the extension problem FPT algorithm and lower bounds Parameterized by treewidth (Tight) combinatorial bounds Open questions Tight algorithmic bounds parameterized by treewidth Is the π( π 10 ) bound tight for 2-CNCF-Coloring? Probably not THANK YOU 1Unless ππβππππ/poly
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