Benjamin Doerr Partial Colorings of Unimodular Hypergraphs Partial
Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Partial Colorings of Unimodular Hypergraphs Overview Introduction Hypergraphs Coloring hypergraphs (discrepancy) Unimodular hypergraphs Partial coloring Partially coloring unimodular hypergraphs Motivation Result Application Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Introduction Hypergraphs Hypergraph: H E = (V; ) V µ set of vertices E: finite 2 V : set of hyperedges j j V =5 vertices j. Ej =4 hyperedges Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Introduction Hypergraphs Hypergraph: H E = (V; ) V µ set of vertices E: finite 2 V : set of hyperedges H f j 2 Eg Induced subhypergraph: = V 0 (V 0 ; E V 0 E ) ) Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Introduction Discrepancy of Hypergraphs !P f¡ g all hyperedges are balanced: Color vertices s. t. 1; +1 : V 2 (v) “ 2 -coloring” (E) : = v E H j j “imbalance of hyperedge E” 2 E disc( ; ) : = max. E (E) H H disc( ) : = min disc( ; ) +1 +1 ¡ H j H¡ j = (E) ); ) 1)=+ ( ; ) = disc( =1¡+ =1(1+ 1) disc( (E) 1=2 = 1 +1 -1 -1 Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Introduction Unimodular Hypergraphs H Def: unimodular iff each induced subhypergraph has discrepancy at most one. … · · · g = ([n]; [i : : : j] 1 i j n ) H · disc( ) 1 j j ) Remark: means E =0 j j even ) (E) j j “perfectly balanced” = E (E) 1 H odd Benjamin. Doerr f j “almost perfect”, “ 1” cannot be avoided The queen of low-discrepancy hypergraphs! Partial Colorings of Unimodular Hypergraphs
Introduction Unimodular Hypergraphs: Examples f Intervals in g [n] : = 1; : : : ; n . £ Rows/Columns in a grid: = V E [m] [n] £f gj 2 g ff g £ j 2 g[f = i j j [n] i [m] Bipartite graphs. Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Introduction Partial Coloring H disc( ) = 1 Observe: is “caused” by the “odd” vertex in odd-cardinality hyperedges. +1 -1 ? ! f¡ g vertices! Plan: Don’t color all : V P 1; 0; +1 (v) = 0 “partial coloring” v vertices with are “uncolored” 2 = (E) (v) ! Hv E , . . . as before f g : V 0 disc( ; ) = 0 Aim: , but “Nice partial coloring” +1 -1 0 doesn’t count! [Beck’s partial coloring method (1981)] Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Partial Colorings of Unimodular Hypergraphs Existence of Nice Partial Colorings? Clearly, not all hypergraphs have nice partial colorings: Complete hypergraphs Projective planes, hypergraphs constructed from Hadamard matrices (proof: the Eigenvalue argument works also for partial colorings) Topic of this talk: Do at least unimodular hypergraphs have nice partial colorings? Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Partial Colorings of Unimodular Hypergraphs Unimodular hgs with no nice partial coloring H H H ff gj 2 g = ([n]; i i [n] ) f j 2 g = ([n]; [i] i [n] ) “singletons” “initial intervals” _ ¡ j ¡ g = ([n]; [i: : j] j i = 2 j i = 4 ) f “intervals of length 3 and 5” No hope for partial coloring? Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Partial Colorings of Unimodular Hypergraphs Sometimes it works: H +1 = ([n]; 0 gj 2 ¡ g i; i + 1; i + 2 i [n 2] ) ff -1 +1 0 “length 3 intervals” -1 Rows and columns in the grid. Uniform unimodular hypergraphs: All hyperedges contain the same number of vertices (needs proof). Question: When are there nice partial colorings? Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Partial Colorings of Unimodular Hypergraphs Result H following two properties are equivalent: The (i) has a perfectly balanced non-trivial (“nice”) partial coloring; !f ¡ g w: V 0; 1=k; : : : (k 1)=k P vertex weights (ii) there an integer k and non-trivial have integral weight w(E) = such that all hyperedges 2 v E w(v) . 3/5 1/5 Remark: The partial in (i) colors at least half of =0 w(u)coloring the vertices with in (ii). 1/5 2/5 Application: “Randomly rounding rationals is as easy as rounding half-integers” [STACS 2007 ] 3/5 Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
Application ¼ all IF: For Ax Ay Bx = By 2 f g x 0; 1=2 n there is a RR y g 0; 1 n [no rounding error w. r. t. totally unimodular 2 f matrix g B] ¼ For all rational there is a RR THEN: y 0; 1 Ay Bx = By 2 f “Proof”: x g 0; 1=k; 2=k; : : : n Bx 2 f¡ gintegral such that B = 0 1; 0; 1 n 2 f coloring: g Exists Partial such that =0 = x ~ 0; 1=2 n x ~i ¡ 1=2 i such that iff = x : x x + (2=k)~ y~ x ~ 2 f g RR of as above, x 0; 1 n Repeat until Benjamin. Doerr such that [low rounding errors w. r. t. matrix A] x Ax 2 f . Partial Colorings of Unimodular Hypergraphs n s. t.
Partial Colorings of Unimodular Hypergraphs Summary H following two properties are equivalent: The (i) has a perfectly balanced non-trivial partial coloring; !f ¡ g w: V 0; 1=k; : : : (k 1)=k P vertex weights (ii) there an integer k and non-trivial have integral weight w(E) = such that all hyperedges 2 v E w(v) . 3/5 1/5 Remark: The partial in (i) colors at least half of =0 w(u)coloring 1/5 the vertices with in (ii). [Open Problem: How many? ] 2/5 Author claims an application. 3/5 Thanks! Benjamin. Doerr Partial Colorings of Unimodular Hypergraphs
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