Random hypergraphs and property B
aa r X i v : . [ m a t h . C O ] F e b RANDOM HYPERGRAPHS AND PROPERTY B
LECH DURAJJAKUB KOZIK
Theoretical Computer Science Department, Faculty of Mathematics and Computer Science,Jagiellonian University, Krak´ow, Poland
DMITRY SHABANOV
Moscow Institute of Physics and Technology, Laboratory of Combinatorial and Geometric Structures,Dolgoprudny, Moscow Region, RussiaNational Research University Higher School of Economics, Faculty of Computer Science, Moscow,Russia
Abstract.
In 1964 Erd˝os proved that (1 + o (1))) e ln(2)4 k k edges are sufficient to build a k -graphwhich is not two colorable. To this day, it is not known whether there exist such k -graphs with smallernumber of edges. Erd˝os’ bound is consequence of the fact that a hypergraph with k / M ( k ) = (1 + o (1)) e ln(2)4 k k randomly chosen edges of size k is asymptotically almost surely not twocolorable. Our first main result implies that for any ε >
0, any k -graph with (1 − ε ) M ( k ) randomlyand uniformly chosen edges is a.a.s. two colorable. The presented proof is an adaptation of the secondmoment method analogous to the developments of Achlioptas and Moore from 2002 who consideredthe problem with fixed size of edges and number of vertices tending to infinity. In the second part ofthe paper we consider the problem of algorithmic coloring of random k -graphs. We show that quitesimple, and somewhat greedy procedure, a.a.s. finds a proper two coloring for random k -graphs on k / O (cid:0) k ln k · k (cid:1) edges. That is of the same asymptotic order as the analogueof the algorithmic barrier defined by Achlioptas and Coja-Oghlan in 2008, for the case of fixed k . Introduction
The smallest number of edges in a k -graph (i.e. k -uniform hypergraph) that is not two colorable isdenoted by m ( k ). Early results by Erd˝os from 60’s [7, 8] determined that the exponential factor of thegrowth of m ( k ) is 2 k (that is log ( m ( k )) ∼ k ). In 1975 Erd˝os and Lov´asz in [9] suggested that perhaps k · k is the correct order of magnitude for m ( k ).Despite a few improvements on the side of the lower bounds (the most recent one by Radhakrishnanand Srinivasan in [12]) the upper bound of Erd˝os from 1964 has not been improved since. He provedin [8] that m ( k ) (1 + o (1)) e ln(2)4 k k . (1) E-mail addresses : [email protected], [email protected], [email protected] .Research of L. Duraj and J. Kozik was partially supported by Polish National Science Center (2016/21/B/ST6/02165).Research of D. Shabanov was supported by grant of the President of Russian Federation no. MD-1562.2020.1 and bythe program ”Leading Scientific Schools”, grant no. NSh- 2540.2020.1.
This bound results from the fact that the random hypergraph with that number of edges, over theset of vertices of size about k /
2, a.a.s. can not be properly two colored. The chosen number of vertices,turns out to give the smallest number of edges in the random construction. Our motivation for takinga closer look on random k -graphs on k / k -graphs with smaller number of edges and discuss the problem of findinga proper coloring efficiently.Analogous problems have been considered in the context of random constraint satisfaction problems,where the size k of constraints/edges was fixed and the number of variables/vertices n was tending toinfinity. In that framework, the straightforward first moment calculation shows that if r > k − ln(2) − ln(2) /
2, then H k ( n, r · n ) is a.a.s. not two colorable. That observation was complemented by Achlioptasand Moore [3] who proved that, if r < k − ln(2) − (1 + ln(2)) / o k (1), then random hypergraph H k ( n, r · n ) is a.a.s. two colorable. This is the first of the papers that directly inspire our developments(the bound itself has been later improved in [5] and [6]).Second result from that area which provides a context for our considerations on efficient coloring,was proved by Achlioptas and Coja-Oghlan in [1]. They discussed the evolution of the space of solutionswhen successive random constraints are added to the instance. They observed that, at some point, theset of solutions, which in certain sense is initially connected, undergoes a sudden change after which itbecomes shattered into exponentially many well separated regions. That behavior has been interpretedas a barrier for effective algorithms. Since that time, for some specific problems, algorithms and theiranalyses were improved up to the threshold of shattering (see e.g. [4]) but none surpassed that barrier.The case of our interests, when n = Θ( k ), is slightly different, and most proofs for fixed k do nottranslate literally. The ideas of the proofs however usually can be applied. As shown by our work, theresulting proofs turn out to be technically simpler. Moreover, since the values like 2 k − are no longerconstants, we were able to obtain the sharp threshold for two colorability of the random hypergraphin many cases. Note that for fixed k , this question is still open (however the results in [5] give verytight bounds). For a similar problem k -SAT with not too slowly growing k , Frieze and Wormald in[10] proved the existence of satisfiability threshold by analogous methods.2. Main results
When discussing two coloring of a hypergraph, we use colors blue and red. We analyze the problemof coloring of random k -graphs with large k . Therefore asymptotic statements shall be understoodwith k → ∞ . We say that a property holds asymptotically almost surely if the probability that it holdsis 1 − o (1). For positive integer b we denote by a b the falling factorial, i.e. a b = Y j =0 ,...,b − ( a − j ) . For positive integers n > k , we define ϕ = ϕ ( n, k ) to be such that( n/ k n k = ϕ ( n, k )2 k . Note that always 2 − k ϕ ( n, k )
1. Moreover, for n = ω ( k / ), we have ϕ ( n, k ) ∼ exp( − k / (2 n )). Inparticular, n = ω ( k ) implies ϕ ( n, k ) ∼ k -graphs. We denote the number of verticesby n , and the number of distinct uniformly sampled edges by m . The resulting probabilistic space isdenoted by H k ( n, m ). In other words H k ( n, m ) can be seen as uniformly chosen k -graph on n verticeswith m edges. For technical convenience we assume that n is always even. Our main interest lies in k -graphs with Θ( k ) vertices, but the presented proof covers much larger range of parameters.2.1. Sharp threshold for two colorability.Theorem 1.
For any ε > and any superlinear function n = n ( k ) satisfying ln( n ) = o ( k ) if c = c ( n, k ) < (1 − ε ) ln(2) /ϕ ( n, k ) , ANDOM HYPERGRAPHS AND PROPERTY B 3 then hypergraph H k ( n, cn k − ) is a.a.s. two colorable (when k goes to infinity). Observe that for ε = 0, the number of edges cn k − is asymptotically equivalent to the upper bound(1). Note also that, by using the first moment argument, it is straightforward to check that if c = c ( n, k ) > (1 + ε ) ln(2) /ϕ ( n, k ) , then hypergraph H k ( n, cn k − ) is not two colorable a.a.s., which means that Theorem 1 provides thesharp threshold for the two colorability of H k ( n, m ) for a growing function k satisfying ln n = o ( k ) andsuperlinear number of vertices.In the proof of Theorem 1 we apply the second moment method along the lines of developmentsof Achlioptas and Moore from [3]. An important modification that allowed to essentially simplify theargument is that we focus only on equitable colorings (following the idea from [11]). The assumptionln( n ) = o ( k ) can be weakened to at least n = o (cid:0) k /k (cid:1) at the expense of more complicated calculations.Similarly, the assumption on superlinear growth of n can be replaced by n > αk for some constant α .It could be dropped completely if the edges of H k ( n, m ) were chosen with replacement.2.2. Algorithms for two coloring.Theorem 2.
For any fixed α < / , superlinear polynomially bounded function n = n ( k ) and m α n ln( k ) k ϕ ( n, k ) − k − , there exists an efficient algorithm that a.a.s. finds a proper two coloring of random k -graph H k ( n, m ) . When the constructive aspects are considered, efficient usually means that the running time isbounded by a polynomial of the input size. It is rather obvious, that the simple procedure that weanalyze satisfy this requirement, as it can be easily seen to work in time O ( nkm ). Moreover, since itdepends only on a small fraction of the actual edges, for some carefully tailored (but hardly natural)representation of the hypergraph, for most inputs its running time can be even sublinear.As far as we know, the problem of algorithmic coloring of k -graphs, when n is polynomially boundedin k , has not been directly addressed before. However some results, obtained for fixed k , can beadapted. Achlioptas, Kim, Krivelevich and Tetali presented in [2] a procedure that finds propercolorings for H k ( n, m ) when the number of edges m is bounded by some function that is O ( n/k · k ).They also argued that their analysis of the procedure is tight up to the constants. Literally, theirproof requires n = ω ( k ), but it can be easily modified to cover also the case when n = Θ (cid:0) k (cid:1) , withthe cost of constants only. The procedure constructs an equitable coloring by coloring consecutivepairs of vertices into distinct colors. The vertices are chosen according to some fixed order as longas there are no edges which are monochromatic so far but have at most three not colored vertices.In such a case it picks two not colored vertices of such an edge. For a properly bounded number ofedges in a random k -graph, a.a.s. the procedure never observes a monochromatic edge with only onenot colored vertex, and hence succeeds in finding a proper coloring. Within the same framework offixed k , Achlioptas and Coja-Oghlan in [1] observed that the shape of the space of proper coloringsof H k ( n, rn ) suddenly changes when r passes value (1 + ε k )2 k − ln( k ) /k . That value was called the shattering threshold . One of the consequences of the change is that the space becomes disconnected(two colorings are considered adjacent if they are of small Hamming distance). Although, above thatthreshold and until m = Θ (cid:0) n k (cid:1) , proper colorings almost surely exist, the problem of finding onebecomes significantly harder. When n is polynomially bounded in k , we observe somewhat analogousbehavior for the number of edges of the order Θ (cid:0) n · ln( k ) /k · k (cid:1) . The analysis of our procedure canbe extended to justify that, for α < /
2, a proper coloring is usually surrounded by a large numberof other proper colorings that are of small Hamming distance. On the other hand, similar argumentscan be used to justify that, when α = ω (1), proper colorings tend to be isolated.3. Proof of Theorem 1
We fix ε > n = n ( k ) and c = c ( n, k ) satisfy the assumptions of Theorem 1. Inthe canonical definition of H k ( n, m ), edges of the hypergraph are sampled without replacement. That RANDOM HYPERGRAPHS AND PROPERTY B causes certain purely technical complications in the calculations. In order to avoid them we alter theprobabilistic space slightly. Let H ′ k ( n, m ) be a random multi-hypergraph in which m edges are sampledwith replacement from the set of all k -subsets of a set of size n . Clearly H ′ k ( n, m ) conditioned on theevent that all edges are distinct is H k ( n, m ). However, for our parameters that is asymptotically almostsure event. Indeed, classical analysis of the birthday paradox problem implies that for m = o (cid:16)(cid:0) nk (cid:1) / (cid:17) the probability of observing two identical edges is o (1). In our case, m is exponential in k (as n = o (2 k )and ϕ ( n, k ) − k ), while (cid:0) nk (cid:1) / > (cid:0) nk (cid:1) k/ is superexponential (since n is superlinear in k ). It meansthat within our parameters any asymptotically almost sure event in H ′ k ( n, m ) is also a.a.s. in H k ( n, m ).From now on we work with H ′ k ( n, m ).Let H be a random hypergraph sampled with distribution H ′ k ( n, m ), X denote the number of properequitable (i.e. every color class consists of exactly n/ H , and X be thenumber of ordered pairs of such colorings. By Payley-Zygmund inequality we getPr[ X > > E [ X ] E [ X ] . We aim at showing that, for large k , hypergraph H is two colorable with high probability so we wouldlike to have E [ X ] E [ X ] ∼ , for k tending to infinity. Since the above quotient is bounded from above by 1 it is enough to provethat(2) E [ X ] E [ X ] o (1) . We have E [ X ] = (cid:18) nn/ (cid:19) − (cid:0) n/ k (cid:1)(cid:0) nk (cid:1) ! m and E [ X ] = n/ X a =0 (cid:18) nn/ (cid:19)(cid:18) n/ a (cid:19) − (cid:0) n/ k (cid:1)(cid:0) nk (cid:1) + 2 (cid:0) ak (cid:1)(cid:0) nk (cid:1) + 2 (cid:0) n/ − ak (cid:1)(cid:0) nk (cid:1) ! m , where a corresponds to the size of the intersection of the sets of red vertices in two equitable colorings,binomial coefficients stand for the number of pairs of equitable colorings at Hamming distance 2 a ,and the expression under the m -th power is the inclusion-exclusion formula for the probability that arandom edge is properly colored by both equitable colorings of a fixed pair with Hamming distance2 a . Denote F ( a ) = (cid:0) n/ a (cid:1) (cid:0) nn/ (cid:1) and G ( a ) = − ( n/ k n k + 2 a k n k + 2 ( n/ − a ) k n k (cid:16) − ( n/ k n k (cid:17) m , so that(3) E [ X ] E [ X ] = n/ X a =0 F ( a ) · G ( a ) . It is natural idea to apply the Laplace method for the above sum – i.e. determine the range of a thatconstitute central interval , prove that properly rescaled sum within that interval is close to a Gaussianintegral and finally show that contributions of the remaining ranges of a are negligible. We generallyfollow this path. We manage however to avoid a lot of technical details.We choose δ = n ln( n ) /k and define the central interval as C = ( δ, n/ − δ ) ∩ N and tail intervals as T = [0 , δ ] ∩ N and T = [ n/ − δ, n/ ∩ N . Note that our assumption on n implies that δ/n = o (1),in particular the central interval is not empty. In the next section we show that the contribution from ANDOM HYPERGRAPHS AND PROPERTY B 5 the central interval to the sum (3) is at most 1 + o (1). Then we show that the contributions from thetails are negligible. That completes the proof of inequality (2) and Theorem 1.3.1. Central region.
We start with showing that G ( a ) is essentially constant in C . Proposition 3.
We have G ( a ) o (1) , uniformly for a ∈ C .Proof. Function G ( a ) is convex and symmetric on the central interval. Therefore is has to be maximizedin the ends of C . We assume that k is large enough that n/ − δ > k . For any a ∈ C , we get G ( a ) G ( δ ) − ϕ k − + 4 ( n/ − δ ) k n k (cid:0) − ϕ k − (cid:1) ! m − ϕ k − + 4 ( n/ k n k ( n/ − δ ) k ( n/ k (cid:0) − ϕ k − (cid:1) m ϕ · − ( k − · (1 − δ/n ) k (cid:0) − ϕ k − (cid:1) ! m exp (cid:18) m · ϕ · − ( k − · (1 − δ/n ) k · (cid:16) − ϕ k − (cid:17) − (cid:19) . We focus on the exponent with the intention of showing that it is o (1). We have m · ϕ · − ( k − · (1 − δ/n ) k · (cid:16) − ϕ k − (cid:17) − = O ( n · exp( − δk/n )) . Since we chose δ = n ln( n ) /k the above expression is o (1), hence G ( a ) G ( δ ) = 1 + o (1). (cid:3) By the above Proposition we have that X a ∈C F ( a ) · G ( a ) (1 + o (1)) · X a ∈C F ( a ) < (1 + o (1)) · n/ X a =0 (cid:0) n/ a (cid:1) (cid:0) nn/ (cid:1) . The last sum, however is well known to be equal to 1. Altogether we get that X a ∈C F ( a ) · G ( a ) o (1) . Tails.
By symmetry of the analyzed functions it is sufficient to consider only one of the tails. Wefocus on T . Functions F ( . ) and G ( . ) are respectively increasing and decreasing on T . Therefore forany a ∈ T , we have F ( a ) · G ( a ) F ( δ ) · G (0) . Now G (0) = (cid:18) − n/ k n k (cid:19) − m = (cid:16) − ϕ k − (cid:17) − m = exp (cid:16) mϕ k − + O (cid:0) m − k ϕ (cid:1)(cid:17) = (1 + o (1)) · exp( c · ϕ · n ) . Note that we used the fact that n = o (cid:0) k (cid:1) in order to obtain m − k ϕ = o (1).In order to bound F ( δ ) we use the well known facts that (cid:0) nn/ (cid:1) > n n and (cid:0) n/ δ (cid:1) < ( n/ H (2 δ/n ) ,where H ( . ) is the binary entropy function. We get F ( δ ) < n · (cid:16) H (2 δ/n ) − (cid:17) n . Combining both bounds we obtain F ( a ) · G ( a ) (1 + o (1)) · n · (exp( c · ϕ − ln(2)(1 − H (2 δ/n ))) n . RANDOM HYPERGRAPHS AND PROPERTY B
Since δ/n = o (1) we also have H (2 δ/n ) = o (1). Therefore, whenever c · ϕ < (1 − ε ) ln(2) the aboveexpression is exponentially small in n . In particular (for sufficiently large n ) it is still o (1) whensummed over T . 4. Algorithmic coloring of random k -graphs Theorem 2 has been stated in terms of uniform random hypergraph H k ( n, m ). However, closelyrelated binomial model allows for much more natural proof. In a random hypergraph H k ( n, p ) every k -subset of the set of vertices of size n is independently added to the hypergraph with probability p .We are going to work with the following technical statement. Proposition 4.
For any fixed α < / , and any superlinear polynomially bounded function n = n ( k ) ,there exists an efficient algorithm that on random k -graph H k ( n, p ) with the expected number of edges m α n ln( k ) k ϕ ( n, k ) − k − a.a.s. finds a proper two coloring. The number of edges in H k ( n, p ) is distributed binomially, and hence it is concentrated. Togetherwith the fact that the property of being two colorable is monotonic, it implies that the statementanalogous to the above proposition is also true for the uniform model H k ( n, m ). Hence the aboveproposition implies Theorem 2. For the rest of this section we work with the binomial model.Let α < / n = n ( k ) satisfy the assumptions of Proposition 4. We define p and q as p · (cid:18) nk (cid:19) = ϕ − q k − , q = 2 α n ln( k ) k , so that the expected number of edges is m as in Proposition 4. Then, the expected number ofmonochromatic edges in an equitable coloring is2 · (cid:18) n/ k (cid:19) · p = q. For a given coloring, an edge is almost monochromatic if all but one of its vertices are of the samecolor. The vertex of that edge with the unique color is called its head . A vertex is called safe if it isnot the head of any almost monochromatic edge.4.1.
Procedure.
Let R , B be a fixed equipartition of the vertex set of H = H k ( n, p ). We initiallycolor the vertices of R red and the vertices of B blue. Since, in our case, the input hypergraph israndom we can work with a fixed partition. If the algorithm were to be applied to a specific instance,it seems more appropriate to start with a random one. The general idea of the procedure is to keeprecoloring safe vertices belonging to monochromatic edges until the coloring becomes proper. Notethat changing the color of a safe vertex does not create new monochromatic edges. Additionally, forconvenience, we want to keep the coloring equitable. Therefore, in every round, we try to repair twomonochromatic edges of different colors by switching the colors of two safe vertices.The main loop of the algorithm is presented as Listing 1. In a single step, it switches colors of twocarefully chosen safe vertices. Procedures PickRedVertex() and
PickBlueVertex() are responsiblefor the choices. The first of them is presented on Listing 2. The other one is symmetric. Globalvariables R and B denote the current partition of the vertices into red and blue. Initially, R = R and B = B . These variables are modified only in the main loop. They are also used implicitly inthe safety checks within choice procedures. Values N R , N B are the numbers of initially red and bluemonochromatic edges, respectively. The value of r (resp. b ) corresponds to the index of currently ormost recently considered initially red (resp. blue) edge. Both these variables are global. Beside themain loop, they are used in the corresponding choice procedures.When the algorithm reaches the point when all initially monochromatic edges are already considered(i.e. r > N R and b > N B ), the main loop ends and the current coloring (that is proper for the input ANDOM HYPERGRAPHS AND PROPERTY B 7 hypergraph) is returned. However the algorithm may fail if, at some point, function
PickRedVertex() or PickBlueVertex() is unable to find a good vertex to be recolored.
Listing 1:
The main procedure R ← R , B ← B , r ← b ← while r N R or b N B do v R ← PickRedVertex() v B ← PickBlueVertex()/* switch colors of vertices */ R ← R \ { v R } ∪ { v B } B ← R \ { v B } ∪ { v R } return ( R , B )We describe only procedure PickRedVertex() , the other one is symmetric. Recall that r is a globalvariable so in each consecutive call of the procedure it starts with the value of r as left by the previouscall. Originally empty list C R contains the vertices that has been already tested for safety. It is usedonly in PickRedVertex() , but we also need its state to be preserved between calls.The procedure first advances into the next red edge and then iterates over its vertices (revealed ina random order) in search for a safe vertex or a vertex that has been recolored before. It may happenthat during the evaluation we run out of red edges (i.e. r exceeds N R ) but we still need to do somerecolorings to get rid of blue ones. In order to deal with this issue we extend the list of red edges – forany r > N R , the r -th red edge is just uniformly chosen k -subset of R . In fact we want the functionto pick a randomly chosen red safe vertex – extending the list of red edges with artificial random onesis just a way of realizing that task.When iterating over the vertices of the current red edge, the procedure performs a few tests. Firstit checks if the current vertex has been observed before (i.e. whether it belongs to C R ). If not, it isappended to C R and checked for safety. If the safety check succeeds, the vertex is returned from thefunction (and hence is picked as the one to be recolored next). Otherwise, we continue iterating overthe vertices of the current edge.If the current vertex has been observed before, it was either recolored or unsafe at that time. Ifit was recolored, then the current edge is no longer red and we can advance to considering the nextedge. This is done by restarting the function (the first instruction is to increase r which points to thecurrent red edge). If the vertex was unsafe at the time of being checked, we treat it as still unsafe andcontinue iterating over the vertices of the current edge (we ignore the possibility that the vertex mightbe safe in the current coloring).It may happen that we reach the end of the current edge without finding safe or recolored vertex.We distinguish two cases here based on the value of r . If r N R , the problematic edge is an actualedge of the hypergraph. In such a case we declare failure. On the other hand, for r > N R , theproblematic edge does not belong to the hypergraph, and has been added artificially to the list. In RANDOM HYPERGRAPHS AND PROPERTY B that case, we can simply ignore the problem and keep looking for another vertex to be recolored, byconsidering next artificial edges.
Listing 2:
Procedure
PickRedVertex() r ← r + 1 // advance to the next edge for j = 1 , . . . , k do v ← reveal the j -th vertex of the r -th red edge if C R does not contain v then append v to C R if v is safe in the current coloring then return v else if v has been recolored then restart the procedure if r N R or C R contains all elements of R then FAIL else restart the procedure4.2. Analysis.
Random hypergraph H k ( n, p ) can be seen as the product of (cid:0) nk (cid:1) independent Bernoullitrials, each with probability of success p . With every such trial, and hence with every k -subset e ofV, we associate the indicator random variable I e that corresponds the outcome of the trial. In orderto keep a clear distinction between potential edges and the actual edges of the hypergraph, instead ofreferring to a potential edge f ⊂ V we refer to its indicator random variable I f .The probability distribution H k ( n, p ) over the k -graphs on n vertices can be also obtained byother constructions. We describe one that can be easily modified to suit our needs. Let N R be arandom variable with binomial distribution B ( (cid:0) n/ k (cid:1) , p ). We first determine the value of N R , thenchose uniformly that number of distinct k -subsets of R and put them in the constructed hypergraphas edges. The edges contained in B are constructed in the same way. The remaining edges, i.e.the ones that are not monochromatic in the initial coloring, are added to the hypergraph just likebefore – according to the values of the corresponding indicators. So far the resulting probabilitydistribution is just the same as in the original construction. It turns out to cause technical difficultiesthat monochromatic edges chosen in such a way are constrained to be distinct and hence they aredependent. For small enough p (like the ones that we are interested in), the dependence is veryweak. To get rid of this constraint we alter the probabilistic space slightly. Random multi-hypergraph H ′ k ( n, p ) is defined by an analogous construction with the only change that the N R (resp. N B ) edgescontained in R (resp. B ) are chosen with repetitions, and hence independently. Clearly, H ′ k ( n, p )conditioned on the event that no edge has been included more than once, is just H k ( n, p ). That eventhappens with asymptotic probability 1. Indeed, birthday paradox problem tells that we need to pickroughly (cid:0) n/ k (cid:1) / independent red edges to observe a repeated one with positive probability. On theother hand N R , which has mean q/
2, almost surely takes much smaller value. This discussion shows,that if some property holds a.a.s. in H ′ k ( n, p ) it also has to hold a.a.s. in H k ( n, p ). From now on wework with H ′ k ( n, p ).The algorithm ends successfully when there are no more monochromatic edges, but it may fail earlierwhen no recoloring candidate is found. We prove below that the algorithm asymptotically almost surelysucceeds. We concentrate on the red edges, as the situation of the blue ones is symmetric. Duringthe safety check, we reveal only the values of indicators I e for k -sets e , for which the checked vertexis the head, i.e. check if our k -graph actually contains corresponding sets as edges. The initiallymonochromatic edges are revealed in a random order. We deliberately altered the probabilistic spaceso that the consecutive red edges are drawn independently and uniformly from all k -subsets of R . Observation 5.
The initial number of red (resp. blue) edges a.a.s. does not exceed q .Proof. The number of initially red edges is distributed binomially with mean q/
2. Chernoff boundshows that probability of having more than q red edges is exponentially small in q . The same boundholds for the number of initially blue edges. (cid:3) ANDOM HYPERGRAPHS AND PROPERTY B 9
Observation 6.
Safety checks on different red (resp. blue) vertices are performed on disjoint sets ofindicators.Proof.
Suppose to the contrary that the indicator of a set e is checked for two red vertices v and w .Without loss of generality the check on w is the later one. But then during v ’s check, w is still red, so e would contain at least two red vertices during the check, which is impossible. (cid:3) The last observation guarantees that each indicator is tested at most once during safety checks ofred vertices. In fact, most of the indicators are tested at most once during the whole evaluation of thealgorithm. However, a small number of indicators may be tested twice, for both colors. Consider suchan indicator I f of some set f , first tested during a safety check of a vertex w , and then again for avertex u of opposite color. Assuming without loss of generality that w was red during the check, and u was blue, the only way for this to happen is that f \ { w } was first entirely blue, and then all thevertices of f except u and w were recolored to red.Observe that it can only harm the algorithm if f is indeed an edge, and thus may render u or w unsafe. We will call such situation – a second safety check on an actual edge – a corrupted safetycheck . But as f must have been initially an almost-monochromatic edge, and then it would have toget almost all its vertices recolored, it is a very unlikely situation, and we will prove in Observation 10that a.a.s. no corrupted safety check will happen.For the sake on analysis, we slightly modify the safety checks to only truly inspect the indicatorsthat were not tested before. For any indicator previously seen, we assume that the check automaticallysucceeds – note that with no corrupted safety checks the assumption must be true. Observation 7.
Independently of the outcomes of the previous checks, the probability of a vertexpassing the modified safety check is at least δ = k − α (1+ o (1)) .Proof. We compute the probability of a successful check on an arbitrary vertex v . It fails if there existsan almost monochromatic edge for which v is the head. There are at most (cid:0) n/ k − (cid:1) candidates for suchan edge, so the probability that none of them shows up is at least (1 − p )( n/ k − ). We have p (cid:18) n/ k − (cid:19) = qϕ − k − (cid:0) n/ k − (cid:1)(cid:0) nk (cid:1) = q k n/ − k + 1) ϕ − k ( n/ k n k = q k ( n − k + 2) = 2 α ln( k ) n ( n − k + 2) = 2 α ln( k )(1 + o (1))) . Using the fact that 1 − p = exp (cid:0) − p + O (cid:0) p (cid:1)(cid:1) = exp ( − p · (1 + o (1))) we obtain:(1 − p )( n/ k − ) = exp (cid:20) − p · (cid:18) n/ k − (cid:19) (1 + o (1)) (cid:21) = k − α (1+ o (1)) . (cid:3) We show next that, during the whole algorithm, on the list of checked vertices C R , there is asignificant fraction of the ones which have passed the safety check (and thus have been recolored). Observation 8.
Let α ′ = (1 / α ) / . Asymptotically almost surely, for every l such that k/ l n/ , the prefix of C R of length l contains at least l · k − α ′ recolored vertices.Proof. Let us first consider a chance for a long series on consecutive failed safety checks – namely, k α ′ failures in a row. Observe that such a sequence can only happen on indicators that were notpreviously checked (otherwise they would automatically suceed). By Observation 6 they are conductedon disjoint sets of vertices and thus are independent. The chance that we observe such a series of failuresis, consequently, (1 − δ ) k α ′ < exp (cid:16) − δ · · k α ′ (cid:17) = exp (cid:16) − k α ′ − α )+ o (1) (cid:17) . The list C R reaches thelength at most n/
2, hence straightforward union bound shows that the probability of observing a longseries of failures is smaller than ( n/
2) exp (cid:16) − k α ′ − α )+ o (1)) (cid:17) = o (1). This implies that every prefix of C R of length l contains at least ⌊ l · · k − α ′ ⌋ recolored vertices.However, since l > k/ l = ω ( k α ′ ), which is sufficient to deduce that, for large enough k , wehave l · k − α ′ ⌊ l · · k − α ′ ⌋ . (cid:3) Observation 9.
The modified procedure asymptotically almost never declares failure.Proof.
The whole procedure fails if one of the choice procedures is unable to find a vertex to berecolored. As usual we focus on
PickRedVertex and show that such situation a.a.s. does not happen.The first possibility for the failure is that the procedure does not find a safe vertex or recoloredvertex in an edge that was initially red. When we reveal the vertices of such an edge e , some of themare already checked. The procedure fails if all unchecked vertices fail the safety check, and no vertex of e from C R was recolored (i.e. safe) when met for the first time. At the time when we start to analyzeedge e , at least one of the following two conditions must hold: either at least k/ e are in C R , or at least k/ k/ e have been checked and added to C R (recall thateach vertex is checked for safety only once, and if it fails it is treated as permanently unsafe.) In thealtered probabilistic space, the choices of the vertices of e are not influenced by other monochromaticedges, so the set of k/ C R is chosen uniformly from k/ C R . Dueto Observation 8 the chance that they are all unsafe is smaller than (1 − k − α ′ ) k/ .The other case is that there are k/ e , all ofthem must not pass the safety check. Observation 6 implies that the chance of this event is at most(1 − δ ) k/ < (1 − k − α ′ ) k/ . Therefore in both cases the chance that the procedure fails while consideringedge e is never bigger than (1 − k − α ′ ) k/ < exp (cid:16) − k − α ′ (cid:17) .By union bound and Observation 5, the procedure fails explicitly with probability at most 2 n/k · ln k · exp (cid:16) − k − α ′ (cid:17) = o (1).Note that once r exceeds N R a failure can only occur when all the vertices of R are already checked.Hence the necessary condition for the procedure to fail at that point is that there are no safe uncheckedvertices at the time when the procedure is called. We show that the probability of that event is verysmall. By the construction of the main procedure, at each time when PickRedVertex is called, thenumber of recolored red vertices does not exceed max( N R , N B ). Both these values are smaller than q a.a.s. It means that, at each time when the procedure is called, list C R contains at most q recoloredvertices. Observation 8 implies that it contains at most ( q + 1) k − α ′ = o ( n ) vertices. Therefore( n/ − o (1)) red vertices are still not checked. The probability that they all turn out to be unsafeis (1 − δ ) ( n/ o (1)) < exp( − δ ( n/ o (1))) which is o (exp( − q )). Since the number of calls to theprocedure is a.a.s. smaller than q , the situation which allows the procedure to fail because all thevertices of R are already checked a.a.s. does not happen. (cid:3) Observation 10.
Asymptotically almost surely, no corrupted safety checks will happen.Proof.
Every time, when a blue vertex fails the safety check, it is the head of at least one edge thatis currently almost red. The tail of every such edge is called a witnessing tail . It is crucial to observethat it is completely irrelevant for the further evaluation of the modified procedure what is the specificcontent of the witnessing tails. Moreover, it is not even necessary to reveal that content to the coloringprocedure. It means that, conditioned on the specific outcomes of the safety checks, the witnessing tailsare distributed uniformly among the ( k − f subsequently gets k − ANDOM HYPERGRAPHS AND PROPERTY B 11
In the final coloring we have at most q vertices recolored to red. For a given witnessing tail, theprobability that k − (cid:0) qk − (cid:1) · ( n/ (cid:0) n/ k − (cid:1) n · q k − ( n/ k − n · q k − ( n/ k − = n · (2 q/n ) k − . The expected number of witnessing tails for every vertex is p · (cid:0) n/ k − (cid:1) = 2 α ln( k )(1 + o (1))) (derivedalready in Observation 7). Therefore the expected final number of witnessing tails is O ( n ln( k )),and by Markov equality, asymptotically almost surely there are no more than n witnessing tails.But then, by simple union bound, the expected number of corrupted safety checks does not exceed n · n · (2 q/n ) k − = o (1). Hence, a.a.s. not even one of them happens. (cid:3) In many places in the presented proofs we used quite rough estimations. It is interesting to observethat we did not lose much. Observation 6 shows that for α > /
2, the fraction of safe vertices among R is o (cid:0) k − (cid:1) . Therefore a random red edge with high probability avoids them all. In consequence, inthese cases, our procedure is very likely to fail. References [1]
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