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Balanced Online Graph Avoidance Games Henning Thomas Master Thesis supervised by Reto Spöhel ETH Zürich TexPoint fonts used in EMF. Read the TexPoint manual.

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Presentation on theme: "Balanced Online Graph Avoidance Games Henning Thomas Master Thesis supervised by Reto Spöhel ETH Zürich TexPoint fonts used in EMF. Read the TexPoint manual."— Presentation transcript:

1 Balanced Online Graph Avoidance Games Henning Thomas Master Thesis supervised by Reto Spöhel ETH Zürich TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AA

2 ETH Zürich – Balanced Online Graph Avoidance Game2 The Balanced Online Graph Avoidance Game Rules: played by one player, called ‘Painter’ starts with the empty graph on n vertices the edges appear in sets of size r u.a.r. and have to be coloured instantly by Painter using each of the available r colors exactly once Painter loses as soon as she creates a monochromatic copy of some fixed forbidden graph F Question: How long can Painter survive in this random graph process?

3 ETH Zürich – Balanced Online Graph Avoidance Game3 Example The balanced online triangle avoidance game with 2 colors on 6 vertices, i.e., F = M, r = 2 : We denote the graph after N steps by G(n,N). G(n,N) is equivalent to a random graph G n,m with m = r ¢ N. We will sometimes also switch to G n,p with p = r ¢ N/n 2. Painter loses in step 7! Step 0: G(6,0)Step 1: G(6,1)Step 2: G(6,2)Step 3: G(6,3)Step 4: G(6,4)Step 5: G(6,5)Step 6: G(6,6)Step 7: G(6,7)

4 ETH Zürich – Balanced Online Graph Avoidance Game4 Previous Results Marciniszyn, Mitsche, Stojaković (2005) proved threshold functions N 0 (n) for a variety of graphs F, e.g. cycles, for the balanced online F-avoidance game with 2 colors, i.e., There exists a strategy such that for N ¿ N 0 (n) we have For N À N 0 (n) we have regardless of Painter’s strategy that

5 ETH Zürich – Balanced Online Graph Avoidance Game5 Previous Results Prakash, Spöhel (2007) proved threshold functions N 0 (n) for a large class of graphs F, including cycles and cliques, for the balanced online F- avoidance vertex-coloring game with r colors.

6 ETH Zürich – Balanced Online Graph Avoidance Game6 Main Result (simplified) For many graphs F, the threshold for the balanced online F avoidance game with r colors is where

7 ETH Zürich – Balanced Online Graph Avoidance Game7 Some Concrete Thresholds Corollary (cycle-avoidance game) The threshold for the balanced online C t -avoidance game with r colors is Corollary (clique-avoidance game) The threshold for the balanced online K t -avoidance game with r ¸ t colors is

8 ETH Zürich – Balanced Online Graph Avoidance Game8 Lower Bound We have to show that there exists a strategy such that Pain- ter a.a.s. (asymptotically almost surely, i.e., with probability tending to 1 for n ! 1 ) survives any N ¿ n 2-1/m r ¤ (F) steps. We introduce the following “smart greedy strategy”: Let In every step take an arbitrary coloring that does not create a monochromatic copy of D. If this is not possible, give up. For simplicity we assume D = F.

9 ETH Zürich – Balanced Online Graph Avoidance Game9 Lower Bound If the strategy stops, the r edges of one step could not be coloured without creating a monochromatic copy of F. An example (F = K 4, r = 3)

10 ETH Zürich – Balanced Online Graph Avoidance Game10 Lower Bound Observation: Each valid coloring corresponds to a perfect matching in this bipartite graph. Hall’s Theorem: We lose in a certain step if and only if there exists C µ {e 1,...., e r } such that each of them is contained in r - |C| + 1 copies of F of the same r - |C| + 1 colors C ¡ (C)

11 ETH Zürich – Balanced Online Graph Avoidance Game11 Lower Bound An example: F = K 4, r = 3 If the smart greedy strategy stops, G(n,N) must contain or

12 ETH Zürich – Balanced Online Graph Avoidance Game12 Lower Bound How often will T appear in G(n,N)? first moment method ) a.a.s. no copy of T in G(n,N) T where

13 ETH Zürich – Balanced Online Graph Avoidance Game13 Lower Bound What about T’ = ? We have

14 ETH Zürich – Balanced Online Graph Avoidance Game14 The Gluing Intuition Why do both structures T and T’ yield the same result? The expected number of a certain structure S around an r- set of edges is S S’S’ T T’T’

15 ETH Zürich – Balanced Online Graph Avoidance Game15 Lower Bound This is still not all! The r copies of F might intersect: Technical Lemma: There is no ‘dangerous structure’ with intersections that is more likely to appear in G(n,N) than or

16 ETH Zürich – Balanced Online Graph Avoidance Game16 Upper Bound We need to show that Painter will close a monochromatic copy of F within any N À n 2-1/m r ¤ (F) steps regardless of her strategy. We apply a two-round approach 1st round: Painter is presented the all edges of the first N/2 steps at once and may colour them offline. 2nd round: Game continues normally, i.e., Painter is presented the remaining r ¢ N/2 edges in sets of size r and has to colour them instantly.

17 ETH Zürich – Balanced Online Graph Avoidance Game17 Upper Bound Proof strategy 1st round: Show that Painter must a.a.s. generate a lot of ‘threats’ for the second round when colouring the edges of the first round 2nd round: Show (with the second moment method) that a.a.s. one of the threats indeed causes Painter to lose the game

18 ETH Zürich – Balanced Online Graph Avoidance Game18 Upper Bound (1st Round) Painter loses in the 2nd round if she is presented Hence, we want the 1st round to create many monochro- matic copies of r ¢ F - “threat” r¢F-r¢F-

19 ETH Zürich – Balanced Online Graph Avoidance Game19 Upper Bound (1st Round) Theorem (Rödl/Rucinski, 1995) Let G be a non-empty graph that is not a star forest. Then G(n,N) a.a.s. has the property that every r-edge-coloring contains  (n v(G) p e(G) ) monochromatic copies of G if where

20 ETH Zürich – Balanced Online Graph Avoidance Game20 Upper Bound (1st Round) apply Rödl/Rucinski to r ¢ F - : For this, we need that r ¢ F - satisfies If F satisfies this condition, Rödl/Rucinski gives us  (n rv(F -) p re(F -) ) monochromatic, say blue, copies of r ¢ F -. At this point, we leave the territory of “for arbitrary graphs F” and enter “for graphs F, for which there exists F - µ F with e F – 1 edges such that m 2 (F - ) · m r ¤ (F)”

21 ETH Zürich – Balanced Online Graph Avoidance Game21 Upper Bound (2nd Round) We show with SMM that Painter is a.a.s. confronted with For each of the M copies of r ¢ F - we introduce an indicator variable Z i for the event that the corresponding r-set of edges is presented Painter during the 2nd round, and set Note that we might have for some i,j.

22 ETH Zürich – Balanced Online Graph Avoidance Game22 Upper Bound (2nd Round) Note that we might have for some i,j. Nevertheless, if Z > 0 a.a.s., then Painter closes at least one monochromatic copy of F. In order to apply the second moment method to Z we need to show We fix the result of the first round and assume M (= number of blue copies of r ¢ F - in G(n,N/2)) =  (n rv(F -) p re(F -) ), which is a.a.s. the case by Rödl/Rucinski.

23 ETH Zürich – Balanced Online Graph Avoidance Game23 Upper Bound (2nd Round) It turns out that if M =  (n rv(F -) p re(F -) ) we have Moreover, we have a.a.s. F F F S1S1 F F F S2S2 D D D J JDJD

24 ETH Zürich – Balanced Online Graph Avoidance Game24 Upper Bound (Summary) After the first round we condition on the events For all J µ S 1 : The number of threat pairs ‘intersecting in a copy of J’ behaves as expected. (is not too large) These events occur a.a.s. and ensure that in the second round. Thus, the second moment method shows that Painter will a.a.s. close a monochromatic copy of F within any steps.

25 ETH Zürich – Balanced Online Graph Avoidance Game25 Main Result Let F be a non-empty graph. Then Painter can a.a.s. survive any steps in the balanced online F- avoidance game. Moreover, if there exists a graph with edges such that, then Painter will a.a.s. lose the game within any steps.

26 ETH Zürich – Balanced Online Graph Avoidance Game26 Graphs not satisfying the 2-round Condition Even for the case F = K 4 and r = 2 our method does not yield a threshold. It seems that whenever the 2-round condition is not satisfied, a better strategy can be obtained by not only avoiding D, but also other graphs (e.g. further subgraphs of D) In the case F = K 4 and r = 2 the strategy looks as follows: avoid with high priority avoid with low priority


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