Presentation is loading. Please wait.

Presentation is loading. Please wait.

Colorings of graphs and Ramsey’s theorem A proper coloring of a graph G is a function from the vertices to a set C of colors such that the ends of every.

Similar presentations


Presentation on theme: "Colorings of graphs and Ramsey’s theorem A proper coloring of a graph G is a function from the vertices to a set C of colors such that the ends of every."— Presentation transcript:

1 Colorings of graphs and Ramsey’s theorem A proper coloring of a graph G is a function from the vertices to a set C of colors such that the ends of every edge have distinct colors. If |C|=k, then we say G is k-colored. The Chromatic number  (G) of a graph G is the minimal number of colors such that G is colorable. If  (G) = 2, then G is a bipartite graph, which has no odd cycle. “Four Color Theorem” (Appel and Haken, 1977) states that if G is planar, then  (G)  4.

2 p2. Applications of graph coloring: Scheduling Final Exams: How can the final exams at a university be scheduled so that no student has two exams at the same time? Frequency Assignments: Suppose no two TV stations can operate within 220km on the same channel. How can the assignment of channels be modeled by graph coloring? Index Register: In efficient compilers the execution of loops is speeded up when frequently used variables are stored temporarily in index registers in the CPU, instead of regular memory. For a given loop, how many index registers are needed?

3 p3.

4 p4. x 1 2 i j l < d? 1 l = d 2 i j j <= d-2 j i H=G-{x} i Is it possible? Can the first i have 2 neighbors with j?

5 p5. Moreover, the component is a simple path from x i to x j in H ij. Since if 2 neighbors of x i in H had the same color j, then the other neighbors of x i in H would have at most d-2 different colors. Then we could recolor x i, which is impossible. Why? Suppose y is the first vertex on the path from x i to x j in C ij with deg  3. Then the neighbors of y in H use at most d-2 colors, so we can recolor y such that x i and x j are not connected in H ij, which is impossible. Why? Thus no such y exists and C ij is a path. Suppose z is in C ij and C ik but it is not x i. Then z has two neighbors with color j and two with color k. Again the neighbor of z in H use at most d-2 colors. Re-coloring z, we have another contradiction. So C ij  C ik ={x i }. i i jk i k k j

6 p6. By the assumption that G has no K d+1 as subgraph, there are 2 neighbors of x, say x 1 and x 2, that are not adjacent. Let a be the neighbor of x 1 with color 2. Interchange the colors 1 and 3 on C 13. Thus a is in C’ 23 and is also in C’ 12. Thus C’ 12  C’ 23  {x 2 }. Contradiction! x1x1 x2x2 x3x3 a C 12 C 23 C 13 x

7 p7. Proof 2 of Brooks Theorem: Problem 3A: Fix d ≥ 3. Let H be a simple graph with all degree ≤ d which cannot be d-colored and which is minimal subject to these properties. (1) Show that H is connected after deleting a vertex. (2) Then show that if partition V(H) into sets X and Y with |Y| ≥ 3, then there are at least 3 vertices a, b, c in Y each of which is adjacent to at least one vertex in X. Proof 2: Let d and H as in Problem 3A. If H is not complete, there are vertices x 1,x n-1,x n so that x 1 is adjacent to x n-1 and x n but the last two are not adjacent. x1x1 xnxn x n-1 Number the rest n-3 vertices such that each x k, k>1 is adjacent to at least one of x i with i< k. I.e. when x 1,...,x k have been chosen k< n-2, choose x k+1 to be any vertex other than x n-1 or x n adjacent to one of x 1,..x k. Why possible?

8 p8. Proof 2 of Brooks Theorem: Once the above is done, d-color the vertices starting at the end of the sequence– assigning x n and x n-1 the same color. x n-1 xnxn x k+1 xkxk x2x2 x1x1 If x k+1,..., x n have been colored properly, then x k connects to at most d-1 of them. Inductively and finally, there is one color available for x 1. Why?

9 Thm 3.2. If the edges of K n are colored red or blue, and r i, i=1,…, n, denotes the number of red edges with vertex i as an endpoint, and if T is the number of monochromatic triangles, then T = Pf: Corollary: i.. r i red edges

10 Ramsey number N( p, q; 2): 至少需要幾人,其中才會有 p 個人彼此都認識或有 q 個人彼此都不認識 ? N( 3, 2; 2) =3. N( 3, 3; 2) =6. Why?

11 Ramsey number N( p, q; 2): 至少需要幾人,其中才會有 p 個人彼此都認識或有 q 個人彼此都不認識 ? Claim: N(p,q;2) ≤ N(p-1,q; 2) + N(p,q-1; 2). Proof: Let n = N(p-1,q; 2) + N(p,q-1; 2). Consider a graph with n vertices and v be one of the vertices. Color the edges of the graph with red and blue colors in an arbitrary way. Red neighbors >= N(p,q-1;2). or Blue neighbors >= N(p-1,q;2). v By induction the red neighbors has p vertices with blue edges only or q-1 vertices with red edges only. Similarly, by induction, the blue neighbors has p-1 vertices with blue edges only or q vertices with red edges only.

12 Ramsey number N( p, q; 2): 至少需要幾人,其中才會有 p 個人 彼此都認識或有 q 個人彼此都不認識 ? By observing the binomial coefficients C(n,k), we know C(n,k)= C(n-1,k) + C(n-1, k-1). By the claim, we have N(p,q;2) ≤ N(p-1,q; 2) + N(p,q-1; 2). We prove Thm 3.4. N(p,q;2) ≤ C(p+q-2, q-1). By induction, assume N(p-1,q; 2) ≤ C(p+q-3, q-1) and N(p,q-1; 2) ≤ C(p+q-3, q-2). Then N(p,q;2) ≤ N(p-1,q; 2) + N(p,q-1; 2) ≤ C(p+q-3, q-1) + C(p+q-3, q-2) = C(p+q-2, q-1)

13

14 p14. " Imagine an alien force, vastly more powerful than us landing on Earth and demanding the value of N(5, 5;2) or they will destroy our planet. In that case, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they asked for N(6, 6;2), we should attempt to destroy the aliens". - Paul Erd ő sPaul Erd ő s

15 p15. p,qp,q – – 4149 – 6156 – 8473 – – – 4958 – 8780 – – – – – 4158 – – – – – – – 6180 – – – – – – – – – – – – 3583≤ – – – – – – – – – – – – 2826 ≤ – – 已知的 Ramsey Numbers

16 KpKp

17 p17.

18

19 p19. Thm 3.7. N(p,p;2)  c p 2 p/2, where c is a constant. Pf. Consider K n and color the edges randomly. For any set S of k vertices let A S be the event that the subgraph on S is monochromatic. Pr[ A S ] = 2 1-C(k,2). Let T be another k-set. A S and A T are dependent iff |S  T|  2, i.e. the subgraphs on S and T share at least one common edge. The degree d is at most C(k,2) C(n, k-2). If C(k,2) C(k,2)C(n, k-2)<1, then none of A S will happen. By some calculation, we have n < c k 2 k/2. This means we will need larger n to have any A S happen. Let k=p. we prove the Thm. □

20 p20. F. R. Ramsey ( ) a logician Thm 3.3. (Ramsey’s theorem 1930) Let r  1 and q i  r, i=1,2,…,s be given. There exists a minimal positive number N(q 1,…,q s ; r) such that: Let S be a set with n elements. Suppose that all C(n,r) r-subsets of S are divided into s mutually exclusive families T 1,…, T s (colors). Then if n  N(q 1,…,q s ; r) there is an i  [s] and some q i -subset of S for which every r-subset of these q i is in T i. We will show the case when s=2. (a) N(p,q;1)= p+q-1, Why? (b) N( p, r; r)=p and N( r, q; r)=q, for p, q  r.

21 p21. By induction on r, we assume it is true for up to r-1. Use induction on p+q, using (b). Define p 1 =N( p-1, q; r) and q 1 =N( p, q-1; r). Let S be a set with n elements, where n  1+ N(p 1, q 1 ; r-1). Let the r-subsets of S be colored with red and blue colors. Let a be an arbitrary element of S and S’=S-{a}. Define a coloring of the (r-1)-subsets of S’ by giving any (r- 1)-subset X  S’ the same color as X  {a}. By ind hyp, S’ either has a subset A of size p 1 such that all its (r-1)-subsets are red or a subset B of size q 1 with all its (r-1)- subsets colored blue. WLOG, suppose the first case happens. Then A has N(p-1, q;r) elements. Proof of Thm 3.3.

22

23 p23. r-1 N(p-1, q; r) elements. N(p,q-1;r) elements. a deg=N(p-1,q;2)+N(p,q-1;2)-1 qp-1 q-1 p Problem 3C: The equality cannot hold, if the right hand side are both even.

24 p24. An application of Ramsey’s Theorem Theorem 3.8: For a given n, there is an integer N(n) such that any collection of N> N(n) points in the plane, no three on a line, has a subset of n points forming a convex n-gon. Proof: (1) 觀察平面上 n 個點,其中任何三點不共線 。 這 n 點 將形成凸 n 邊型若且唯若 其中任意四點形成凸四邊型 。 (2) 令 N(n)=N(n,n;3) 。 然後將這 N(n) 個點由 1 編號並將任意三點形成的三角型塗上紅色如 果這三點的編號由小到大是順時鐘方向;否則 塗成藍色。由 N(n,n;3) 的定義知存在 n 個點 , 其中任意三個點形 成的三角型都是紅色的。

25 p25. 考慮這 n 個點中的任意 4 點,試問可否產生下列情形 ? 也就是說點 d 是否可以落在一紅色的三角型 ? 其中假設 a < b < c 。 由圖可知 a < d < c , 因為三角型 adc 為紅色。同理 a < b < d 。 但由此可推論出 b < d < c ,這表示 三角型 bdc 應當塗成藍色。 故得到矛盾。所以得知不可能有上述的情況發生, 也就是說任意四點都會形成凸四邊型。 由 (1) 得知這 n 個點會形成一凸 n 邊型。 a b c d


Download ppt "Colorings of graphs and Ramsey’s theorem A proper coloring of a graph G is a function from the vertices to a set C of colors such that the ends of every."

Similar presentations


Ads by Google