Download presentation

Presentation is loading. Please wait.

Published bySilvester Banks Modified over 2 years ago

1
From Topological Methods to Combinatorial Proofs of Kneser Graphs Daphne Der-Fen Liu 劉 德 芬 Department of Mathematics California State University, Los Angeles

2
Kneser Graphs Given positive integers n ≥ 2k, the Kneser Graph KG(n, k) has the vertex set of all k-element subsets of [n] = {1, 2, 3, …, n}. Two vertices A and B are adjacent if A and B are disjoint, A∩B = .

3
Example: KG(5, 2) is Petersen graph 1 2 1 3 1 5 1 4 3 5 3 4 2 5 2 4 2 3 4 5

4
Lovász Theorem [1978] (KG(n, k)) = n – 2k + 2. The proof was by the Borsuk-Ulam theorem, which can be proved by Tucker Lemma [1946] (with extremely fine “triangulations” of S n ) Tucker Lemma Borsuk-Ulam Lovász Thm Matoušek [2004] Combinatorial proof !

5
Circular Chromatic Number For positive integers p, q, with p ≥ 2q, a (p, q)-coloring c for a graph G is a mapping c: V(G) → {0, 1, 2, …, p-1} such that if uv E(G) then q ≤ |c(u) – c(v)| ≤ p - q. Circular chromatic number of G is: χ c (G) = inf {p/q : G admits a (p,q)-coloring} Note, a (p, 1)-coloring is a proper coloring.

6
Johnson-Holroyd-Stahl Conjecture [1997] c (KG(n, k)) = n – 2k + 2. Zhu Surveys Yes, for even n [Meunier ‘05] and [Simonyi and Tardos ‘06] ? Yes, for sufficiently large n [Hajiabolhassan and Zhu, 2003] Known result: For any graph G, c (G) = (G).

7
Chen Theorem [2011] Johnson-Holroyd-Stahl Conjecture [1997] c (KG(n, k)) = n – 2k + 2. Because c (G) = (G), so Lovász Thm Chen Theorem Proof was by Ky Fan Lemma [1954]

8
Tucker Lemma Borsuk-Ulam Lovász Thm Matoušek [2004] Fan Lemma Chen Theorem A combinatorial proof modified from Prescott and Su [2005] (2) (1) (3) This talk Chang, L. Zhu [2012]

9
Use 1, 2, -1, -2 to label vertices so that it is antipodal on the boundary, Always exists a complementary edge (sum 0) ! and avoid edges with label sum = 0. IMPOSSIBLE !! 1 1 1 -2 2 X

10
Same conclusion for other symmetric triangulations of a square D 2 : N X N 3 X 3 These are all symmetric triangulation of D 2

11
Tucker Lemma [1946] Given an array of N 2 elements, in N rows and N columns (N > 1), and given four labels 1,-1, 2, -2, let each element of the array be assigned one of the four labels in such a way that each pair of antipodal elements on the boundary of the array is assigned a pair of labels whose sum is zero; then there is at least one pair of adjoining elements of the array that have labels whose sum is zero. Can be extended to N k

12
D2D2 D2 D2 S1S1 D3D3 D3 D3 S2S2 Special Triangulations of S n

13
Alternating Simplex of a labeling Let K be a triangulation of S n. Let : V(K) { 1, 2, …, m} be a labeling A simplex with d vertices is positive alternating if labels of its vertices can be expressed as { k 1, – k 2, k 3, – k 4 ……, (-1) d-1 k d }, where 1 ≤ k 1 < k 2 < k 3 < k 4 … < k d m. Negative alternating is similar, except the leading label is negative, – k 1, k 2, – k 3 ……

14
Ky Fan Lemma [1956] Let K be a barycentric derived subdivision of the octahedral subdivision of the n-sphere S n. Let m be a positive integer. Label each vertex of K with one of 1, 2, ….., m, so that: The labels assigned to any two antipodal vertices of K have sum 0 Any 1-simplex in K have labels sum ≠ 0 Then there exist an odd number of positive alternating n-simplices in K. (Hence, m > n) Antipodal labeling No complementary edges

15
Boundary of the First Barycentric Subdivision of D n 0,0 1,0 1,1 1,-1 0,-1 For each simplex, the vertices can be ordered as V 1, V 2, … such that: If the i-th coordinate of V i is non-zero, say z, then the i-th coordinate of all V j, j > i, must be z. -1,-1 -1,0 0,1 -1,1

16
D2D2 D2 D2 S1S1 D3D3 D3 D3 S2S2 Special Triangulations of S n

17
Example (Recall) 1 1 1 -2 2 X D2 S1D2 S1 By Key Fan Lemma, 1,0 1,1 There is an odd number of positive alternating 1-simplex (an edge).

18
Useful Essence of Fan Lemma Barycentric subdivision of S n-1 R n Fix n. A (nontrivial) signed n-sequence is: A = (a 1, a 2, a 3,...., a n ), each a i { 0, +, - }, and at least one a i ≠ 0. A can be expressed by A = (A +, A-), where A + = { i : a i = + }, A - = { i : a i = - }. Opposite of A: - A = (A-, A + ). Example: A = ( -, 0, +, +, +) = ( {3, 4, 5}, {1} ) - A = ( +, 0, -, -, - ) = ( {1}, {3, 4, 5} )

19
Simplices in the Triangulations of S n-1 A d-set consists of d elements from n : = {A 1, A 2, ….., A d }, A 1 < A 2 < ….. < A d Let n denote all non-trivial signed n-sequences. Let A, B n, A = (A +, A - ), B = (B +, B - ). Denote A ≤ B, if A + B + and A - B - Denote |A| = | A + | + | A - | Dimension ( ) = d = | |. Note, A + A - = and |A| 1.

20
Antipodal labeling Let : n → { 1, 2, … m}. is sign-preserving: (- A) = - (A), A n Complementary pair: A < B n, (A)+ (B) = 0. Positive alternating d-sets: is a d-set, ( ) = { k 1, - k 2, k 3, ….., (-1) d-1 k d }, 0 < k 1 < k 2 < k 3 < ….. < k d ≤ m.

21
Fan Lemma applied to the 1 st barycentric subdivision of octahedral subdivision of S n-1 Assume : n → { 1, 2, … m } is sign- preserving without complementary pairs. Then there exist an odd number of positive alternating n-sets. Consequently, m n. Octahedral Tucker Lemma: Assume : n → { 1, 2, … (n-1)} is sign- preserving. Then there exists some complementary pair.

22
Proof of Fan Lemma: Construct a graph G Vertices: is a d or (d-1)-sets, max( ) = d: Type I: is an agreeable alternating (d-1)-set, d 2. Type II: is an agreeable almost alternating d-set. Type III: is an alternating d-set. Edges: ’ is and edge if all below are true: (1) < ’ and | | = | ’| - 1 (2) is alternating (positive or negative) (3) ( ’ ) = sign ( ) (4) max( ’) = | ’ |

23
Claim All vertices in G have degree 2, except {(+,0, …,0)}, {(-,0, …, 0)} and alternating n-sets which are of degree 1. So, G consists of disjoint paths. The negative of each path is also a path in G. So, there are an even number of paths in G. The two ends of a path are not opposite sets. By symmetry, there are an odd number of positive alternating n-sets. Q.E.D. Finished (1)

24
Lovász Theorem (KG(n, k)) = n – 2k + 2. Claim: (KG(n, k)) ≤ n – 2k + 2. Define a (n–2k+2)-coloring c of KG(n, k) by: 1 2 3 4 5.… n – 2k +1n – 2k +2.…. n -1 n 2k – 1

25
Assume (KG(n, k)) ≤ n – 2k + 1 Let c be a (n-2k+1)-coloring for KG(n, k) using colors in { 2k-1, 2k, 2k+1, …, n-1 }. Define : n → { 1, 2, … (n-1)} by: where c(U) = max {c(W): W U and |W| = k}.

26
Define : n → { 1, 2, … (n-1)} by: where c(U) = max {c(W): W U and |W| = k}. is sign-preserving: (- A) = - (A), A n No complementary pairs. If (A) = - (B) & A < B, then c(A’) = c (B’), for some A’ A +, B’ B -. Impossible as A + B - = , so A’, B’ adjacent. Contradicting Fan Lemma, as (n -1) < n. Lovász Thm Fished (2)

27
Alternative Kneser Coloring Lemma Chen [JCTA, 2011], Chang-L-Zhu [ JCTA, 2012] Suppose c is a proper (n-2k+2)-coloring for KG(n, k). Then [n] can be partitioned into: [n] = S T {a 1, a 2, …., a n-2k+2 }, where |S| = |T| = k- 1, and c(S {a i })= c(T {a i })= i, i =1, 2,..., n-2k+2. By Alternative Kneser Coloring Lemma, Chen Theorem can be proved easily.

28
Claim: (KG(n, k)) ≤ n – 2k + 2. Define a (n–2k+2)-coloring c of KG(n, k) by: 1 2 3 4 5.… n – 2k +1n – 2k +2.…. n -1 n 2k – 1 S T {a n - 2k+2 } {a 1, a 2, …., a n -2k+1 } a i = i, i = 1, 2, …, n – 2k+1 Example (Recall)

29
Alternative Kneser Coloring Lemma Any proper (n-2k+2)-coloring c for KG(n, k), [n] = S T {a 1, a 2, …., a n-2k+2 }, where |S| = |T| = k-1, and c(S {a i })= c(T {a i })= i, i =1, 2,..., n-2k+2. In KG(n, k), the sets S {a i }, T {a i }, i = 1,..., n–2k+2, Induce a complete bipartite graph K n- 2k+2, n- 2k+2 minus a perfect matching which used all colors called a colorful K n- 2k+2, n- 2k+2

30
Alternative Kneser Coloring Lemma Any proper (n-2k+2)-coloring c for KG(n, k), then [n] = S T {a 1, a 2, …., a n-2k+2 }, where | S | = | T | = k – 1, and S {a1}S {a1} S {a2}S {a2} S {a n-2k+2 } : : c (S {a i }) = c (T {a i } ) = i, i = 1, 2,..., n – 2k+ 2. T {a1}T {a1} T {a2}T {a2} T {a n-2k+2 } : : 11 2 2 n-2k+2 : : : :

31
Proof of Alternative Kneser Coloring Theorem Let c be a proper (n-2k+2)-coloring of KG(n, k) using colors from { 2k-1, 2k-2,…, n }. Define : n → { 1, 2, … n } by: where c(U) = max {c(W): W U and |W| = k}. Let be a linear order on subsets of [n] such that if |U| < |W| then U W.

32
is sign-preserving: (- A) = - (A), A n No complementary pairs. As seen before. By Fan Lemma, there exist an odd number of positive alternating n-sets.

33
Important Claim Let be a positive alternating n-set. Then ( ) = {1, - 2, 3, - 4, …, (-1) n-1 n } : A 1 < A 2 <.... < A n | A i | = i, for all i = 1, 2, …, 2k-2 |A + 2k-2 | = |A - 2k-2 | = k-1 a 2k-1, a 2k, … a n [n] \ (A + 2k-2 A - 2k-2 ), so that c (A + 2k-2 {a 2k-1, a 2k+1, … a 2i+1 } ) = 2i +1, i c (A - 2k-2 {a 2k, a 2k+2, … a 2i } ) = 2j, j

34
Proof (continue) Let = {A n : |A+| = |A-| = k-1}. By Claim every positive alternating n-set contains exactly one element from . For every A , let (A, ) = # positive alternating n-sets containing A. By Fan Lemma, A (A, ) is odd. So (Z, ) 1 (mod 2) for some Z .

35
Proof (continue) By Fan Lemma, A (A, ’) is odd. Define ’ : n → { 1, 2, … n } by: ’(Z) = - (Z); and ’(A) = (A) if A Z. ’ is sign-preserving without complementary pairs Since A Z (A, ) = A Z (A, ’), and (- Z, ) = (Z, ’) = 0, So we have, (Z, ) (- Z, ’) 1 (mod 2)

36
Let Z = (Z+, Z-) = (S, T). Then - Z = (T, S). Apply Claim to both , ’ we get: Let , ’ be the positive alternating n-sets for, ’, containing Z and – Z, respectively. a 2k-1, a 2k, … a n [n] \ (S T ), c ( S {a 2k-1, a 2k+1, … a 2i -1 } ) = c( T {b 2k-1, b 2k+1, … b 2i-1 } ) = 2i – 1, i b 2k-1, b 2k, … b n [n] \ (T S ), so that c ( S {a 2k, a 2k+2, … a 2i } ) = c ( T {b 2k, b 2k, … b 2i } ) = 2i, i

37
Hence, c (S { a 2k-1 }) = c(T { b 2k-1 }) = 2k – 1. So, a 2k-1 = b 2k-1 By induction, a i = b i and c (S { a i }) = c(T { b i }) = i for all i. This completes the proof of Alternative Kneser Coloring Theorem. Fished (3)

38
Thanks to Pen-An Chen for great results iii Happy Birthday Professor Chang iii THANK YOU for being a Great Mentor !! THANK YOU ALL !! Thanks to Xuding Zhu for excellent lecture note Thanks to the Conference Committee

Similar presentations

OK

6.896: Topics in Algorithmic Game Theory Lecture 8 Constantinos Daskalakis.

6.896: Topics in Algorithmic Game Theory Lecture 8 Constantinos Daskalakis.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on methods of gene transfer in plants Download ppt on power steering Ppt on applied operational research pdf Ppt on brand equity Download ppt on political parties class 10 Ppt on business plan with example Ppt on case study of microsoft Perspective view ppt on android Ppt on tunnel diode construction Scrolling message display ppt online