Generalizing the degree sequence problem

Transcription

1 Mddlebury College March 2009 Arzona State Unversty Dscrete Mathematcs Semnar

2 The degree sequence problem Problem: Gven an nteger sequence d = (d 1,...,d n ) determne f there exsts a graph G wth d as ts sequence of degrees. If such a G exsts then d s sad to be graphc, and G s called a realzaton.

3 An example Is d = (3,3,3,3,3,3) graphc?

4 An example Is d = (3,3,3,3,3,3) graphc? Havel (1955) and Hakm (1962) gave an algorthm to decde. (3,3,3,3,3,3) (2,2,2,3,3) = (3,3,2,2,2) (2,1,1,2) = (2,2,1,1) (1,0,1) = (1,1,0) (0,0) As (0,0) s graphc, so s the gven.

5 An example Is d = (3,3,3,3,3,3) graphc? Havel (1955) and Hakm (1962) gave an algorthm to decde. (3,3,3,3,3,3) (2,2,2,3,3) = (3,3,2,2,2) (2,1,1,2) = (2,2,1,1) (1,0,1) = (1,1,0) (0,0) As (0,0) s graphc, so s the gven. To construct a realzaton, work backwards usng smple edge augmentatons.

6 Erdős-Galla crteron Theorem [Erdős, Galla (1960)] A nonncreasng sequence of nonnegatve ntegers d = (d 1,...,d n ) (n 2) s graphc f, and only f, n =1 d s even and for each nteger k, 1 k n 1, k d k(k 1) + =1 n =k+1 mn{k,d }.

7 Erdős-Galla crteron Theorem [Erdős, Galla (1960)] A nonncreasng sequence of nonnegatve ntegers d = (d 1,...,d n ) (n 2) s graphc f, and only f, n =1 d s even and for each nteger k, 1 k n 1, k d k(k 1) + =1 n =k+1 mn{k,d }. The degrees of the frst k vertces are absorbed wthn k-subset and the degrees of remanng vertces. A necessary condton whch s also suffcent!

8 Theorem (Erdős, Galla) For a graphc d, n =1 d 2(n 1) f and only f there exsts a connected G realzng d.

9 Theorem (Erdős, Galla) For a graphc d, n =1 d 2(n 1) f and only f there exsts a connected G realzng d. Proof: (Suffcency) If there exsts a connected realzaton then G contans a spannng tree. Thus G has n 1 edges and so n =1 d 2(n 1).

10 Theorem (Erdős, Galla) For a graphc d, n =1 d 2(n 1) f and only f there exsts a connected G realzng d. Proof: (Suffcency) If there exsts a connected realzaton then G contans a spannng tree. Thus G has n 1 edges and so n =1 d 2(n 1). (Necessty) Pck the realzaton of d wth the fewest number of components. If ths number s 1, then we are done. Otherwse one of the components contans a cycle. Performng a smple edge-exchange allows us to move to a realzaton wth fewer components.

11 For a subgraph F, d s sad to be potentally F-graphc f there exsts a realzaton of d contanng F.

12 For a subgraph F, d s sad to be potentally F-graphc f there exsts a realzaton of d contanng F. (2,2,2,2,2,2) s potentally K 3 -graphc.

15 Problem Gven a subgraph F, determne the least even nteger m s.t. Σd m d s potentally F-graphc. Denote m by σ(f, n).

16 Erdős, Jacobson, Lehel Conjecture Conjecture (EJL ) For n suffcently large, σ(k t,n) = (t 2)(2n t + 1) + 2. Lower bound arses from consderng: + ÃØ ¾ d = ((n 1) t 2,(t 2) n t+2 ) ÃÒ Ø ¾

17 Erdős, Jacobson, Lehel Conjecture Conjecture settled: t = 3 Erdős, Jacobson, & Lehel(1991), t = 4 Gould, Jacobson, & Lehel(1999), L & Song(1998), t = 5 L & Song(1998), t 6 L, Song, & Luo(1998) t 3 S.(2005), Ferrara, Gould, S. (2009+) - purely graph-theoretc proof. Theorem For n suffcently large, σ(k t,n) = (t 2)(2n t + 1) + 2.

18 Sketch of our proof Uses nducton on t. Erdős-Galla guarantees enough vertces of hgh degree. Uses noton of an edge-exhange. Edge-exchange allows us to place desred subgraph on vertces of hghest degree and buld K t from smaller clque guaranteed by nductve hypothess.

19 Extendng the EJL-conjecture to an arbtrary graph F Let F be a forbdden subgraph. Let α(f) denote the ndependence number of F and defne: and u := u(f) = V (F) α(f) 1, s := s(f) = mn{ (H) : H F, H = α(f) + 1}. Consder the followng sequence, π(f,n) = ((n 1) u,(u + s 1) n u ).

20 A General Lower Bound If F s a subgraph of F then σ(f,n) σ(f,n) for every n. Let σ(π) denote the sum of the terms of π. Proposton (Ferrara, S. - 09) Gven a graph F and n suffcently large then, σ(f,n) max{σ(π(f,n)) + 2 F F } (1) = max{n(2u(f ) + s(f ) 1) F F } (2)

21 Proof of Lower Bound Proof: Let F F be the subgraph whch acheves the max. Consder, + K u(f ) + An s(f ) regular graph on n u(f ) vertces. u(f ) = V (F ) α(f ) 1 s(f ) = mn{ (H) : H F, H = α(f ) + 1}

22 A Stronger Lower Bound Let v (H) be the number of vertces of degree n H. Let M (H) denote the set of nduced subgraphs on α + 1 vertces wth v (H) > 0. For all, s α 1 defne: m = mn M (H){vertces of degree at least } n s = m s 1 and n = mn{m 1,n 1 } Fnally, set δ α 1 = n α 1 and for all, s α 2 defne δ = n n +1 and π (F,n) = ((n 1) u,(u + α 1) δ α 1,(u + α 2) δ α 2,... (u + s) δs,(u + s 1) n u Σδ ).

23 An Example Let F = K 6,6. Then u(k 6,6 ) = = 5 and s(k 6,6 ) = 4. m 4 = 3 and m 5 = 2 n 4 = m 4 1 = 2 and n 5 = mn{m 5 1,n 4 } = 1 δ 5 = n 5 = 1 and δ 4 = n 5 n 4 = 1 Thus, π (K 6,6,n) = ((n 1) 5,10,9,8 n 7 )

24 A Stronger Lower Bound Theorem (Ferrara, S. - 09) Gven a graph F and n suffcently large then, σ(f,n) max{σ(π (F,n)) + 2 F F }

25 When Does Equalty Hold? clques complete bpartte graphs Chen, L, Yn 04; Gould, Jacobson, Lehel 99; L, Yn 02 complete multpartte graphs Chen, Yn 08; G. Chen, Ferrara, Gould, S. 08; Ferrara, Gould, S. 08 matchngs Gould, Jacobson, Lehel 99 cycles La 04 (generalzed) frendshp graph Ferrara, Gould, S. 06, (Chen, S., Yn 08) clque mnus an edge La 01; L, Mao, Yn 05 dsjont unon of clques Ferrara 08

26 Conjecture Gven a graph F and n suffcently large then, σ(f,n) = max{σ(π (F,n)) + 2 F F } Conjecture (weaker verson) Gven a graph F, let ǫ > 0. Then there exsts an n 0 = n 0 (ǫ,f) such that for any n > n 0 σ(f,n) max{(n(2u(f ) + d(f ) 1 + ǫ) F F }.

27 Conjecture Gven a graph F and n suffcently large then, σ(f,n) = max{σ(π (F,n)) + 2 F F } Conjecture (weaker verson) Gven a graph F, let ǫ > 0. Then there exsts an n 0 = n 0 (ǫ,f) such that for any n > n 0 σ(f,n) max{(n(2u(f ) + d(f ) 1 + ǫ) F F }. Conjecture (strong form) holds for graphs wth ndependence number 2 (Ferrara, S. - 09)

28 An example of the generalzed problem Is the followng graphc? < V,d,D >=< {V 1,V 2 },(5 4,3 8 ), [ ] >

29 An example of the generalzed problem Is the followng graphc? < V,d,D >=< {V 1,V 2 },(5 4,3 8 ), [ ] >

30 An example of the generalzed problem Is the followng graphc? < V,d,D >=< {V 1,V 2 },(5 4,3 8 ), [ ] > V 1 V 2

33 Let d = (d v 1 1,dv 2 2,...,dv k k ) where v = V and so V s the set of vertces of degree d. Let V = {V 1,...,V k }. Let D = (d j ) be a k k matrx, wth d j denotng the number of edges between V and V j ; d s the number of edges contaned entrely wthn V.

34 Let d = (d v 1 1,dv 2 2,...,dv k k ) where v = V and so V s the set of vertces of degree d. Let V = {V 1,...,V k }. Let D = (d j ) be a k k matrx, wth d j denotng the number of edges between V and V j ; d s the number of edges contaned entrely wthn V. Jont degree-matrx graphc realzaton problem Gven < V,d,D >, decde whether a smple graph G exsts such that, for all, each vertex n V has degree d, and, for j, there are exactly d j edges between V and V j, whle, for all, there are exactly d edges contaned n V.

35 Amanatds, Green and Mhal (AGM) have shown that the followng natural necessary condtons for a realzaton to exst are also suffcent. The condtons are: Degree feasblty: 2d + Σ j [k],j d j = v d, for all 1 k, and Matrx feasblty: D s a symmetrc matrx wth non-negatve ntegral entres, d j v v j, for all 1 k, and d ( v 2), for all 1 k.

36 AGM s algorthmc proof Algorthm rests on a balanced degree nvarant. It starts wth the empty graph and adds one edge at a tme whle keepng the dfference between any two vertex degrees n a gven V to at most 1. Whle there exsts some,j such that d j s not satsfed the algorthm adds an edge between V and V j.

37 AGM algorthm G p M M j N N j V V j

38 AGM algorthm G p M M j N u v N j V V j

39 AGM algorthm G p+1 M M j N u v N j V V j

40 AGM algorthm G p M v M j N u N j V V j

41 AGM algorthm G p M v M j N u N j V V j

42 AGM algorthm G p+1 M N u V v V j M j N j

43 AGM algorthm G p M u v M j N N j V V j

44 AGM algorthm G p M u v M j N N j V V j

45 AGM algorthm G p+1 M u v M j N x y N j V V j x may equal y

46 AGM algorthm G p M N u v V

47 AGM algorthm G p+1 M N u v V

48 AGM algorthm G p M N v u V

49 AGM algorthm G p+1 M N v u If N =1 V

50 AGM algorthm G p M N v u V If N > 1

51 AGM algorthm G p+1 M N v u V If N > 1

52 AGM algorthm G p M N v u V

53 AGM algorthm G p M N v u V If N =1

54 AGM algorthm G p+1 M N v u V If N =1

55 AGM algorthm G p M N v u If N >1 V

56 AGM algorthm G p+1 M v u If N >1 N V

57 Theorem (Jont Degree-Matrx Realzaton Theorem - AGM) Gven < V,d,D >, f degree and matrx feasblty hold, then a graph G exsts that realzes < V, d, D >. Furthermore, such a graph can be constructed n tme polynomal n n.

58 Can one prove our conjectures usng the Jont Degree-Matrx Theorem?

59 Can we prove our conjectures usng the AGM-result? Is d = (10 6,4 8 ) potentally K 6 -graphc?

60 Can we prove our conjectures usng the AGM-result? Is d = (10 6,4 8 ) potentally K 6 -graphc? If a realzaton of d exsts that contans a copy of K 6, then, t s known, a realzaton exsts n whch the copy of K 6 les on the vertces of hghest degree.

61 Can we prove our conjectures usng the AGM-result? Is d = (10 6,4 8 ) potentally K 6 -graphc? If a realzaton of d exsts that contans a copy of K 6, then, t s known, a realzaton exsts n whch the copy of K 6 les on the vertces of hghest degree. So, by Theorem of AGM, we must smply construct a D for whch degree and matrx feasblty hold and for whch, n ths example, d 11 = ( 6 2) = 15.

62 Can we prove our conjectures usng the AGM-result? Is d = (10 6,4 8 ) potentally K 6 -graphc? If a realzaton of d exsts that contans a copy of K 6, then, t s known, a realzaton exsts n whch the copy of K 6 les on the vertces of hghest degree. So, by Theorem of AGM, we must smply construct a D for whch degree and matrx feasblty hold and for whch, n ths example, d 11 = ( 6 2) = 15. The value of d 11 forces, by degree feasblty, d 12 = d 21 = 30. In turn, by degree feasblty agan, we get d 22 = 1. It s now easy to check that matrx feasblty holds. Thus, d has a realzaton contanng a copy of K 6.

63 Summary Proved the EJL-conjecture Generalzed the EJL-conjecture and proved a specfc case Jont Degree-Matrx Theorem appears to be a useful tool. Can we use t to prove the generalzed EJL-conjecture?

64 Summary Proved the EJL-conjecture Generalzed the EJL-conjecture and proved a specfc case Jont Degree-Matrx Theorem appears to be a useful tool. Can we use t to prove the generalzed EJL-conjecture? Thank you!