Generalizing the degree sequence problem


 Nathaniel Harris
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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 ErdősGalla 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 ErdősGalla 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 ksubset 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 edgeexchange allows us to move to a realzaton wth fewer components.
11 For a subgraph F, d s sad to be potentally Fgraphc f there exsts a realzaton of d contanng F.
12 For a subgraph F, d s sad to be potentally Fgraphc f there exsts a realzaton of d contanng F. (2,2,2,2,2,2) s potentally K 3 graphc.
13 For a subgraph F, d s sad to be potentally Fgraphc f there exsts a realzaton of d contanng F. (2,2,2,2,2,2) s potentally K 3 graphc.
14 For a subgraph F, d s sad to be potentally Fgraphc 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 Fgraphc. 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 graphtheoretc 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. ErdősGalla guarantees enough vertces of hgh degree. Uses noton of an edgeexhange. Edgeexchange 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 EJLconjecture 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
31 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
32 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 degreematrx 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 nonnegatve 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 DegreeMatrx 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 DegreeMatrx Theorem?
59 Can we prove our conjectures usng the AGMresult? Is d = (10 6,4 8 ) potentally K 6 graphc?
60 Can we prove our conjectures usng the AGMresult? 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 AGMresult? 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 AGMresult? 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 EJLconjecture Generalzed the EJLconjecture and proved a specfc case Jont DegreeMatrx Theorem appears to be a useful tool. Can we use t to prove the generalzed EJLconjecture?
64 Summary Proved the EJLconjecture Generalzed the EJLconjecture and proved a specfc case Jont DegreeMatrx Theorem appears to be a useful tool. Can we use t to prove the generalzed EJLconjecture? Thank you!
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