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Realization polytopes for the degree sequence of a graph Michael D. Barrus Department of Mathematics Brigham Young University CanaDAM 2013 June 12, 2013 M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 1 /

  1. Realization polytopes for the degree sequence of a graph Michael D. Barrus Department of Mathematics Brigham Young University • CanaDAM 2013 June 12, 2013 M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 1 / 23

  2. A fractional excursion It is possible to go to a graph theory conference and to ask oneself, at the end of every talk, What is the fractional analogue? What is the right definition? — Scheinerman and Ullman, Fractional Graph Theory M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 2 / 23

  3. A fractional excursion It is possible to go to a graph theory conference and to ask oneself, at the end of every talk, What is the fractional analogue? What is the right definition? — Scheinerman and Ullman, Fractional Graph Theory ( 2 , 2 , 2 ) , ( 2 , 1 , 1 ) , ( 1 , 2 , 1 ) , ( 1 , 1 , 2 ) , ( 1 , 1 , 0 ) , ( 1 , 0 , 1 ) , ( 0 , 1 , 1 ) , ( 0 , 0 , 0 ) M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 2 / 23

  4. A fractional excursion It is possible to go to a graph theory conference and to ask oneself, at the end of every talk, What is the fractional analogue? What is the right definition? — Scheinerman and Ullman, Fractional Graph Theory ( 2 , 2 , 2 ) , ( 2 , 1 , 1 ) , ( 1 , 2 , 1 ) , ( 1 , 1 , 2 ) , ( 1 , 1 , 0 ) , ( 1 , 0 , 1 ) , ( 0 , 1 , 1 ) , ( 0 , 0 , 0 ) U.N. Peled and M.K. Srinivasan. The polytope of degree sequences. Linear Algebra and its Applications, 114/115:349–377 (1989). M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 2 / 23

  5. b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b Keeping track of realizations ( 2 , 2 , 2 , 1 , 1 ) M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 3 / 23

  6. Keeping track of realizations 2 0) ∈ R 10 (0, 1, 0, 1, 1, 1, 0, 0, 0, 1 3 12 13 14 15 23 24 25 34 35 45 4 5 M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 4 / 23

  7. Keeping track of realizations 2 0) ∈ R 10 (0, 1, 0, 1, 1, 1, 0, 0, 0, 1 3 12 13 14 15 23 24 25 34 35 45 4 5 Simple graph realizations of ( 2 , 2 , 2 , 1 , 1 ) must satisfy x 12 + x 13 + x 14 + x 15 = 2 x 12 + x 23 + x 24 + x 25 = 2 x 13 + x 23 + x 34 + x 35 = 2 x 14 + x 24 + x 34 + x 45 = 1 x 15 + x 25 + x 35 + x 45 = 1 x ij ∈ { 0 , 1 } M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 4 / 23

  8. Keeping track of realizations 2 0) ∈ R 10 (0, 1, 0, 1, 1, 1, 0, 0, 0, 1 3 12 13 14 15 23 24 25 34 35 45 4 5 Simple graph realizations of ( 2 , 2 , 2 , 1 , 1 ) must satisfy x 12 + x 13 + x 14 + x 15 = 2 x 12 + x 23 + x 24 + x 25 = 2 x 13 + x 23 + x 34 + x 35 = 2 x 14 + x 24 + x 34 + x 45 = 1 x 15 + x 25 + x 35 + x 45 = 1 0 ≤ x ij ≤ 1 M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 4 / 23

  9. A polytope Given d , let S ( d ) be the polytope in R ( n 2 ) defined by Degree conditions n � for 1 ≤ i ≤ n x ij = d i j = 1 j � = i Hypercube bounds 0 ≤ x ij ≤ 1 for 1 ≤ i , j ≤ n M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 5 / 23

  10. A polytope Given d , let S ( d ) be the polytope in R ( n 2 ) defined by Degree conditions n � for 1 ≤ i ≤ n x ij = d i j = 1 j � = i Hypercube bounds 0 ≤ x ij ≤ 1 for 1 ≤ i , j ≤ n What are the vertices? M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 5 / 23

  11. Two polytopes 2 0) ∈ R 10 (0, 1, 0, 1, 1, 1, 0, 0, 0, 1 3 12 13 14 15 23 24 25 34 35 45 4 5 Simple graph realizations of ( 2 , 2 , 2 , 1 , 1 ) must satisfy x 12 + x 13 + x 14 + x 15 = 2 x 12 + x 23 + x 24 + x 25 = 2 x 13 + x 23 + x 34 + x 35 = 2 x 14 + x 24 + x 34 + x 45 = 1 x 15 + x 25 + x 35 + x 45 = 1 0 ≤ x ij ≤ 1 S ( d ) : bounded by hyperplanes R ( d ) : convex hull M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 6 / 23

  12. S ( 2 , 2 , 2 , 1 , 1 ) (1 , 0 , 0 , 1 , 1 , 0 , 0 , 1 , 0 , 0) (1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 , 1 , 0) (1 , 1 , 0 , 0 , 0 , 0 , 1 , 1 , 0 , 0) (1 , 1 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 1) (0 , 1 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 0) (1 , 1 , 0 , 0 , 0 , 1 , 0 , 0 , 1 , 0) (0 , 1 , 1 , 0 , 1 , 0 , 1 , 0 , 0 , 0) Here, R ( d ) = S ( d ) , i.e., vertices correspond exactly to simple graph realizations. M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 7 / 23

  13. b b b b b b b b b b b b S ( 1 , 1 , 1 , 1 , 1 , 1 ) n � x ij = 1 , 0 ≤ x ij ≤ 1 j = 1 j � = i M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 8 / 23

  14. b b b b b b b b b b b b S ( 1 , 1 , 1 , 1 , 1 , 1 ) n � x ij = 1 , 0 ≤ x ij ≤ 1 j = 1 j � = i ( 1 , 1 , 1 , 1 , 1 , 1 ) has 15 realizations, but S ( 1 , 1 , 1 , 1 , 1 , 1 ) has 25 vertices, so R ( d ) � = S ( d ) . M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 8 / 23

  15. b b b b b b b b b b b b S ( 1 , 1 , 1 , 1 , 1 , 1 ) n � x ij = 1 , 0 ≤ x ij ≤ 1 j = 1 j � = i ( 1 , 1 , 1 , 1 , 1 , 1 ) has 15 realizations, but S ( 1 , 1 , 1 , 1 , 1 , 1 ) has 25 vertices, so R ( d ) � = S ( d ) . A non-integral vertex: ( 0 , 0 , 0 , 1 / 2 , 1 / 2 , 1 / 2 , 1 / 2 , 0 , 0 , 1 / 2 , 0 , 0 , 0 , 0 , 1 / 2 ) M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 8 / 23

  16. b b b b b b b b b b b b S ( 1 , 1 , 1 , 1 , 1 , 1 ) n � x ij = 1 , 0 ≤ x ij ≤ 1 j = 1 j � = i ( 1 , 1 , 1 , 1 , 1 , 1 ) has 15 realizations, but S ( 1 , 1 , 1 , 1 , 1 , 1 ) has 25 vertices, so R ( d ) � = S ( d ) . A non-integral vertex: ( 0 , 0 , 0 , 1 / 2 , 1 / 2 , 1 / 2 , 1 / 2 , 0 , 0 , 1 / 2 , 0 , 0 , 0 , 0 , 1 / 2 ) Each fractional edge has value 1 / 2. M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 8 / 23

  17. b b b b b b Questions ( 0 , 0 , 0 , 1 / 2 , 1 / 2 , 1 / 2 , 1 / 2 , 0 , 0 , 1 / 2 , 0 , 0 , 0 , 0 , 1 / 2 ) What determines whether the vector of a fractional realization is a vertex of S ( d ) ? M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 9 / 23

  18. b b b b b b Questions ( 0 , 0 , 0 , 1 / 2 , 1 / 2 , 1 / 2 , 1 / 2 , 0 , 0 , 1 / 2 , 0 , 0 , 0 , 0 , 1 / 2 ) What determines whether the vector of a fractional realization is a vertex of S ( d ) ? Which degree sequences, like ( 2 , 2 , 2 , 1 , 1 ) , satisfy S ( d ) = R ( d ) , i.e., the vertices of S ( d ) correspond exactly to simple graph realizations of d ? M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 9 / 23

  19. b b b b b b b b b b b b b b b b Vertices of S ( d ) ? ( 1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 , 1 , 0 ) ( 2 / 3 , 1 , 0 , 1 / 3 , 2 / 3 , 2 / 3 , 0 , 0 , 1 / 3 , 1 / 3 ) ( 0 , 0 , 0 , 1 / 2 , 1 / 2 , 1 / 2 , 1 / 2 , 0 , 0 , 1 / 2 , 0 , 0 , 0 , 0 , 1 / 2 ) M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 10 / 23

  20. b b b b b b b b b b b b b b b b Vertices of S ( d ) ? ( 1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 , 1 , 0 ) ( 2 / 3 , 1 , 0 , 1 / 3 , 2 / 3 , 2 / 3 , 0 , 0 , 1 / 3 , 1 / 3 ) ( 0 , 0 , 0 , 1 / 2 , 1 / 2 , 1 / 2 , 1 / 2 , 0 , 0 , 1 / 2 , 0 , 0 , 0 , 0 , 1 / 2 ) 2 / 3 1 / 3 M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 10 / 23

  21. b b b b b b b b b b b b b b b b Vertices of S ( d ) ? ( 1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 , 1 , 0 ) ( 2 / 3 , 1 , 0 , 1 / 3 , 2 / 3 , 2 / 3 , 0 , 0 , 1 / 3 , 1 / 3 ) ( 0 , 0 , 0 , 1 / 2 , 1 / 2 , 1 / 2 , 1 / 2 , 0 , 0 , 1 / 2 , 0 , 0 , 0 , 0 , 1 / 2 ) 2 / 3 1 / 3 Theorem Given a graphic sequence d, a point in S ( d ) is a vertex of S ( d ) if and only if the non-integral edges in the corresponding realization form a disjoint union of odd cycles. When this is the case, there are an even number of these cycles, and each non-integral edge has value 1 / 2 . M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 10 / 23

  22. When can/can’t this happen? ( 1 , 1 , 1 , 1 , 1 , 1 ) vs ( 2 , 2 , 2 , 1 , 1 ) M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 11 / 23

  23. When can/can’t this happen? ( 1 , 1 , 1 , 1 , 1 , 1 ) vs ( 2 , 2 , 2 , 1 , 1 ) For d not admitting disjoint odd cycles of 1 / 2 -edges, S ( d ) = R ( d ) , i.e., vertices of S ( d ) correspond exactly to simple graph realizations of d . M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 11 / 23

  24. When can/can’t this happen? ( 1 , 1 , 1 , 1 , 1 , 1 ) vs ( 2 , 2 , 2 , 1 , 1 ) For d not admitting disjoint odd cycles of 1 / 2 -edges, S ( d ) = R ( d ) , i.e., vertices of S ( d ) correspond exactly to simple graph realizations of d . In every realization, each edge is either 100% there or not there — no halfways about it. We call such d forceful sequences , and we call their realizations forceful graphs . How do we recognize forceful sequences/graphs? M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 11 / 23

  25. What happens in nonforceful graphs For odd k , ℓ : k ℓ vertices vertices 1 1 1 1 0 1 1 1 1 1 1 2 2 1 1 1 1 1 M. D. Barrus (BYU) Realization polytopes for degree sequences June 12, 2013 12 / 23

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