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The Subtree Polynomial Lucas Mol Joint work with Jason Brown - - PowerPoint PPT Presentation

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I NTRODUCTION C OUNTING S UBTREES C OMPLEX S UBTREE R OOTS R EAL S UBTREE R OOTS O PEN P ROBLEMS The Subtree Polynomial Lucas Mol Joint work with Jason Brown (Dalhousie University) CanaDAM 2019 Graph Polynomials Minisymposium I NTRODUCTION C

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

The Subtree Polynomial

Lucas Mol Joint work with Jason Brown (Dalhousie University) CanaDAM 2019 – Graph Polynomials Minisymposium

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

PLAN

INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

PLAN

INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

THE SUBTREE POLYNOMIAL

Let T be a tree (or forest), and let S be the collection of all subtrees (connected subgraphs) of T.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

THE SUBTREE POLYNOMIAL

Let T be a tree (or forest), and let S be the collection of all subtrees (connected subgraphs) of T. ◮ Define the subtree polynomial of T by ΦT(x) =

n

  • S∈S

x|V(S)|.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

THE SUBTREE POLYNOMIAL

Let T be a tree (or forest), and let S be the collection of all subtrees (connected subgraphs) of T. ◮ Define the subtree polynomial of T by ΦT(x) =

n

  • S∈S

x|V(S)|. ◮ Alternatively, ΦT(x) =

n

  • k=1

sk(T)xk, where sk(T) is the number of subtrees of T of order k.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T: ΦT(x) =

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T: ΦT(x) = 5x

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T: ΦT(x) = 5x + 4x2

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T: ΦT(x) = 5x + 4x2 + 4x3

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T: ΦT(x) = 5x + 4x2 + 4x3 + 3x4

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T: ΦT(x) = 5x + 4x2 + 4x3 + 3x4 + x5

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

GRAPH PARAMETERS ENCODED

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

GRAPH PARAMETERS ENCODED

◮ ΦT(1) is the number of subtrees of T.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

GRAPH PARAMETERS ENCODED

◮ ΦT(1) is the number of subtrees of T.

◮ An important topological index, inversely correlated to the Wiener index.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

GRAPH PARAMETERS ENCODED

◮ ΦT(1) is the number of subtrees of T.

◮ An important topological index, inversely correlated to the Wiener index.

◮ The quantity MT = Φ′

T(1)

ΦT(1) is the mean subtree order of T.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

GRAPH PARAMETERS ENCODED

◮ ΦT(1) is the number of subtrees of T.

◮ An important topological index, inversely correlated to the Wiener index.

◮ The quantity MT = Φ′

T(1)

ΦT(1) is the mean subtree order of T.

◮ First studied by Jamison.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

GRAPH PARAMETERS ENCODED

◮ ΦT(1) is the number of subtrees of T.

◮ An important topological index, inversely correlated to the Wiener index.

◮ The quantity MT = Φ′

T(1)

ΦT(1) is the mean subtree order of T.

◮ First studied by Jamison.

◮ Less obvious: −ΦT(−1) is the independence number of T (Jamsion, 1987).

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

THE LOCAL SUBTREE POLYNOMIAL

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

THE LOCAL SUBTREE POLYNOMIAL

Let T be a tree with vertex v, and let Sv be the collection of subtrees of T containing v.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

THE LOCAL SUBTREE POLYNOMIAL

Let T be a tree with vertex v, and let Sv be the collection of subtrees of T containing v. ◮ The local subtree polynomial of T at v is given by ΦT,v(x) =

  • S∈Sv

x|V(S)|.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

THE LOCAL SUBTREE POLYNOMIAL

Let T be a tree with vertex v, and let Sv be the collection of subtrees of T containing v. ◮ The local subtree polynomial of T at v is given by ΦT,v(x) =

  • S∈Sv

x|V(S)|. Evidently, we have: ΦT(x) = ΦT,v(x) + ΦT−v(x).

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) =

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) = x

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) = x + x2

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) = x + x2 + 2x3

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) = x + x2 + 2x3 + 2x4

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) = x + x2 + 2x3 + 2x4 + x5

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) = x + x2 + 2x3 + 2x4 + x5 ΦT−v(x) = 4x + 3x2 + 2x3 + x4

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

AN EXAMPLE

Consider the following tree T with vertex v: v ΦT,v(x) = x + x2 + 2x3 + 2x4 + x5 ΦT−v(x) = 4x + 3x2 + 2x3 + x4 Notice: Many of the “large” subtrees of T contain v, while many

  • f the “small” subtrees of T do not contain v.
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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

PLAN

INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

COUNTING SUBTREES

◮ Let sk(T) denote the number of subtrees of T of order k, i.e., ΦT(x) =

n

  • k=1

sk(T)xk.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

COUNTING SUBTREES

◮ Let sk(T) denote the number of subtrees of T of order k, i.e., ΦT(x) =

n

  • k=1

sk(T)xk. ◮ Among all trees of order n, the star has the largest number

  • f subtrees, while the path has the smallest number of

subtrees.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

COUNTING SUBTREES

◮ Let sk(T) denote the number of subtrees of T of order k, i.e., ΦT(x) =

n

  • k=1

sk(T)xk. ◮ Among all trees of order n, the star has the largest number

  • f subtrees, while the path has the smallest number of

subtrees. ◮ In fact, Jamison demonstrated that for any tree T of order n ≥ 5 not isomorphic to Pn or K1,n−1, sk (Pn) < sk(T) < sk

  • K1,n−1
  • for all 3 ≤ k ≤ n − 1.
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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

COUNTING SUBTREES

◮ Let sk(T) denote the number of subtrees of T of order k, i.e., ΦT(x) =

n

  • k=1

sk(T)xk. ◮ Among all trees of order n, the star has the largest number

  • f subtrees, while the path has the smallest number of

subtrees. ◮ In fact, Jamison demonstrated that for any tree T of order n ≥ 5 not isomorphic to Pn or K1,n−1, sk (Pn) < sk(T) < sk

  • K1,n−1
  • for all 3 ≤ k ≤ n − 1.

◮ We give a short proof of this result.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

CSIKV´

ARI’S GENERALIZED TREE SHIFT

Let T be a tree, and let v1 and v2 be non-leaf vertices of T connected by a path whose internal vertices all have degree 2.

T1 T2 q v1 v2

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

CSIKV´

ARI’S GENERALIZED TREE SHIFT

Let T be a tree, and let v1 and v2 be non-leaf vertices of T connected by a path whose internal vertices all have degree 2.

T1 T2 q v1 v2

The following tree T ′ is said to be obtained from T by a generalized tree shift:

T1 T2 q + 1

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A POSET

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A POSET

◮ Let Tn be the set of trees

  • f order n.
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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A POSET

◮ Let Tn be the set of trees

  • f order n.

◮ Say T T ′ if T ′ can be

  • btained from T by a

sequence of generalized tree shifts.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A POSET

◮ Let Tn be the set of trees

  • f order n.

◮ Say T T ′ if T ′ can be

  • btained from T by a

sequence of generalized tree shifts. ◮ is a partial order on Tn.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A POSET

◮ Let Tn be the set of trees

  • f order n.

◮ Say T T ′ if T ′ can be

  • btained from T by a

sequence of generalized tree shifts. ◮ is a partial order on Tn. ◮ The path Pn is the minimum element, and the star K1,n−1 is the maximum element.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A POSET

◮ Let Tn be the set of trees

  • f order n.

◮ Say T T ′ if T ′ can be

  • btained from T by a

sequence of generalized tree shifts. ◮ is a partial order on Tn. ◮ The path Pn is the minimum element, and the star K1,n−1 is the maximum element.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree of order n ≥ 4, and let T ′ be a tree

  • btained from T by a generalized tree shift. For every

k ∈ {3, . . . , n − 1}, sk(T) < sk(T ′).

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree of order n ≥ 4, and let T ′ be a tree

  • btained from T by a generalized tree shift. For every

k ∈ {3, . . . , n − 1}, sk(T) < sk(T ′).

T1 T2 q v1 v2

T :

T1 T2

T ′ :

q + 1

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree of order n ≥ 4, and let T ′ be a tree

  • btained from T by a generalized tree shift. For every

k ∈ {3, . . . , n − 1}, sk(T) < sk(T ′).

T1 T2 q v1 v2

T :

T1 T2

T ′ :

q + 1

Sketch of Proof: Show that ΦT ′(x) − ΦT(x) = 1 x · q

  • i=0

xi

  • ·
  • ΦT1,v1(x) − x
  • ·
  • ΦT2,v2(x) − x
  • .
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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

PLAN

INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

SUBTREE ROOTS

If ΦT(z) = 0, then z ∈ C is called a subtree root of T.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

SUBTREE ROOTS

If ΦT(z) = 0, then z ∈ C is called a subtree root of T.

Re(z) Im(z)

The subtree roots of all trees of order at most 14.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

BOUNDING THE MODULUS

Theorem: Let T be a tree. If ΦT(z) = 0, then |z| ≤ 1 +

3

√ 3.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

BOUNDING THE MODULUS

Theorem: Let T be a tree. If ΦT(z) = 0, then |z| ≤ 1 +

3

√ 3. ◮ This bound is tight! The only tree with a subtree root of modulus 1 +

3

√ 3 is the star K1,3.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree. If ΦT(z) = 0, then |z| ≤ 1 +

3

√ 3.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree. If ΦT(z) = 0, then |z| ≤ 1 +

3

√ 3. Proof idea: Use the fact that ΦT(z) = ΦT,v(z) + ΦT−v(z).

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree. If ΦT(z) = 0, then |z| ≤ 1 +

3

√ 3. Proof idea: Use the fact that ΦT(z) = ΦT,v(z) + ΦT−v(z). By the reverse triangle inequality, |ΦT(z)| ≥ |ΦT,v(z)| − |ΦT−v(z)|.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree. If ΦT(z) = 0, then |z| ≤ 1 +

3

√ 3. Proof idea: Use the fact that ΦT(z) = ΦT,v(z) + ΦT−v(z). By the reverse triangle inequality, |ΦT(z)| ≥ |ΦT,v(z)| − |ΦT−v(z)|. So it suffices to show that if |z| > 1 +

3

√ 3, then |ΦT,v(z)| > |ΦT−v(z)|.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Theorem: Let T be a tree. If ΦT(z) = 0, then |z| ≤ 1 +

3

√ 3. Proof idea: Use the fact that ΦT(z) = ΦT,v(z) + ΦT−v(z). By the reverse triangle inequality, |ΦT(z)| ≥ |ΦT,v(z)| − |ΦT−v(z)|. So it suffices to show that if |z| > 1 +

3

√ 3, then |ΦT,v(z)| > |ΦT−v(z)|. This is proven using a rather technical strong induction. The key is a good lower bound on |ΦT,v(z)|.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Lemma: Let T be a tree of order n with vertex v. If |z| ≥ 2, then |ΦT,v(z)| ≥ |z| · (|z| − 1)n−1 .

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Lemma: Let T be a tree of order n with vertex v. If |z| ≥ 2, then |ΦT,v(z)| ≥ |z| · (|z| − 1)n−1 . Key ingredient of proof:

v v1 v2 vk

T2 T1 Tk . . .

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Lemma: Let T be a tree of order n with vertex v. If |z| ≥ 2, then |ΦT,v(z)| ≥ |z| · (|z| − 1)n−1 . Key ingredient of proof:

v v1 v2 vk

T2 T1 Tk . . . ΦT,v(z) = z ·

k

  • i=1
  • 1 + ΦTi,vi(z)
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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

PLAN

INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

REAL SUBTREE ROOTS

◮ The subtree polynomial has positive coefficients

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

REAL SUBTREE ROOTS

◮ The subtree polynomial has positive coefficients

◮ ⇒ no subtree roots in (0, ∞)

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

REAL SUBTREE ROOTS

◮ The subtree polynomial has positive coefficients

◮ ⇒ no subtree roots in (0, ∞)

◮ The subtree polynomial has no complex roots of modulus greater than 1 +

3

√ 3

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

REAL SUBTREE ROOTS

◮ The subtree polynomial has positive coefficients

◮ ⇒ no subtree roots in (0, ∞)

◮ The subtree polynomial has no complex roots of modulus greater than 1 +

3

√ 3

◮ ⇒ no subtree roots in

  • −∞, −1 −

3

√ 3

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

REAL SUBTREE ROOTS

◮ The subtree polynomial has positive coefficients

◮ ⇒ no subtree roots in (0, ∞)

◮ The subtree polynomial has no complex roots of modulus greater than 1 +

3

√ 3

◮ ⇒ no subtree roots in

  • −∞, −1 −

3

√ 3

  • ◮ Question: What about the interval (−1 −

3

√ 3, 0)?

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0).

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0). Sketch of Proof:

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0). Sketch of Proof: Let T be a tree.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0). Sketch of Proof: Let T be a tree. ◮ Claim: Φ′

T(x) > 0 for all x ∈ (−1, 0).

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0). Sketch of Proof: Let T be a tree. ◮ Claim: Φ′

T(x) > 0 for all x ∈ (−1, 0).

◮ Since ΦT(0) = 0, the fact that ΦT(x) < 0 for x ∈ [−1, 0) follows by MVT.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0). Sketch of Proof: Let T be a tree. ◮ Claim: Φ′

T(x) > 0 for all x ∈ (−1, 0).

◮ Since ΦT(0) = 0, the fact that ΦT(x) < 0 for x ∈ [−1, 0) follows by MVT. ◮ Useful identity: xΦ′

T(x) =

  • v∈V(T)

ΦT,v(x)

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0). Sketch of Proof: Let T be a tree. ◮ Claim: Φ′

T(x) > 0 for all x ∈ (−1, 0).

◮ Since ΦT(0) = 0, the fact that ΦT(x) < 0 for x ∈ [−1, 0) follows by MVT. ◮ Useful identity: xΦ′

T(x) =

  • v∈V(T)

ΦT,v(x) ◮ Claim: ∀v ∈ V(T), x ∈ (−1, 0) ⇒ ΦT,v(x) ∈ (−1, 0).

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-FREE SUBINTERVAL

Theorem: No tree has a subtree root in the interval [−1, 0). Sketch of Proof: Let T be a tree. ◮ Claim: Φ′

T(x) > 0 for all x ∈ (−1, 0).

◮ Since ΦT(0) = 0, the fact that ΦT(x) < 0 for x ∈ [−1, 0) follows by MVT. ◮ Useful identity: xΦ′

T(x) =

  • v∈V(T)

ΦT,v(x) ◮ Claim: ∀v ∈ V(T), x ∈ (−1, 0) ⇒ ΦT,v(x) ∈ (−1, 0). ◮ Claim is proved using induction and the nice recursive formula for ΦT,v(x): ΦT,v(x) = x ·

k

  • i=1
  • 1 + ΦTi,vi(x)
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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-DENSE SUBINTERVAL

Theorem: The closure of the collection of all real subtree roots contains the interval [−2, −1].

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

A ROOT-DENSE SUBINTERVAL

Theorem: The closure of the collection of all real subtree roots contains the interval [−2, −1]. Proof technique: We show that the collection of real subtree roots of the following family of trees is dense in [−2, −1]:

a b

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

PLAN

INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

Find a tighter bound on the subtree roots of all trees of order n.

Re(z) Im(z)

Our bound: 1 +

3

√ 3

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

Find a tighter bound on the subtree roots of all trees of order n.

Re(z) Im(z)

Our bound: 1 +

3

√ 3 Conjectured bound: 1 +

n−1

√ n − 1

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

Find a tighter bound on the subtree roots of all trees of order n.

Re(z) Im(z)

Our bound: 1 +

3

√ 3 Conjectured bound: 1 +

n−1

√ n − 1 A stronger conjecture

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

CLOSURE

Does the closure of the collection of subtree roots contain the entire annulus 1

2 ≤

  • z + 1

2

  • ≤ 3

2?

Re(z) Im(z)

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

◮ Consider “subtrees” of graphs.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

◮ Consider “subtrees” of graphs.

◮ Roots of the subtree polynomial still appear to be bounded in modulus by 1 +

3

√ 3.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

◮ Consider “subtrees” of graphs.

◮ Roots of the subtree polynomial still appear to be bounded in modulus by 1 +

3

√ 3.

Re(z) Im(z)

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

A question of Jamison:

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

A question of Jamison: ◮ For which trees is the sequence of coefficients s2(T), s3(T), . . . , sn(T) unimodal?

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

A question of Jamison: ◮ For which trees is the sequence of coefficients s2(T), s3(T), . . . , sn(T) unimodal? ◮ In particular, is this sequence unimodal when T has no vertices of degree 2?

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

OPEN PROBLEMS

A question of Jamison: ◮ For which trees is the sequence of coefficients s2(T), s3(T), . . . , sn(T) unimodal? ◮ In particular, is this sequence unimodal when T has no vertices of degree 2?

◮ Fact: This sequence is NOT log-concave for all such trees.

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INTRODUCTION COUNTING SUBTREES COMPLEX SUBTREE ROOTS REAL SUBTREE ROOTS OPEN PROBLEMS

Thank you!