Trees, Tanglegrams, and Tangled Chains

Transcription

1 Trees, Tanglegrams, and Tangled Chains Sara Billey University of Washington Based on joint work with: Matjaž Konvalinka and Frederick (Erick) Matsen IV arxiv: WPI, September 12, 2015

2 Outline Background Formulas for Trees, Tanglegrams and Tangled Chains Algorithms for random generation Open Problems

3 Rooted Binary Trees B n = set of rooted inequivalent binary trees with n leaves B n 1, 1, 1, 2, 3, 6, 11, 23, 46, 98,...

4 Rooted Binary Trees B n = set of rooted inequivalent binary trees with n leaves B n 1, 1, 1, 2, 3, 6, 11, 23, 46, 98,... Examples. (1), (2), (3) represent the unique rooted binary trees for n = 1, 2, 3 respectively.

5 Rooted Binary Trees B n = set of rooted inequivalent binary trees with n leaves B n 1, 1, 1, 2, 3, 6, 11, 23, 46, 98,... Examples. (1), (2), (3) represent the unique rooted binary trees for n = 1, 2, 3 respectively. B 4 = {((1)(3)), ((2)(2))}, B 5 = {((1)((1)(3))), ((1)((2)(2))), ((2)(3))}, ((1)(((1)((1)((1)(3))))(((2)(2))(((1)(3))((2)(3)))))) is in B 20. B 20 = 293, 547

6 Catalan objects C n = set of plane rooted binary trees with n leaves C n 1, 1, 2, 5, 14, 42,... Example. ((1)(2)) and ((2)(1)) are distinct as plane trees.

7 Automorphism Groups of Rooted Binary Trees Let T B n rooted binary tree with n leaves. A(T ) is the automorphism group of T given a canonical labeling of its leaves. Example. T = ((1)((2)(2))) generated by 3 involutions [1, 3, 2, 4, 5], [1, 2, 3, 5, 4], [1, 4, 5, 2, 3] (2 3) (4 5) (2 4)(3 5) A(T ) = 2 3 = 8.

8 Tanglegrams Defn. An (ordered binary rooted) tanglegram of size n is a triple (T, w, S) where S, T B n and w S n. Two tanglegrams (T, w, S) and (T, w, S ) are equivalent provided T = T, S = S and w A(T )wa(s). T n = set of inequivalent tanglegrams with n leaves t n = T n 1, 1, 2, 13, 114, 1509, 25595, ,... Example. n = 3, t 3 = 2

9 Tanglegrams Case n = 4, t 4 = 13 :

10 Enumeration of Tanglegrams Questions.(Matsen) How many tanglegrams are in T n? How does t n grow asymptotically? First formula.:. t n = S B n T B n 1 A(T )wa(s) w S n This formula allowed us to get data up to n = 10. Sequence wasn t in OEIS.

11 Motivation to study tanglegrams Cophylogeny Estimation Problem in Biology.: Reconstruct the history of genetic changes in a host vs parasite or other linked groups of organisms. Tanglegram Layout Problem in CS.: Find a drawing of a tanglegram in the plane with planar embeddings of the left and right trees and a minimal number of crossing (straight) edges in the matching. Eades-Wormald (1994) showed this is NP-hard. Tanglegrams appear in analysis of software development in CS.

12 Main Enumeration Theorem Thm 1. The number of tanglegrams of size n is t n = λ summed over binary partitions of n. ( ) l(λ) 2 i=2 2(λ i + + λ l(λ) ) 1, z λ Defn. A binary partition λ = (λ 1 λ 1...) has each part λ k = 2 j for some j N. Defn. z λ = 1 m 0 2 m 1 4 m2 (2 j ) m j m 0!m 1!m 2! m j! for λ = 1 m 0 2 m 1 4 m 2 8 m3.

13 The numbers z λ are famous! Defn. More generally, z λ = 1 m 1 2 m 2 3 m3 j m j m 1!m 2!m 2! m j! for λ = 1 m 1 2 m 2 3 m3. Facts.: 1. The number of permutations in S n of cycle type λ is n! z λ. 2. If v S n has cycle type λ, then z λ is the size of the stabilizer of v under the conjugation of S n on itself. 3. For fixed u, v S n of cycle type λ, z λ = #{w S n wvw 1 = u }. 4. The symmetric function h n (X) = λ p λ (X) z λ.

14 Main Enumeration Theorem Thm. The number of tanglegrams of size n is t n = λ ( ) l(λ) 2 i=2 2(λ i + + λ l(λ) ) 1, z λ summed over binary partitions of n and z λ. Example. The 4 binary partitions of n = 4 are λ : (4) (22) (211) (1111) z λ : ! ! 1 4 4! t 4 = = 13

15 Corollaries Cor 1. t n = c2 n 1 n! 4 n 1 µ z µ l(µ) i=1 n(n 1) (n µ +1), µi 1 j=1 (2n 2(µ 1+ +µ i 1 ) 2j 1) 2 summed is over binary partitions µ with all parts equal to a positive power of 2 and µ n. Cor 2.: As n gets large, t n n! e n 1 πn 3. Cor 3.: There is an efficient recurrence relation for t n based on stripping off all copies of the largest part of λ. We can compute t 4000 exactly.

16 Second Enumeration Theorem Thm 2. The number of binary trees in B n is b n = λ summed over binary partitions of n. ( ) l(λ) i=2 2(λ i + + λ l(λ) ) 1, z λ

17 Second Enumeration Theorem Thm 2. The number of binary trees in B n is b n = λ summed over binary partitions of n. ( ) l(λ) i=2 2(λ i + + λ l(λ) ) 1, z λ Question. What if the exponent k is bigger than 2? t(k, n) = λ l(λ) i=2 ( 2(λ i + + λ l(λ) ) 1 ) k z λ.

18 Tangled Chains Defn. A tangled chain of size n and length k is an ordered sequence of binary trees with complete matchings between the leaves of neighboring trees in the sequence. Thm 3. The number of tangled chains of size n and length k is t(k, n) = λ l(λ) i=2 ( 2(λ i + + λ l(λ) ) 1 ) k z λ.

19 Outline of Proof of Theorem 1 t n = S B n T B n 1 A(T )wa(s) w S n

20 Outline of Proof of Theorem 1 For S, T fixed t n = S B n T B n 1 A(T )wa(s) w S n A(T )wa(s) = A(T ) A(S) A(T ) wa(s)w 1

21 Outline of Proof of Theorem 1 For S, T fixed t n = S B n T B n A(T )wa(s) = w S n A(T ) wa(s)w 1 = 1 A(T )wa(s) w S n A(T ) A(S) A(T ) wa(s)w 1 w S n u A(T ) v A(S) χ[u = wvw 1 ]

22 Outline of Proof of Theorem 1 For S, T fixed t n = S B n T B n A(T )wa(s) = w S n A(T ) wa(s)w 1 = = u A(T ) v A(S) 1 A(T )wa(s) w S n A(T ) A(S) A(T ) wa(s)w 1 w S n u A(T ) v A(S) w S n χ[u = wvw 1 ] χ[u = wvw 1 ]

23 Outline of Proof of Theorem 1 For S, T fixed t n = S B n T B n A(T )wa(s) = w S n A(T ) wa(s)w 1 = = u A(T ) v A(S) 1 A(T )wa(s) w S n A(T ) A(S) A(T ) wa(s)w 1 w S n u A(T ) v A(S) w S n χ[u = wvw 1 ] = A(T ) λ A(S) λ z λ λ n χ[u = wvw 1 ] where A(T ) λ = {w A(T ) type(w) = λ}. Only binary partitions occur!

24 Outline of Proof of Main Theorem t n = = S B n T B n S B n T B n λ 1 A(T )wa(s) w S n A(T ) λ A(S) λ z λ A(T ) A(S)

25 Outline of Proof of Main Theorem t n = = S B n T B n S B n T B n λ z λ = λ 1 A(T )wa(s) w S n A(T ) λ A(S) λ z λ A(T ) A(S) A(T ) λ A(T ) T B n 2

26 Outline of Proof of Main Theorem To show: t n = = S B n T B n S B n T B n λ z λ = λ A(T ) λ = A(T ) T B n via the recurrence 1 A(T )wa(s) w S n A(T ) λ A(S) λ z λ A(T ) A(S) A(T ) λ A(T ) T B n ( ) l(λ) i=2 2(λ i + + λ l(λ) ) 1 =: q λ z λ 2q λ = q λ/2 + Conclusion: t n = z λ q 2 λ. λ 1 λ 2 =λ 2 q λ 1q λ 2

27 Random Generation of Tanglegrams Input: n Step 1: Pick a binary partition λ n with prob z λ q 2 λ /t n. Step 2: Choose T and u A(T ) λ uniformly by subdividing λ = λ 1 λ 2 according to the recurrence for q λ. Similarly, choose S and v A(T ) λ uniformly by subdividing. Step 3: Among the z λ permutations w such that u = wvw 1, pick one uniformly. Output: (T, w, S).

28 Random Generation of a Permutation in A(T ) Input: Binary tree T B n with left and right subtrees T 1 and T 2. If n = 1, set w = (1) A(T ), unique choice. Otherwise, recursively find w 1 A(T 1 ) and w 2 A(T 2 ) at random. If T 1 T 2, set w = w 1 w 2. If T 1 = T 2, choose either w = w 1 w 2 or w = πw 1 w 2 with equal probability. Here π = (1 k)(2 (k + 1))(3 (k + 3)) (k n) where k = n/2 flips the labels on the leaves of the two subtrees. Output: Permutation w A(T ).

29 Random Generation of Tanglegrams:Step 2 Step 2: Choose T and u A(T ) λ uniformly by subdividing λ = λ 1 λ 2 according to the recurrence for q λ. Input: λ n. If n = 1, output T =, u = (1) A(T ), unique choice. Otherwise, pick a subdivision of λ from two types {(λ/2, λ/2)} {(λ 1, λ 2 ) : λ 1 λ 2 = λ} with probability proportional to q λ/2 + q λ 1q λ 2 = 2q λ.

30 Random Generation of Tanglegrams:Step 2 Step 2: Choose T and u A(T ) λ uniformly by subdividing λ = λ 1 λ 2 according to the recurrence for q λ. Type 1: (λ/2, λ/2). Use the algorithm recursively to compute T 1 B n/2 and a permutations u 2 A(T 1 ) λ/2. Uniformly at random, generate anther permutation u 1 A(T 1 ). Set T = (T 1, T 1 ), u = πu 1 πu 1 1 πu 2. Type 2: (λ 1, λ 2 ). Use the algorithm recursively to compute trees T 1, T 2 and permutations u 1 A(T 1 ) λ 1 u 2 A(T 2 ) λ 2. Switch if necessary so T 1 T 2. Set Output: (T, u). T = (T 1, T 2 ), u = u 1 u 2.

31 Random Generation of Tanglegrams:Step 2 Example If λ = (6, 4), then λ = 10, λ/2 = (3, 2) and π = (1 6)(2 7)(3 8)(4 9)(5 10). If then all in cycle notation. w 1 = (1 4)(2 5)(3) and w 2 = (6 9 7)(8 10) w = πw 1 πw 1 1 πw 2 = ( )( ),

32 Review: Random Generation of Tanglegrams Input: n Step 1: Pick a binary partition λ n with prob z λ q 2 λ /t n. Step 2: Choose T and u A(T ) λ uniformly by subdividing λ = λ 1 λ 2 according to the recurrence for q λ. Similarly, choose S and v A(T ) λ uniformly by subdividing. Step 3: Among the z λ permutations w such that u = wvw 1, pick one uniformly. Output: (T, w, S).

33 Random Tanglegrams: n=10

34 Random Tanglegrams: n=20

35 Random Tanglegrams: n=30

36 Random Tanglegrams: n=50

37 Random Tanglegrams: n=100

38 Positivity and symmetric functions go hand in hand with enumeration. This is a story that began with an enumeration question and via work of Gessel now connects to symmetric functions, plethysm of Schur functions, and Kronecker coefficients.

39 Open Problems 1. Is there a closed form or functional equation for T (x) = t n x n like there is for binary trees B(x)? B(x) = x ( ) B(x) 2 + B(x 2 ) 2. Is there an efficient algorithm for depth first search on tanglegrams? 3. Can one describe the lex minimal permutations in the double cosets A(T )\S n /A(S) for S, T B n?