Key words. binary tree, Fitch s algorithm, maximum parsimony tree, DNA/RNA sequences, probability

Transcription

1 SIAMJ.DISCRETE MATH. c 997 Society for Industrial and Applied Mathematics Vol. 0, No. 3, pp , August THE LENGTH OF A LEAF COLORATION ON A RANDOM BINARY TREE A. M. HAMEL AND M. A. STEEL Abstract. An assignment of colors to objects induces a natural integer weight on each tree that has these objects as leaves. This weight is called parsimony length in biostatistics and is the basis of the maximum parsimony technique for reconstructing evolutionary trees. Equations for the average value (over all binary trees) of the parsimony length of both fixed and random colorations are derived using generating function techniques. This leads to asymptotic results that extend earlier results confined to just two colors. A potential application to DNA sequence analysis is outlined briefly. Key words. binary tree, Fitch s algorithm, maximum parsimony tree, DNA/RNA sequences, probability AMS subject classifications. 05C05, 05A5, 9D0 PII. S Introduction. Let, n, denote the set of (unrooted) trees with n leaves (vertices of degree ) labeled,,...,n and with all remaining vertices unlabeled and of degree 3. Such trees, which we will simply call binary trees, are useful representations of evolutionary relationships in taxonomy. In that case, the set [n] ={,,...,n} represents the extant taxa being classified, while the remaining vertices in the tree represent ancestral taxa. It is often convenient to represent the (global) ancestral taxon of all these taxa by a root vertex obtained by subdividing an edge (the most ancient edge) of the tree. Let R(n), n >, denote the set of all such edge-rooted binary trees on leaf set [n]. We define R() as the singleton set consisting of an isolated (root) vertex labeled. Note for n the bijection ψ : R(n ), where, if T,ψ(T) is the edge-rooted binary tree which results when leaf n and its incident edge are deleted, as in Figure. Edge subdivision also gives a bijection, as in Figure. We let ψ : {(T,e):T, e E(T)} R(n), := and R(n) := R(n) for n andn, respectively. Since E(T ) =n 3, for each T, it follows (from ψ and ψ ) that, for n 3, () R(n) =(n 3) =(n 3)!!=3 5 (n 3) (n )! = n (n )!, Received by the editors July, 994; accepted for publication (in revised form) May 3, 996. The first author of this paper was supported by a postdoctoral fellowship from the Natural Sciences and Engineering Research Council of Canada. Department of Mathematics and Statistics, University of Canterbury, Christchurch, New Zealand (physamh@cantua.canterbury.ac.nz, m.steel@math.canterbury.ac.nz). 359

2 360 A. M. HAMEL AND M. A. STEEL Fig.. Bijections between rooted and unrooted binary trees. a result dating back at least as far as 870 to a paper by Schröder [9]. applying Stirling s formula to R(n), Thus, by () R(n) n! π n n 3/. (A definition of asymptotic ( ) appears at the beginning of section 3.) Let χ be a coloration of [n] byasetcof r colors. For example, in phylogenetic analysis each site j in a collection of n aligned DNA/RNA sequences (where r = or 4) gives a coloration χ = χ j of [n] for which χ j (i) is the actual nucleotide (when r = 4) or its purine/pyrimidine classification (when r = ) that occurs at site j in the ith sequence. Given a tree T in orr(n) and a coloration χ of [n] letl(t,χ) be the minimal number of edges of T that need to be assigned differently colored ends in order to extend χ to a coloration of all the vertices of T (any such minimizing extension is called a minimal extension of χ for T ). The number l(t,χ) is called the parsimony length of χ on T, and it is the basis of the widely used maximum parsimony technique for reconstructing evolutionary trees from aligned genetic sequences. This approach selects the tree(s) T which minimizes (minimize) the sum of l(t,χ j ) over all sites j in the sequences this sum is the length of T on the sequences. Such a tree a maximum parsimony tree requires the fewest mutations to account for the variations in the aligned sequences. The aim of this paper is to develop analytic techniques that would allow the length of the maximum parsimony tree on the original sequences to be compared with the average length of all binary trees on either (i) the original sequences or (ii) randomized versions of the original sequences (i.e., sequences generated randomly with the same expected frequencies of colors as the original sequences, as in Steel, Lockhart, and Penny []). These two average values are obtained by evaluating, respectively, certain functions µ n and µ n (which we describe in section ) at each sequence site and summing up the resulting values across the sites. An asymptotic formula for µ n is described in section 3 and since, as we show, µ n and µ n are asymptotically equivalent, this provides an asymptotic formula for µ n as well. Our results exploit some special properties of the generating functions which count various classes of leaf labeled trees

3 PARSIMONY LENGTH OF A RANDOM BINARY TREE 36 Fig.. A rooted tree with sets assigned to vertices by the parsimony operation. according to their parsimony length. In this sense the exact and asymptotic analyses complement and extend the approaches of Carter et al. [] and Butler [], respectively, both of which analyzed similar systems of generating functions with just two colors (although the problems these authors considered were slightly different from ours). First we describe a convenient technique for computing l(t,χ) known as the (first pass of) Fitch s algorithm (Fitch [3], Hartigan [5]). If T, subdivide any edge of T to obtain a tree T R(n). Note that l(t,χ)=l(t,χ). Now direct all edges of T away from the root and recursively assign nonempty subsets of colors to the vertices of T beginning with the leaves and progressing toward the root, as follows: () leaf i [n] is assigned the singleton set {χ(i)}, () once the descendants of vertex v have both been assigned sets A, B, then assign vertex v the set A B, where is the (nonassociative, binary) parsimony operation defined on C φ, { A B if A B φ, A B = A B if A B = φ. The set assigned to the root of T is called the root set. (In the case T R() the root set is just {χ()}.) These concepts are illustrated in Figure. A fundamental property of this procedure is the following. Lemma. (Hartigan [5]). l(t,χ) is the number of times an empty intersection (option in the above description of ) is encountered in this assignment of sets of colors to the vertices of T. Furthermore, the root set is precisely the set of those colors that occur in at least one minimal extension of χ for T. We will use both of these properties in section. Notation. () For convenience, we write x to denote (x,x,...,x r ), and x a to denote the monomial x a xa...xar r, a! to denote a!a!...a r!. () We also write [x a ]f(x) to denote the coefficient of x a xa...xar r in f(x),

4 36 A. M. HAMEL AND M. A. STEEL as in Goulden and Jackson [4]. (3) C = {α,...,α r } will denote the set of colors which are assigned to the elements of the set [n] ={,...,n}.ifa i = χ (α i ),i=...r,wesayχis of type a =(a,...,a r ). Thus, a i 0and r i= a i = n.. Calculations (exact). The aim of this paper is to calculate the two averages that we now define. Definition (µ n and µ n). Let µ n (a) be the average, over all trees T,of the length of a fixed coloration of [n] of type a on T. For probability distribution φ = (φ,φ,...,φ r ), φ 0,φ 0,...,φ r 0, r i= φ i =,letµ n(φ)be the average, over all trees T, of the expected length of a random coloration of [n] on T. In this random coloration each element of [n] is independently assigned color α i with probability φ i. Note that µ n(φ) is the average, over all trees T, of n l(t,χ) φ χ(j), χ j= and so (3) µ n(φ) = a ( ) n φ a µ n (a). a Here and elsewhere a summation over a ranges over all nonnegative r-tuples a,...,a r with r i= a i = n. Also, note that µ n and µ n are symmetric functions in a,...,a r and φ,...,φ r, respectively. The following generating function forms the basis for our calculations. For φ A C,letT A (x,z)= f A (a,l) a,l a! x a z l, where f A (a,l) is the number of trees in R(n), n, of parsimony length l 0 and root set A for a fixed r-coloration of [n] oftypea. By Lemma., the set {T A (x,z), A C} satisfies the system of simultaneous quadratic equations described in Steel [0], (4) T A (x,z)= (B,C):B C=A T B(x,z)T C (x,z)+ (B,C): B C= B C=A z T B(x,z)T C (x,z)+δ A (x), where (5) δ A (x) = { xi if A = {α i }, 0 if A >. For r = this system can be treated by the multivariate Lagrange inversion formula (Goulden and Jackson [4]) to give an explicit closed-form expression for f A (a,l) see Carter et al. [], Steel [0]. Theorem. (exact formulae). Let T A (x) :=T A (x,). (i) µ n (a) = a! R(n) [xa ] (A,B):A B= ( T A(x)T B (x) r i= x i ).

5 PARSIMONY LENGTH OF A RANDOM BINARY TREE 363 (ii) µ n(φ) = n! R(n) [xn ] (A,B):A B= T A(φx)T B (φx)( x), where φx =(φ x,...,φ r x). (iii) µ n(φ) =µ n (φ)+ i,a: α i A Proof of Theorem.. Let φ i (n )! R(n ) [xn ]T A (φx). (6) R(x,z):= A T A (x,z), R(x) :=R(x,). First observe that, from (4), we have the fundamental identity (7) R(x,z)= R (x,z)+(z ) (B,C): B C= T B(x,z)T C (x,z)+ r x i. i= Putting z = in (7), we obtain (8) so that (9) R(x) = r R (x)+ x i, i= r R(x) = x i. i= In particular, (0) and so () Let () Then, from (7), R(φx) = x, [x n ]R(φx) = R(n). n! Q(x) := z R(x,z) z=. Q(x) =Q(x)R(x)+ (B,C): B C= T B(x)T C (x).

6 364 A. M. HAMEL AND M. A. STEEL Hence Applying (9) gives (3) Q(x) = Q(x) = (B,C):B C= (B,C):B C= T B(x)T C (x) ( R(x)). ( T B(x)T C (x) r i= x i ). Now from () a![x a ]Q(x) is the sum l lf(a,l), where f(a,l)= A f A(a,l), the total number of trees T R(n) of length l for a coloration χ of [n] oftypea. Thus a! R(n) [xa ]Q(x) is the average length over all trees in R(n) of the length of χ. However, each edge rooting of a binary tree leads to an identical parsimony length (i.e., the position of the root is irrelevant to the length), so this quantity is also the average length over all trees in of the length of χ, which in view of (3) establishes part (i). (ii) Applying part (i) to (3) we obtain (4) µ n(φ) = a ( n a ) φ a a! R(n) [xa ]F(x), where F (x) = (A,B):A B= T A(x)T B (x)( r i= x i). Rewriting (4), we have µ n(φ) = n! R(n) a φ a [x a ]F(x)= n! R(n) [xn ]F(φx), as required. (iii) For any tree T R(m) with root vertex ρ and subject to a random coloration of [m] according to φ, lets(t ) denote the (random variable) root set of T (as defined in section ). Suppose T andχis a coloration of [n]. By the bijection ψ : R(n ) and Lemma. (rooting T on the edge incident with the leaf labeled n), we have (5) where l(t,χ)=l(ψ(t),χ )+δ(t,χ), { if χ(n) S(ψ(T )), δ(t,χ)= 0 otherwise, and where χ is the restriction of χ to [n ]. Let µ(t ) denote the expected value of l(t,χ)forarandomχ(generated according to φ). Then from (5) we have (6) Now (7) µ(t )=µ(ψ(t)) + Prob[δ(T,χ)=]. Prob[δ(T,χ)=]=Prob[χ(n) S(ψ(T ))] = φ i Prob[S(ψ(T )) = A]. i,a: α i A

7 PARSIMONY LENGTH OF A RANDOM BINARY TREE 365 Also, by definition, (8) µ n(φ) = T µ(t ), while (9) µ n (φ) = B(n ) T B(n ) µ(t )= = T B(n ) T (n 3)µ(T ) µ(ψ(t )). Thus, combining (6) (9) we have (0) Now () µ n(φ) =µ n (φ)+ i,a: α i A T Also, n![x n ]T A (φx) = a Prob[S(ψ(T )) = A] = φ i T R(n ) ( ) n a φ a f A (a), and so n![x n ]T A (φx) = a φ a = T R(n) = T R(n) = T R(n) Thus, the term on the right of () is just χ:χhas type a a Prob[S(ψ(T )) = A]. T R(n ) T R(n) S(T,χ)=A χ:χ has type a and S(T,χ)=A χ:s(t,χ)=a φ a Prob[χ] Prob[S(T,χ)=A]. (n )! R(n ) [xn ]T A (φx), Prob[S(T )=A]. which, together with (0), establishes part (iii), thereby completing the proof of Theorem.. 3. Calculations (asymptotic). In this section we obtain asymptotic results concerning µ n(φ) andµ n (a). Theorem 3. below shows that µ n(φ) andµ n (a)are asymptotically equivalent since they both grow linearly with n, and their difference (when φ = n a) is bounded by a term of order n. The theorem also provides a prescription for calculating, in principle, their asymptotic values by solving a system of simultaneous quadratic equations involving real numbers. In the case of two colors this can be done analytically, but generally numerical techniques would seem to be

8 366 A. M. HAMEL AND M. A. STEEL required. However, in the case of equifrequency colorations (φ i = r ) the resulting system is considerably simpler, being of dimension r rather than r, and we solve this for r 4 in Corollary 3.. A second corollary provides a biology-oriented application. f(n) We adopt the standard notation f(n) g(n) if lim n g(n) = and f(n) =O(g(n)) if f(n) g(n) is bounded as n. Theorem 3. (asymptotic formulae). (i) µ n(φ) µ n, where µ = µ (φ) is given by µ = t A t B (A,B):A B= and where the numbers t A = T A (x) x=, A Cform the unique nonnegative φ solution to the simultaneous system t A = (B,C):B C=A t Bt C + δ A(φ) with δ A given by (5). (ii) For r =colors, µ = 3 ( 3φ φ ). (iii) µ n (a) nµ (φ) for φ = n a. Indeed, µ n(a) µ n(φ) n(r )/ for all n. Proof of Theorem 3.. (i) We first recall a special case of Lemma (i) of Meir, Moon, and Mycielski [6]: suppose F (x) andg(x) are power series in x and that [x n ]F (x) =O(ρ n n 3 ), and F (ρ) 0. Then [x n ]G(x) bρ n n, () Taking G(x) =( x) [x n ]F(x)G(x) F(ρ)[x n ]G(x). we have from () that (3) [x n ]G(x) = R(n+) n! =(n ) R(n) n! π ρ n n, where ρ =.NowtakeF(x)=T A(φx)T B (φx) for any pair of nonempty sets A, B C. Since for all C C,C the power series T C (φx) has all nonnegative coefficients, we have [x n ]F (x) = [x n ]T A (φx)t B (φx) [x n ] T C (φx) C

9 PARSIMONY LENGTH OF A RANDOM BINARY TREE 367 =[x n ]R(φx) from (6) =[x n ](R(φx) x) from (8) = R(n) from () n! ρ n n 3 π for ρ = from (). Thus [x n ]F (x) =O(ρ n n 3 ), and so, from (3) and the fact that F has nonnegative coefficients (so that F (ρ) 0), we can apply () to Theorem. (ii) to deduce that µ n = n! R(n) n (A,B):A B= (A,B):A B= T A(φx)T B (φx) x= [x n ]( x) T A ( φ ) T B ( φ ), as claimed. The prescribed system for t A := T A ( φ) follows from (4) by putting z = and x = φx. Now t A = T A ( φ) is clearly nonnegative. Thus, applying induction on A to the simultaneous system described in (i) shows that, moreover, t A > 0 for all A. Wenow show that there is only one such solution to this system. Let t =[t A ]andt =[t A ] be two solutions of the system of simultaneous quadratic equations described in Theorem 3., with t A,t A >0 for all A. Wewishtoshowthat t=t. First, note that A t A = A t A = since A t A = ( A t A) +. Let ɛ = A t A t A.Wehave ɛ= t B t C t Bt C A B,C: B C=A t B t C t Bt C A A B,C: B C=A B,C: B C=A t B t C t C + t C t B t B (since t B t C t Bt C t B t C t B t C + t B t C t Bt C ) = t B t C t C + t C t B t B B,C B,C ) = ɛ ( B = ɛ. t B + C t C It follows that both inequalities in the above derivation must, in fact, be equalities. In particular, the second inequality becomes an equality only if, for all B,C : B C = A, t B t C lies between t Bt C and t B t C. Since t and t both have positive coordinates

10 368 A. M. HAMEL AND M. A. STEEL we either have or Thus,ifwelet t B t B and t C t C t B t B and t C t C whenever B C = A. A := we have A 0 with equality precisely if (t B t B)(t C t C) B,C: B C=A (4) t B = t B or t C = t C for each (B,C) : B C=A. Now t A = B,C: t B t C + B C=A δ A(φ) and similarly for t A, so expanding out A we obtain A =t A δ A (φ)+t A δ A (φ) B,C: B C=A (t Bt C + t B t C), so A = t A + t A A A A B,C(t Bt C + t Bt C) ( )( ( )( ) = t B t C ) t B t C B C B C =0. It follows that A = 0 for all A (since A 0 for all A and A A = 0). From (4) this implies that for any pair B,C we have t B = t B or t C = t C. In particular, taking B = C we obtain that t B = t B,andsot=t,asclaimed. (ii) This result follows from part (iii) of Theorem 3., and the analogous result for µ n (a) from Moon and Steel [7]. However, it can also be derived more directly from Theorem 3. (i). We have, for C = {α, β}, (5) µ =T {α} T {β}, where T {α} = T {α} ( φ), T {β}=t {β} ( φ), and T {α,β} = T {α,β} ( φ) satisfy the system T {α} = T {α} + T {α}t {α,β} + φ, T {β} = T {β} + T {β}t {α,β} + φ, T {α,β} = T {α,β} + T {α}t {β}. Butler [] solved this system, and from his equation (6) we have T {α} = 3 ( +3φ + P), T {β} = 3 ( 3φ + P),

11 PARSIMONY LENGTH OF A RANDOM BINARY TREE 369 where P = 3φ φ, and from this we can obtain µ directly from (5). (iii) We first claim that (6) µ n (a) µ n (a ) a a, where denotes the l norm on R r. Since the components of a and a both sum to n, a a =kfor some integer k. In this case we can find two colorations χ, χ of [n] oftypesa,a, respectively, and such that χ and χ agree on all but k elements of [n]. Now, for any binary tree T, it is easily checked that l(t,χ ) l(t,χ)+k since any minimal extension of χ for T produces an extension χ of T by just changing the colors of the (at most k) leavesoft for which χ and χ disagree (and thereby increasing the number of edges of T with differently colored ends by at most k). Although χ may not be a minimal extension of χ for T, we nevertheless obtain the claimed inequality. Conversely, l(t,χ) l(t,χ )+k,andso l(t,χ) l(t,χ ) k, which, upon averaging over all binary trees, gives which establishes (6). Now from (3), µ n (a) µ n (a ) k, µ n = E[µ n (A)], the expected value of µ n (A), where A =(A,...,A r ) is drawn from a multinomial distribution with parameters n and φ,...,φ r. Then µ n (a) µ n = E[µ n (a) µ n (A)] E[ µ n (a) µ n (A) ] E[ a A ] from (6) (7) = r E[ a i A i ]. i= Now A i has a binomial distribution with parameters n and φ i, and since E[A i ]=nφ i = a i, we have, applying the convex version of Jensen s inequality (Rényi [8]), E[ a i A i ] E[(a i A i ) ] = Var[A i ] = nφ i ( φ i ), so that, from (7), µ n (a) µ n(φ) r i= nφi ( φ i ), which, by the concave version of Jensen s inequality, is at most n(r )/.

12 370 A. M. HAMEL AND M. A. STEEL Corollary 3. (equifrequency colorations). Suppose φ i = r Then µ = where the t i satisfy the system (j,k):j+k r; j,k t i = (j,k) r! j!k!(r j k)! t jt k, π ijkt j t k + δ i, r for i =,...,r. for i =,...,r and where δ i, = if i = and 0 otherwise and where π ijk is the number of sets of sizes j and k which, under the parsimony operation (*), give a specific root set of size i for the tree; i.e., ( i j) if j + k = i, ( π ijk = r i )( r j ) j i k i if j, k i, 0 else. Examples. Forr=wehave t = t +t t + 4, t = t +t, which gives t = 6,t = 6, and so from Corollary 3. we obtain µ = 3, which agrees with Theorem 3. (ii). For r> it seems necessary to solve the system {t i } by numerical methods. For r = 3 and 4, the equations become t = t +t t +t t 3 +t + 6, t =t + t +t t 3, and t 3 =3t t + t 3 t = t +3t t +3t +3t t 3 +3t t 3 +t t 4 + 8, t =t + t +t t 3 +t 3+t t 4, t 3 =3t t + t 3+t 3 t 4, t 4 =4t t 3 +3t + t 4, respectively, and we find that t =(0.4855, , ) and µ =0.474 for r = 3 and t =(0.7656, , , ) and µ = for r =4. As a second and biologically-oriented application of Theorem 3., let us, as in section, regard a collection of n aligned DNA sequences of length c as a collection

13 PARSIMONY LENGTH OF A RANDOM BINARY TREE 37 χ,...,χ c of r-colorations of [n] (for r = or 4). T for this data; that is, Let l(t ) denote the length of l(t )= c l(χ j,t), j= and let l be the average value of l(t )over. We also consider a randomized version of l as follows. Let l (T ) be the expected length of a given binary tree T on sequences randomly generated by assigning each of the c sites in sequence i a color α j with probability φ i j, as in Steel, Lockhart, and Penny []. Let l denote the average value of l (T )over. Finally, let P(n) denote the set of partitions of n into at most r-parts (thus P(n) ={(p,...,p r ):p p p r 0, r i= p i = n}). Corollary 3.. Asymptotically (as n ), (i) l n p P(n):N(p)>0 µ ( n p ) N ( p ), where N(p) is the number of sites j for which the type of χ j, arranged in decreasing order, gives partition p. (ii) l cnµ (φ), where φ = n n φ i. i= Proof of Corollary 3.. (i) l = = = = c j= T l(t ) T j= c µ n (a (j) ), j= T c l(χ j,t) l(χ j,t) where a j is the type of χ j.thus l= p P(n):N(p)>0 µ n (p)n(p), and the result now follows from parts (i) and (iii) of Theorem 3.. (ii) (8) l = ce [µ n (A)],

14 37 A. M. HAMEL AND M. A. STEEL where E denotes expectation at a single site in the probability space described above. Since A j is a sum of n independent (but not necessarily identical) 0 random variables D ij, with Prob[D ij =]=φ i j,wehavethat (9) n A pφ, where p denotes convergence in probability (as n )and φ= n φ i. n i= Now from (3), (6), and (9) it can be checked that ( ) n µ n n A (30) n µ n(φ) p 0 as n. Also, by Theorem 3., parts (i) and (iii), respectively, n µ n(φ) µ (φ) 0, µ n (A) ( ) n n µ n n A p 0 as n ;thus (3) as n. Now, from (8), µ n (A) n p µ (φ) [ ] l = cne µn (A), n which, together with (3), establishes part (ii). REFERENCES [] J. P. Butler, Fractions of trees with given root traits; the limit of large trees, J. Theoret. Biol., 47 (990), pp [] M. Carter, M. D. Hendy, D. Penny, L.A. Székely, and N. C. Wormald, On the distribution of lengths of evolutionary trees, SIAM J. Discrete Math., 3 (990), pp [3] W. M. Fitch, Towards defining the course of evolution: Minimum change for a specific tree topology, Syst. Zool., 0 (97), pp [4] I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiley, New York, 983. [5] J. A. Hartigan, Minimum mutation fits to a given tree, Biometrics, 9 (973), pp [6] A. Meir, J. W. Moon, and J. Mycielski, Hereditary finite sets and identity trees, J. Combin. Theory Ser. B, 35 (983), pp [7] J. W. Moon and M. A. Steel, A limiting theorem for parsimoniously bicoloured trees, Appl. Math. Lett., 6 (993), pp [8] A. Rényi, Probability Theory, North Holland, Amsterdam, 970. [9] E. Schröder, Vier combinatorische Probleme, Zeitschrift für Mathematik und Physik, 5 (870), pp [0] M. A. Steel, Decompositions of leaf-colored binary trees, Adv. in Appl. Math., 4 (993), pp. 4. [] M. A. Steel, P. J. Lockhart, and D. Penny, Confidence in evolutionary trees from biological sequence data, Nature, 364 (993), pp