Phylogenetic Trees Lecture 2

.

Phylogenetic Trees

Lecture 2

Based on: Durbin et al Section 7.3, 7.4, 7.8

2

The Four Points ConditionTheorem: A set M of L objects is additive iff any subset of four objects can be labeled i,j,k,l so that:

d(i,k) + d(j,l) = d(i,l) +d(k,j) ≥ d(i,j) + d(k,l)

We call {{i,j},{k,l}} the “split” of {i,j,k,l}.

The four point condition doesn’t provides an algorithm to construct a tree from distance matrix, or to decide whether there is such a tree.

The first methods for constructing trees for additive sets used neighbor joining methods:

3

Constructing additive trees:The neighbor joining problem

Let i, j be neighboring leaves in a tree, let k be their parent, and let

m be any other vertex.

The formula

shows that we can compute the distances of k to all other leaves.

This suggest the following method to construct tree from a

distance matrix:

1. Find neighboring leaves i, j in the tree,

2. Replace i, j by their parent k and recursively construct a tree T

for the smaller set.

3. Add i, j as children of k in T.

)],(),(),([),( jidmjdmidmkd 2

1

4

Neighbor Finding

How can we find from distances alone a pair of nodes which are neighboring leaves?

Closest nodes aren’t necessarily neighboring leaves.

AB

CD

Next we show one way to find neighbors from distances.

5

Neighbor Finding: Saitou & Nei method

Theorem [Saitou & Nei] Assume all edge weights are positive. If D(i, j) is minimal (among all pairs of leaves), then i and j are neighboring leaves in the tree.

ij

kl

m

T1T2

is a leaf

For a leaf , let ( , ).im

i r d i m Definition: Let , be leaves Then

( , ) ( 2) ( , ) ( )where is the number of leaves in

i j

i jD i j L d i j r r

L T

The proof is rather involved !

6

Neighbor Joining Algorithm Set L to contain all leavesIteration: Choose i, j such that D(i, j) is minimal Create new node k, and set

Remove i, j from L, and add kTerminate:

when |L| =2 , connect two remaining nodes

1( , ) ( ( , ) ( , ) ( , )) (for some )

2( , ) ( , ) ( , )

1for each node , ( , ) ( ( , ) ( , ) ( , ))

2

d i k d i j d i m d j m m

d j k d i j d i k

m d k m d i m d j m d i j

ij

k

m

7

Saitou & Nei’s Idea:

is a leaf

For a leaf , let ( , ).im

i r d i m

D12 = (a+c+d) – (1/3)(a+b + a+c+d + a+c+e+f+ a+c+e+g + d+c+a + d+c+b + d+e+f +

d+e+g)

D13 = (a+b) – (1/3)(a+b + a+c+d + a+c+e+f+ a+c+e+f + b+a + b+c+d + b+c+e+f +

b+c+e+g)

Hence D12 - D13 = (4/3) c

2

5•

a

b

c d

e

fg

1

3 4•

• •

•2

1

LLet (i, j) = d(i, j) – (ri + rj)

“ L-2 ” is crucial!

D

8

Saitou & Nei’s proof

Notations used in the proof :

p(i, j) = the path from vertex i to vertex j; P(D,C) = (e1, e2, e3) = (D, E, F, C)

For a vertex i, and an edge e=(i , j):Ni(e) = |{k : e is on p(i, k), k is a leave}|.e.g. ND(e1) = 3, ND(e2) = 2, ND(e3) = 1NC(e1) = 1

A B

CD

e1

e3

e2

EF

9

Saitou & Nei’s proof: Crucial Observation

( , )

( , )

For leaves , connected by a path ( , ,.., , ):

( )[ ( ) ( )]

( 2)[ ( , ) ( , )] ( )[ ( ) ( )]

i j i je p i j

i je p l k

i j i l k j

r r lth e N e N e

L d i l d k j lth e N e N e

i

j

kl

Rest of T is a leaf

Observe that ( , ) ( ) ( ), i im e E

r d i m lth e N e

10


Proof of Theorem: Assume for contradiction that D(i, j) is minimized for i, j which are not neighboring leaves.Let (i, l, ..., k, j) be the path from i to j. let T1 and T2 be the subtrees rooted at k and l which do not contain edges from P(i,j) (see figure).

ij

kl

T1T2

Notation: |T| = #(leaves in T).

11

Saitou & Nei’s proofCase 1: i or j has a neighboring leaf. WLOG j has a neighbor leaf m.A. D(i,j) - D(m,j)=(L-2)(d(i,j) - d(j,m) ) – (ri+rj) + (rm+ rj)

=(L-2)(d(i,k)-d(k,m) )+rm-ri

B. rm-ri ≥ (L-2)(d(k,m)-d(i,l)) + (4-L)d(k,l)

i j

kl

mT2

Substituting B in A:D(i,j) - D(m,j) ≥

(L-2)(d(i,k)-d(i,l)) + (4-L)d(k,l) = 2d(k,l) > 0,

contradicting the minimality assumption.

(since for each edge eP(k,l), Nm(e) ≥ 2 and Ni(e) L-2)

12


Case 2: Not case 1. Then both T1 and T2 contain 2 neighboring leaves.WLOG |T2| ≥ |T1| . Let n,m be neighboring leaves in T1. We shall prove that D(m,n) < D(i,j), which will again contradict the minimality assumption.

i j

kl

mn

p

T1

T2

13


i j

kl

mn

p

T1

T2

A. 0 ≤ D(m,n) - D(i,j)= (L-2)(d(m,n) - d(i,j) ) + (ri+rj) – (rm+rn)

B. rj-rm< (L-2)(d(j,k) – d(m,p)) + (|T1|-|T2|)d(k,p)C. ri-rn < (L-2)(d(i,k) – d(n,p)) + (|T1|-|T2|)d(l,p)

Adding B and C, noting that d(l,p)>d(k,p):D. (ri+rj) – (rm+rn) < (L-2)(d(i,j)-d(n,m)) + 2(|T1|-|T2|)d(l,p)

Substituting D in the right hand side of A:D(m,n ) - D(i,j)< 2(|T1|-|T2|)d(l,p) ≤ 0, as claimed. QED

14

A simpler neighbor finding methodSelect an arbitrary node r. For each pair of labeled nodes (i, j) let C(i, j) be

defined by the following figure:

C(i,j)

i

j

r

Claim: Let i, j be such that C(i, j) is maximized.Then i and j are neighboring leaves.

)],(),(),([),( jidrjdridjiC 2

1

15

Neighbor Joining Algorithm Set M to contain all leaves, and select a root r. |M|=L If L =2, return tree of two vertices

Iteration: Choose i, j such that C(i, j) is maximal Create new vertex k, and set

remove i, j, and add k to M Recursively construct a tree on the smaller set, then add i, j as

children on k, at distances d(i,k) and d(j,k).

ij

k

m

)],(),(),([),(

),(),(),(

)],(),(),([),(

jidmjdmidmkdm

kidjidkjd

rjdridjidkid

2

1 , nodeeach for

2

1

16

Complexity of Neighbor Joining Algorithm

Naive Implementation:Initialization: Θ(L2) to compute the C(i, j)’s.

Each Iteration: O(L) to update {C(i, k): i L} for the new node k. O(L2) to find the maximal C(i, j).

Total of O(L3).

ij

k

m

17

Complexity of Neighbor Joining Algorithm

Using Heap to store the C(i, j)’s :

Initialization: Θ(L2) to compute and heapify the C(i,j)’s.

Each Iteration: O(1) to find the maximal C(i,j). O(L logL) to delete {C(m,i), C(m,j)} and add C(m,k) for

all vertices m.

Total of O(L2 log L).

(implementation details are omitted)

18

Ultrametric trees as special weighted trees

Definition: An Ultrametric tree is a rooted weighted tree all of whose leaves are at the same depth. Edge weights can be represented by the distances of internal vertices from the leaves.

E.g., the tree produced by UPGMA.

Note: each internal vertex has at least two children

8

A E D CB

5

3

3

0:

3333

2

5

5

3

19

Ultrametric trees A more recent (and more efficient) way for constructing and identifying additive trees.Idea: Reduce the problem to constructing trees by the “heights” of the internal nodes. For leaves i, j, D(i, j) represent the “height” of the common ancestor of i and j.

AE

D C

B

8

5

3

3

20

Ultrametric Trees Definition: T is an ultrametric tree for a symmetric positive real

matrix D (called ultrmetric matrix) if:1. The leaves of T correspond to the rows and columns of D2. Internals nodes have at least two children, and the Least Common

Ancestor of i and j is labeled by D(i, j).3. The labels decrease along paths from root to leaves

A B C D E

A 0 8 8 5 3

B 0 3 8 8

C 0 8 8

D 0 5

E 0AE

D C

B

8

5

3

3

21

We will study later the following question:

Given a symmetric positive real matrix D,

Is there an ultrametric tree T for D?

Centrality of Ultrametric Trees

But first we show ultrametric trees can be used to construct trees for additive sets and other related problems.

22

Use the labels to define weights for all internal edges in the natural way.For this, consider the labels of leaves to be 0. We get an additive ultrametric tree whose height is the label of the root.

E

D C

B

8

5

3

3

2

53

A

3 3

5

3

3

Transforming Ultrametric Trees to Weighted Trees

Note that in this tree all leaves are at the same height. This is why it is called ultrametric.

23

Transforming Weighted Trees to Ultrametric Trees

A weighted Tree T can be transformed to an ultrametric tree T’ as follows:

Step 1: Pick a node k as a root, and “hang” the tree at k.

a

b

c

d

2

23

4

1

a

b

c d

2

13

4 2

k=a

24


Step 2: Let M = maxid(i, k). M is taken to be the height of T’.Label the root by M, and label each internal node j by M - d(k, j). “ k ” is the root.

a

b

c

d

2

23

4

1

a

b

c d

2

13

4 2

9

7

4

k = a, M = 9

25


Step 3: “Stretch” edges of leaves so that they are all at distance M from the root

M = 9

a

b

c d

2

13

4 2

9

7

4

(9)

(6)

(2)(0)

a

b

c d

7

9

7

4

2

3

4

9

4( M-d(k,i) )

k

mi

26

Re-constructing Weighted Trees from Ultrametric Trees

M = 9

Weight of an internal edge is the difference between the labels (heights) its endpoints. Assume that the distance matrix D = [d(i, j)] of the original unrooted tree is given.Weights of an edge to leaf i is obtained by subtracting “M - d(k, i)” from its current weight.

a

b

c d

7(-6)

9

7

4

4

9 (-9)

4(-2)

a

b

c d

1

2

3

4

0

2

k

m

i

(M–d(k,m))–(M–d(k,i)) = d(i,m)

27

How D’ is constructed from D

a

b

c d

2

13

4 2

9

7

1( , ) ( ( , ) ( , ) ( , ))

2(Here, a, b, c)

d k m d i k d j k d i j

k i j

D’(i, j) should be the height of the Least Common Ancestor of i and j in T’, the ultrametric tree hanged at k.

Let M = maxi d(i, k) and m is the LCA of i and j.Thus, D’(i, j) = M - d(k, m), where d(k, m) is computed by:

k

m

i

j

Note that this can be computed without the additive tree!

28

The transformation of D to D’

a b c d

a 9 9 9

b 7 7

c 4

d

a b c d

a 3 9 7

b 8 6

c 6

d

Distance matrix D

a

b

c d

2

13

4 2

Ultrametric matrix D’

a

b

c d

9

7

4

M=9

T T’

29

Identifying Ultrametric Trees

Definition: A distance matrix D is ultrametric if for each 3 indices i, j, k

D( i, j ) ≤ max {D( i, k ), D( j, k )}.

(i.e., there is a tie for the maximum value)

Theorem (U): D has an ultrametric tree iff it is ultrametric.

(to be proved later)

30

Theorem: D is an additive distance matrix if and only if D’ is an ultrametric matrix.

Note that the construction of D’ is independent of the additive tree.

Proof. ( ) Use the conversion from an additive tree to an ultrametric tree and Theorem (U) .

( ) Use Theorem (U) and the conversion from an ultrametric tree to an additive tree and check that the additive tree indeed realizes the distance matrix.

31

Solving the Additive Tree Problem by the Ultrametric Problem: Outline

We solve the additive tree problem by reducing it to the ultrametric problem as follows:

1. Given an input matrix D = D(i, j) of distances,

transform it to a matrix D’= D’(i, j) , where D’(i,j) is

the height of the Least Common Ancestor of i and j in

the corresponding ultrametric tree T’. (If not

ultrametric, then the input matrix is not additive!)

2. Construct the ultrametric tree, T’, for D’.

3. Reconstruct the additive tree T from T’.

32

LCA and distances in Ultrametric Tree

Let LCA(i, j) denote the lowest common ancestor of leaves i and j. Let D(i, j) be the height of LCA(i, j), and dist(i,j) be the distance from i to j.

Claim: For any pair of leaves i, j in an ultrametric tree:

D(i, j)= 0.5 dist(i, j).A B C D E

A 0 8 8 5 3

B 0 3 8 8

C 0 8 8

D 0 5

E 0A E D

CB

8

5

33

33

Identifying Ultrametric Distances

Definition: A distance matrix D of dimension L by L is ultrametric iff for each 3 indices i, j, k :

D( i, j ) ≤ max { D( i, k ), D( j, k ) }.

j k

i 9 6

j 9

Theorem(U): The following conditions are equivalent for an LL symmetric matrix D:

1. D is ultrametric

2. There is an ultrametric tree of L leaves such that for each pair of leaves i, j :

D(i, j) = height(LCA(i, j)) = ½ dist(i, j).

Note: D(i, j) ≤ max {D(i, k), D(j, k)} is easier to check than the 4 points condition. Therefore the theorem implies that ultrametric sets are easier to characterize then an additive sets.

34

Properties of ultrametric matrix used in the Proof of the Theorem (U)

Definition: Let D be an L by L matrix, and let S {1,...,L}.

D[S] is the submatrix of D consisting of the rows and columns with indices from S.

Claim 1: D is ultrametric iff for every S {1,...,L}, D[S] is ultrametric.

Claim 2: If D is ultrametric and maxi,jD(i, j)=m, , then m appears in every row of D above the row where the max occurs.

j k

? ?

j m

One of the “?” Must be m

35

Ultrametric tree Ultrametric matrix

There is an ultrametric tree s.t. D(i, j) = ½ dist(i, j).

D is an ultrametric matrix: By properties of Least Common Ancestors in trees

ijk

D(k, i) = D(j, i) ≥ D(k, j)

36

Ultrametric matrix Ultrametric tree

Proof of D is an ultrametric matrix D has an ultrametric tree :

By induction on L, the size of D.Basis: L= 1: T is a leaf

L= 2: T is a tree with two leaves

0 9

0

0

i

j

i j

i

i

9

ji

37

Induction step

Inductive Hyp.: Assume that it’s true for 1, 2, … , L-1.

Induction step: L > 2. Let m = m1 be the maximum distance.

Let Si ={l: D(1, l) = mi}, and { S1 , S2 , … Sk } form a partition of the leaves into k classes. (note: |S1| > 0)

By Claim 1, D[Si], i = 1, 2, …, k are all ultrametric and

hence we can construct tree T1 for S1, rooted at m and trees Ti for Si with root labeled mi < m for i = 2, …, k.

(if mi = 0 then Ti is a leaf).

38

Notice that on any ultrametric tree the path from the root to the leave “1” must have exactly k+1 nodes, where k is the number of classes.

Each node on this path must be labeled by one of the distinct entries in row 1, and those labels must appear in decreasing order on the path.

1 2 3 4 5 6 7 8

1234..

0 4 3 4 6 4 3 60 4 2 6 1 4 6

13,7

2,4,6

5,8

6

4

3

T1

T2

T3

39

Correctness Proof

By Inductive Hypothesis, Ti ’s are all ultrmetric trees, and

we assemble them along the path from the root to leave “1” to form the tree T.

To prove that T is an ultrametric tree for D, need to check that D(i, j) is the label of the LCA of i and j in T.

If i and j are in the same subtree, this holds by induction;

otherwise the label of the node that the higher tree attaches to the path, which is the LCA, is indeed D(i, j) .

QED

40

Complexity AnalysisLet f (L) be the time complexity for L×L matrix.

f (1)= f (2) = constant. For L > 2: Constructing S1 and S2: O(L). Let |S1| = k, |S2| = L-k.

Constructing T1 and T2: f (k) + f (L-k).

Joining T1 and T2 to T: Constant.

Thus we have:

f (L) ≤ maxk[ f (k) + f (L-k)] +cL, 0 < k < L.

f (L) = cL2 satisfies the above.

Need an appropriate data structure!

41

Recall: identifying Additive Trees via Ultrametric trees

We solve the additive tree problem by reducing it to the

ultrametric problem as follows:

1. Given an input matrix D = D(i, j) of distances,

transform it to a matrix D’= D’(i, j), where D’(i, j) is

the height of the LCA of i and j in the

corresponding ultrametric tree T’.

2. Construct the ultrametric tree, T’, for D’.

3. Reconstruct the additive tree T from T’.

42

How D’ is constructed from D

D’(i, j) should be the height of the Least Common Ancestror of i and j in T’, the ultrametric tree hanged at k:

Thus, D’(i,j) = M - d(k, m), where d(k, m) is computed by:

a

b

c d

2

13

4 2

9

7

1( , ) ( ( , ) ( , ) ( , ))

21

(For =a, =b, =c, '(b,c) 9 (3 9 8) 72

d k m d i k d j k d i j

k i j D

43

The transformation D D’ T’T

a b c d

a 0 9 9 9

b 0 7 7

c 0 4

d 0

a b c d

a 0 3 9 7

b 0 8 6

c 0 6

d 0

D

a

b

c d

2

13

4 2

D’

a

b

c d

9

7

4

M=9

T T’

Phylogenetic Trees Lecture 2

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