A Study on Measuring Distance between Two Trees 阮夙姿教授 Advisor: 阮夙姿教授林陳輝 Presenter : 林陳輝

A Study on Measuring Distance between Two Trees

Advisor: 阮夙姿教授阮夙姿教授Presenter : 林陳輝林陳輝

CSIE, National Chi Nan University 2

OutlineIntroduction

Problem definition

Related workThe metric and algorithms

Mixture distanceBasic algorithmThe modified algorithm

Mixture - matching distanceMixture - matching distance

Conclusions and Future work


Introduction

Evolutionary tree

Comparing trees

Comparing trees is not easy

-Phylogenetic tree, wikipedia


Mixture tree

taxa

Time

S.-C. Chen and B. G. Lindsay, “Building Mixture Trees from Binary Sequence Data,” Biometrika, 2006.


Problem definition

1111

99 88

11 33 55 77

A B C D E F G H

v1

v2 v3

v4v5 v6

v7

•The leaves are associating taxas

•There is a time parameter on every internal node


OutlineIntroduction

Problem definition






Related workPath difference metric

dp(T1, T2) = ||d(T1) – d(T2)||2

d(Ti) is a vector that contains all pair leaves distance of

Ti.

M. A. Steel and D. Penny, “Distributions of Tree Comparison Metrics – Some New Results,” Syst. Biol. 42(2):126-141, 1993.


Related workNodal metric

In full binary trees, the complexity is O(n3).In complete binary trees, the complexity is O(n2 log n). John Bluis and Dong-Guk Shin, “Nodal Distance Algorithm: Calculating a Phylogenetic Tree Comparison Metric,” Proc. of the 3rd IEEE Symposium on BioInformatics and BioEngineering, 87- 94, 2003

leaves. are for ,) ,() ,(Distance21

yx,yxDyxD TT


Related work

Matching distanceP. W. Diaconis and S. P. Holmes, “Matchings and Phylogenetic Trees.," Proc. Natl Acad Sci U S A, Vol. 95, No. 25, pp. 14600~14602, 1998.

The algorithm for matching distanceG. Valiente, A Fast Algorithmic Technique for Comparing Large Phylogenetic Trees," SPIRE, pp. 370~375, 2005.


Matching Representation

1 2

3 4

5 6

0

0

0

0

07 8

9 10

11

{1,2} {5,6} {3,7} {4,8} {9,10}


Matching distance

{1,2} {5,6} {3,7} {4,8} {9,10}

{1,3} {4,6} {2,7} {5,8} {9,10}

The distance is 2

3 4

5 6

8

9 10

7

1 2

2 5

4 6

8

9 10

7

1 3

11 11T1

T2

T1

T2


OutlineIntroduction

Problem definition




Conclusion and Future work


Mixture distance and algorithmsDefinition:

pTi (x, y) is time parameter of the LCA of leaves x, y

leaves. are for ,),(),(Distance21

yx,yxpyxp TT

99

11 33

A B C D

v1

v3v2

99

22 33

A BC D

v1

v3v2


Distance conditions

The distance from an object to itself is zero.

The distance from A to B is the same as the distance from B to A.

The Triangle Inequality holds true.

- J. Felsenstein, Inferring phylogenies. Sunderland, MA: Sinauer Associates, 2004.


Algorithm

C(n, 2)

Algorithmic idea: grouping

Full binary tree99

11 33

A B C D

v1

v2

88

44

11

A B C D

v1

v2

v3v3

AB: |8 – 1| = 7

AC: |8 – 9| = 1

AD: |8 – 9| = 1

BC: |4 – 9| = 5

BD: |4 – 9| = 5

CD: |1 – 3| = 2

Distance = 21

leaves. are for ,),(),(Distance21

yx,yxpyxp TT


99

77 88

22 33 44 55

A B C D E F G H

v1

v2 v3

v4 v5 v6v7

T199

66 88

11 33 44 55

HG FA B CD E

v1

v2 v3

v4 v5 v6v7

T2

Algorithm


99

HG FA B CD E

T2

Red:1 Green:1

99

7788

22 33 44 55

A B C D E F G H

v1

v2v3

v4 v5 v6v7

Red:0 Green:1

Red:1 Green:0

Red:0 Green:1

Red:1 Green:0

66 88

11 33 44 55

v1

v2 v3

v4 v5 v6v7

Red:1Green:1

Red:2 Green:2

T1

|pT1(v1) － pT2

(v6)| × (1 × 1+0 × 0) = |9 － 4| × (1*1+0*0) =

5

|pT1(v1) － pT2

(v7)| × (0 × 0+1 × 1) = |9 － 5| × (0*0+1*1) =

4

|pT1(v1) － pT2

(v3)| × (1 × 1+1 × 1) = |9 － 8| × (1*1+1*1) =

2


T2

99

66 88

11 33 44 55

HG FA B CD E

v1

v2 v3

v4 v5 v6v7

Red:0 Green:1

Red:0Green:1

99

77 88

22 3344 55

A B C D E F G H

v1

v2 v3

v4 v5 v6v7

T1

Red:1 Green:0

Red:1 Green:0

Red:0 Green:0

Red:0 Green:0

Red:0Green:2

Red:2Green:0

|pT1(v2) － pT2

(v2)| × (2 × 0 + 0 × 0) = |7 － 6| × (2 × 0 + 0 × 0) =

0|pT1(v2) － pT2

(v3)| × (0 × 1 + 0 × 1) = |7 － 8| × (0 × 1 + 0 × 1)

= 0|pT1(v2) － pT2

(v1)| × (2 × 2 + 0 × 0) = |7 － 9| × (2 × 2 + 0 × 0) =

8

Red:2Green:2

CTLab

2/0--2/0--1/1 |- 2/0--0/0


Complexity analysis

For every internal node of T1, coloring all leaves

needs O(n).

Counting distance in T2 needs O(n).

The time complexity is O(n2).


The modified algorithm

Boost up the basic algorithm

Too much empty color information


T2

99

66 88

11 33 44 55

HG FA B CD E

v1

v2 v3

v4 v5 v6v7

Red:0 Green:1

Red:0Green:1

99

77 88

22 3344 55

A B C D E F G H

v1

v2 v3

v4 v5 v6v7

T1

Red:1 Green:0

Red:1 Green:0

Red:0 Green:0

Red:0 Green:0

Red:0Green:2

Red:2Green:0

|pT1(v2) － pT2

(v2)| × (2 × 0 + 0 × 0) = |7 － 6| × (2 × 0 + 0 × 0) =

0|pT1(v2) － pT2

(v3)| × (0 × 1 + 0 × 1) = |7 － 8| × (0 × 1 + 0 × 1)

= 0|pT1(v2) － pT2

(v1)| × (2 × 2 + 0 × 0) = |7 － 9| × (2 × 2 + 0 × 0) =

8

Red:2Green:2

Empty color information

CTLab

2/0--2/0--1/1 |- 2/0--0/0


T2

99

66 88

11 33 44 55

HG FA B CD E

v1

v2 v3

v4 v5 v6v7

T2

99

88

11

A B CD

v1

v3

v4


The modified algorithm

Finding LCA in constant time with O(n) preprocessing

MA Bender, MIF Colton, The LCA Problem Revisited, Proc. LATIN, 2000

2-way merge problemR.C.T. Lee, S. S. Tseng, R.C. Chang and Y. T. Tsai, Introduction to the Design and Analysis of Algorithms. McGraw-Hill Education, 2005


9

7 8

2 3 4 5

HG FA B CD E

v1

v2 v3

v4 v5 v6v7

T2

9

6 8

1 3 4 5

A B C D E F G H

v1

v2 v3

v4 v5 v6v7

T1

1 2

3

4 5

6

7

8 9

10

11 12

13

14

15

1 2 45 8 911 12


9

7 8

2 3 4 5

HG FA B CD E

v1

v2 v3

v4 v5 v6

v7

T2

1 2

45 8 911 12

1, 2 11, 12 5,84, 9

13 v4 |1 – 2| (1 1 + 0 0) = 19

6 8

1 3 4 5

A B C D E F G H

v1

v2 v3

v4 v5 v6v7

T1

1 2

3

4 5

6

7

8 9

10

11 12

13

14

15

1 2


9

7 8

2 3 4 5

HG FA B CD E

v1

v2 v3

v4 v5 v6

v7

T2

45 8 9

11 12

1, 2 11, 12 5,84, 9

1, 2, 11, 12 4, 5, 8, 9

1, 2, 4, 5, 8, 9, 11, 12

|9 – 7| (2 2 – 0 0) = 8

9

6 8

1 3 4 5

A B C D E F G H

v1

v2 v3

v4 v5 v6v7

T1

1 2

3

4 5

6

7

8 9

10

11 12

13

14

15

9

1 5

v1

v4

3 13v7

11 121 2

1 2

15

HGA B


Complexity analysis

To reconstruct subtree of T1 is in linear time

Counting distance in reconstructed subtree needs O(m).

The height of complete binary tree is O(logn)

The total complexity is O(nlogn) in complete binary tree.


OutlineIntroduction

Problem definition

Related worksThe metric and algorithms





Mixture-matching distance

Distance =

i is matching distance between T1 and T2.

PTm denotes the product of all time parameter in Tm

2 ,1 , and ,for , /1 mnPPiPP mnmn TTTT


9

7 8

2 3 4 5

HG FA B CD E

T2

9

6 8

1 3 4 5

A B C D E F G H

T1

1 2 3 4 5 6 7 8

9 10 11 12

13 14

15

1 2 4 58

9 11 10

367

12

13 14

15

{1, 2} {3, 4} {5, 6} {7, 8} {9,10} {11, 12} {13, 14}

{1, 2} {3, 6} {4, 5} {7, 8} {9,12} {10, 11} {13, 14}

Distance = 1 - (25920 / 60480) + 2 ≒ 2.571

604801 TP

259202 TP

T1

T2


0

1

∞

The sameNo different leaves

i

i transposition

Distance

Distance = 1 - (25920 / 60480) + 2 ≒ 2.571

The time complexity is O(n)

2 ,1 , and ,for , /1 mnPPiPP mnmn TTTTDistance =


OutlineIntroduction

Problem definition

Related worksThe metric and algorithms





Conclusions

Metric ConsiderenceTime complexity

Full binary tree

Complete binary tree

Path difference metric Structure N/ANodal distance Structure O(n3) O(n2logn)

Mixture distanceStructure and

time parameterO(n2) O(nlogn)

Matching distance Structure O(n)

Mixture-matching distance

Structure and

time parameterO(n)


Future work

Improve the time complexity

Extend to k - ary trees

Add mutation point

Thanks for Your Listening.

Documents

A Study on Measuring Distance between Two Trees 阮夙姿 教授 Advisor: 阮夙姿 教授 林陳輝 Presenter : 林陳輝

A Study on Measuring Distance between Two Trees 阮夙姿教授 Advisor: 阮夙姿教授林陳輝 Presenter : 林陳輝