tree distance algorithm

Workshop ontree distance

By Hector Francofrancoph at tcd dot ie

Trinity College of Dublin

summary

1. Levenshtein algorithm & Sub-string matching

2. The tree edit distance

3. Basic tree concepts

4. Maps

5. Tai maps

6. Tree distance

7. Zhang Shasha algorithm & alignment

summary

• Levenshtein algorithm & Sub-string matching• The tree edit distance• Basic tree concepts• Maps• Tai maps• Tree distance• Zhang Shasha algorithm & alignment

String distance metrics: Levenshtein• Edit-distance metrics

– Precursor of tree distance.– Distance is shortest sequence of edit commands that

transform s to t. (meaning the sequence of edit command that sum less cost of mapping s to t)

– Simplest set of operations:• Copy/map character from s over to t, (cost 0)• Delete a character in s (cost 1)• Insert a character in t (cost 1)• Substitute one character for another (cost 1)

– This is “Levenshtein distance”

Levenshtein distance - example

• distance(“William Cohen”, “Willliam Cohon”)

W I L L I A M _ C O H E N

W I L L L I A M _ C O H O N

C C C C I C C C C C C C S C

0 0 0 0 1 1 1 1 1 1 1 1 2 2

SDomain:

TRange:

op

cost

alignment

Computing Levenshtein distanceD(i,j) = score of best alignment from s1..si to t1..tj

= min

D(i-1,j-1) + d(si,tj) //subst/copyD(i-1,j)+1 //insertD(i,j-1)+1 //delete

(simplify by letting d(c,d)=0 if c=d, 1 else)

also let D(i,0)=i (for i inserts) and D(0,j)=j

Computing Levenshtein distance - 3

D(i,j)= min


= D(s,t)

C O H E N

0 1 2 3 4 5

M 1 1 2 3 4 5

C 2 1 2 3 4 5

C 3 2 2 3 4 5

O 4 3 2 3 4 5

H 5 4 3 2 3 4

N 6 5 4 3 3 3

The yellow row and column, correspond to the row 0 and column 0 of the table, and they are initiated in increasing order.

For (int x = 0; x<size(target), x++) D(0,x) = x;For (int x = 0; x<size(source), x++) D(x,0) = x;d(si,tj) represents the cost of change the letter si into the letter tj, where if the letter is the same the cost will be 0 and if is a different letter the cost will be 1.

Computing Levenshtein distance - 3

D(i,j)= min


C O H E N

0 1 2 3 4 5

M 1 1 2 3 4 5

C 2 1 2 3 4 5

C 3 2 2 3 4 5

O 4 3 2 3 4 5

H 5 4 3 2 3 4

N 6 5 4 3 3 3

T1 T2 Cost

M - 1

C - 1

C C 0

O O 0

H H 0

- E 1

N N 0

T1 T2 Cost

M - 1

C C 0

C - 1

O O 0

H H 0

- E 1

N N 0

Practice

G O O D

0 1 2 3 4

G 1 0 1 2 3

O 2 1 0 1 2

D 3 2 1 1 1

G O O DG O DG O D

TRYTRY

Sub string matching

• What is the best matching substring of S in T?

u n i v e r s

0 0 0 0 0 0 0

n 1 0 1 1 1 1 1

i 2 1 0 1 2 2 2

e 3 2 1 1 1 2 3

r 4 3 2 2 2 1 2

Look for the minimum value

“niver”

Cost 0 for delete positions before the sub-string

summary


Edit distance between trees• Tai 1979 introduce a criterion for matching nodes between tree

representations, and Zhang and Shasha 1989 develop an algorithm that find an optimal matching tree solution for a given pair of trees.

• For this algorithm it’s considered left to right order: what means that the order of the children of each node is important and ancestor: what means that the ancestor of each node is important.

• For match or convert form one tree into another there are three operations allowed, as deleting a node, inserting a node and replace or changing a node.

• When a node n is deleted, all its children are attached to the parent of n, in a insertion, it happen the opposite, it can heritage nodes as it’s own children, and a changing only affects the label of the node without any changes on the tree morphology.

summary


Post-order traversal of trees

• To traverse a non-empty binary tree in postorder, perform the following operations recursively at each node:

• 1 Traverse the left subtree.• 2 Traverse the right subtree.• 3 Visit the node .

practice yourself

1

5

32

4

Ancestors:

Left most descendentFunction l(x) give as the most left descendent of the node x

Key roots x is a key root if:

)()(,| xlklxkk =>¬∃

summary


Mappings

M

H

S

I I

S

S

B

Domain: Range:

ΛDeleted: in the domain:

Changed

Exact match

Inserted: in the range

Mappings

M

H

S

I I

S

S

B

Domain: Range:

ΛTransformation = { 1,2 } , { 2, Λ } , { 3,1 } , { 4,4 } , {Λ,3 }Note: this is NOT a tai map.

1

12

2

33

4 4

TR: Transformations

M: Map

The setsM

H

SI I

S

S

B

Domain: Range:

Λ1

12

2

33

4 4

C: changeCost =1

EX: exact matchCost = 0

I: InsertionCost = 1

D: deletionCost= 1

TR = EX + C + I + D M = EX + C

Mappings

TR = { 1,2 } ,{ 2, Λ } , { 3,1 } , { 4,4 } , {Λ,3 }

M

H

S

I I

S

S

B

Domain: Range:

Λ1

12

2

33

4 4

• Sets:

C: relabeled (a,b) { 1,2 }

Ex: exact match (c,c) { 3,1 } , { 4,4 }

I: insertions (Λ,b) {Λ,3 }

D: deletions (a,Λ) { 2, Λ }

Mappings more formal

• Let V (T) denote the set of nodes of a tree T• S is the Source tree• T is the Target tree.• M mapping

summary


Tai mapping: a restricted map

M={(a,b), (c,d)}

1 One to one node

2 Sibling order preserved:brothers do not change order

3 Ancestor order preserved: by change are not new ancestors

1 a=c iff b=d

6 a<c iff b<d

7 Anc(a,c) iff anc (b,d).

c f

b d e

a

• Sample of tai mapping:

• M= {(c,f),(a,e)}

practice

.

Ancestry!multiple!

Possible tai mapping

Sibling order

1

2

34

3

21

5

4 1

5

32

4

summary


Definition : Tree distance

• There can be multiple possible tai mappings between two trees, and at least there is one with the smallest cost.

• “the tree-distance is the cost of the least expensive Tai mapping”

• Cost = |I| + |D| + |C|

Practice: prove the “triangle inequality”

TR: Transformations

M: Map

C: changeCost =1



D: deletionCost= 1

TR: Transformations

M: Map

C: changeCost =1



D: deletionCost= 1

Prove: γ (M3) ≤ γ(M1)+ γ(M2) | M3 = M1*M2Clue: γ (M0) = |C0|+|I0|+|D0|

PracticeAll possible combinations of the set m1 and m2 in m3 gives same or less cost combined, than alone.

SET1 COST SET2 COST SET3 COST DIFF

EX 0 EX 0 EX 0 0

EX 0 C 1 C 1 0

EX 0 D 1 D 1 0

C 1 EX 0 C 1 0

C 1 C 1 C/EX 1/0 -1/-2

C 1 D 1 D 1 -1

I 1 EX 0 I 1 0

I 1 C 1 I 1 -1

I 1 D 1 None 0 -2

D 1 None 0 D 1 0

D 1 I 1 C/EX 1/0 -1/-2

None 0 I 1 I 1 0

summary


Zhang shasha algorithm

• finds the tree distance (cost of the least costly Tai mapping)

• Important concepts:• Tree table -> tscore• Forest table -> fscore

Forest Distance

• Forest distance == F(), Tree distance T().• Cost function δ = F and T

Tree Tree Tree

Forest

Tree

δ( , )+γ( )

Tree distance

δ( , )=min

δ( , )+γ( )

δ( , )+γ( )

delete

add

change

Forest distance

δ( , ) =min

δ( , )+γ( )

δ( , )+γ( )delete

add

δ( , )+ δ( , )

δ( , )+γ( )

Why this is not allowed?Clue: check tai mapping restrictions

δ( , ) =min

δ( , )+γ( )

δ( , )+γ( )delete

add

change

δ( , )+γ( )δ( , )=change

Algoritm: psedudo code

• Preprocessing() // – Get most left node and key root for each node.

• For s:=1 to |keyroots(t1)|for t:= 1 to |keyroots(t2)|

i=keyroots(t1)[s];j=keyroots(t2)[t];Treedist(i,j);

end• Return tdist[i,j];

Algoritm: 2

Treedist(pos1,pos2) {bound1=pos1-left1[pos1]+2;bound2=pos2-left2[pos2]+2;

fdist= new int[bound1,bound2];fdist[0][0] = 0; for(i=1,i<bound1;i++)

fdist[i][0]= fdist[i-1][0] + c[i][0] for(i=1,i<bound1;i++)

fdist[0][i]= fdist[0][i-1] + c[0][1]

prep

are

fore

st t

able

Algoritm: 3

For(k=left[pos1],i=1;k<=pos1;k++,i++)For(l=left[pos1],j=1;l<=pos1;l++,j++)

if((left1[k]==left1[pos1]&&(left2[l]==left2[pos2])){ // if both are trees, then tree distance

/// then:

Fdist[i][j]=MIN(

fdist[i-1][j]+c[0][l]

fdist[i][j-1]+c[ k][0]

fdist[i-1][j-1]+c[k][l]

Tdist[k][l]=fdist[i][j];

Algoritm: 4

}else{ /// else:

M=left1[k]-left1[pos1];

N= left2[l]-left2[pos2];

Fdist[i][j]=MIN(

fdist[i-1][j]+c[0][l]

fdist[i][j-1]+c[ k][0]

fdist[m][n]+tdist[k][l];

)

}

Sample• There is a permanent tree distance table, and

a dynamic forest distance table.• Let’s follow the algorithm to solve this

problem:• Color means different labels.

6

5

2

1

3 4

6

2 4

1 3

5

t1 t26

6

5

Sample

• Position:• Left1 array = • Left2 array =• LR_keyroots1 =• LR_keyroots2 =

6

5

2

1

3 4

6

2 4

1 3

5

t1 t2

0 1 2 3 4 5 6

Nan 1 1 3 4 1 1

Nan 1 1 3 3 5 1

Nan 0 0 1 1 0 1

Nan 0 0 0 1 1 1

Most left


6

5

2

1

3 4

6

2 4

1 3

5

t1 t2

0 1 2 3 4 5 6

Nan 1 1 3 4 1 1

Nan 1 1 3 3 5 1

Nan 0 0 1 1 0 1

Nan 0 0 0 1 1 1

Key roots


6

5

2

1

3 4

6

2 4

1 3

5

t1 t2

0 1 2 3 4 5 6

Nan 1 1 3 4 1 1

Nan 1 1 3 3 5 1

Nan 0 0 1 1 0 1

Nan 0 0 0 1 1 1

Atomic cost: c

• Measure of change the label node n to the node m.

• First row and column corresponds to the cost of delete/insert a node with such label.

6

5

2

1

3 4

6

2 4

1 3

5

t1 t2

- 1 1 1 1 1 1

1 0 1 0 1 0 1

1 1 0 1 0 1 1

1 0 1 0 1 0 1

1 0 1 0 1 0 1

1 1 0 1 0 1 1

1 1 0 1 0 1 1

1

2

3

4

5

6

1 2 3 4 5 6

ΛΛ

Tree distance table

• First row and column are unless , it just simplify the code.

• The other cell are suppose to be start by 0, but we use N = Nan to illustrate that we only use the cells with value.

• Matrix 6 *6 as the number of nodes

• Each position x means the sub-tree T[l(x) ..x).

6

5

2

1

3 4

6

2 4

1 3

5

t1 t2

- - - - - - -

- N N N N N N

- N N N N N N

- N N N N N N

- N N N N N N

- N N N N N N

- N N N N N N

0 1 2 3 4 5 6

01 23 456

Treedist(3,4)

Calculations:

• Bound1 = 3-3+2=2• Bound2 = 4-3+2=3• Fdist = new int[2][3]• Prepare forest distance

table

Forest distance

- - - - - - -

- N N N N N N

- N N N N N N

- N N N N N N

- N N N N N N

- N N N N N N

- N N N N N N

Treedist(3,4) step2Calculations:

• k = 3, i = 1• L = 3, j = 1• Is a tree? -> yes

• Set the value in Forest distance table and in tree distance table

• L++, j++

Forest distance0 1 2

1 0

- - - - - - -

- N N N N N N

- N N N N N N

- N N 0 N N N

- N N N N N N

- N N N N N N

- N N N N N N

T( , )=min3 3

F(Λ, )+ γ( )=23

F( ,Λ)+ γ( )=2F(Λ, Λ)+γ( )=0

3

3

3

3

It is a tree so:2.We can look in

diagonal3.We must copy the value in tree table


• k = 3, i = 1• L = 4, j = 2• Is a tree? -> yes

• Set the value in Forest distance table and in tree distance table

• L++, j++

Forest distance0 1 2

1 0 1

- - - - - - -

- N N N N N N

- N N N N N N

- N N 0 1 N N

- N N N N N N

- N N N N N N

- N N N N N N

T( , )=min3

F(Λ, )+ γ( )=3F( , )+ γ( )=1F(Λ, ) +γ( )=2

3

3

4

4

4

3

4

3

3

3

Treedist(3,5) step1

Calculations:


table

Forest distance

- - - - - - -

- N N N N N N

- N N N N N N

- N N 0 1 N N

- N N N N N N

- N N N N N N

- N N N N N N


• k = 3, i = 1• L = 5, j = 1• Is a tree? -> yes

Forest distance

T( , )=min3 5

F(Λ, )+ γ( )=25

F( ,Λ)+ γ( )=2F(Λ, Λ)+γ( )=0

3

3

5

5 - - - - - - -

- N N N N N N

- N N N N N N

- N N 0 1 0 N

- N N N N N N

- N N N N N N

- N N N N N N

0 1

1 0

Treedist(3,6) step1

Calculations:


table

Forest distance0 1 2 3 4 5 6

1

- - - - - - -

- N N N N N N

- N N N N N N

- N N 0 1 0 N

- N N N N N N

- N N N N N N

- N N N N N N

Treedist(3,6) step2

Calculations:

• k = 3, i = 1• L = 1, j = 1• Is a tree? -> yes


1 0

- - - - - - -

- N N N N N N

- N N N N N N

- 0 N 0 1 0 N

- N N N N N N

- N N N N N N

- N N N N N N

T( , )=min3

F(Λ, )+ γ( )=2F( , Λ )+ γ( )=2F(Λ, Λ)+γ( )=0

3

3

1

1

1

1

Treedist(3,6) step3

Calculations:

• k = 3, i = 1• L = 2, j = 2• Is a tree? -> yes


1 0 1

- - - - - - -

- N N N N N N

- N N N N N N

- 0 1 0 1 0 N

- N N N N N N

- N N N N N N

- N N N N N N

T( , )=min3

F(Λ, )+ γ( )=2F( , )+ γ( )=1F(Λ, )+γ( )=2

3

3

2

2

2

1

2

1

1

1

Treedist(3,6) step4

Calculations:

• k = 3, i = 1• L = 4, j = 4• Is a tree? -> NO

• M = 0• N = 2


1 0 1 2

- - - - - - -

- N N N N N N

- N N N N N N

- 0 1 0 1 0 N

- N N N N N N

- N N N N N N

- N N N N N N

T( , )=min3

F(Λ, )+ γ( )=4F( , )+ γ( )=2F(Λ, )+T( , )=2

3

3

221

3

21

3

21

21

3 3

It is NOT a tree so:2.We can NOT look

in diagonal

Treedist(3,6) step5

Calculations:

• k = 3, i = 1• L = 3, j = 3• Is a tree? -> NO

• M = 0• N = 2


1 0 1 2 3

- - - - - - -

- N N N N N N

- N N N N N N

- 0 1 0 1 0 N

- N N N N N N

- N N N N N N

- N N N N N N

T( , )=min3

F(Λ, )+ γ( )=5F( , )+ γ( )=3F(Λ, )+T( , )=3

3

3

2

21

2 41 3

2 41 3

21 3

43

3

Size of forest

• Tree distance (3,4)-> starting in [3,3] (most left).F(3..3,3..3),F(3..3,3..4)

Key roots6

5

2

1

3 4

t1

2 4

1 3

t2


• Tree distance (4,4)-> starting in [4,3] (most left).

F(4..4,3..3),F(4..4,3..4)

Key roots6

5

2

1

3 4

t1

2 4

1 3

t2


• Tree distance (4,4)-> starting in [4,3] (most left).

F(4..4,3..3),F(4..4,3..4)• Tree distance (6,4)-> starting in [1,3] (most left).

F(4..4,3..3),F(4..4,3..4)

Key roots6

5

2

1

3 4

t1

2 4

1 3

t2


Key roots6

5

2

1

3 4

t1

2 4

1 3

t2

• Tree distance (6,4)-> starting in [1,3] (most left).F(1..1,3..4), treeF(1..2,3..4), treeF(1..3,3..4), forestF(1..4,3..4), forestF(1..5,3..4), treeF(1..6,3..4), tree

Key roots 6

5

2

1

3 4

t1

2 4

1 3

t2

2

1

3 4 We need the result of the matching for the sub-trees nodes 3 and 4, and it happen that hey are key roots!

δ( , ) =min

δ( , )+γ( )

δ( , )+γ( ) delete

add

δ( , )+ δ( , )

Practice:

• Calculate the tree distance between this two trees and observe the similarity with Levenshtein algorithm. (Only 1 key root.)

G

o

o

d

G

o

d

Forest table

= tree table

= levenshtein table.

Tree distance

Time:

Space:

Calculate the alignment

Tree distance- - - - - - -

- 0 1 0 1 0 5

- 1 0 1 0 1 4

- 0 1 0 1 0 5

- 0 1 0 1 0 5

- 4 3 4 3 4 2

- 5 4 5 4 5 3

Forest distance 6*60 1 2 3 4 5 6

1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 2 1 2

5 4 3 2 3 2 2

6 5 4 3 4 3 3

The algorithm can be extended in order to get the alignment:All tables are need it.This talbes corresponds to the same example as we looked before.



- 0 1 0 1 0 5

- 1 0 1 0 1 4

- 0 1 0 1 0 5

- 0 1 0 1 0 5

- 4 3 4 3 4 2

- 5 4 5 4 5 3


1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 2 1 2

5 4 3 2 3 2 2

6 5 4 3 4 3 3

(6,6)



- 0 1 0 1 0 5

- 1 0 1 0 1 4

- 0 1 0 1 0 5

- 0 1 0 1 0 5

- 4 3 4 3 4 2

- 5 4 5 4 5 3


1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 2 1 2

5 4 3 2 3 2 2

6 5 4 3 4 3 3

(6,6),



- 0 1 0 1 0 5

- 1 0 1 0 1 4

- 0 1 0 1 0 5

- 0 1 0 1 0 5

- 4 3 4 3 4 2

- 5 4 5 4 5 3


1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 2 1 2

5 4 3 2 3 2 2

6 5 4 3 4 3 3

(6,6), (4,5)



- 0 1 0 1 0 5

- 1 0 1 0 1 4

- 0 1 0 1 0 5

- 0 1 0 1 0 5

- 4 3 4 3 4 2

- 5 4 5 4 5 3


1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 2 1 2

5 4 3 2 3 2 2

6 5 4 3 4 3 3

(6,6), (4,5), (2,2)



- 0 1 0 1 0 5

- 1 0 1 0 1 4

- 0 1 0 1 0 5

- 0 1 0 1 0 5

- 4 3 4 3 4 2

- 5 4 5 4 5 3


1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 2 1 2

5 4 3 2 3 2 2

6 5 4 3 4 3 3

(6,6), (4,5), (2,2), (1,1).

THANKS FOR YOUR ATENTION

tree distance algorithm

Education