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Tree distance algorithm
- 1. Workshop on tree distance By Hector Franco francoph at tcd dot ie Trinity College of Dublin
- 6. Computing Levenshtein distance D(i,j) = score of best alignment from s1..si to t1..tj = min D(i-1,j-1) + d(si,tj) //subst/copy D(i-1,j)+1 //insert D(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
- 7. Computing Levenshtein distance - 3 D(i,j)= min D(i-1,j-1) + d(si,tj) //subst/copy D(i-1,j)+1 //insert D(i,j-1)+1 //delete 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. = 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
- 8. Computing Levenshtein distance - 3 D(i,j)= min D(i-1,j-1) + d(si,tj) //subst/copy D(i-1,j)+1 //insert D(i,j-1)+1 //delete 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
- 9. Practice G O O D G O D G O D TRY TRY 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
- 15. practice yourself 1 5 3 2 4
- 16. Ancestors:
- 17. Left most descendent Function l(x) give as the most left descendent of the node x
- 18. Key roots x is a key root if:
- 20. Mappings M H S I I S S B Domain: Range: Deleted: in the domain: Changed Exact match Inserted: in the range
- 21. Mappings M H S I I S S B Domain: Range: Transformation = { 1,2 } , { 2, } , { 3,1 } , { 4,4 } , { ,3 } Note: this is NOT a tai map. 1 1 2 2 3 3 4 4
- 22. The sets TR: Transformations M: Map C: change Cost =1 EX: exact match Cost = 0 I: Insertion Cost = 1 D: deletion Cost= 1 TR = EX + C + I + D M = EX + C M H S I I S S B Domain: Range: 1 1 2 2 3 3 4 4
- 30. Practice: prove the “triangle inequality” Prove: (M3) (M1)+ (M2) | M3 = M1*M2 Clue: (M0) = |C0|+|I0|+|D0| TR: Transformations M: Map C: change Cost =1 EX: exact match Cost = 0 I: Insertion Cost = 1 D: deletion Cost= 1 TR: Transformations M: Map C: change Cost =1 EX: exact match Cost = 0 I: Insertion Cost = 1 D: deletion Cost= 1
- 31. Practice All 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
- 35. Tree distance ( , )+ ( ) ( , )= min ( , )+ ( ) ( , )+ ( ) delete add change
- 36. Forest distance ( , ) = min ( , )+ ( ) ( , )+ ( ) delete add ( , )+ ( , )
- 37. Why this is not allowed? Clue: check tai mapping restrictions ( , )+ ( ) ( , ) = min ( , )+ ( ) ( , )+ ( ) delete add change
- 38. ( , )+ ( ) ( , )= change
- 74. THANKS FOR YOUR ATENTION