1. Comparative RNA Structural Analysis

2. Overview Comparative RNA Structural Analysis
Method 1: Align, then fold
Method 2: Fold, then compare

3. Overview Comparative RNA Structural Analysis
Method 1: Align, then fold
Method 2: Fold, then compare

4. Comparative RNA Structural Analysis Problem Definition
Input: A set of sequences with assumed structural similarities.
Output: Alignment, and common structural elements.

5. Possible approaches Homologous RNA sequences 1 Sequence alignment
Aligned Sequences
Fold
alignments
Aligned Structures

6. Possible approaches Homologous RNA sequences 1 2 Fold Sequence
AUCCCCGUAUCGAUC
CUCGGCGUAUCGGUC 1 2
Fold
Sequences
Sequence
alignment
Homologous RNA
secondary Structures
Aligned Sequences
Structure
Alignment
Fold
alignments
Aligned Structures

7. Simultaneous Fold and Alignment
Possible approaches
Homologous RNA sequences
AUCCCCGUAUCGAUC
CUCGGCGUAUCGGUC 1 3 2
Fold
Sequences
Sequence
alignment
Sankoff
Simultaneous Fold and Alignment
Homologous RNA
secondary Structures
Aligned Sequences
Structure
Alignment
Fold
alignments
Aligned Structures

8. Align, then fold First step: multiple alignment
We want to use an algorithm we know to fold our aligned sequences.
How can we modify Nussinov algorithm to fold multiple alignments?
A C G T G G A G A A C G G
A C C C T A A A G G G G A
T A T A G C A A T T A T C
C G G A T T A G T T C C G
G A T T G G A C G A A T A
G G G C T A A A T G C C A

9. Align, then fold We need a new scoring function
Scoring a base pair is different than scoring a pair of columns in our alignment.
Using the new scoring function, we can apply Nussinov algorithm on the converted input (with slight changes).

10. Covariation Columns that “change together” construct a stem
A C G U G G A G A A C G G
A C C C U A A A G G G G A
U A U A G C A A U U A U C
C G G A U U A G U U C C G
G A U U G G A C G A A U A
G G G C U A A A U G C C A

11. The Mixy algorithm
For each column 𝑖 in the alignment, define 𝑓 𝑖 𝑥 , 𝑥∈ 𝐴,𝑈,𝐶,𝐺 to be 𝑥’s frequency in column 𝑖.
A C G U G A A C G G
A C C C U G G G G G
A A U A G U U A U G
2 3
𝑓 2 𝐶 =

12. The Mixy algorithm
For each column 𝑖 in the alignment, define 𝑓 𝑖 𝑥 , 𝑥∈ 𝐴,𝑈,𝐶,𝐺 to be 𝑥’s frequency in column 𝑖.
For each 𝑖 and 𝑗, define 𝑓 𝑖,𝑗 𝑥,𝑦 , 𝑥,𝑦∈{𝐴,𝑈,𝐶,𝐺} to be the frequency of 𝑥 in column 𝑖 and 𝑦 column 𝑗 on the same sequence.
A C G U G A A C G G
A C C C U G G G G G
A A U A G U U A U G
2 3
𝑓 2,9 𝐶,𝐺 =
𝑓 2,9 𝐴,𝐺 =

13. The Mixy algorithm
For each column 𝑖 in the alignment, define 𝑓 𝑖 𝑥 , 𝑥∈ 𝐴,𝑈,𝐶,𝐺 to be 𝑥’s frequency in column 𝑖.
For each 𝑖 and 𝑗, define 𝑓 𝑖,𝑗 𝑥,𝑦 , 𝑥,𝑦∈{𝐴,𝑈,𝐶,𝐺} to be the frequency of 𝑥 in column 𝑖 and 𝑦 column 𝑗 on the same sequence.
Clearly, if 𝑥 and 𝑦 are independent, 𝑓 𝑖,𝑗 𝑥,𝑦 𝑓 𝑖 𝑥 ∗ 𝑓 𝑗 𝑦 ≈1.
A C G U G A A C G G
A C C C U G G G G G
A A U A G U U A U G
2 3
𝑓 2,9 𝐶,𝐺 =
𝑓 2,9 𝐴,𝐺 =

14. The Mixy algorithm
Now, to measure mutual information between columns 𝑖 and 𝑗 we’ll define: 𝐻 𝑖,𝑗 = 𝑥,𝑦 𝑓 𝑖,𝑗 𝑥,𝑦 log 2 𝑓 𝑖,𝑗 𝑥,𝑦 𝑓 𝑖 𝑥 ∗ 𝑓 𝑗 𝑦
A C G U G A A C G G
A C C C U G G G G G
A A U A G U U A U G
𝑓 2,9 𝐶,𝐺 log 2 𝑓 2,9 𝐶,𝐺 𝑓 2 𝐶 ∗ 𝑓 9 𝐺 + 𝑓 2,9 𝐴,𝑈 log 2 𝑓 2,9 𝐴,𝑈 𝑓 2 𝐴 ∗ 𝑓 9 𝑈
𝐻 2,9 =
= 2 3 ∗ log ∗ ∗ log ∗ = 2 3 ∗ log ∗log⁡(3)=
2 3 ∗ ∗1.58=0.526

15. The Mixy algorithm
Now, to measure mutual information between columns 𝑖 and 𝑗 we’ll define: 𝐻 𝑖,𝑗 = 𝑥,𝑦 𝑓 𝑖,𝑗 𝑥,𝑦 log 2 𝑓 𝑖,𝑗 𝑥,𝑦 𝑓 𝑖 𝑥 ∗ 𝑓 𝑗 𝑦
𝑓 1,10 𝐴,𝐺 log 2 𝑓 1,10 𝐴,𝐺 𝑓 1 𝐴 ∗ 𝑓 10 𝐺 = 3 3 ∗ log ∗ =1∗0=0
A C G U G A A C G G
A C C C U G G G G G
A A U A G U U A U G
𝐻 1,10 =

16. The Mixy algorithm
Now, to measure mutual information between columns 𝑖 and 𝑗 we’ll define: 𝐻 𝑖,𝑗 = 𝑥,𝑦 𝑓 𝑖,𝑗 𝑥,𝑦 log 2 𝑓 𝑖,𝑗 𝑥,𝑦 𝑓 𝑖 𝑥 ∗ 𝑓 𝑗 𝑦
𝑓 3,7 𝐺,𝐴 log 2 𝑓 3,7 𝐺,𝐴 𝑓 3 𝐺 ∗ 𝑓 7 𝐴 + 𝑓 3,7 𝐶,𝐺 log 2 𝑓 3,7 𝐶,𝐺 𝑓 3 𝐶 ∗ 𝑓 7 𝐺 +
𝑓 3,7 𝑈,𝑈 log 2 𝑓 3,7 𝑈,𝑈 𝑓 3 𝑈 ∗ 𝑓 7 𝑈 + 𝑓 3,7 𝐴,𝐶 log 2 𝑓 3,7 𝐴,𝐶 𝑓 3 𝐴 ∗ 𝑓 7 𝐶 =
=4∗ 1 4 ∗ log ∗ =1∗𝑙𝑜𝑔 4 =2
A C G U G A A C G G
A C C C U G G G G G
A A U A G U U A U G
A A A A G U C U U G
𝐻 3,7 =

17. The Mixy algorithm 0≤ 𝐻 𝑖,𝑗 ≤2
Now, to measure mutual information between columns 𝑖 and 𝑗 we’ll define: 𝐻 𝑖,𝑗 = 𝑥,𝑦 𝑓 𝑖,𝑗 𝑥,𝑦 log 2 𝑓 𝑖,𝑗 𝑥,𝑦 𝑓 𝑖 𝑥 ∗ 𝑓 𝑗 𝑦
0≤ 𝐻 𝑖,𝑗 ≤2
A C G U G A A C G G
A C C C U G G G G G
A A U A G U U A U G
A A A A G U C U U G
Higher value means that columns 𝑖 and 𝑗 are correlated
Lower value means that columns 𝑖 and 𝑗 are not correlated

18. Overview Comparative RNA Structural Analysis
Method 1: Align, then fold
Method 2: Fold, then compare

19. Ordered rooted tree representation
Shapiro, 1988:
nodes - elements of secondary structure (hairpin loop, bulge, internal loop or multi-loop).
edges - base-paired (stem) regions.

20. Ordered rooted tree representation
Shapiro, 1988:
nodes - elements of secondary structure (hairpin loop, bulge, internal loop or multi-loop).
edges - base-paired (stem) regions.
Zhang, 1998:
nodes - unpaired bases (leaves) or paired bases (internal nodes). Each node is labeled with a base or a pair of bases.
edges - connecting consecutive stem base-pairs or a leaf base with the last base-pair in the corresponding stem.

21. Problem definition
The subtree isomorphism problem [Matula, 1968,1978]:
Given a pattern tree P and a text tree T,
find a subtree of T which is isomorphic to P,
In other words: find if some subtree of T is identical in structure to P
The subtree homeomorphism problem [Chung, 1987, Reyner, 1977, Pinter et al., 2004]:
Similar to isomorphism problem, where degree-2 nodes can be deleted from the text tree.

22. Subtree homeomorphism problem
Let P and 𝑇 be two ordered, rooted trees.
Let 𝑡 be a subtree of 𝑇, rooted at node 𝑣∈𝑇
A mapping 𝛼: P → t is a one-to-one matching of a node of P to a node of 𝑡.
The mapping must preserve the ancestor relations of the nodes and their relative order.
The subtree homeomorphism score of a mapping, denoted S(𝛼,v), is:
S(𝛼,v)
node-to-node similarity score function
𝑢∈𝑃, 𝑣∈𝑡
edge-to-edge similarity score function
euP, evt
The penalty of deleting a
degree-2-node from T
The penalty for deleting any other node in T

23. Subtree homeomorphism problem
Given P and 𝑇, we want to find a subtree 𝑡 in T such that the score S(𝛼,v) is maximal
How can we do that?
Ho can we solve this problem efficiently?
Dynamic programming!

24. Subtree homeomorphism problem
Isomorphism
Homeoomorphism

25. Rooted Ordered Subtree Isomorphism
Given trees 𝑃 and 𝑇, and the scoring table below, compute Labeled Ordered Rooted Subtree Isomorphism of 𝑃 and 𝑇.
No deletions are allowed from 𝑃
Only deletions of complete subtrees from 𝑇 are allowed, with penalty = 0 𝑃 𝑇 b e f c d a c’ f’ a' e' b' d' g' h’ 4 1 3

26. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b e f c d a
Rows are post ordered 𝑃 nodes
Columns are post ordered 𝑇 nodes 4 1 3

27. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b e f c d a
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

28. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c d a
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

29. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c d a
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

30. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c d a
ℎ𝑒𝑖𝑔ℎ𝑡 𝑐 >ℎ𝑒𝑖𝑔ℎ𝑡( 𝑏 ′ )
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

31. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
ℎ𝑒𝑖𝑔ℎ𝑡 𝑐 >ℎ𝑒𝑖𝑔ℎ𝑡( 𝑏 ′ )
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

32. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

33. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

34. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 4 1 3

35. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 4 1 3

36. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 4 1 3 −∞

37. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 4 1 3 −∞

38. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 𝟎 4 1 3 −∞

39. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 𝟎 4 1 3 −∞

40. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 𝟎 4 1 3 1 −∞

41. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 𝟎 𝟎 𝟎 4 1 3 1 4 −∞ 4 1 1

42. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 𝟎 𝟎 𝟎 4 1 3 1 4 −∞ 4 1 1

43. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ d a
Small DP table e‘ f‘ g‘ e f 𝟎 𝟎 𝟎 4 1 3
𝑆 𝑐, 𝑐 ′ =3+5=8 1 4 −∞ 4 1 1

44. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
Small DP table e‘ f‘ g‘ e f 𝟎 𝟎 𝟎 4 1 3
𝑆 𝑐, 𝑐 ′ =3+5=8 −∞ 1 4 4 1 1

45. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

46. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
Small DP table h‘ e f 𝟎 4 1 3 4 −∞ 1

47. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
Small DP table h‘ e f 𝟎 4 1 3 4 −∞ 1

48. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

49. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
𝑑𝑒𝑝𝑡ℎ 𝑐 >𝑑𝑒𝑝𝑡ℎ(𝑎′)
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

50. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
𝑑𝑒𝑝𝑡ℎ 𝑐 >𝑑𝑒𝑝𝑡ℎ(𝑎′)
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

51. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
ℎ𝑒𝑖𝑔ℎ𝑡 𝑎 >ℎ𝑒𝑖𝑔ℎ𝑡( 𝑐 ′ )
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

52. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
ℎ𝑒𝑖𝑔ℎ𝑡 𝑎 >ℎ𝑒𝑖𝑔ℎ𝑡( 𝑐 ′ )
𝑆 𝑢,𝑣 =
𝑖𝑓 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑙𝑒𝑎𝑣𝑒𝑠
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 4 1 3

53. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
Small DP table b‘ c‘ d‘ b c d 4 1 3

54. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a
Small DP table b‘ c‘ d‘ b c d 4 1 3 4 4 3 −∞ 8 −∞ 3 3 4

55. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a 20
Small DP table b‘ c‘ d‘ b c d 4 1 3 4 4 3
𝑆 𝑐, 𝑐 ′ =4+16=20 −∞ 8 −∞ 3 3 4

56. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a 20
Where is the solution? 4 1 3

57. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a 20 4 1 3

58. Rooted Ordered Subtree Isomorphismf c d a 𝑃 𝑇 c’ f’ a' e' b' d' g' h’ b’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a 20 e‘ f‘ g‘ e f b‘ c‘ d‘ b c d 𝟎 4 1 3 4 4 3 1 4 −∞ 8 −∞ 4 1 1 3 3 4

59. Running time complexityb’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a 20
If 𝑃 has m nodes and 𝑇 has 𝑛 node
There are 𝑛𝑚 cells in the large DP table
In the worst case – for each cell we will compute a small DP table with 𝑚𝑛 cells
Resulting in 𝑂( 𝑛 2 𝑚 2 ) running time
Is there a tighter bound? 𝑃 𝑇 b e f c d a c’ f’ a' e' b' d' g' h’

60. Running time complexityb’ e‘ f‘ g‘ c‘ h‘ d‘ a' b 4 1 3 e f c −∞ 8 d a 20
If 𝑃 has m nodes and 𝑇 has 𝑛 node
Each node 𝑢 in 𝑃 will be in a small DP table only when its father is compared to a node in 𝑇
A father of a node in P is compared at most 𝑛 times
⟹𝑂(𝑚𝑛)
Symmetrically, for a node 𝑣 in T
Overall: 𝑂 𝑚𝑛+𝑚𝑛+𝑚𝑛 =𝑂(𝑚𝑛) 𝑃 𝑇 b e f c d a c’ f’ a' e' b' d' g' h’
Large DP
Small DP for a node in P
Small DP for a node in T