Suffix Trees

1
Suffix Trees

• Lecture 12 October 11, 2005
• Algorithms in Biosequence Analysis
• Nathan Edwards - Fall, 2005

2
Suffix Trees

• Finish proof of Ukkonens O(m) construction
• Generalized suffix trees
• First applications of suffix trees

3
Example

• Suffix tree of S xabxa

x
b

a
a
x
6
b

a
b
x
5

x

a
4
a


3
2
1
4
Example

• Implicit suffix tree of S xabxa

x
b
a
a
x
b
a
b
x
x
a
a
3
1
2
5
Ukkonens O(m) construction high-level

• Construct T1
• For i 1 to m 1 Phase i1 make Ti1 from
  Ti For j 1 to i 1 Extension j Find
  path from root labeled Sji Extend path with
  S(i1)

6
Suffix extension rules

• Extension of Sji ß with S(i1)
• ß ends at a leaf ? add S(i1) to leaf edge label
• No path at end of ß starts with S(i1)? add new
  leaf edge at end of ß with label
  S(i1). create new node at end of ß if
  necessary
• A path at end of ß starts with S(i1)? do nothing

7
Example

• Implicit suffix tree of S axabx

b
a
x
x
x
a
a
4
b
b
x
b
x
1
x
3
2
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Example

• Implicit suffix tree of S axabxbRule 1

b
a
x
x
x
a
a
b
b
b
4
x
b
x
b
x
b
1
b
3
2
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Example

• Implicit suffix tree of S axabxbRule 2

b
a
x
x
x
b
a
a
b
b
b
5
4
x
b
x
b
x
b
1
b
3
2
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Example

• Implicit suffix tree of S axabxbRule 3

b
a
x
x
x
b
a
a
b
b
b
5
4
x
b
x
b
x
b
1
b
3
2
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Suffix links

• Given
• String xa, character x, string a
• Internal node v with node-label xa
• If there is a node s(v) with node-label a, then v
  ? s(v) is called a suffix link.
• If a is empty, s(v) is the root.

12
Suffix links

• Corollary If implicit suffix tree Ti has an
  internal node v with path label xa,Ti must have a
  node s(v) with path-label a.

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Skip/Count Edge Traversal
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Summary

• Given node v with label xa, suffix link points to
  node s(v) with label a.
• Suffix links skip/count edge traversal
• find ß Sji given ß0 Sj-1i
• accomplish all extensions of a phase in O(m) time

15
Problem!

• Total length of the edge labels could be ?(m2)
• Construct suffix tree on S abcdefghijklmnopqrs
  tuvwxyz
• Total length of edge labels is 2627/2
• Need output size to be O(m) too!
• Use indices instead of substrings on edges

16
Example

• Suffix tree of S xabxa

x
b

a
a
x
6
b

a
b
x
5

x

a
4
a


3
2
1
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Example

• Suffix tree of S xabxa
• Note that we must now keep S in memory!

6,6
1,2
2,2
6
3,6
6,6
5
6,6
3,6
3,6
4
3
2
1
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Suffix extension rules

• Extension of Sji ß with S(i1)
• ß ends at a leaf ? change label (j,i) to (j,i1)
• No path at end of ß starts with S(i1)? add new
  leaf edge at end of ß with label
  (i1,i1). create new node at end of ß if
  necessary
• A path at end of ß starts with S(i1)? do nothing

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Rule 3 ends the phase

• Extension of Sji ß with S(i1)
• If Rule 3 applies, then ß is already followed by
  S(i1) in Ti
• Some other suffix Sj0i, j lt j0, of S1i
  starts with ß S(i1),
• So we know Sj01i is in Ti
• Therefore Sj1i is already followed by S(i1)
  in Ti
• End the phase once Rule 3 applies
• The remaining do nothings are done for free!

20
Applying Rule 1 is easy

• Once Sji ends at a leaf (Rule 1), every
  extension of suffix j ends at the same leaf
• Sji1, Sji2,
• Instead of changing label (j,i) to (j,i1) for
  each such leaf edge, we make its label (j,e)
• e is a global whose value is i1 in phase i1
• One change to e updates all leaf edges

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Where are the leaf nodes?

• Notice that once we create a leaf j with Rule 2,
  suffix j always ends at a leaf.
• Therefore, all extensions until Rule 3 applies
  result in leaf-edges.
• So, leaf extensions (Rule 1) for all Rule 1 and 2
  extensions of previous iterations can be done
  implicitly, in constant time.

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Where are the leaf nodes?
Rules 1 2
Rule 3
1 1 1 1 1 1 1 1 1 1 2 2 2 1 2
3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
Phase 1
Phase 2
Phase 3
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Where are the leaf nodes?
Rules 1 2
Rule 3
1 1 1 1 1 1 1 1 1 1 2 2 2 1 2
3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
Phase 1
Phase 2
Phase 3
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Where are the leaf nodes?
Rules 1 2
1 1 1 1 1 1 1 1 1 1 2 2 2 1 2
3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2
Phase 1
Phase 2
Phase 3
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Single Phase Algorithm

• Phase i1 make Ti1 from Ti Add S(i1)
• Increment e to i1
• Compute explicit extensions from p until Rule 3
  applies
• Set p to position where Rule 3 applied.
• Position of suffix Spi in tree at end of phase
  i1, is the position of suffix Spi1 required
  for start of phase i2.

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Running time analysis

• m phases, phase i extensions overlap phase i1
  extensions in at most one position.
• 2m explicit extensions
• Explicit extension takes constant time plus
  number of node-skips after following suffix link.
• Similar node-depth argument as before
• node-depth doesnt change between phases
• decrease node-depth at most 2(explicit-extns)
• max node-depth at most m

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Post-processing

• Convert implicit suffix tree to suffix tree in
  O(m) time
• Add terminator symbol to S
• Replace e with S
• O(m) time, visit each leaf edge.

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Generalized Suffix Trees

• It is unusual to have a single long string to
  analyze.
• Usually have a set of strings
• Even the genome is a set of chromosome sequences!
• Generalized suffix trees store all suffixes of a
  set of strings
• leaves index string number position
• Build suffix tree of S S1S2S3..._at_Sk

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Generalized Suffix Trees

• Synthetic suffixes can be misleading
• Remove them by changing edge label
• Can run Ukkonens algorithm on each string in
  turn
• Just find prefix of S2 that is already in tree
• Resume algorithm with next character extension
• Edge labels might be from different strings
• Use 1 never-match character, instead of many
  unique characters
• Encode absolute position or relative position and
  string index?

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Applications

• Longest common substring
• Lightweight longest common substring
• Maximal repeats
• Maximal unique matches (MUMs)

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Longest common substring

• Longest substring contained in S1 and S2.
• Construct generalized suffix tree of S1, S2
• Each leaf represents a suffix of S1, S2, or both.
• Mark each node of suffix tree with
• 1 if all leaves in its sub-tree are suffixes of
  S1
• 2 if all leaves in its sub-tree are suffixes of
  S2
• 0 if a mixture of suffixes of S1 and S2
• Longest common substring is represented by the
  node with mark 0 of maximum string-depth.
• Why must LCS end at a node?

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Longest common substring

• Leaves are easy to mark
• Mark internal nodes, bottom upif all its
  children are from S1 mark with 1if all its
  children are from S2 mark with 2otherwise, mark
  with 0.
• Post-order traversal to mark all nodes, and find
  deepest 0 mark in tree.
• O(S1S2) time

33
Lightweight LCS

• Build suffix tree for S1
• Initialize all nodes of tree with label 0
• Match the initial characters of S2 from the root
  of the tree
• Given mismatch at character S2(i1)
• find node v at or above end of S21i
• Label(v) i

34
Lightweight LCS

• Given match from the root, between S2ji and
  mismatch at character S2(i1)
• find node v at or above end of S2ji
• Label(v) i-j1
• Jump to suffix link s(v) and use skip/count edge
  traversal to find S2j1i in tree
• Re-start character comparisons in tree with
  S2(i1)
• Treat end of S2 like a mismatch.
• Do tree traversal to find largest label

35
Lightweight LCS

• O(S1S2) time
• Very similar to Keyword-Tree automaton
• Build suffix tree for smaller of S1 and S2

36
Maximal Repeats

• Given a string S, find all substrings with at
  least two copies in S, that cannot be extended to
  the left or right
• S AAGATATGATAGGATC
• Maximal repeats GATA, AG, GAT

37
Maximal Repeats

• Can be determined in linear time
• If a is a maximal repeat, then there is an
  internal node with label a.
• Why?
• Therefore, there are at most m S maximal
  repeats.
• Why?

38
Maximal Repeats

• So, which internal nodes represent maximal
  repeats?
• Define the left-character of a suffix Sjm to
  be character S(j-1).
• An internal node is called left-diverse if at
  least two leaves in its sub-tree have different
  left-characters

39
Maximal Repeats

• Theorem
• The node-label a of internal node v is a maximal
  repeat if and only if v is left diverse.
• Left-diverse nodes can be marked using a
  post-order traversal of tree, as in longest
  common substring

40
Maximal unique match

• Useful for anchoring larger alignments
• MUMmer Delcher, ., Salzberg, 99, 02, 04
• MUMs occur exactly once in each of S1 and S2 and
  cant be extended in either direction and still
  be a match
• Maximal repeats in generalized suffix tree of S1
  and S2 that have exactly two leaves, one from S1
  and one from S2