Building Suffix Trees in O(m) time

1
Building Suffix Trees in O(m) time

• Weiner had first linear time algorithm in 1973
• McCreight developed a more space efficient
  algorithm in 1976
• Ukkonen developed a simpler to understand variant
  in 1995
• This is what we will focus on

2
Implicit Suffix Trees

• An implicit suffix tree for string S is a tree
  obtained from the suffix tree for S by
• removing from all edge labels
• removing any edges that now have no label
• removing any node that does not still have at
  least two children
• Some suffixes may no longer be leaves
• An implicit suffix tree for prefix S1..i of S
  is similarly defined based on the suffix tree for
  S1..i
• Ii will denote the implicit suffix tree for
  S1..i

3
Example

• Implicit tree for xabxa from tree for xabxa
• xabxa, abxa, bxa, xa, a,

b
x
a

x
a
6
a
b

x

b
5

x
a
a
4


3
2
1
4
Remove

• Remove
• xabxa, abxa, bxa, xa, a,

b
x
a

x
a
6
a
b

x

b
5

x
a
a
4


3
2
1
5
Remove after

• Remove
• xabxa, abxa, bxa, xa, a

b
x
a
x
a
6
a
b
x
b
5
x
a
a
4
3
2
1
6
Remove unlabeled edges

• Remove unlabeled edges
• xabxa, abxa, bxa, xa, a

b
x
a
x
a
6
a
b
x
b
5
x
a
a
4
3
2
1
7
Remove unlabeled edges

• Remove unlabeled edges
• xabxa, abxa, bxa, xa, a

b
x
a
x
a
a
b
x
b
x
a
a
3
2
1
8
Remove interior nodes

• Remove internal nodes with only one child
• xabxa, abxa, bxa, xa, a

b
x
a
x
a
a
b
x
b
x
a
a
3
2
1
9
Final implicit tree

• Remove internal nodes with only one child
• xabxa, abxa, bxa, xa, a

b
x
x
a
a
a
b
b
x
x
a
a
3
2
1
10
Basic Structure of Algorithm

• Initialization
• I1 has one edge labeled S(1)
• For i 1 to m-1 (build Ii1)
• For j 1 to i1
• Find location of string Sj..i in tree
• Extend to incorporate character S(i1)
• Expand final implicit tree to make full suffix
  tree

11
Order of operations visualization

• S xabxacdefghixabcab
• i1 1
• x
• i1 2
• extend x to xa
• a
• i1 3
• extend xa to xab
• extend a to ab
• b

12
Extension Rules

• Case 1 Sj..i ends at a leaf
• Add character S(i1) to end of label on leaf edge
• Case 2 Not a leaf, but no path from end of
  Sj..i location continues with S(i1)
• Split a new leaf edge for character S(i1)
• May need to create an internal node if Sj..i
  ends in the middle of an edge
• Case 3 Sj..i1 is already in the tree
• No update

13
Visualization

• Implicit tree for axabxb from tree for axabx

b
Rule 1 at a leaf node
Rule 3 already in tree
Rule 2 add a leaf edge (and an interior node)
14
Observations

• Once Sj..i is located in the tree, extending to
  accommodate S(i1) is constant time
• Making Ukkonens algorithm O(m2)
• Finding the Sj..i locations in the suffix trees
  quickly when explicit computation is needed

15
Edge Label Representation

• Potential Problem
• Size of edge labels may be W(m2)
• Example
• S abcdefghijklmnopqrstuvwxyz
• Total length is Sjltm1 j m(m1)/2
• Similar problem can happen when the length of the
  string is arbitrarily larger than the alphabet
  size
• Solution
• Label edges with pair of indices indicating
  beginning and end of positions of the substring
  in S

16
Full edge label illustration

• String S xabxa

b
x
a

x
a
6
a
b

x

b
5

x
a
a
4


3
2
1
17
Compact edge label illustration

• String S xabxa

(1,2) or (4,5)?
(2,2)
(6,6)
(3,6)
6
(6,6)
5
(6,6)
(3,6)
(3,6)
4
3
2
1
18
Modified Extension Rules

• Rule 2 new leaf edge
• label new leaf edge (i1, i1)
• Rule 1 leaf edge extension
• label had to be (p,i) before extension
• given rule 2 above and an induction argument
• now will be (p, i1)
• Rule 3 still nothing needs to be done

19
Suffix Links

• Consider the two strings a and xa
• Suppose some internal node v of the tree is
  labeled with xa and another node s(v) in the tree
  is labeled with a
• Then the edge (v,s(v)) is a suffix link
• Do all internal nodes (the root is not considered
  an internal node) have suffix links?

20
Suffix Link Lemma

• If a new internal node v with path-label xa is
  added to the current tree in extension j of some
  phase i1, then
• the path labeled a already exists at an internal
  node of the tree or
• the internal node labeled a will be created in
  the extension of j1 or
• string a is empty and s(v) is the root

21
Proof of Suffix Link Lemma

• A new internal node is created only by extension
  rule 2
• This means there are two distinct suffixes of
  S1..i1 that start with xa
• xaS(i1) and xacb where c is not S(i1)
• This means there are two distinct suffixes of
  S1..i1 that start with a
• aS(i1) and acb where c is not S(i1)
• Thus, if a is not empty, a will label an internal
  node once extension j1 is processed which is the
  extension of a

22
Using suffix links to speed up location of Sj..i

• S1..i must end at a leaf since it is the
  longest string in implicit tree Ii
• Keep a pointer to this leaf in all cases and
  extend according to rule 1
• Locating Sj1..i from Sj..i which is at node
  w
• If w is an internal node, set v to w
• Otherwise, set v parent(w)
• If v is the root, you must traverse from root to
  find Sj1..i
• If not, go to s(v) and begin search for remaining
  portion of Sj..i from there
• Remaining portion is the label of the edge we
  traversed up
• (see figure 6.5 on page 100)

23
Skip/count Trick

• Problem Moving down from s(v) naively takes time
  proportional to the number of characters compared
• Solution
• At each node, only compare the first character in
  an edge label to the next character to be checked
• Then, use the number of characters on that edge
  to update search in constant time
• Running time is now proportional to the number of
  nodes in the path searched rather than the number
  of characters
• See Figure 6.6 on page 102

24
O(m2) argument

• node-depth of v number of nodes on path from
  root to node v
• Lemma For any suffix link (v, s(v)) traversed in
  Ukkonens algorithm, at that moment, nd(v) lt
  nd(s(v))1
• If xb is an ancestral internal node of v where b
  is not empty, then it has a suffix link to a node
  with path-label b
• See Figure 6.7 on page 103

25
O(m2) argument

• Lemma Any phase takes O(m) time with skip/count
  trick
• Proof
• Decrements to node depth at most 2m
• i1 lt m extensions per phase
• Walking up decreases node depth at most 1
• Suffix link traversal decreases node depth at
  most 1
• At most 3m downward edge traversal
• Max node depth is m
• None are negative
• Each downward traversal increases depth by at
  least 1

26
Observation

• Making Ukkonens algorithm O(m)
• Implicit computations of many extensions
• Need to take argument for a single phase and
  extend to multiple phases

27
Rule 3

• Suppose suffix extension rule 3 applies to
  Sj..i1.
• This means Sj..i1 already appears in the
  implicit suffix tree as a prefix of a larger
  suffix (and is thus a substring of S1..i)
• Then it applies to Sk..i1 for k gt j.
• Clearly, Sk..i1 must also be a substring of
  S1..i and thus must be in the tree
• Thus, stop a phase once the first application of
  rule 3 occurs
• All future rule 3 extensions are done implicitly,
  not explicitly

28
Implicit expansion of leaf nodes

• Once a leaf, always a leaf
• It will always be extended using rule 1
• Implicit expansion of leaf nodes
• When a leaf edge is created in phase i1, instead
  of labeling it with (p, i1), label it with (p,e)
• e is a global index that is set to i1 once in
  each phase
• In later phases, we will not need to explicitly
  extend this leaf but rather can implicitly extend
  it by incrementing e once in its global location
• How can we easily identify leaf nodes to avoid
  explicitly expanding them in later phases?

29
Avoiding leaf nodes

• For phase i, let last(i) denote the last
  extension of phase i that is not by rule 3
• Observation
• All the suffixes Sj..i for 1 lt j lt last(i)
  end at a leaf node
• All the extensions for 1 lt j lt last(i) in phase
  i1 and higher are rule 1 extensions by previous
  slide
• Therefore, last(i1) gt last(i)
• Trick
• In phase i1, only explicitly compute extensions
  for last(i)1 up till first rule 3 extension is
  found

30
Single phase algorithm

• Phase i1
• Increment e to i1 (implicitly extending all
  existing leaves)
• Explicitly compute successive extensions starting
  at last(i)1 and continuing until a rule 3
  extension or no more extensions needed
• Exact location is known given next step in
  previous phase
• Set last(i1) appropriately (last rule 1 or 2
  extension)
• Observation
• Phase i and i1 share at most 1 explicit
  extension
• In all phases, there will be at most 2m extensions

31
Visualization of explicit extensions

• S xabxacdefghixabcab
• Phase 1 extend x (rule 1)
• Phase 2 extend a (rule 2)
• Phase 3 extend b (rule 2)
• Phase 4 extend x (rule 3)
• Phase 5 extend x (rule 3)
• Phase 6 extend x (rule 2), extend a (rule 2)
  extend c (rule 2)
• Phase 7 extend d (rule 2)

32
O(m) argument

• Lemma All phases take O(m) time
• Proof
• Similar to previous node-depth argument
• Key new observation
• From one phase to the next, we either start at
  root or we work with same node we ended with last
  time so the node depth is identical
• With 2m total extensions at most, we get O(m)
  upward and downward traversals of links

33
Finishing up

• Convert final implicit suffix tree to a true
  suffix tree
• Add using just another phase of execution
• Now all suffixes will be leaves
• Replace e in every leaf edge with m
• Just requires a traversal of tree which is O(m)
  time