Suffix Trees

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Suffix Trees
Construction and Applications
João Carreira 2008
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Outline

• Why Suffix Trees?
• Definition
• Ukkonen's Algorithm (construction)?
• Applications

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Why Suffix Trees?
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Why Suffix Trees?

• Asymptotically fast.

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Why Suffix Trees?

• Asymptotically fast.
• The basis of state of the art data structures.

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Why Suffix Trees?

• Asymptotically fast.
• The basis of state of the art data structures.
• You don't need a Phd to use them.

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Why Suffix Trees?

• Asymptotically fast.
• The basis of state of the art data structures.
• You don't need a Phd to use them.
• Challenging.

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Why Suffix Trees?

• Asymptotically fast.
• The basis of state of the art data structures.
• You don't need a Phd to use them.
• Challenging.
• Expose interesting algorithmic ideas.

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Definition
Suffix Tree for an m-character string

• m leaves numbered 1 to m

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Definition
Suffix Tree for an m-character string

• m leaves numbered 1 to m
• edge-label vs node-label

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Definition
Suffix Tree for an m-character string

• m leaves numbered 1 to m
• edge-label vs node-label
• each internal node has at least two children

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Definition
Suffix Tree for an m-character string

• m leaves numbered 1 to m
• edge-label vs node-label
• each internal node has at least two children
• the label of the leaf j is S j..m

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Definition
Suffix Tree for an m-character string

• m leaves numbered 1 to m
• edge-label vs node-label
• each internal node has at least two children
• the label of the leaf j is S j..m
• no two edges out of the same node can have
  edge-labels
• beginning with the same character

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Definition Example
String xabxac Length (m) 6 characters Number of
Leaves 6 Node 5 label ac
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Implicit vs Explicit

• What if we have axabx ?

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Ukkonen's Algorithm
suffix tree construction
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Ukkonen's Algorithm
suffix tree construction

• Text S 1..m
• m phases
• phase j is divided into j extensions

• In extension j of phase i 1
• find the end of the path from the root labeled
  with substring S j..i
• extend the substring by adding the character S(i
  1) to its end

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Extension Rules

• Rule 1 Path ß ends at a leaf. S(i 1) is added
  to the end of the label on that leaf edge.

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Extension Rules

• Rule 2 No path from the end of ß starts with
  S(i 1), but at least one labeled path
  continues from the end of ß.

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Extension Rules

• Rule 3 Some path from the end of ß starts with
  S(i 1), so we do nothing.

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Ukkonen's Algorithm
suffix tree construction
Complexity
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Ukkonen's Algorithm
suffix tree construction
Complexity

• m phases

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Ukkonen's Algorithm
suffix tree construction
Complexity

• m phases
• phase j -gt j extensions

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Ukkonen's Algorithm
suffix tree construction
Complexity

• m phases
• phase j -gt j extensions
• find the end of the path of substring ß
  O(ß) O(m)?

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Ukkonen's Algorithm
suffix tree construction
Complexity

• m phases
• phase j -gt j extensions
• find the end of the path of substring ß
  O(ß) O(m)?
• each extension O(1)?

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Ukkonen's Algorithm
suffix tree construction
Complexity

• m phases
• phase j -gt j extensions
• find the end of the path of substring ß
  O(ß) O(m)?
• each extension O(1)?

O(m3)?
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First make it run, then make it run fast.
Brian Kernighan
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Suffix Links
Definition

• For an internal node v with path-label xa, if
  there is another node s(v), with
• path-label a, then a pointer from v to s(v) is
  called a suffix link.

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Suffix Links
Lemma

• If a new internal node v with path label xa is
  added to the current tree in extension
• j of some phase, then either the path labeled a
  already ends at an internal node
• or an internal at the end of the string a will be
  created in the next extension
• of the same phase.

If Rule 2 applies
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Suffix Links
Lemma

• If a new internal node v with path label xa is
  added to the current tree in extension
• j of some phase, then either the path labeled a
  already ends at an internal node
• or an internal at the end of the string a will be
  created in the next extension
• of the same phase.

• If Rule 2 applies
• S j..i continues with c ? S(i 1)?

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Suffix Links
Lemma

• If a new internal node v with path label xa is
  added to the current tree in extension
• j of some phase, then either the path labeled a
  already ends at an internal node
• or an internal at the end of the string a will be
  created in the next extension
• of the same phase.

• If Rule 2 applies
• S j..i continues with c ? S(i 1)?
• S j 1..i continues with c.

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Single Extension Algorithm
Extension j of phase i 1
1. Find the first node v at or above the end of
S j - 1..i that either has a suffix link
from it or is the root. Let ? denote the string
between v and the end of S j 1..i .
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Single Extension Algorithm
Extension j of phase i 1
1. Find the first node v at or above the end of
S j - 1..i that either has a suffix link
from it or is the root. Let ? denote the string
between v and the end of S j 1..i .
2. If v is the root, follow the path for S j..i
(as in the naive algorithm). Else traverse the
suffix link and walk down from s(v) following
the path for string ?.
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Single Extension Algorithm
Extension j of phase i 1
1. Find the first node v at or above the end of
S j - 1..i that either has a suffix link
from it or is the root. Let ? denote the string
between v and the end of S j 1..i .
2. If v is the root, follow the path for S j..i
(as in the naive algorithm). Else traverse the
suffix link and walk down from s(v) following
the path for string ?.
3. Using the extension rules, ensure that the
string S j..i S(i1) is in the tree.
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Single Extension Algorithm
Extension j of phase i 1
1. Find the first node v at or above the end of
S j - 1..i that either has a suffix link
from it or is the root. Let ? denote the string
between v and the end of S j 1..i .
2. If v is the root, follow the path for S j..i
(as in the naive algorithm). Else traverse the
suffix link and walk down from s(v) following
the path for string ?.
3. Using the extension rules, ensure that the
string S j..i S(i1) is in the tree.
4. If a new internal w was created in extension j
1 (by rule 2), then string a must end at
node s(w), the end node for the suffix link from
w. Create the suffix link (w, s(w)) from w to
s(w).
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Node Depth
The node-depth of v is at most one greater than
the node depth of s(v).
xß
xß
ß
ß
xa
a
xa
a
x?
x?
?
?
Node depth 4
Node depth 3
equal node-depth 3
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Skip/count Trick

• ? number of characters in an edge
• Directly implemented edge traversal O(?)?

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Skip/count Trick

• ? number of characters in an edge
• Directly implemented edge traversal O(?)?

• Jump from node to node.
• K number of nodes in a path
• Time to traverse a path O(K)?

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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof
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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof

• There are i 1 m extensions in phase i 1

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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof

• There are i 1 m extensions in phase i 1
• In a single extension, the algorithm walks up at
  most one edge, traverses one suffix link,
• walks down some number of nodes, applies the
  extension rules and may add a suffix link.

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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof

• There are i 1 m extensions in phase i 1
• In a single extension, the algorithm walks up at
  most one edge, traverses one suffix link,
• walks down some number of nodes, applies the
  extension rules and may add a suffix link.
• The up-walk decreases the current node-depth by
  at most one.

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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof

• There are i 1 m extensions in phase i 1
• In a single extension, the algorithm walks up at
  most one edge, traverses one suffix link,
• walks down some number of nodes, applies the
  extension rules and may add a suffix link.
• The up-walk decreases the current node-depth by
  at most one.
• Each suffix link traversal decreases the
  node-depth by at most another one.

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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof

• There are i 1 m extensions in phase i 1
• In a single extension, the algorithm walks up at
  most one edge, traverses one suffix link,
• walks down some number of nodes, applies the
  extension rules and may add a suffix link.
• The up-walk decreases the current node-depth by
  at most one.
• Each suffix link traversal decreases the
  node-depth by at most another one.
• Each down-walk moves to a node of greater depth.

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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof

• There are i 1 m extensions in phase i 1
• In a single extension, the algorithm walks up at
  most one edge, traverses one suffix link,
• walks down some number of nodes, applies the
  extension rules and may add a suffix link.
• The up-walk decreases the current node-depth by
  at most one.
• Each suffix link traversal decreases the
  node-depth by at most another one.
• Each down-walk moves to a node of greater depth.
• Over the entire phase the node-depth is
  decremented at most 2m times.

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Ukkonen's Algorithm

• Using the skip/count trick
• any phase of Ukkonen's algorithm takes O(m) time.

Proof

• There are i 1 m extensions in phase i 1
• In a single extension, the algorithm walks up at
  most one edge, traverses one suffix link,
• walks down some number of nodes, applies the
  extension rules and may add a suffix link.
• The up-walk decreases the current node-depth by
  at most one.
• Each suffix link traversal decreases the
  node-depth by at most another one.
• Each down-walk moves to a node of greater depth.
• Over the entire phase the node-depth is
  decremented at most 2m times.
• No node can have depth greater than m, so the
  total increment to current node-depth
• (down walks) is bounded by 3m over the entire
  phase.

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Ukkonen's Algorithm

• m phases
• 1 phase O(m)?

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Ukkonen's Algorithm

• m phases
• 1 phase O(m)?

O(m2)?
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First make it run fast, then make it run faster.
João Carreira
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Edge-Label Compression

• A string with m characters has m suffixes.

• If edge labels are represented with characters,
  O(m2) space is needed.

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Edge-Label Compression

• A string with m characters has m suffixes.

• If edge labels are represented with characters,
  O(m2) space is needed.

To achieve O(m) space, each edge-label
(p, q)?
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Two more tricks...
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Rule 3 is a show stopper
If rule 3 applies in extension j, it will also
apply in all further extensions until the end of
the phase.
Why?
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Rule 3 is a show stopper
If rule 3 applies in extension j, it will also
apply in all further extensions until the end of
the phase.
Why?

• When rule 3 applies, the path labeled S j..i
  must continue with character S(i 1), and
• so the path labeled S j 1..i does also,
  and rule 3 again applies in extensions j1...i1.

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Rule 3 is a show stopper

• End any phase i 1 the first time rule 3 applies.

• The remaining extensions are said to be done
  implicitly.

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Once a leaf always a leaf

• Leaf created gt always a leaf in all successive
  trees.
• No mechanism for extending a leaf edge beyond
  its current leaf.

• Once there is a leaf labeled j, extension rule 1
  will always apply to extension j
• in any sucessive phase.

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Once a leaf always a leaf

• Leaf created gt always a leaf in all successive
  trees.
• No mechanism for extending a leaf edge beyond
  its current leaf.

• Once there is a leaf labeled j, extension rule 1
  will always apply to extension j
• in any sucessive phase.

(p, e)?
Leaf Edge Label
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Single Phase Algorithm
In each phase i
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Single Phase Algorithm
During construction
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Implicit to Explicit
One last phase to add character O(m)?
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Suffix Trees are a Swiss Knife
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Applications
Exact String Matching
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Applications
Exact String Matching
Preprocessing O(m)?
Search O(n k)?
Three ocurrences of string aw.
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Applications
And much more..

• Longest common substring O(n)?
• Longest repeated substring O(n)?
• Longest palindrome O(n)?
• Most frequently occurring substrings of a
  minimum length O(n)?
• Shortest substrings occurring only
  once O(n)?
• Lempel-Ziv decomposition O(n)?
• .....

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Biology easily has 500 years of exciting
problems to work on.
Donald Knuth
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web.ist.utl.pt/joao.carreira
Questions?