Suffix trees

Trie• A tree representing a set of strings.

ab

c

e

e

f

d b

f

e g

{ aeef ad bbfe bbfg c }

Trie (Cont)• Assume no string is a prefix of

anothera

bc

e

e

f

d b

f

e g

1) Each edge is labeled by a letter,

2) No two edges outgoing from the same node are labeled the same.

3) Each string corresponds to a leaf.

Compressed Trie • Compress unary nodes, label edges by

stringsa

bc

e

e

f

d b

f

e g

a

bbf

c

eefd

e g

Suffix tree Given a string s a suffix tree of s is a compressed trie of all suffixes of s

To make these suffixes prefix-free we add a special character, say $, at the end of s

Suffix tree (Example) Let s=abab, a suffix tree of s is a compressed trie of all suffixes of s=abab${

$ b$ ab$ bab$ abab$ }

ab

ab

$

ab$

b

$

$

$

Note that a suffix tree has O(n) nodes n = |s| (why?)

Trivial algorithm to build a Suffix tree

Put the largest suffix in

Put the suffix bab$ in

abab$

abab

$

ab$

b

Put the suffix ab$ in

ab

ab

$

ab$

b

ab

ab

$

ab$

b

$

Put the suffix b$ in

ab

ab

$

ab$

b

$

ab

ab

$

ab$

b

$

$

Put the suffix $ in

ab

ab

$

ab$

b

$

$

ab

ab

$

ab$

b

$

$

$

We will also label each leaf with the starting point of the corres. suffix.

ab

ab

$

ab$

b

$

$

$

12

ab

ab

$

ab

$

b

3

$ 4

$

5

$

AnalysisTakes O(n2) time to build.

It is possible to construct it in O(n) time

But, how come? does it take O(n) space ?

$

Linear space ? • Consider the string aaaaaabbbbbb

a

c

bbbbbb$

a b

a

a

a

b

b

b

b

b$

bbbbbb$

bbbbbb$

bbbbbb$

bbbbbb$

$

$

$

$

$

abbbbbb$

To use only O(n) space encode the edge-labels

a

c

bbbbbb$

a b

a

a

a

b

b

b

b

b$

bbbbbb$

bbbbbb$

bbbbbb$

(7,13)

$

$

$

$

$

• Consider the string aaaaaabbbbbb

$

abbbbbb$

To use only O(n) space encode the edge-labels

(1,1)

c

(7,13)

(7,13)

(7,13)

(7,13)

(7,13)

• Consider the string aaaaaabbbbbb

(6,13)

(1,1)

(1,1)

(1,1)

(1,1)

(13,13)

(7,7)

(13,13)

(13,13)

(13,13)

(12,13)

(13,13)

(7,7)

(7,7)

(7,7)

(7,7)

What can we do with it ?Exact string matching:Given a text T, |T| = n, preprocess it

such that when a pattern P, |P|=m, arrives you can quickly decide if it occurs in T.

We may also want to find all occurrences of P in T

Exact string matchingIn preprocessing we just build a suffix tree in O(n) time

12

ab

ab

$

ab$

b

3

$ 4

$

5

$

Given a pattern P = ab we traverse the tree according to the pattern.

12

ab

ab

$

ab$

b

3

$ 4

$

5

$

If we did not get stuck traversing the pattern then the pattern occurs in the text. Each leaf in the subtree below the node we reach corresponds to an occurrence.By traversing this subtree we get all k occurrences in O(m+k) time

Generalized suffix tree Given a set of strings S a generalized suffix tree of S is a compressed trie of all suffixes of s S

To associate each suffix with a unique string in S add a different special char to each s

Generalized suffix tree (Example)

Let s1=abab and s2=aab here is a generalized suffix tree for s1 and s2

{ $ # b$ b# ab$ ab# bab$ aab# abab$ }

1

2

a

b

ab

$

ab$

b

3

$

4

$

5$

1

b#

a

2

#

3

#4

#

So what can we do with it ? Matching a pattern against a database of strings

Longest common substring (of two strings)Every node with a leaf

descendant from string s1 and a leaf descendant from string s2 represents a common substring.

A maximal common substring corresponds to such node.

1

2

a

b

ab

$

ab$

b

3

$

4

$

5$

1

b#

a

2

#

3

#4

#

Find such node with largest “string depth”

Lowest common ancetorsA lot more can be gained from the suffix tree if we preprocess it so that we can answer LCA queries on it

Why?The LCA of two leaves represents the longest common prefix (LCP) of these 2 suffixes

1

2

a

b

ab

$

ab$

b

3

$

4

$

5$

1

b#

a

2

#

3

#4

#

Lowest common ancestors

Write an Euler tour of the tree3

4

51

672

4 1 7 1 3 5 6 5 33 1231

2

3

46

9

10

8

57

11

12

0

LCA(1,5) = 3

Shallowest node

3

4

51

672

4 1 7 1 3 5 6 5 33 123

1

2

3

46

9

10

8

57

11

12

0

2 1 2 1 0 1 2 1 00 110

minimum

Node id

Level

Range minimum3

4

51

672

4 1 7 1 3 5 6 5 33 123

1

2

3

46

9

10

8

57

11

12

0

2 1 2 1 0 1 2 1 00 110

minimum

Preprocess an array, such that given i,j you can find the minimum in [i,j] fastReduction takes linear time

Trivial algorithms for RMQ• O(n) space, O(n) query time

• O(n2) space, O(1) query time

Less trivial algorithms to RMQ

• Try to use O(nlog(n)) space to do a query in O(1) time..

• Simpler: O(n + sqrt(n)) to do a query in sqrt(n)

Lowest common ancetorsA lot more can be gained from the suffix tree if we preprocess it so that we can answer LCA queries on it

Why?The LCA of two leaves represents the longest common prefix (LCP) of these 2 suffixes

1

2

a

b

ab

$

ab$

b

3

$

4

$

5$

1

b#

a

2

#

3

#4

#

Finding maximal palindromes

• A palindrome: caabaac, cbaabc• Want to find all maximal palindromes

in a string s

Let s = cbaaba

The maximal palindrome with center between i-1 and i is the LCP of the suffix at position i of s and the suffix at position n-i+2 of sr

Maximal palindromes algorithm

Prepare a generalized suffix tree for s = cbaaba$ and sr = abaabc#

For every i, find the LCA of suffix i of s and suffix n-i+2 of sr

3

a

a b

a

baaba$

b

3

$

7$

b

7

#c

1

6a b

c #

5

2 2

a $

c #

a

5

6

$

4

4

1

c #

a $

$abc #

c #

Let s = cbaaba$ then sr = abaabc#

AnalysisO(n) time to identify all palindromes

Suffix trees for compression

aabcbbabbsb (2,5) bbabbabcbbabbabbsb…S[1..n] =

Compression aabcbbabbsbabcbbbbabbabcbbabbabbsb…

aabcbbabbsb (2,5) bbabbabcbbabbabbsb…S[1..n] =

Let priori=S[i..i+Li-1] = S[si..si+Li-1] be this prefix

Compression

Let Li be the length of the longest prefix of S[i..n] that is a substring of S[1..i-1]Let si the previous position of this prefix

aabcbbabbsbabcbbbbabbabcbbabbabbsb…

aabcbbabbsb (2,5) bbabbabcbbabbabbsb…

S[1..n]=

Compressionprior12

(s12,L12)

i=12

aabcbbabbsbabcbbbbabbabcbbabbabbsb…

For (i=1 ; i<=n ;;){Compute (si,Li) if Li>1 {output(si,Li); i=i+Li}; else {output(S[i]); i=i+1}}

aabcbb

(2,2)

bs

(6,4)(? si,Li) איך לחשב את

“Ziv-Lempel” compression

aabcbbabbsbabbbbbabbabcbbabbabbsb…

Implementation using suffix tree

Before compression:

• Build a suffix tree T for S.

• For each node v, compute cv : – the smallest leaf index in v’s subtree.– the starting position of the leftmost copy of the

substring that labels the path from the root to v.

• O(n) time.

leaf i

root

S[i...n]vstring-len)v( + cv ≥ i

computing (si,Li):

icv

string-len)v(

Let v be the “first” node such that:

set si ← cv

set Li ← i-cv

v2

v1

Example

S = abababab 1 2 3 4 5 6 7 8

1

1

1

12

2

2

a

a

a

a

a

a

a

b

b

b

bb

b

b

b

$

$

$

$

$$

$ $18 76 54 32

i=1 Li=0 a

i=2 Li=0 b

i=3 Li=2 si=1 )1,2(

i=5 Li=4 si=1 )1,4(

Suffix array• We loose some of the functionality

but we save space.Let s = ababSort the suffixes lexicographically: ab, abab, b, babThe suffix array gives the indices of the suffixes in sorted order

2 0 3 1

How do we build it ?• Build a suffix tree• Traverse the tree in in-order,

lexicographically picking edges outgoing from each node and fill the suffix array.

• O(n) time• Can also build it directly

ExampleLet S = mississippi

iippiissippiississippimississippipi

7

4

1

0

9

8

6

3

10

5

2

ppisippisisippissippississippi

L

R

Let P = issa

M

How do we search for a pattern ?

• If P occurs in T then all its occurrences are consecutive in the suffix array.

• Do a binary search on the suffix array

• Takes O(mlogn) time• Can also do it in O(m+log(n)) with an

additional array

Computing the suffix array• Can do it in linear time without

constructing the suffix tree ?

Divide into triples

$ y a b b a d a b b a d o

abb ada bba do$

Divide into triples

$ y a b b a d a b b a d o

abb ada bba do$

$ y a b b a d a b b a d o

bba dab bad o$$

Radix sort triplets

$ y a b b a d a b b a d o

abb ada bba do$

$ y a b b a d a b b a d o

bba dab bad o$$

abb

ada

bba

do$

bad

dab

o$$

Change the alphabet

$ y a b b a d a b b a d o

abb ada bba do$

$ y a b b a d a b b a d o

bba dab bad o$$

abb

ada

bba

do$

bad

dab

o$$

1

2

3

4

5

6

7

1 2 4 6

4 5 3 7

Sort the suffixes

$ y a b b a d a b b a d o

abb ada bba do$

$ y a b b a d a b b a d o

bba dab bad o$$

1 2 4 6

4 5 3 7

1 2 4 62 4 64 664 5 3 75 3 73 77

Recursively (problem size decreases to 2/3 the original)

abb ada bba do$ bba dab bad o$$

1 2 4 6 4 5 3 7

0 1 2 3 4 5 6 7

0 - 123645371 - 24645372 - 4645373 - 645374 - 45375 - 5376 - 377 - 7

0 - 123645371 – 24645376 – 374 - 45372 - 4645375 – 5373 - 645377 - 7

Go back to original problem abb ada bba do$ bba dab bad o$$

1 2 4 6 4 5 3 70 - 123645371 - 24645372 - 4645373 - 645374 - 45375 - 5376 - 377 - 7

0 - 123645371 – 24645376 – 374 - 45372 - 4645375 – 5373 - 645377 - 7

$ y a b b a d a b b a d o

1 2 3 4 5 6 7 8 9 10 11 120

0/1 1/4 2/7 3/10 4/2 5/5 6/8 7/11

Go back to original problem abb ada bba do$ bba dab bad o$$

1 2 4 6 4 5 3 70 - 123645371 - 24645372 - 4645373 - 645374 - 45375 - 5376 - 377 - 7

0 - 123645371 – 24645376 – 374 - 45372 - 4645375 – 5373 - 645377 - 7

$ y a b b a d a b b a d o

1 2 3 4 5 6 7 8 9 10 11 120

0/1 1/4 2/7 3/10 4/2 5/5 6/8 7/11

1 4 2 6 5 3 7 8

Sort the remaining third

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

(b, 2) (a, 5) (a, 7)(y, 1)

(b, 2)(a, 5) (a, 7) (y, 1)

36 9 0

1 2 3 4 5 6 7 8 9 10 11 120

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

(b, 2) (a, 5) (a, 7)(y, 1)

(b, 2)(a, 5) (a, 7) (y, 1)

36 9 0

1 2 3 4 5 6 7 8 9 10 11 120

1 4 8 2 7 5 10 11

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

36 9 0

1 4 8 2 7 5 10 11

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

36 9 0

4 8 2 7 5 10 11

6

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

39 0

4 8 2 7 5 10 11

6 4

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

39 0

8 2 7 5 10 11

6 4 9

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

3 0

8 2 7 5 10 11

6 4 9 3

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

0

8 2 7 5 10 11

6 4 9 3 8

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

0

2 7 5 10 11

6 4 9 3 8 2

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

0

7 5 10 11

6 4 9 3 8 2 7

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

0

5 10 11

6 4 9 3 8 2 7 5

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

0

10 11

6 4 9 3 8 2 7 5

Merge

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

6 4 9 3 8 2 7 5 10 11 0

summary

$ y a b b a d a b b a d o1 4 2 6 5 3 7 8

1 2 3 4 5 6 7 8 9 10 11 12

1

0

6 4 9 3 8 2 7 5 10 11 0

When comparing to a suffix with index 1 (mod 3) we compare the char and break ties by the ranks of the following suffixes

When comparing to a suffix with index 2 (mod 3) we compare the char, the next char if there is a tie, and finally the ranks of the following suffixes