Time-Space Trade-Offs for Longest Common Extensions Philip Bille 1 , - - PowerPoint PPT Presentation

Time-Space Trade-Offs for Longest Common Extensions

Philip Bille1, Inge Li Gørtz1, Benjamin Sach2, and Hjalte Wedel Vildhøj1

1Technical University of Denmark, DTU Informatics, {phbi,ilg,hwvi}@imm.dtu.dk 2University of Warwick, Department of Computer Science, sach@dcs.warwick.ac.uk

CPM 2012, Helsinki July 4, 2012

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The Longest Common Extension Problem

Definition

Problem: Preprocess a string T of length n to support LCE queries:

◮ LCE(i, j) = The length of the longest common prefix of the suffixes

starting at position i and j in T. Example

T = b a n a n a s

LCE(2, 4) = ?

1 2 3 4 5 6 7

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The Longest Common Extension Problem

Definition

Problem: Preprocess a string T of length n to support LCE queries:

◮ LCE(i, j) = The length of the longest common prefix of the suffixes

starting at position i and j in T. Example

T = b a n a n a s

LCE(2, 4) = ?

a n a n a s a n a s

1 2 3 4 5 6 7

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The Longest Common Extension Problem

Definition

Problem: Preprocess a string T of length n to support LCE queries:

◮ LCE(i, j) = The length of the longest common prefix of the suffixes

starting at position i and j in T. Example

T = b a n a n a s

LCE(2, 4) = 3

a n a n a s a n a s

1 2 3 4 5 6 7

4 / 56

The Longest Common Extension Problem

Definition

Problem: Preprocess a string T of length n to support LCE queries:

◮ LCE(i, j) = The length of the longest common prefix of the suffixes

starting at position i and j in T. Example

T = b a n a n a s

LCE(2, 5) = 0

a n a n a s n a s

1 2 3 4 5 6 7

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The Longest Common Extension Problem

Definition

Problem: Preprocess a string T of length n to support LCE queries:

◮ LCE(i, j) = The length of the longest common prefix of the suffixes

starting at position i and j in T. Example

T = b a n a n a s

LCE(2, 5) = 0

a n a n a s n a s

1 2 3 4 5 6 7 ◮ We assume that the input is given in read-only memory and is not

included in the space complexity.

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j LCE(i, j) =

1 2 3 4 5 6 7

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j LCE(i, j) = 1

1 2 3 4 5 6 7

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j LCE(i, j) = 2

1 2 3 4 5 6 7

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j LCE(i, j) = 3

1 2 3 4 5 6 7

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j Time: O(n) Space: O(1) LCE(i, j) = 3

1 2 3 4 5 6 7

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j Time: O(n) Space: O(1) LCE(i, j) = 3

1 2 3 4 5 6 7

#2: Store the suffix tree

2

nas

4

s na

6

s a

1

bananas

3

nas

5

s na

7

s

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j Time: O(n) Space: O(1) LCE(i, j) = 3

1 2 3 4 5 6 7

#2: Store the suffix tree

NCA(2, 4) 2

nas

4

s na

6

s a

1

bananas

3

nas

5

s na

7

s

LCE(i, j) = |NCA(i, j)| = 3

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j Time: O(n) Space: O(1) LCE(i, j) = 3

1 2 3 4 5 6 7

#2: Store the suffix tree

NCA(2, 4) 2

nas

4

s na

6

s a

1

bananas

3

nas

5

s na

7

s

Time: O(1) Space: O(n) LCE(i, j) = |NCA(i, j)| = 3

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Two Simple Solutions

#1: Store nothing

T = b a n a n a s

i j Time: O(n) Space: O(1) LCE(i, j) = 3

1 2 3 4 5 6 7

#2: Store the suffix tree

NCA(2, 4) 2

nas

4

s na

6

s a

1

bananas

3

nas

5

s na

7

s

Time: O(1) Space: O(n) LCE(i, j) = |NCA(i, j)| = 3

Trade-off?

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Our Results

Time: O (n) Space: O (1) Time: O (1) Space: O (n)

Trade-off? Less space Faster

Store nothing Store suffix tree

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Our Results

Time: O (n) Space: O (1) Time: O

• τ log
• LCE(i,j)

τ

• Space:

O n

τ

• Time:

O (τ) Space: O

• n

√τ

• Time:

O (1) Space: O (n)

Trade-off? Less space Faster

Randomized Deterministic Store nothing Store suffix tree

Trade-off parameter τ, 1 ≤ τ ≤ n

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i j

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i j

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i j

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i j

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i j

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i j

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i j

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Difference Covers

A difference cover modulo τ is a set of integers D ⊆ {0, 1, . . . , τ − 1} such that for any distance d ∈ {0, 1, . . . , τ − 1}, D contains two elements separated by distance d modulo τ. Ex: The set D = {1, 2, 4} is a difference cover modulo 5. d 1 2 3 4 i, j 1, 1 2, 1 1, 4 4, 1 1, 2

1 2 4 3 1 4 2 3

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

D D D D

Difference Covers

A difference cover modulo τ is a set of integers D ⊆ {0, 1, . . . , τ − 1} such that for any distance d ∈ {0, 1, . . . , τ − 1}, D contains two elements separated by distance d modulo τ. Ex: The set D = {1, 2, 4} is a difference cover modulo 5. d 1 2 3 4 i, j 1, 1 2, 1 1, 4 4, 1 1, 2

1 2 4 3 1 4 2 3

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A Deterministic Solution

Idea: Store a subset of the n suffixes in a compacted trie.

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

D D D D

Lemma (Colbourn and Ling1)

For any τ, a difference cover modulo τ of size at most √ 1.5τ + 6 can be computed in O(√τ) time.

Analysis

Time: O(τ) Space: O(#stored suffixes) = O n

τ |D|

• = O
• n

√τ

• 1C. J. Colbourn and A. C. Ling. Quorums from difference covers. Inf. Process. Lett.

75(1-2):9–12, 2000

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A Randomized Solution (Monte Carlo)

Rabin-Karp Fingerprints

Let p be a sufficiently large prime and choose b ∈ Zp uniformly at random. φ(S) =

|S|

• k=1

S[k]bk mod p .

T = d b c a a b c a b c a a b c a c

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

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A Randomized Solution (Monte Carlo)

Rabin-Karp Fingerprints

Let p be a sufficiently large prime and choose b ∈ Zp uniformly at random. φ(S) =

|S|

• k=1

S[k]bk mod p .

T = d b c a a b c a b c a a b c a c = 3 1 2 0 0 1 2 0 1 2 0 0 1 2 0 2

φ(T[2 . . . 7]) = 1b1 + 2b2 + 0b3 + 0b4 + 1b5 + 2b6 mod p

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

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A Randomized Solution (Monte Carlo)

Rabin-Karp Fingerprints

Let p be a sufficiently large prime and choose b ∈ Zp uniformly at random. φ(S) =

|S|

• k=1

S[k]bk mod p .

T = d b c a a b c a b c a a b c a c = 3 1 2 0 0 1 2 0 1 2 0 0 1 2 0 2

φ(T[2 . . . 7]) = 1b1 + 2b2 + 0b3 + 0b4 + 1b5 + 2b6 mod p

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Crucial property: With high probability φ is collision-free on substrings of T, i.e., φ(S1) = φ(S2) iff S1 = S2.

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A Randomized Solution (Monte Carlo)

Rabin-Karp Fingerprints

Let p be a sufficiently large prime and choose b ∈ Zp uniformly at random. φ(S) =

|S|

• k=1

S[k]bk mod p .

T = d b c a a b c a b c a a b c a c = 3 1 2 0 0 1 2 0 1 2 0 0 1 2 0 2

φ(T[2 . . . 7]) = 1b1 + 2b2 + 0b3 + 0b4 + 1b5 + 2b6 mod p

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Crucial property: With high probability φ is collision-free on substrings of T, i.e., φ(S1) = φ(S2) iff S1 = S2. Also important: φ(T[i . . . j + 1]) can be computed from φ(T[i . . . j]) in O(1) time.

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A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

S

φ

′′

φ

′

Observation: If S is block aligned we can compute φ(S) in O(1) time. Otherwise, the time needed is O(τ).

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A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

S

φ

′′

φ

′

Observation: If S is block aligned we can compute φ(S) in O(1) time. Otherwise, the time needed is O(τ).

35 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

36 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

37 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• 38 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• 39 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• 40 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• 41 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• ×

×

42 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• ×

×

• 43 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• ×

×

• ×

×

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A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• ×

×

• ×

×

• 45 / 56

A Randomized Solution (Monte Carlo)

How to answer a query

Idea: Store fingerprints of suffixes starting at every τ’th position in T. T =

Blocks of τ chars

i j

• ×

×

• ×

×

• Analysis

Time: Only O(log( LCE

τ )) fingerprint comparisons each taking time O(τ).

Hence query time O

• τ log

LCE

τ

• .

Space: O n

τ

• .

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A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing?

47 / 56

A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing? Challenge: Doing this quickly while using O( n

τ ) space.

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A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing? Challenge: Doing this quickly while using O( n

τ ) space.

Observation: Whenever we compare two fingerprints, we can ensure that one of them is of the form T[jτ . . . jτ + τ · 2ℓ − 1] for some ℓ, j

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A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing? Challenge: Doing this quickly while using O( n

τ ) space.

Observation: Whenever we compare two fingerprints, we can ensure that one of them is of the form T[jτ . . . jτ + τ · 2ℓ − 1] for some ℓ, j . . . this cuts down the number of fingerprints we need to check!

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A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing? Challenge: Doing this quickly while using O( n

τ ) space.

Observation: Whenever we compare two fingerprints, we can ensure that one of them is of the form T[jτ . . . jτ + τ · 2ℓ − 1] for some ℓ, j . . . this cuts down the number of fingerprints we need to check! General idea: For each ℓ ≥ 0 in increasing order, check that for all i, j, φ(T[i . . . i + τ·2ℓ − 1]) = φ(T[jτ . . . jτ + τ·2ℓ − 1]) iff T[i . . . i + τ·2ℓ − 1] = T[jτ . . . jτ + τ·2ℓ − 1] T =

φ(T[i . . . i + τ · 2ℓ − 1])

T =

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A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing? Challenge: Doing this quickly while using O( n

τ ) space.

Observation: Whenever we compare two fingerprints, we can ensure that one of them is of the form T[jτ . . . jτ + τ · 2ℓ − 1] for some ℓ, j . . . this cuts down the number of fingerprints we need to check! General idea: For each ℓ ≥ 0 in increasing order, check that for all i, j, φ(T[i . . . i + τ·2ℓ − 1]) = φ(T[jτ . . . jτ + τ·2ℓ − 1]) iff T[i . . . i + τ·2ℓ − 1] = T[jτ . . . jτ + τ·2ℓ − 1] T =

φ(T[i . . . i + τ · 2ℓ − 1]) = φ(T[jτ . . . jτ + τ · 2ℓ − 1])

T =

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A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing? Challenge: Doing this quickly while using O( n

τ ) space.

Observation: Whenever we compare two fingerprints, we can ensure that one of them is of the form T[jτ . . . jτ + τ · 2ℓ − 1] for some ℓ, j . . . this cuts down the number of fingerprints we need to check! General idea: For each ℓ ≥ 0 in increasing order, check that for all i, j, φ(T[i . . . i + τ·2ℓ − 1]) = φ(T[jτ . . . jτ + τ·2ℓ − 1]) iff T[i . . . i + τ·2ℓ − 1] = T[jτ . . . jτ + τ·2ℓ − 1] T =

φ(T[i . . . i + τ · 2ℓ − 1]) = φ(T[jτ . . . jτ + τ · 2ℓ − 1]) φ(T[i . . . i + τ · 2ℓ−1 − 1])

?

= φ(T[jτ . . . jτ + τ · 2ℓ−1 − 1])

T =

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A Randomized Solution (Las Vegas)

Question: Can we verify that φ is collision free during preprocessing? Challenge: Doing this quickly while using O( n

τ ) space.

Observation: Whenever we compare two fingerprints, we can ensure that one of them is of the form T[jτ . . . jτ + τ · 2ℓ − 1] for some ℓ, j . . . this cuts down the number of fingerprints we need to check! General idea: For each ℓ ≥ 0 in increasing order, check that for all i, j, φ(T[i . . . i + τ·2ℓ − 1]) = φ(T[jτ . . . jτ + τ·2ℓ − 1]) iff T[i . . . i + τ·2ℓ − 1] = T[jτ . . . jτ + τ·2ℓ − 1] T =

φ(T[i . . . i + τ · 2ℓ − 1]) = φ(T[jτ . . . jτ + τ · 2ℓ − 1]) φ(T[i + τ · 2ℓ−1 . . . i + τ · 2ℓ − 1])

?

= φ(T[jτ + τ · 2ℓ−1 . . . jτ + τ · 2ℓ − 1])

T =

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Conclusions

We gave three time-space trade-offs for LCE on a single string:

◮ A deterministic solution

◮ O(τ) query time ◮ O(n/√τ) space (even during preprocessing) ◮ O(n2/√τ) preprocessing time

◮ A Monte-Carlo solution

◮ O (τ log (LCE(i, j)/τ)) query time (correct with high prob.) ◮ O(n/τ) space (even during preprocessing) ◮ O(n) preprocessing time.

◮ A Las-Vegas solution

◮ O (τ log (LCE(i, j)/τ)) query time (correct with certainty) ◮ O(n/τ) space (even during preprocessing) ◮ O(n log n) preprocessing time with high prob. 55 / 56

Conclusions

We gave three time-space trade-offs for LCE on two strings:

◮ A deterministic solution

◮ O(τ) query time ◮ O(n/τ + m/√τ) space (even during preprocessing) ◮ O(nm/√τ) preprocessing time

◮ A Monte-Carlo solution

◮ O (τ log (LCE(i, j)/τ)) query time (correct with high prob.) ◮ O((n + m)/τ) space (even during preprocessing) ◮ O(n) preprocessing time.

◮ A Las-Vegas solution

◮ O (τ log (LCE(i, j)/τ)) query time (correct with certainty) ◮ O((n + m)/τ) space (even during preprocessing) ◮ O(n log n) preprocessing time with high prob. 56 / 56