Time-Space Trade-Offs for the Longest Common Substring Problem - - PowerPoint PPT Presentation
Time-Space Trade-Offs for the Longest Common Substring Problem Tatiana Starikovskaya 1 and Hjalte Wedel Vildhj 2 1 Moscow State University, Department of Mechanics and Mathematics, tat.starikovskaya@gmail.com 2 Technical University of Denmark,
Tatiana Starikovskaya1 and Hjalte Wedel Vildhøj2
1Moscow State University, Department of Mechanics and Mathematics,
tat.starikovskaya@gmail.com
2Technical University of Denmark, DTU Compute, hwv@hwv.dk
CPM 2013, Bad Herrenalb, Germany June 17, 2013
1 / 27
Definition
Problem: Given T1, T2, . . . , Tm of total length n. Compute the longest substring, which appears in at least 2 ≤ d ≤ m strings. Example
T1 = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
T2 = a c a c c t a c c c t a g T3 = a c t a g t a a t g c a t
2 / 27
Definition
Problem: Given T1, T2, . . . , Tm of total length n. Compute the longest substring, which appears in at least 2 ≤ d ≤ m strings. Example
T1 = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
T2 = a c a c c t a c c c t a g T3 = a c t a g t a a t g c a t d = 3 ⇒ LCS = c t a g
3 / 27
Definition
Problem: Given T1, T2, . . . , Tm of total length n. Compute the longest substring, which appears in at least 2 ≤ d ≤ m strings. Example
T1 = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
T2 = a c a c c t a c c c t a g T3 = a c t a g t a a t g c a t d = 3 ⇒ LCS = c t a g d = 2 ⇒ LCS = c t a c c
4 / 27
A patented solution
5 / 27
1
acctaccctag$
7
c t a g $
10
$
3
a c c c t a g $ t c c
12
$
6
ctacct$
1
gctagctacct$ g a
2
a c c t a c c c t a g $
8
ctag$
11
$
4
ccctag$
9
g$ a t c
12
$
5
ctag$
8
t$ c c
10
ccctag$
4
g$ g a t c
13
$
7
cct$
3
gctacct$ cta
2
gctagctacct$ g
13
$
6
ctag$
9
t$ cc
11
ccctag$
5
g$ g a t
Build Generalized Suffix Tree
T1 = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
T2 = a c a c c t a c c c t a g
6 / 27
1
acctaccctag$
7
c t a g $
10
$
3
a c c c t a g $ t c c
12
$
6
ctacct$
1
gctagctacct$ g a
2
a c c t a c c c t a g $
8
ctag$
11
$
4
ccctag$
9
g$ a t c
12
$
5
ctag$
8
t$
cc
10
ccctag$
4
g$ g
a t c
13
$
7
cct$
3
gctacct$ cta
2
gctagctacct$ g
13
$
6
ctag$
9
t$ cc
11
ccctag$
5
g$ g a t
Build Generalized Suffix Tree
T1 = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
T2 = a c a c c t a c c c t a g
7 / 27
1
acctaccctag$
7
c t a g $
10
$
3
a c c c t a g $ t c c
12
$
6
ctacct$
1
gctagctacct$ g a
2
a c c t a c c c t a g $
8
ctag$
11
$
4
ccctag$
9
g$ a t c
12
$
5
ctag$
8
t$
cc
10
ccctag$
4
g$ g
a t c
13
$
7
cct$
3
gctacct$ cta
2
gctagctacct$ g
13
$
6
ctag$
9
t$ cc
11
ccctag$
5
g$ g a t
Space: Θ(n)
T1 = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
T2 = a c a c c t a c c c t a g
8 / 27
Question
Can the LCS problem be solved (deterministically) in O
space and O
time for 0 ≤ ε ≤ 1?
Our Answer
Yes if 0 ≤ ε ≤ 1
For two strings (d = m = 2), the problem can be solved in: Time: O
Space: O
for any 0 < ε ≤ 1
3.
In the general case (2 ≤ d ≤ m), the problem can be solved in: Time: O
O
for any 0 ≤ ε < 1
3.
9 / 27
When the LCS is long
Idea: Preprocess a sparse sample of the n suffixes for LCP queries. T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28 10 / 27
When the LCS is long
Idea: Preprocess a sparse sample of the n suffixes for LCP queries. T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
DCτ DCτ DCτ DCτ DCτ DCτ
Difference Covers
A difference cover modulo τ is a set of integers DCτ ⊆ {0, 1, . . . , τ − 1} such that for any distance d ∈ {0, 1, . . . , τ − 1}, DCτ contains two elements separated by distance d modulo τ. Ex: The set DCτ = {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
11 / 27
When the LCS is long
Idea: Preprocess a sparse sample of the n suffixes for LCP queries. T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
DCτ DCτ DCτ DCτ DCτ DCτ ◮ Number of sampled suffixes: O
n
τ |DCτ|
n
√τ
12 / 27
When the LCS is long
Idea: Preprocess a sparse sample of the n suffixes for LCP queries. T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
◮ Number of sampled suffixes: O
n
τ |DCτ|
n
√τ
◮ The LCS is the LCP of two suffixes.
13 / 27
When the LCS is long
Idea: Preprocess a sparse sample of the n suffixes for LCP queries. T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
◮ Number of sampled suffixes: O
n
τ |DCτ|
n
√τ
◮ The LCS is the LCP of two suffixes. ◮ If |LCS| ≥ τ one of the first τ characters of the LCS is sampled in
both strings.
14 / 27
When the LCS is long
Idea: Preprocess a sparse sample of the n suffixes for LCP queries. T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
◮ Number of sampled suffixes: O
n
τ |DCτ|
n
√τ
◮ The LCS is the LCP of two suffixes. ◮ If |LCS| ≥ τ one of the first τ characters of the LCS is sampled in
both strings.
◮ Hence the LCS corresponds to a pair (p∗ 1, p∗ 2) maximizing
lcp
15 / 27
When the LCS is long
Idea: Preprocess a sparse sample of the n suffixes for LCP queries. T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
RB(11)= (g c t a c)R = c a t c g
◮ Number of sampled suffixes: O
n
τ |DCτ|
n
√τ
◮ The LCS is the LCP of two suffixes. ◮ If |LCS| ≥ τ one of the first τ characters of the LCS is sampled in
both strings.
◮ Hence the LCS corresponds to a pair (p∗ 1, p∗ 2) maximizing
lcp
16 / 27
When the LCS is long
How to compute the pair (p∗
1, p∗ 2) faster than O
n2
τ
T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
SAτ = 14 21 17 26 6 1 16 22 11 12 19 24 4 27 7 2 9 [ , , , , , , , , , , , , , , , , ] LCPτ = 3 1 2 2 1 2 1 2 3 4 1 1 [ , , , , , , , , , , , , , , , ] SAR
τ =
14 1 17 21 26 6 16 22 11 19 12 24 4 2 27 7 9 [ , , , , , , , , , , , , , , , , ] LCPR
τ =
1 1 4 3 2 4 1 3 2 1 2 4 [ , , , , , , , , , , , , , , , ]
Main observation: lcp(T[p∗
1..], T[p∗ 2..]) ∈ [ℓmax − τ + 1; ℓmax], so we can
ignore all pairs with lcp values smaller than ℓmax − τ + 1.
17 / 27
When the LCS is long
How to compute the pair (p∗
1, p∗ 2) faster than O
n2
τ
T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
SAτ = 14 21 17 26 6 1 16 22 11 12 19 24 4 27 7 2 9 [ , , , , , , , , , , , , , , , , ] LCPτ = 3 1 2 2 1 2 1 2 3 4 1 1 [ , , , , , , , , , , , , , , , ] SAR
τ =
14 1 17 21 26 6 16 22 11 19 12 24 4 2 27 7 9 [ , , , , , , , , , , , , , , , , ] LCPR
τ =
1 1 4 3 2 4 1 3 2 1 2 4 [ , , , , , , , , , , , , , , , ]
Main observation: lcp(T[p∗
1..], T[p∗ 2..]) ∈ [ℓmax − τ + 1; ℓmax], so we can
ignore all pairs with lcp values smaller than ℓmax − τ + 1.
18 / 27
When the LCS is long
How to compute the pair (p∗
1, p∗ 2) faster than O
n2
τ
T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
SAτ = 14 21 17 26 6 1 16 22 11 12 19 24 4 27 7 2 9 [ , , , , , , , , , , , , , , , , ] LCPτ = 3 1 2 2 1 2 1 2 3 4 1 1 [ , , , , , , , , , , , , , , , ] SAR
τ =
14 1 17 21 26 6 16 22 11 19 12 24 4 2 27 7 9 [ , , , , , , , , , , , , , , , , ] LCPR
τ =
1 1 4 3 2 4 1 3 2 1 2 4 [ , , , , , , , , , , , , , , , ]
Main observation: lcp(T[p∗
1..], T[p∗ 2..]) ∈ [ℓmax − τ + 1; ℓmax], so we can
ignore all pairs with lcp values smaller than ℓmax − τ + 1.
19 / 27
When the LCS is long
How to compute the pair (p∗
1, p∗ 2) faster than O
n2
τ
T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
SAτ = 14 21 17 26 6 1 16 22 11 12 19 24 4 27 7 2 9 [ , , , , , , , , , , , , , , , , ] LCPτ = 3 1 2 2 1 2 1 2 3 4 1 1 [ , , , , , , , , , , , , , , , ] SAR
τ =
14 1 17 21 26 6 16 22 11 19 12 24 4 2 27 7 9 [ , , , , , , , , , , , , , , , , ] LCPR
τ =
1 1 4 3 2 4 1 3 2 1 2 4 [ , , , , , , , , , , , , , , , ]
Main observation: lcp(T[p∗
1..], T[p∗ 2..]) ∈ [ℓmax − τ + 1; ℓmax], so we can
ignore all pairs with lcp values smaller than ℓmax − τ + 1.
20 / 27
When the LCS is long
How to compute the pair (p∗
1, p∗ 2) faster than O
n2
τ
T = a
1
g
2
g
3
c
4
t
5
a
6
g
7
c
8
t
9
a
10
c
11
c
12
t
13
$1
14
a
15
c
16
a
17
c
18
c
19
t
20
a
21
c
22
c
23
c
24
t
25
a
26
g
27
$2
28
SAτ = 14 21 17 26 6 1 16 22 11 12 19 24 4 27 7 2 9 [ , , , , , , , , , , , , , , , , ] LCPτ = 3 1 2 2 1 2 1 2 3 4 1 1 [ , , , , , , , , , , , , , , , ]
O(τ) Analysis (sketch): O(τ) rounds each using O(n/√τ) time and space: Time: O (n√τ) Space: O (n/√τ) Time: O
Space: O
0 < ε ≤ 1
2.
τ = n2ε
21 / 27
When the LCS is shorter than τ
T1 = 2τ Si
◮ The LCS is a substring of one of the strings of length 2τ. ◮ Build the generalized suffix tree for a batch Si of strings of total
length O( n
√τ ). ◮ Traverse the suffix tree with T2 in O(n) time to find the node of
greatest string depth.
◮ Repeat for all O(√τ) batches.
22 / 27
When the LCS is shorter than τ
T1 = 2τ Si
◮ The LCS is a substring of one of the strings of length 2τ. ◮ Build the generalized suffix tree for a batch Si of strings of total
length O( n
√τ ). ◮ Traverse the suffix tree with T2 in O(n) time to find the node of
greatest string depth.
◮ Repeat for all O(√τ) batches.
Time: O (n√τ) Space: O (n/√τ) Time: O
Space: O
0 ≤ ε ≤ 1
3.
τ = n2ε τ = O(n/√τ)
23 / 27
When the LCS is long
Challenge: The difference cover property only holds for pairs. T = T1 T2 T3 T4 LCS LCS LCS
24 / 27
When the LCS is long
Challenge: The difference cover property only holds for pairs. T = T1 T2 T3 T4 LCS LCS LCS τ 5 4 1 5 3 2
25 / 27
When the LCS is long
Challenge: The difference cover property only holds for pairs. T = T1 T2 T3 T4 LCS LCS LCS τ 5 4 1 5 3 2 Algorithm: Extract the maximum head until we have d − 1 distinct
Computing the next element in a list can be done in O(log n(log2 n + d)). Extracting it costs O(√τ). At most O(d√τ) extractions. Time: O
26 / 27
Results
For two strings (d = m = 2), the LCS problem can be solved in: Time: O
Space: O
for any 0 < ε ≤ 1
3.
In the general case (2 ≤ d ≤ m), the LCS problem can be solved in: Time: O
O
for any 0 ≤ ε < 1
3.
Open Problems
Can the generalized solution be improved? Can the trade-off interval of
2? Can the problem be solved in
O(n1+ε) time and O(n1−ε) space for any 0 ≤ ε ≤ 1?
27 / 27