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An Optimal Algorithm for An Optimal Algorithm for Online Square DetectionOnline Square Detection

Gen-Huey Chen, Jin-Ju Hong, Hsueh-I LuGen-Huey Chen, Jin-Ju Hong, Hsueh-I Lu

National Taiwan UniversityNational Taiwan University

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OutlineOutlineThe definitions of the square detection problem The definitions of the square detection problem

and the online square detection problemand the online square detection problemThe techniques of the algorithm in [Cro86] for tThe techniques of the algorithm in [Cro86] for t

he square detection problemhe square detection problemOur algorithm for the online square detection prOur algorithm for the online square detection pr

oblemoblemConclusionConclusion

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Square Detection ProblemSquare Detection ProblemSquare: a nonempty string of the form XXSquare: a nonempty string of the form XXE.g. “a b c a b c” is a square.E.g. “a b c a b c” is a square.

“ “a b c a b c a” is not a square.a b c a b c a” is not a square.

Input: a string Input: a string SSSquare detection problem:Square detection problem:

Is there a square in Is there a square in SS??

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Online Square Detection ProbleOnline Square Detection Problemm

Leung, Peng, and Ting in COCOON’04Leung, Peng, and Ting in COCOON’04Input: a string Input: a string SSLet Let mm be the unknown smallest integer s.t. be the unknown smallest integer s.t. SS[1..[1..

mm] contains a square.] contains a square.Online square detection problem:Online square detection problem:

Determine Determine mm as soon as as soon as SS[[mm] is read.] is read.An An OO((mm log log22mm)-time algorithm )-time algorithm [LPT04][LPT04]

An An OO((mm log logββ)-time algorithm in our paper)-time algorithm in our paper

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Algorithm in [Cro86] forAlgorithm in [Cro86] forSquare Detection ProblemSquare Detection Problem

forfor kk = 1 = 1 toto pp // // pp: # of blocks: # of blocks

{ { ifif a square ends in a square ends in BBii thenthen returnreturn YES; } YES; }

returnreturn NO; NO;

B1 B2 B3 B4 . . . Bp

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ff-factorization-factorizationLet Let ddkk denote the starting position of the denote the starting position of the kk-th bloc-th bloc

k k BBkk..

BBkk is is SS[[ddkk]] if if SS[[ddkk] does not occur before ] does not occur before ddkk, or , or the the

longest prefix of longest prefix of SS[[ddkk....nn] that occurs before ] that occurs before ddkk..

1 2 3 4 5 6 7 8 9 10 111 2 3 4 5 6 7 8 9 10 11 ……E.g. E.g. SS = a a a b b a b a b a a … = a a a b b a b a b a a …

BB11 BB22 BB33 BB44 BB55 BB66

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ff-factorization (cont.)-factorization (cont.)

A square ending in A square ending in BBkk is centered either in is centered either in BBkk-1-1

or in or in BBkk..

. . . Bk-1 Bk

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Square Ending in the Square Ending in the kk-th Block-th BlockCase 1. The square is entirely in the Case 1. The square is entirely in the kk-th block.-th block.

Case 2. The square begins in the (Case 2. The square begins in the (kk-1)-st block.-1)-st block.Case 2.1. The square is centered in the (Case 2.1. The square is centered in the (kk-1)-st block.-1)-st block.

Case 2.2. The square is centered in the Case 2.2. The square is centered in the kk-th block.-th block.

Case 3. The square begins before the (Case 3. The square begins before the (kk-1)-st block -1)-st block and centered in the (and centered in the (kk-1)-st or -1)-st or kk-th block.-th block.

…

…

…

…

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Our Algorithm for OnlineOur Algorithm for OnlineSquare Detection ProblemSquare Detection Problem

forfor ii = 1 = 1 toto nn // // n n = |= |S|S|

{ compute the { compute the ff-factorization of -factorization of SS[1..[1..ii];];

ifif a square ends at a square ends at SS[[ii] ] thenthen returnreturn ii; }; }

returnreturn NO-SQUARE; NO-SQUARE;

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Square Ending at Square Ending at SS[[ii]] in in BBkk

Case 1. The square is entirely in the Case 1. The square is entirely in the kk-th block.-th block.

Case 2. The square begins in the (Case 2. The square begins in the (kk-1)-st block.-1)-st block.Case 2.1. The square is centered in the (Case 2.1. The square is centered in the (kk-1)-st block.-1)-st block.

Case 2.2. The square is centered in the Case 2.2. The square is centered in the kk-th block.-th block.

Case 3. The square begins before the (Case 3. The square begins before the (kk-1)-st block -1)-st block and centered in the (and centered in the (kk-1)-st or -1)-st or kk-th block.-th block.

…

…

…

…

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LL((ii11, , ii22, , ii)-square)-square::

RR((ii11, , ii22, , ii)-square)-square::

S

i1 c i2 i

i1 c < i2

S

i1 ci2 i

i2 c < i

j

i1 j < i2

j

i1 j < i2

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Square Ending at Square Ending at SS[[ii]] in in BBkk

Case 1. The square is entirely in the Case 1. The square is entirely in the kk-th block.-th block.

Case 2. The square begins in the (Case 2. The square begins in the (kk-1)-st block.-1)-st block.Case 2.1. The square is centered in the (Case 2.1. The square is centered in the (kk-1)-st block.-1)-st block.

Case 2.2. The square is centered in the Case 2.2. The square is centered in the kk-th block.-th block.

Case 3. The square begins before the (Case 3. The square begins before the (kk-1)-st block -1)-st block and centered in the (and centered in the (kk-1)-st or -1)-st or kk-th block.-th block.

…

…

…

…

LL((ddkk-1-1, , ddkk, , ii)-square :)-square :

RR((ddkk-1-1, , ddkk, , ii)-square :)-square :

RR(1, (1, ddkk-1-1, , ii)-square :)-square :

dk-1 dk i

1 dk-1 i

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Our Algorithm for OnlineOur Algorithm for OnlineSquare Detection ProblemSquare Detection Problem

forfor ii = 1 = 1 toto nn // // n n = |= |S|S|

{ compute the { compute the ff-factorization of -factorization of SS[1..[1..ii];];

let let SS[[ii] belong to ] belong to BBkk;;

ifif an an LL((ddkk-1-1, , ddkk, , ii)-square)-square is detected is detected thenthen returnreturn ii;;

ifif an an RR((ddkk-1-1, , ddkk, , ii)-square)-square is detected is detected thenthen returnreturn ii;;

ifif an an RR(1, (1, ddkk-1-1, , ii)-square)-square is detected is detected thenthen returnreturn ii;;

}}

returnreturn NO-SQUARE; NO-SQUARE;

amortizedO(logβ)time

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Longest Common ExtensionsLongest Common ExtensionsFor positions For positions ii11ii22ii33 in in SS

XXRR((ii11, , ii22, , ii33)): longest common right extension of : longest common right extension of

positions positions ii11 and and ii22 with boundary with boundary ii33

1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

E.g. E.g. SS = a b a b b a b a b a = a b a b b a b a b a

XXLL((ii22, , ii33, , ii11)): longest common left extension of p: longest common left extension of p

ositions ositions ii22 and and ii33 with boundary with boundary ii11

XXRR(3, 8, 10) = 2(3, 8, 10) = 2XXLL(4, 9, 2) = 3(4, 9, 2) = 3

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Head Extension Function: Head Extension Function: XXRR(1, (1, jj, ,

ii))If the string If the string SS is read character by character, in is read character by character, in

the the ii-th iteration, for all -th iteration, for all jjii, , XXRR(1, (1, jj, , ii) can be c) can be c

omputed in omputed in OO(1) time with totally (1) time with totally OO((ii)-time pr)-time preprocessing.eprocessing.

1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

E.g. E.g. SS = a b a b b a b a b a = a b a b b a b a b a

XXRR(1,(1,jj,10) ,10) 10 0 2 0 0 4 0 3 0 110 0 2 0 0 4 0 3 0 1

We call We call XXRR(1, (1, jj, , ii) ) the head extension functionthe head extension function

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LL((ii11, , ii22, , ii))-square-square

Y Z Y ZS

i1 j i2 i

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[ML84] [ML84] SS has an has an LL((ii11, , ii22, , ii)-square)-square if and only if ther if and only if ther

e is an index e is an index jj with with ii11jj<<ii22 such that such that XXRR((jj, , ii22, , ii)) = = ||SS[[ii

22....ii]|]| and and XXLL((jj-1, -1, ii22-1, -1, ii11)) + + XXRR((jj, , ii22, , ii)) ||SS[[jj....ii22-1]|-1]|..

LL((ii11, , ii22, , ii))-square-square

Y Z Y ZS

i1 j i2 i

S[1..i-1] contains no square.

=

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Detecting Detecting LL((ddkk-1-1, , ddkk, , ii))-squares-squares

Let Let zz((jj)) = = ||SS[[jj....ddkk-1]|-1]|--XXLL((jj-1,-1,ddkk-1,-1,ddkk-1-1) ) for all for all jj in in BBkk-1-1

In the In the ii-th iteration: is there an index -th iteration: is there an index jj in in BBkk-1-1 s.t. s.t. XXRR

((jj, , ddkk, , ii)) = = zz((jj))??

Y Z Y =Z ?S

dk-1 j dk i

z(j)

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In the In the ddkk-th iteration -th iteration (preprocessing)(preprocessing)

Compute Compute zz((jj)) for all for all jj in in BBkk-1-1

Build the suffix tree of Build the suffix tree of BBkk-1-1$$

For all For all uu, compute, compute

min{min{zz((jj)| )| jj ↔ a leaf in ↔ a leaf in uu’s subtree}’s subtree}

Y Z YS

dk-1 j dk i

z(j)

u

z(j)O(|Bk-1|logβ) time

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In the In the ii-th iteration-th iteration

If |If |SS[[ddkk....ii]| equals the value stored in ]| equals the value stored in uu

a square ends at position a square ends at position ii

Y Z Y =Z ?S

dk-1 j dk i

z(j)

u

z(j)

S[dk..i]

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RR((ii11, , ii22, , ii))-square-square

Y Z Y ZS

i1 i2 j i

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RR((ii11, , ii22, , ii))-square-square

[ML84] [ML84] SS has an has an RR((ii11, , ii22, , ii)-square)-square if and only if the if and only if the

re is an index re is an index jj with with ii22<<jj<<ii such that such that XXRR((ii22, , j+1j+1, , ii)) = = ||

SS[[jj+1..+1..ii]|]| and and XXLL((ii22-1, -1, jj, , ii11)) + + XXRR((ii22, , jj, , ii)) ||SS[[ii22....jj]|]|..

Y Z Y ZS

i1 i2 j i

S[1..i-1] contains no square.

=

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Detecting Detecting RR((ddkk-1-1, , ddkk, , ii))-square-square

Let Let zz((jj)) = = ||SS[[ddkk....jj]|]|--XXLL((ddkk-1,-1,jj,,ddkk-1-1) ) for all for all jj in in BBkk

Insert the position Insert the position jj into the set of into the set of jj++zz((jj))For all For all jj in the set of in the set of ii, , XXRR((ddkk, , jj+1, +1, ii)) = = zz((jj))??

Y Z Y =Z ?S

dk-1 dk j i

z(j)set of j+z(j)

insert jamortizedO(logβ) time

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Computing Computing XXLL((ddkk-1, -1, jj, , ddkk-1-1))

||SS[[gg,,ddkk-1]| = min( |-1]| = min( |SS[[ddkk-1-1....ddkk-1]|, -1]|, ||SS[[ddkk....jj]|]| ) )

For all For all vv with with ggvv<<ddkk, , XXLL((vv, , ddkk-1, -1, gg)) can be compute can be compute

d in d in OO(1) time using the technique of computing the (1) time using the technique of computing the head extension function.head extension function.

Y Z YS

dk-1 dk j i

g

v

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Computing Computing XXLL((ddkk-1, -1, jj, , ddkk-1-1)) (cont.)(cont.)

Let Let FF((jj)) denote the longest suffix of denote the longest suffix of SS[[ddkk....jj]] that is al that is al

so a substring of so a substring of SS[[gg....ddkk-1]-1]

XXLL((ddkk-1,-1,jj,,ddkk-1-1)) = | = |FF((jj)| if )| if yy==ddkk-1-1

min( min( ||FF((jj)|)|, , XXLL((yy,,ddkk-1,-1,gg)) ) otherwise ) otherwise

Y Z YS

dk-1 dk j i

g

y

F(j)

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Time ComplexityTime Complexity

forfor ii = 1 = 1 toto nn // // n n = |= |S|S|

{ compute the { compute the ff-factorization of -factorization of SS[1..[1..ii];];

let let SS[[ii] belong to ] belong to BBkk;;

ifif an an LL((ddkk-1-1, , ddkk, , ii)-square)-square is detected is detected thenthen returnreturn ii;;

ifif an an RR((ddkk-1-1, , ddkk, , ii)-square)-square is detected is detected thenthen returnreturn ii;;

ifif an an RR(1, (1, ddkk-1-1, , ii)-square)-square is detected is detected thenthen returnreturn ii;;

}}

returnreturn NO-SQUARE; NO-SQUARE;

amortizedO(logβ)time

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ConclusionConclusionEach of those Each of those OO(log(logββ) terms comes from the tr) terms comes from the tr

aversal in a suffix tree of a string with aversal in a suffix tree of a string with OO((ββ) dis) distinct characters.tinct characters.

Expected time: Expected time: OO((mm))Is it possible to reduce the running time to worIs it possible to reduce the running time to wor

st-case st-case OO((mm) time for a general alphabet?) time for a general alphabet?