Complexity of the path avoiding forbidden pairs problem revisited

0.41 MB·English

by  Jakub Kováč

Complexity of the path avoiding forbidden pairs problem revisited
JakubKova´cˇ
DepartmentofComputerScience,ComeniusUniversity,Mlynska´Dolina,
84248Bratislava,Slovakia
Abstract
LetG =(V,E)beadirectedacyclicgraphwithtwodistinguishedverticess,t,andletF beasetofforbiddenpairsof
1
vertices. WesaythatapathinG issafe,ifitcontainsatmostonevertexfromeachpair{u,v} ∈ F. GivenG and F,
1
thepathavoidingforbiddenpairs(PAFP)problemistofindasafes–tpathinG.
0
2 We systematically study the complexity of different special cases of the PAFP problem defined by the mutual
positionsoffobiddenpairs. Fixonetopologicalordering≺ofvertices;wesaythatpairs{u,v}and{x,y}aredisjoint,
c
e ifu≺v≺ x≺y,nested,ifu≺ x≺y≺v,andhalving,ifu≺ x≺v≺y.
D The PAFP problem is known to be NP-hard in general or if no two pairs are disjoint; we prove that it remains
NP-hard even when no two forbidden pairs are nested. On the other hand, if no two pairs are halving, the problem
5
isknowntobesolvableincubictime. WesimplifyandimprovethisresultbyshowinganO(M(n))timealgorithm,
2
whereM(n)isthetimetomultiplytwon×nbooleanmatrices.

]
M Keywords: path,forbiddenpairs,NP-hard,dynamicprogramming
D
.
s
c
1. Introduction
[

2 LetG =(V,E)beadirectedgraphwithtwodistinguishedverticess,t∈V andletF ⊆V×V beasetofforbidden
v pairs of vertices. We say that a path π is safe, if it does not contain any forbidden pair, i.e., π contains at most one
6
vertexfromeachpair{u,v} ∈ F. GivenGandF,thepathavoidingforbiddenpairsproblem(henceforthPAFP)isto
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findasafe s–tpathinG. Inthispaper,westudythecomplexityofdifferentspecialcasesoftheproblemondirected
9
3 acyclicgraphs.
.
1
1.1. Motivation
1
1 ThePAFPproblemwasfirststudiedbyKrauseetal.[1]andSrimaniandSinha[2]motivatedbydesigningtest
1
cases for automatic software testing and validation. We can represent a program as a directed graph where vertices
:
v representsegmentsofcodeandedgesrepresenttheflowofcontrolfromonecodesegmentintoanother. Thegoalisto
i coverthisgraphwith s–tpathscorrespondingtodifferenttestcases. However,notallpathscorrespondtoexecutable
X
sequencesintheprogram.ThereforeKrauseetal.[1]introducedforbiddenpairswhichidentifythemutuallyexclusive
r
a codesegmentsandformulatedthePAFPproblem. Unfortunatelly,asshownbyGabowetal.[3],theproblemisNP-
hardevenfordirectedacyclicgraphs.
Adifferentmotivationcamefrombioinformaticsandtheproblemofpeptidesequencingviatandemmassspec-
trometry. Peptidesarepolymerswhichcanbethoughofasstringsovera20characteralphabetofaminoacidsand
thesequencingproblemistodeterminetheaminoacidsequenceofagivenpeptide. Tothisend,manycopiesofthe
peptidearefragmentedandthemassofthefragmentsismeasured(veryprecisely)bymassspectrometer. Theresult
oftheexperimentisamassspectrumwhereeachpeakcorrespondstomassofsomeprefixorsomesuffixoftheamino
acidsequence,orisanoise. Thespectrumisthencomparedagainstadatabaseofknownfragmentweights.
Emailaddress:[email protected](JakubKova´cˇ)
PreprintsubmittedtoDiscreteAppliedMathematics December30,2011
Chenetal.[4]suggestedthefollowingformulationofthepeptidesequencingproblem: Letuscreateaspectrum
graphwithtwovertices p ands foreachpeakw withweightsw(p)=w −1andw(s)=W−w +1,whereW isthe
i i i i i i i
weightofthewholepeptide. Weaddanedgefromxtoyifthedifferencebetweenweightsw(y)−w(x)equalsthetotal
massofsomeknownsequenceofaminoacids. Thus,pathsinthisgraphcorrespondtoaminoacidsequences. Paths
goingthrough p correspondtow beingaweightofsomeprefixandsimilarly,pathsgoingthrough s correspondto
i i i
w beingaweightofsomesuffix. (Pathsgoingthroughneither p nors correspondtow beinganoise.) However,w
i i i i i
cannotbeapreffixweightandasuffixweightatthesametime,so{p,s}willformaforbiddenpairforeachi. Thisis
i i
averyspecialcaseofthePAFPproblemindirectedacyclicgraphswherealltheforbiddenpairsarenestedandChen
etal.[4]showedthatitispolynomiallysolvable.
ThePAFPproblemondirectedacyclicgraphsalsoaroseinacompletelydifferentapplicationinbioinformatics–
genefindingusingRT-PCRtests[5]. Inthisapplication,wehaveasocalledsplicinggraphwhereverticesrepresent
non-overlapping segments of the DNA sequence, length of a vertex is the number of nucleotides in this segment,
andedge(u,v)indicatesthatsegmentvimmediatelyfollowssegmentuinsomegenetranscript. Thus, pathsinthis
splicinggraphcorrespondtoputativegenes. Theproblemistoidentifythetruegeneswithahelpofinformationfrom
RT-PCRexperiments.
Without going into biology details, let us define a (simplified) result of an RT-PCR experiment as a triple t =
(u,v,(cid:96)),whereu,v∈V aretwoverticesand(cid:96)isthelengthofaproduct. Letπbeapathgoingthroughuandvinthe
splicinggraph;ifthelengthoftheu–vsubpathisequalto(cid:96),wesaythatπexplainstestt,otherwise,itisinconsistent
withtestt. WecandefineascoreofapathπwithrespecttoasetoftestsT asasumofthescoresofallofitsvertices
andedges,plusabonusBforeachexplainedtestfromT,andminusapenaltyPforeachinconsistenttest. Thegene
findingwithRT-PCRtestsproblemistofindan s–t pathwiththehighestscoreinthegivensplicinggraphG witha
setofRT-PCRtestsT.
Notethatifwesetalllengthstoanunattainablevalue,say−1,andwesetahigh(infinite)penaltyPforinconsistent
tests,webasicallygetthePAFPproblem. Thus,thePAFPproblemisatthecoreofgenefindingwithRT-PCRtests
and the latter problem inherits all NP-hardness results for the PAFP problem. On the positive side, we have shown
in our previous work [5] that some polynomial solutions for special cases of the PAFP problem can be extended to
pseudo-polynomialalgorithmsforthegenefindingproblem.
1.2. Previousresults
AsshownbyGabowetal.[3],thePAFPproblemisNP-hardingeneral,butseveralspecialcasesarepolynomially
solvable. Yinnone[6]studiedthePAFPproblemunderskewsymmetryconditionswhereforeachtwoforbiddenpairs
{u,u(cid:48)},{v,v(cid:48)} ∈ F, if there is an edge from u to v, there is also an edge from v(cid:48) to u(cid:48). He proved that under such
conditions, the problem is polynomially equivalent to finding an augmenting path with respect to a given matching
andthuspolynomiallysolvable.
Fordirectedacyclicgraphs,wehavealreadymentionedthatthenestedcaseissolvableinpolynomialtime[4];Kol-
manandPangra´c[7]wereabletodeviseapolynomialalgorithmifthesetofforbiddenpairshasawell-parenthesized
orahalvingstructure(seePreliminaries).
Recently,approximabilityandparameterizedcomplexityofthePAFPproblemhavebeenstudied:Weadd1tothe
objectivefunctiontodisallowazerocostsolutions–otherwisetheproblemistriviallyinapproximable. Hajiaghayi
etal.[8]showedthateventhenthereisaconstantc > 0suchthatminimizing1+thenumberofforbiddenpairson
an s–t pathisnotc·n-approximable. Bodlaenderetal.[9]studiedthePAFPproblemonundirectedgraphs. When
parameterized by the vertex cover ofG = (V,E), the problem is W[1]-hard (the proof also carries over to directed
acyclicgraphs). Ontheotherhand,whenparameterizedbythevertexcoverofH =(V,F)(whereedgesareforbidden
pairs), the problem is fixed parameter tractable (FPT), but has no polynomial kernel unless NP ⊆ coNP/poly. The
problemisalsoFPTwhenparameterizedbythetreewidthofG∪H.
1.3. Contributionsandroadmap
Inthispaper,wesystematicallystudydifferentspecialcasesofthePAFPproblemondirectedacyclicgraphs. In
thenextsection,weintroducethedifferentspecialcasesbasedonmutualpositionsofforbiddenpairs. InSection3,
weprovethatthePAFPproblemisNP-hardevenifthesetofforbiddenpairshasorderedstructureandinSections4
and 5, we improve upon the results of Chen et al. [4] and Kolman and Pangra´c [7] for the nested, halving, and
well-parenthesizedforbiddenpairs.
2
Table1:ComplexityofthePAFPproblemforitsdifferentspecialcases;nandmdenotethenumberofverticesandedgesofG,respectively;O(nω)
isthecomplexityofbooleanmatrixmultiplication,ω<2.3727[10,11].
AllowedForbiddenPairs
Problem Complexity Example
disjoint nested halving
generalproblem (cid:88) (cid:88) (cid:88) NP-hard[3]
overlappingstructure × (cid:88) (cid:88) NP-hard[7]
ordered (cid:88) × (cid:88) NP-hard[new]
well-parenthesized (cid:88) (cid:88) × O(n3)[7],O(nω)[new]
halving × × (cid:88) O(n5)[7],O(nω+1)[new]
nested × (cid:88) × O(nm)[4],O(nω)[new]
disjoint (cid:88) × × O(n+m)[trivial]
2. Preliminaries
LetG =(V,E)beadirectedacyclicgraphandletF bethesetofforbiddenpairs. AsalreadynoticedbyYinnone
[6]andKolmanandPangra´c[7],wemayassumethateveryvertexexceptforsandtbelongstoexactlyoneforbidden
pair,i.e.,(cid:83)F =V −{s,t}. Thisissimplybecauseifvertexvdoesnotbelongtoanyforbiddenpair,wecanremoveit
andreplaceall2-edgepathsu,v,wbyadirectedge(u,w). Ontheotherhand,ifvbelongstok > 1forbiddenpairs,
wecanreplaceitbyadirectedpathoflengthkandmovetheendsofforbiddenpairstodifferentverticesonthispath.
Todefinespecialcasesofinterest, wefixonetopologicalorderingofvertices. Wesaythatvertexuisbeforeor
precedesv,u≺v,ifuprecedesvinthislinearorder. Letusdenotetheforbiddenpairs{f, f(cid:48)}fori=1,...,k,where
i i
f ≺ f(cid:48)and f ≺ f ≺···≺ f ,i.e.,weorderthembypositionoftheleftmemberofthepair.
i i 1 2 k
Werecognizethreepossibletypesofmutualpositionofpairs{u,v}and{x,y}(withoutlossofgenerality,letu≺v,
x≺y,andu≺ x): disjoint(u,v≺ x,y;seeFig.1(a)),nested(u≺ x,y≺v;seeFig.1(b)),andhalving(u≺ x≺v≺y;
see Fig. 1(c)). All the special cases are obtained by restricting the set of forbidden pairs F to only certain types of
mutualpositions(seeTable1). Thisgivesus23 =8cases,fromwhichthese6classesarenon-trivialandinteresting:
(a) disjointpairs (b) nestedpairs (c) halvingpairs
Figure1:Differentmutualpositionsoftwoforbiddenpairs.
1. generalcase–therearenoconstraintsonthepositionsofpairs;
2. overlappingstructure1–everytwoforbiddenpairsoverlap(theymaybenestedorhalving,butnotdisjoint);as
aconsequence, f ≺ f ≺···≺ f ≺ f(cid:48) ≺ f(cid:48) ≺···≺ f(cid:48) forsomepermutationσ;
1 2 k σ(1) σ(2) σ(k)
3. ordered – there may be disjoint and halving pairs, but no two forbidden pairs are nested; as a consequence
f ≺ f ≺···≺ f and f(cid:48) ≺ f(cid:48) ≺···≺ f(cid:48);
1 2 k 1 2 k
4. well-parenthesized–theremaybedisjointandnestedpairs,butnotwopairsarehalving;thiscasedeservesits
namesinceifwewrite( and) forthei-thpair,wegetawell-parenthesizedsequence;
i i
5. halving–everytwopairshalveeachother; f ≺ f ≺···≺ f ≺ f(cid:48) ≺ f(cid:48) ≺···≺ f(cid:48);
1 2 k 1 2 k
6. nested–thereareonlynestedpairsi.e.,theverticesinforbiddenpairsareordered f ≺ f ≺ ... ≺ f ≺ f(cid:48) ≺
1 2 k k
···≺ f(cid:48) ≺ f(cid:48);thisisaspecialcaseofthewell-parenthesizedcase.
2 1
1notethatthisspecialcaseisreferedtoashalvingstructurebyKolmanandPangra´c[7];wereservetheterm“halving”forsetswhereeverytwo
pairshalveeachother
3
ThepreviousworkandourownresultsaresummarizedinTable1.
Forcompletenessandasawarm-up,weincludeourownproofofNP-hardnessofthePAFPprobleminthegeneral
andoverlappingcase. ThisproofisalsosimplerthantheonegivenbyKolmanandPangra´c[7].
Theorem1. ThePAFPproblemisNP-hard,evenwhenthesetofforbiddenpairshasoverlappingstructure.
Proof. By reduction from 3-SAT: Let φ = (cid:86) φ be a formula over m variables x ,...,x , with n clauses φ =
1≤i≤n i 1 m i
((cid:96) ∨(cid:96) ∨(cid:96) ),whereeachliteral(cid:96) iseither x or¬x . WewillconstructgraphG andasetofforbiddenpairs F
i,1 i,2 i,3 i,j k k
suchthatthereisans–tpathavoidingpairsinF ifandonlyifφissatisfiable.
x1 x2 x3 xm
···
···
s t
··· ···
¬x1 ¬x2 ¬x3 ··· ¬xm φ1 φ2 φ3 ··· φn
Figure2:InputforthePAFPproblemfortheformulaφ1∧φ2∧···∧φn.Alledgesaredirectedfromlefttoright.
G consistsoftwoparts: Thefirstpartcontainsavertexforeachvariable x anditsnegation¬x (seeFig.2). A
k k
path traversing this first part corresponds to a truth assignment of variables where the visited vertices are true. The
secondpartcontainsavertexforeachliteral(cid:96) (seeFig.2). Forbiddenpairsconnectingeveryliteralfromthefirst
i,j
parttoeveryoccurenceofitsnegationinthesecondpartofGwillensurethatwecanonlygothrough“true”vertices.
Thusan s–tpathavoidingF existsifandonlyifeveryclauseissatisfied. Sinceeveryforbiddenpairstartsinthefirst
partandendsinthesecondpart,allpairsoverlap.
3. Orderedforbiddenpairs
In this section, we turn to a seemingly more restricted version of the PAFP problem, allowing only disjoint and
halvingforbiddenpairs. Thisspecialcasehasnotbeenstudiedbefore.
Theorem2. ThePAFPproblemisNP-hard,evenwhenthesetofforbiddenpairsisordered.
Proof. We will prove the claim by reduction from 3-SAT. Let φ be a logical formula over m variables x ,...,x ,
1 m
whichisaconjunctionofnclausesφ ∧···∧φ ,whereφ = ((cid:96) ∨(cid:96) ∨(cid:96) )andeachliteral(cid:96) iseither x or¬x .
1 n i i,1 i,2 i,3 i,j k k
WewillconstructgraphGwithalinearorder≺onitsverticesandanorderedsetofforbiddenpairsF suchthatthere
isans–tpathavoidingpairsinF ifandonlyifφissatisfiable.
Graph G consists of several blocks B and B of 2m vertices shown in Fig. 3(a), (b). The blocks are connected
(cid:96)
togetherasoutlinedinFig.3(c).Anyleft-to-rightpaththroughtheblockBnaturallycorrespondstoatruthassignment
ofthevariablesand,sinceB hasanisolatedvertex¬(cid:96),apaththroughblockB correspondstoanassignmentwhere
(cid:96) (cid:96)
(cid:96) is true. A clause gadget consists of three such blocks, each corresponding to one literal. Any s–t path must pass
throughoneofthethreeblocks,andthuschooseanassignmentthatsatisfiestheclause. TheforbiddenpairsinF will
enforcethattheassignmentofthevariablesisthesameinallblocks. Thisisdonebyaddingaforbiddenpairbetween
allliterals(cid:96)(cid:48)intheB -blockswiththeircounterparts¬(cid:96)(cid:48)inthepreviousandthefollowingB-block.
(cid:96)
TheorderofliteralsinaB-blockis¬x ≺ x ≺¬x ≺···≺ x ,whiletheorderinaB -blockis x ≺¬x ≺ x ≺
1 1 2 m (cid:96) 1 1 2
··· ≺ ¬x . Letvi ≺ vi ≺ vi ≺ ··· betheorderofverticesingraphGi. AzippingoperationtakesgraphsG1,G2,G3
m 1 2 3
andproducesanewgraphG1∪G2∪G3withverticesorderedv1 ≺v2 ≺v3 ≺v1 ≺v2 ≺v3 ≺···. Theclausegadgets
1 1 1 2 2 2
areproducedbyzippingthethreeblockscorrespondingtotheirliterals. Ifwedonotallowmultipleforbiddenpairs
startingorendinginthesamevertex,wecansubstituteverticesinGforshortpathsasinFig.3(d). Itiseasytocheck
thatundersuchlinearorder,notwopairsinF arenested.
4
¬x1 ¬x2 ¬x3 ··· ¬xm x1 x2 x3 ··· ‘ ··· xm
x1 x2 x3 ··· xm ¬x1 ¬x2 ¬x3 ··· ¬‘ ··· ¬xm
(a) Block B–verticesofthisgraphcorrespondto (b) BlockB(cid:96)issimilartoaB-block,exceptthattheorderof
positive and negative literals; a path through this verticesisdifferentandvertex¬(cid:96)isisolated. Thus,apath
blockcorrespondstoatruthassignmentofthevari- throughB(cid:96)correspondstoanassignmentwhere(cid:96)istrue.
ables.
φ φ φ
1 2 n
B B B
‘1,1 ‘2,1 ‘n,1
s B B B B B t
‘1,2 ‘2,2 · · · ‘n,2
B B B
‘1,3 ‘2,3 ‘n,3
(c) ConstructionofGfromtheblocksandzippedblockscorrespondingtotheclauses.Forbiddenpairsenforcethat
theassignmentofvariablesisthesameinallblocks.
forbiddenpairs
xk xk+1 ··· B‘k,1
···
¬xk ¬xk+1
¬xk ¬xk+1 ··· xk xk+1 ··· B‘k,2
··· ···
xk xk+1 ¬xk ¬xk+1
blockB xk xk+1 ··· B‘k,3
···
¬xk ¬xk+1
clauseφk
(d) AnenlargedviewofgraphGshowingblockB,thefollowingblocksforclauseφkandthewaythey
areconnectedbyforbiddenpairs.Notethatnotwoforbiddenpairsarenested.
Figure3:ConstructionofthegraphGfora3-SATformulaφ.Alledgesaredirectedfromlefttoright.
4. Well-parenthesizedforbiddenpairs
ThefirstpolynomialalgorithmforthePAFPproblemwithwell-parenthesizedforbiddenpairswasgivenbyKol-
manandPangra´c[7]. Theiralgorithmusesthreerulesforreducingtheinputgraph:
1. contractionofavertex–ifvdoesnotappearinanyforbiddenpair,removeitandaddadirectedge(u,w)for
everypairofedges(u,v),(v,w);
2. removalofanedge–ifedgee∈ E∩F joinstwoverticesthatmakeupaforbiddenpair,removeefromE;
3. removal of a forbidden pair – if (u,v) ∈ F is a forbidden pair, but there is no path from u to v, remove (u,v)
fromF.
Thesethreerulesarealternatelyappliedtotheinputgraphuntilweendupwithvertices sandtonly–eitherjoined
byanedgeordisconnected–whichisatrivialproblem.
AsimpleimplementationofthisapproachgivesanO(n2m)algorithm. Usingfastmatrixmultiplication,thetime
complexitycanbereducedtoO(nω+1)≈O(n3.373)andusingadynamicdatastructurefor“findingpathsanddeleting
edgesindirectedacyclicgraphs”byItaliano[12],itcanbereducedstilltoO(n3).
HerewedescribeourownO(n3)algorithm,itsadvantagesbeingsimplicity,extensibility,andimprovability: The
algorithmdoesnotuseanyadvanceddatastructures. Itcanbeeasilyextendedtosolveproblemssuchas
• findans–tpathpassingtheminimumnumberofforbiddenpairsor
5
• givenagraphwherealledgeshavescoresandtherearebonusesorpenaltiesforsome(well-parenthesized)pairs
ofvertices,findans–tpathwithmaximumscore(thisproblemwasconsideredbyKova´cˇ etal.[5]).
Itseemsunlikelythattheseproblemscanbesolvedusingtheformerapproach(becauseofrule2). Furthermore,our
algorithm can be improved using the Valiant’s technique and fast matrix multiplication algorithms [13, 14] or the
Four-Russianstechnique[15]. Notethatthereductiontomatrixmultiplicationisnotonlyoftheoreticalinterest,since
therearefastandpracticalhardware-basedsolutionsformultiplyingtwomatrices[16,17].
Theorem3. ThePAFPproblemwithwell-parenthesizedforbiddenpairscanbesolvedinO(n3)time.
Proof. Wemodifytheinputgraphsothatnotwoforbiddenpairsstartorendinthesamevertex. LetP[u,v]betrueif
asafeu–vpathexistsandlet J[u,v]betrueifthereisaforbiddenpair(q,v) ∈ F,u ≺ q ≺ v,andthereisasafeu–v
pathsuchthatthefirstedgejumpsoverq.
ThevaluesofPand J canbefoundbydynamicprogramming: Itiseasytocompute J[u,v](ifwealreadyknow
P[w,v]forallu ≺ w (cid:22) v)byinspectingtheneighboursofuandconversely,wecanalsocompute P[u,v]efficiently
usingthetable J: Ifnoforbiddenpairendsinvorvertexuis“inside”theforbiddenpair(q,v) ∈ F, wejustsearch
theneighboursofvforavertexthatcouldbepenultimateontheu–vpath. Otherwise, let(q,v) ∈ F beaforbidden
pairsuchthatu ≺ q ≺ v. Supposethatasafeu–vpathexistsandletwbethelastvertexonthispathbeforeq. Then
P[u,w] and J[w,v] are both true. Conversely, if P[u,w] and J[w,v] are true for some w ≺ q, by concatenating the
correspondingpaths,wegetasafeu–vpath: Thepathobviouslyavoidsallforbiddenpairsbeforeorafterq(fromthe
definitionofP[u,w]andJ[w,v])andtherearenoforbiddenpairshalving(q,v).
Thus,P[s,t]canbecomputedincubictimeusingthefollowingtworecurrences:
(cid:40) (cid:87)
P[w,v] ifu≺qand(q,v)∈ F isaforbiddenpair (1)
J[u,v]= (u,w)∈E,q≺w
undefined otherwise
 true ifu=v
P[u,v]=  (cid:87)(cid:87)falus(cid:22)ew≺v,((Pw[,vu)∈,EwP]∧[u,Jw[w],v]) iiifff((nuqo,,vvfo))ri∈bsiFaddfieosnrabpifdaodirrebenidnpddaesinri,npuvai≺orrq(q≺,vv)∈ F forq≺u ((23))
u(cid:22)w≺q
This algorithm can be further improved to O(nω) time by using fast boolean matrix multiplication. The proof
is actually simple thanks to the work of Zakov et al. [14] that simplified and generalized the Valiant’s technique
[13]. TheyintroduceagenericproblemcalledInsideVectorMultiplicationTemplate(VMT)whichcanbesolvedin
subcubictime. AproblemisconsideredanInsideVMTproblemifitfulfillsthefollowingrequirements:
1. Thegoaloftheproblemistocomputeforeveryi, jaseriesofinsidepropertiesβ1 ,β2 ,...,βK.
i,j i,j i,j
(cid:76) (cid:16) (cid:17)
2. Let 1 ≤ k ≤ K, and let µk be a result of a vector multiplication of the form µk = βk(cid:48) ⊗βk(cid:48)(cid:48) , for
i,j i,j q∈(i,j) i,q q,j
some1 ≤ k(cid:48),k(cid:48)(cid:48) ≤ K. Assumethatthefollowingvaluesareavailable: µk ,allvaluesβk(cid:48) for1 ≤ k(cid:48) ≤ K and
i,j i(cid:48),j(cid:48)
(i(cid:48), j(cid:48))(cid:40)(i, j)andallvaluesβk(cid:48) for1≤k(cid:48) <k. Then,βk canbecomputedino(n)time.
i,j i,j
3. In the multiplication variant that is used for computing µk , the ⊕ operation is associative, and the domain of
i,j
elementscontainsazeroelement. Inaddition,thereisamatrixmultiplicationalgorithmforthismultiplication
variant,whoserunningtimeM(n)overtwon×nmatricessatisfiesM(n)=o(n3).
Theorem 4 (Zakov et al. [14]). For every Inside VMT problem there is an algorithm whose running time is o(n3).
Inparticular, let M(n)bethecomplexityofthematrixmultiplicationusedandsupposethatβk canbecomputedin
i,j
Θ(1) time in item 2 of the definition above. Then the time complexity is Θ(M(n)logn), if M(n) = O(n2logkn); and
Θ(M(n)),ifM(n)=Ω(n2+ε)forε>2and4M(n/2)≤d·M(n)forsomed <1andsufficientlylargen.
Corollary1. ThePAFPproblemwithwell-parenthesizedforbiddenpairscanbesolvedinO(nω)time,where2<ω<
2.3727istheexponentinthecomplexityofthebooleanmatrixmultiplication.
6
Proof. WeformulateoursolutionfromTheorem3asanInsideVMTproblem. Thegoalistocomputeinsideproper-
tiesA,J,α,β,P,P(cid:100)•,andP•(cid:100). Properties J andP correspondtothedynamicprogrammingtablesfromtheproof
u,v u,v
ofTheorem3,otherpropertiesareauxilliary.PropertyAistheadjacencymatrixofgraphGanditisconstant(A =1
u,v
ifandonlyif(u,v)∈ E). Propertiesα,βareusedtostorethepartialresultsfromcases(2)and(3)inthecomputation
of P[u,v]. Finally,theauxiliaryproperties P(cid:100)• and P•(cid:100) canbecomputedfrom Pinconstanttimeandaredefinedas
follows:
P(cid:100)• = P ∧(q≺w) if(q,v)∈ F,elsefalse P•(cid:100) = P ∧(w≺q) if(q,v)∈ F,elsefalse
w,v w,v w,v w,v
NowwecanrewritethecomputationofJ andP usingbooleanvectormultiplicationasfollows:
u,v u,v
(cid:87) P[w,v] (cid:32) J =(cid:76) (A ⊗P(cid:100)•) (1’)
(u,w)∈E,q≺w u,v w∈(u,v) u,w w,v
(cid:87) P[u,w] (cid:32) α =(cid:76) (P ⊗A ) (2’)
u(cid:22)w(cid:22)v,(w,v)∈E u,v w∈(u,v) u,w w,v
(cid:87) (P[u,w]∧J[w,v]) (cid:32) β =(cid:76) (P•(cid:100) ⊗J ) (3’)
u(cid:22)w≺q u,v w∈(u,v) w,v w,v
PropertyP canbecomputedfromα andβ inconstanttime.
u,v u,v u,v
5. Theothercasesandconcludingremarks
NotethattheO(nω)algorithmforwell-parenthesizedforbiddenpairsalsoimprovesupontheresultbyChenetal.
[4]forthenestedcase. Itremainsanopenproblemwhetherthereisamoreefficientalgorithmforthenestedcase.
AnO(nω+1)timealgorithmforhalvingforbiddenpairsisachievedbyarefinedversionofthealgorithmgivenby
KolmanandPangra´c[7].Recallthatinthiscase,theinputgraphGconsistsoftwoparts:alltheforbiddenpairsstartin
thefirstpart,andendinthesecondpartinthesameorder. Letusdenotetheverticesinthefirstparts≺ x ≺···≺ x
1 n
andverticesinthesecondparty ≺ ··· ≺ y ≺ t,where{x,y}areforbiddenpairs. Wemayassumethatallvertices
1 n i i
areaccessiblefromsandthattisaccessiblefromeveryvertex.
Ifthereisadirectedgefromstothesecondpartorifthereisanedgefromthefirstparttot,asafes–tpathexists
trivially. Otherwise,wereducethehalvingcasetoninstancesofthenestedcase. Therewillbeasafes–tpathinGif
andonlyifthereisasafes–t(cid:48)pathinatleastoneoftheproducedinstances.
First,removeallthe(x,y )edges,addanewterminalvertext(cid:48),andreversethedirectionofalledgesinthesecond
i j
part ofG. Note that in this new order, s ≺ x ≺ ··· ≺ x ≺ t ≺ y ≺ ··· ≺ y ≺ t(cid:48), the forbidden pairs are nested.
1 n n 1
The k-th instance is obtained by adding edges (x ,t) and (y ,t(cid:48)) for each edge (x ,y ), so there is a safe s–t path
k (cid:96) k (cid:96)
s,...,x ,y ,...,tintheoriginalgraphGifandonlyifthereisasafes–t(cid:48)paths,...x ,t,...,y ,t(cid:48)inthenewgraph.
k (cid:96) k (cid:96)
Itremainsanopenproblemwhetheramoreefficientalgorithmexists.
Acknowledgements.
The autor would like to thank Bronˇa Brejova´ for many constructive comments. The research of Jakub Kova´cˇ
is supported by APVV grant SK-CN-0007-09 Marie Curie Fellowship IRG-231025 to Dr. Bronˇa Brejova´, Come-
niusUniversitygrantUK/121/2011,andbyNationalScholarshipProgramme(SAIA),SlovakRepublic. Preliminary
versionofthisworkappearedinKova´cˇ etal.[5].
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