Complexity results for scheduling tasks in fixed intervals on two types of machines

Complexity results for scheduling tasks in fixed intervals on

two types of machines

Citation for published version (APA):

Nakajima, K., Hakimi, S. L., & Lenstra, J. K. (1982). Complexity results for scheduling tasks in fixed intervals on two types of machines. SIAM Journal on Computing, 11(3), 512-520. https://doi.org/10.1137/0211040

DOI:

10.1137/0211040

Document status and date: Published: 01/01/1982 Document Version:

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COMPLEXITY RESULTS FOR SCHEDULING TASKS

IN

FIXED

INTERVALS

ON

TWO

TYPES

OF

MACHINES*

K. NAKAJIMA,t S. L. HAKIMI,, AND J. K. LENSTRA

Abstract. Supposethatnindependent tasks aretobe scheduledwithoutpreemption on anunlimited

number ofparallelmachines of twotypes: inexpensiveslow machinesand expensive fastmachines.Each

task requires a given processing time ona slow machine or agivensmaller processingtime on a fast machine.Wemaketwodifferentfeasibility assumptions: (a)each task hasaspecified processing interval, thelength ofwhich isequaltothe processingtimeonaslow machine;(b)each taskhasaspecified starting

time.Foreitherproblem type, wewish to find a feasiblescheduleof minimum total machine cost. Itis

shown that bothproblems are NP-hardinthe strong sense. Theseresults arecomplemented by polynomial algorithms forsomespecial cases.

Keywords, _{parallel machines,}tasks, release dates, deadlines, computational complexity, NP-hardness, polynomial algorithm

1. Introduction. Webeginby considering the following problem.

Suppose

there are ntasks

T,

T,

and an unlimited number of identicalparallelmachines. Each task

_T.

requiresagivenprocessing timep and istobe executedwithout interruption

between agivenrelease date

_r

and a given deadline

_d

_r

+

pi.The tasksare independent in the sense that there are no precedence constraints between them. Each machine can execute any task, but no more than one at a time. The problem is to find the minimum number ofmachines neededtoexecute all tasksaswell as acorresponding schedule of the tasks on the machines.

Thisproblemisknown asthe"fixedjob scheduleproblem"

[6]

andas the "channel assignment problem"

[8], [9],

[10].

Ithas applications in such diverse areasasvehicle

scheduling

[2],

[15],

machine scheduling

[6,

[8],

and computer wiring

[8], [9], [10].

As

a special case ofDilworth’s chaindecompositionproblem, it is solvable in

O(n

2)

time by the staircase rule of Ford and Fulkerson

[3,

p.

65]

andby the step-function methodof GertsbakhandStern

[6].

Hashimoto and

Stevens

[9],

[10]

presentedsome interesting graph theoretical approaches to the problem and proposed an

O(n2.)

algorithm, for which Kernighan, Schweikert and Persky

[12]

gave an

O(n

log

n)

implementation. Recently,

Gupta,

Lee and

Leung

[8]

independently developed a

different

O(n

log

n)

algorithm and also showed that any solution method for the problem requiresf(nlog

n)

time.

In

thispaperwe willconsider a natural generalization of thisproblemwhichhas potential applications in the scheduling areas mentioned above. Again, there are n

independent tasks T1," ’,

Tn,

but there are two types of machines: slow machines

ofcost C and

_fast

machines ofcost C

>

C

.

Each task

_T.

requires a processingtime pi on a slow machine or

q(<p)

on a fast machine and is to be executed without

interruption between its release date

_r

anditsdeadline

_d

_=r

+

p. It is assumed that all numerical problem data are integers.

In

a feasible schedule, the tasks assigned *Receivedby theeditorsJuly30,1979, andin final revisedform September9, 1981.Thisresearch wassupportedinpart by theNational Science Foundationunder grant ENG79-09724.

f ComputerScienceDivision,DepartmentofElectricalEngineering,TexasTech University, Lubbock,

Texas79409. FormerlyatDepartment of ElectricalEngineering and Computer Science, Northwestern University,Evanston,Illinois60201.

tDepartmentof ElectricalEngineering andComputerScience, Northwestern University,Evanston,

Illinois60201.

MathematischCentrum,Amsterdam, the Netherlands. 512

SCHEDULING TASKS IN FIXED INTERVALS 513

to slow machines have to start at their release dates in order to meet their

dead-lines.

For

the tasks

_T

assigned to fast machines, we make two different feasibility assumptions:

(a) VST

(variablestartingtimes):

_T.

maystart atanytime intheinverval

[r.,

dj qj];

(b)

FST

(fixed

startingtimes):

_T.

hasto start at time

_r.

A

schedule using m slowmachineandmrfastmachines has totalcost

rnSC +

rnrC

.

For

either problem type,we wish to find a feasible schedule ofminimum total cost.

In

2 we show that the VST problem is NP-hard

[1], [4],

[5], [11],

even if all release dates areequal.

In

3weextend our techniquestoprove that theFSTproblem

is NP-hard in the case of arbitrary release dates; the case of equal release dates is

trivially solvable in

O(n)

time. The NP-hardness results are

"strong" [4], [5]

in the sense thatthey hold even withrespect to aunary encoding of the data; this implies that there exists nopseudopolynomial algorithm for these problems unless f9

W.

In

4 and 5 we considerthe special case that q 1, ] 1,...,n.

We

present

O(n

log

n)

algorithms for theVSTproblemwithequal release dates and for theFST

problemwitharbitrary releasedates, respectively.

TABLE

Summary_ofcomplexity results

p.arbitrary q.arbitrary

r.

arbitrary NP-hard ( 2) Open VST rjequal NP-hard ( 2) O(nlogn) ( 4) rjarbitrary NP-hard O(nlogn) FST ( 3) O(n) ( 5) O(n)

r.

equal (3) (3)

These results represent an almost complete complexity classification of the

problemclass under consideration, as demonstrated by Table 1. 2. NP-hardness of theVSTproblem.

THEOREM 1. The VST problemis NP-hardin the strongsense,even

_if

allrelease

dates areequal.

Our

proofholds for the case that

Cr/C

3 and

pffq

3, ] 1,.

,

n.Theorem

1 dominates apreviousresult,stating that the

VST

problemisNP-hard inthe strong sense if the release dates are arbitrary,

Cr/C

is anarbitraryconstantbetween 1 and

7,andpffqi 4,j

1,...,

n

[17].

Proof of

Theorem 1.

We

have to show that a problem which is known to be

NP-complete in the strong sense is (pseudopolynomially) reducible to the

VST

problem.

Our

starting pointwillbethe following problem

[5,

p.224,

[SP15]]:

3-PARTITION: Given a set $

{1,..., 3t}

and positive integers al,’ ’,a3t,b with

4X-b

<

_a

<

1/2b,/"

S,

and

s

_a

tb,does thereexist apartition of Sinto disjoint 3-element subsetsSl,

,

St

such that

Y’4s,

ai b, 1,

,

t?

Given any instance of 3-PARTITION, we construct, in (pseudo-) polynomial time,acorresponding instance of theVSTproblemwithequal release dates as follows"

1. The costcoefficients are definedby

C

1, Cr 3. 2. There are 4t tasks:

a-tasks

T.,j S,

with

r

0,

p

6a,

q

2a,

We claim that 3-PARTITION has a solution if and only if there exists a feasible

schedulewithtotalcost at most

C*

3t.

Suppose

that3-PARTITION hasa solution{$1, ’,

St}.

Itispossibleto construct a feasible schedule for all tasks on fast machines

M(,...,

M,

r as follows (cf. Fig.

1)"

foreach e

{1,.

,

t},

machine

M

processes the three tasks

T.,

] Si, in non’de-creasing order of

q

value in the interval

[0, 2b],

and the task

_T

b in

[2b, 3b];

note

thatthe startingtimeof each task falls withinthe requiredinterval. Thetotalcostof thisschedule isequalto tC

r=

C*.

Conversely, suppose that there exists a feasible schedule withtotalcost at most

C*=

3t. No slow machine can process more than one task. No fast machine can

process more than four tasks, since the completion time of the fourth task will be larger than 2b and the starting time ofa fifth task should be no larger than 2b. Let

there be m slow machines and mr fast machines.

We

have, by the hypothesis,

m

+

3mr<-3tand,by the above arguments,m >-4t-4m

r.

Thefirstinequality implies thatmr-<_ _{and the}twotogether imply thatmr>_- t.

We

conclude thatmr t. Itfollows that there arenoslowmachinesand fast machines, each processing four tasks.

Instance of 3-PARTITION"

,t 3; b 25; ] 2 3 4 5 6 7 8 9

t, 7 7 7 8 8 8 9 10 11 Solution’ {{1,2,9}, {3,4,8},{5,6,7}}

Corresponding VST schedule on t fast machines"

FXG. 1. Illustrationo[thetranfformationinTheorem 1.

None of these fast machines can processmore than one b-task, since otherwise

the completion time of the fourth task would be larger than 3b.

It

follows that the

ithfastmachineprocessesexactlyoneb-task andthree a-tasks

T,/"

E$i, with

s,

q’

-<

2b.Since

Y.s

q’

2tb,wehave

_Y.s,

q

2b, 1,.

.,

t.Thecollection{$1,"

,

St}

constitutesasolutionto3-PARTITION. F1

3. NP-hardnessofthe

FST

problem.

THEOREM2. The FSTproblemisNP-hardin thestrongsense.

THEOREM3. The FST problemissolvablein

O(n

time

_if

all release datesareequal. OurNP-hardnessproofholds for the case that

Cr/C

(t

+

2)/(t

+

1)

andpi/qi

z,] 1,...,n, where and z are input variables. Theorem 2 is. still true if

Cr/C

isan arbitraryconstantbetween2 and 3 and

p/q

2,

]

1,.

.,

n

[16];

the proofof thisfurther refinement is quiteinvolved.Theorem3 shows that the NP-hardness result cannot beextendedtothe case of equal releasedates, unless

SCHEDULING TASKS IN FIXED INTERVALS 515

Proof of

Theorem 2.

We

will start from the following strongly NP-complete problem

[5,

p.224,

[SP17]]:

NUMERICAL

MATCHING WITH TARGET

SUMS. Givena set$

{1,

,

t}

and positive integers al,"’,a,, bl,... ,bt, c,"’,ct with

Yis(ai+bi)=Y.gsCi,

do thereexistpermutationsa

and/3

of$suchthata(i)

+

b,)

ci, S?

We may assume without loss of generality that a <...

<

at,b <’"

<

bt

and

ca <

<

ct.Further,wewill assumethatforanyinstance of thisproblem thereexists apositive integer z such that

z

<aa

<

<at

<2z

<ba

<

<bt

<3z

<ca

<

<ct

<5z.

(If

this doesnothold, thendefinez max

{at

+

1,

bt +

1}

and set ag ag

+

z,

bi

be +

2z,cg cg

+

3z,

S.) We

will usethenotation$’=

_{1,.

,

t-

1}.

Givenany instance of

NUMERICAL

MATCHING

WITH TARGET

SUMSwe

construct,in(pseudo-)polynomial time,acorresponding instanceof the

FST

problem

asfollows:

1. Thecostcoefficients aredefinedby C

+

1, Cr

+

2. 2. Thereare

2tz+

tasks:

a-tasks

T,

$, b-tasks

Tg,

h

S,

S, c-tasks

TT,

S,

d-tasks

Ti,

h

S’,

6

S,

with

r

O,

with

_ri

ah, with

r

ci, with

ri

2z

+

zbi, P zai, q ai,

pbhi

zbi,

qi

bi,

Z3 2

pi=3

q=3z,

d 3 d 2

phi Z qhi Z

We claimthat NUMERICAL MATCHING

WITH

TARGETSUMS has a solution ifand onlyifthereexists a feasibleschedulewithtotalcost at most

C*

3

+

z

+

t.

Suppose

that thematching problemhasa solution

(a,/3).

Itispossibletoconstruct afeasible schedule for all taskson fast

machines

M{,

$, and

z-

slow machines

M,g,

he

S’,

e

S,

asfollows

(cf.

Fig.

2)"

for each e $,machine

M{

processesthetasks

b

_2]

T(g),

T(i)(i), Ti

in the intervals

[0, a(i)], [a(i),

a(i)

+ bt3(/)],

[ci,ci

+

3z

(note

that

a(i)

+

b,(i) ci), and each of the t-1 machinesMShi, h

S’,

processes one of the t-1 tasks

T,i,

h

S-{a(-(i))},

in

[ah,

ah

+

zbi] and one of the t-1 tasks

T’i,

h

S’,

in

[2z

+

zbi,2z

+

zbi

+

z

3]

(notethat ah

< 2Z).

The total cost of thisschedule is equal to

tCf

+

(t

2-t)C

C*.

Conversely, suppose that there exists afeasible schedule with total cost at most

C*.

Wemake the following propositions.

PROPOSITION 1.

Two

a-tasksarenotassignedtothesamemachine.

Proof.

Each a-taskisprocessedduring theinterval

[0,

z

].

PROPOSITION 2.

Two

b-tasksarenotassignedtothesame machine.

Proof.

Eachb-task is processedduring the interval

[2z,

3z].

PROPOSITION 3.

Two

c- ord-tasksarenotassignedtothesamemachine.

Proof.

Eachc- ord-taskis processedduring theinterval

[32

"2

+

z, 3z2

+

32’].

PROPOSITION4.

An

a-taskandab-taskarenotassignedtothesameslowmachine.

Proof.

On a slow machine, each a- or b-task is processed during the interval

[2z-l,z+z].

PROPOSITION5.

A

b-taskandac-taskarenotassignedtothesameslowmachine.

Proof.

On a slow machine, each b- or c-task is processed during the interval

[5z

1, 2z

+

2z

+

1].

All tasks are assignedto at most machines,since

(t

z

+

1)C >

C*.

Propositions

1, 2 and 3 imply that there are exactly 2 machines, each processing at most one

Instance of NUMERICAL

MACHING

WITH TARGET SUMS" t 3; z 4; i Solution-2 3 5 6 7 91011 14 16 18 2 3 2 3

Corresponding FST schedule on t fast machines and

Z2_Z

slow

machines-0 5 14 0 6 16 0 7 6 7 5 5 62

c’48

T 64

c’48

T 2 18 66 42 44 108 4344 45 48 47 48

%’

_II

49 52 50 52

T11

"64

T2

"64

-]

108

T22

"64 112 1 2 116 116

FIG. 2. Illustration_ofthetransformationinTheorem 2.

most fast ones,since

(t

2-

_t-1)C

₊

_(t

₊

_1)cf>

_C*.

_{Propositions4and 5 imply that}

there are exactly fast machines, each processing one a-task, one b-task and one

c-task; hence, there are exactly 2- slow machines, each processing one b-task and

oned-task.

We

denote the fast machines by

M,(,

E

S,

and the

t=-t

slow machines by

MShi,

he

S’,

eS. It may be assumed that

_T7

is assigned to

M(,

e

S,

and

Thd

to Mhi, h

S’,

S. There exists a permutation a of S such that

T<i)

is assigned to

Mfi,

isS.

Letusdefinethesizeof

Taxi

as bi,itsprocessingtime on afast machine. Thesize

of a b-task on

M/f

is at most ci-a(i), and the size of a b-task on

Mhi

is at most

[(2z

+

zbi-a)/zJ

bi.

Thesum of these upper bounds over allmachines isequal to

s

(c- a(i))

+

s’,

s

b

,s

b, which is the totalsizeof all b-tasks. It follows thatalltheseupperbounds are actually achieved.

More

explicitly, for each s

S,

there

exists anindex/3(i) Ssuchthat

T(i),<i)

isassignedtoM,

f.,

and there exists anindex

SCHEDULING TASKS IN FIXED INTERVALS 517

M

hi,h

S’,

while

b

Tv(i)g is assigned to afast machine. Thisimplies that the functions

/3

and 3’ arepermutations ofS withy(/3(i))= a(i),

S.

Since

Tb(g)t(g

leaves no idle time between

T(i

and

T

on

M,

we have a(i

+

bt(g ci,

S.

The pair

(a,/3)

constitutes asolutiontothe matching problem.

Proof

ol

Theorem 3.

In

theFSTproblemwithequalrelease dates, each task has

to start at the same time and therefore each machine can process at most one task.

It follows that an optimal schedule uses n slow machines andhas total cost nC

s.

It isconstructedin

O(n)

time.

4.

A

well-solvablecase of theVSTproblem.

THEOREM4.

In

thecase thatqj 1, 1,.

,

n, the VST problemissolvable in

O(n

log

n)

time

_if

allrelease datesareequal.

The complexity of the VST problem with all qj 1 and arbitrary release dates remainsunresolved(cf. Table

1).

Proof of

Theorem 4.

In

the

VST

problemwithequalreleasedates,aslow machine canprocessat most onetask butafastmachinemaybe abletoprocessmore thanone. Letusassume that there arem fast machines, with 0 _<-m

<=

n, and let

X,

denote themaximum numberout ofthe n unit-time tasks that can be completedintime on these machines.

A

schedule usingm fastmachines hasto use n-X,, slow machines;

its totalcost is equal to C, mC

+(n-X,)C

.

Itfollows that an optimal schedule has totalcostmino__<,,_<_

{C.,}.

For each given value of m, the number X,, and acorresponding schedule on

rh

fastmachines canbefoundby an

O(n

log

n)

algorithm from Lawler

[14], [7,

p.

295].

Straightforward application of this algorithm forrn 0,..

,

n would yield an overall optimal schedule in

O(n

2

log

n)

time.

However,

all

Xo,

’,

Xn

togethercanbe determinedbyan

O(n

log

n)

algorithm,

whichconstructsaschedule on n fast machineswiththe property that,for anyvalue of m, the partialscheduleonthefirstrnmachines isanoptimalscheduleonm machines

[13].

This algorithm considers the tasks in order of nondecreasing deadlines and assigns each task to the machine with lowest index onwhich it canbe completed in

time.

A

formalstatement is as follows.

VST

ALGORITHM (only

_fast

machines, allqi 1, allri

O)

Initialize. Reorder the tasks in suchawaythat d -<

_=<

d,;set

do

-.

Intro-duce an array x ofsize n and set x,, 0, rn 1,.

.,

n

Ix,,

tasks have been assignedtomachine

Mr,,,

].

Introduce anarray/x of size n

[T

willbe assignedto

Mt,

]. Set

rn 1.

Iterate.

for j 1 ton do

begin

set m -ifdi_l

_<di

then 1 elseif

x.

<

_d.

then m elsem

+

1; set_tzi m,

xm

xm +

1

end.

Finalize. Set

Xo*-

0;for m 1 ton doset

X,

X,_

+

x,.

It can be shown that

X,

is the maximumnumber of tasks that can be completed in time on m fast machines, for m=0,...,n

[13].

The algorithm requires

O(n

log

n)

time to order the tasks, and

O(n)

time to construct the schedule and to determine the values

Xo,’",X,. It

follows that an overall optimal schedule is obtained in

O(n

log

n)

time. []

Note.

Since x,,

_->x,+,

rn 1,..., n-l,X, is a concave function of m, so that

C,

isconvex.

A

similarobservation will beexploited in the next section.

5.

A

well-solvablecase of the

FST

problem.

THEOREM 5.

In

thecase thatqi

1,/"

1,.

,

n, the FST problemis solvablein

O(n

log

n)

time.

The assumption thatallq 1 istoostrong" an analysis of theproofbelow shows thatour algorithm is applicable in the more generalsituation that theq/are bounded from above by the minimum length of the interval between two different adjacent release dates.Although thisrestriction stilllimitsthe practical valueofourresult,we

feel thatthe insight gained might be usefulinthe design of approximation algorithms forthe general

FST

problem.

Proof

of

Theorem 5. The development of our algorithm will proceed along the same lines as in the previous section. First, we will assume that there are rn fast machines and we will determine an optimal set of tasks to be scheduled on these

machines.

Next,

we will compute the minimum number of slow machinesneeded to execute the remaining tasks. Finally,we willdescribe an efficient methodtofind the optimal value of m.

We

start by representing the problem data in a convenient way.

Suppose

that the release dates assume k different values 1,’",

r-k

with _{1<’’" <k. For} ] 1,...,k, there are

_n

tasks

T1/,..

,

T,

with release dates r0. rnji

?i

and deadlines

dl>-...>-d,j. We

have n

==1

_n

and define

n’=max<=<__k{n}.

This

representation can be obtained by sorting the release dates and the deadlines in

O(n

log

n)

timeand applyingabucketsort

[1]

toorder the taskswiththesame release date accordingtodeadlines in

O(n)

time.

Letus nowassumethat thereare rn fast machines

M(,.

,

M,

with0_-< rn -<

n’.

For

_f=

1,...,k, each of these machines can process exactly one of the tasks

T,...,

_T.

It is obviously optimal to assign

_Tit

to

M

for j 1,..., k and i=

1,...,

min

{n,

m},

so that the remaining tasks will be as short as possible. Let

-,

denote the set of tasks that are not

assigned

to the rn fast machines, where

’o

{T1,

,

T,}

and

,,

,

and let

_lm

denote theminimumnumber of slowmachines

needed to execute these tasks.

A

schedule using rn fast machines has total cost

Cr

mCf+

I,C

s.

Itfollows that an optimal schedule uses

m*

fast machines, where

Cm.

mino=<,=<,,

{

C,,

}.

For

each given value of m, the number l,, anda corresponding schedule of the tasks inff, on

Im

slow machines can be found in

O(n

log

n)

time. This problemhas

alreadybeen discussed inthe first two paragraphs of 1. The following algorithm is a slightmodification ofthe channel assignment algorithm of

Gupta,

Lee and

Leung

[8];

for simplicity, it isstated for thecasethatm 0.

FST ALGORITHM (only slow machines)

Initialize. Reorder the tasks in suchawaythat

_r

<-

_<=

r,;determinea permuta-tion 8 of

{1,...,

n}

such that

d8(1)=<

<-

_ds(,).

Introduceastack$ of

size n and push machineindices 1,...,n onto $in such awaythatrn ison top of rn

+

1,rn 1,.

.,

n 1. Introduce an arrayh ofsize n

[T

willbe assignedto

M

A

Set/"

1 1

Ierate.

while j

-_

n do

if

_r

<

thenbeginset

_A

<-- top element of

S;

pop S; set

<--]

+

1 end

elsebeginpush

_A

ontoS;set <--

+

1 end. Finalize. Set

_lo

<-- maxi_<,

{Ai}.

It can be shown that

lo

is the minimum number of slow machines needed to executealltasks. Thealgorithmrequires

O(n

log

n)

time toorder thetasks,and

O(n)

SCHEDULING TASKS IN FIXED INTERVALS 519

time to construct the schedule and to compute the value

_lo.

Since the release dates and the deadlines havealreadybeensorted,each application ofthisalgorithm requires only

O(n)

time. Straightforward computation of l, for m=0,...,

n’

would yield an overall optimal schedulein

O(n

logn

+

n’n)

O(n

2)

time.

However,

it will be shown below that C, is a convex function of m, and this

propertycan be exploitedto arrive at an

O(n

log

n)

algorithm. The convexity of C,

impliesthat,ifC,

<

C,/x,then

m*

{0,

,

m},

and else

m*

{m

+

1,...,

n’}.

Thus,

m*

can be found by a bisection search as follows" for m

[n 1,

compute C, and C,/1, reduce the domain of

m*

by a factor of two by eliminating either

[0,

m]

or

[m+

1,

n’],

and repeat theprocedureonthe remaininginterval.The optimal value of

m isfound inat most

[log2 (n’+

1)]

iterations.

The entire algorithm requires

O(n

log

n)

time to sortthe release dates and the

deadlinesand,foreach ofO(log

n’)

valuesofm,

O(n)

timetocompute C,. Itfollows that

an

overalloptimal scheduleisobtainedin

O(n

log

n)

time.

Itremains to beshown that

C,

is aconvexfunction ofm. Since

_Cm

mC

+

l,C

s,

wehavetoprove thatl, is convex,orequivalently that

(1)

l,_l-l,>-l,-l,+l, re=l,..., n’-l.

We

define the degree

of

overlap ofthe set

such that [rj, dj). Let

X,(t)

denotethedegreeof overlap of ff, at andx,_l(t)the degree ofoverlapof

T,-I

T,

at t,i.e.,X,-l(t) x,_l(t)-X,(t).

It

isknown

[9]

that

(2)

l, maxt

{X,,(t)},

m 0,...,

n’.

Since the number oftasks

_T

,-1-

’,

and thelengths of their intervals

[r, d)

do

notincrease asmincreases, it isalsotruethat

(3)

Xm-l(t)

>--X,(t)

all t, m 0,...,

n’-

1.

Defining

t

such

thatXm(tm)

maxt

{Xm(t)},

m 0," ",

n’,

and applying

(2),

werewrite

(1)

as

X,-l(t,-l)-X,(t,) >-_X,(t,)-X,+(t,/x).

We

haveforthe left-handsidethat

Xm-l(tm-1)-Xm(tm)

X,-l(t,-l)-X,-l(tm)

+

Xm-l(tm)

>-_x,-l(t,).

Similarly,wehave for the right-handside that

X,(t,)--Xm+l(t,+l)

X,+l(t,)+

x,(t,)-X,+l(t,/l) <-_x,(t,).

Application of

(3)

for t, nowimplies the validity of

(1).

Thiscompletesthe proof

ofTheorem5.

Note. By

means

ot

ingenious counting techniques, the above algorithm for

com-puting a single value l, can be extended to an

O(n

log

n)

algorithm for computing all lo,’’’, ln, together

[13];

whenthe data have alreadybeen sorted,itrequires only

O(n)

time,asbefore.

A

similarresult hasbeenusedin the previoussection.

Acknowledgment. The authors gratefully acknowledgeconstructive suggestions byB.

J. Lageweg.

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