Deterministic Blind Rendezvous in Cognitive Radio Networks

0.15 MB·

by  Sixia Chen

Deterministic Blind Rendezvous
in Cognitive Radio Networks
SixiaChen AlexanderRussell
Department ofComputerScience and Engineering
UniversityofConnecticut
4
1 AbhishekSamanta Ravi Sundaram
0
CollegeofComputerScience
2
NortheasternUniversity
n
a
J January 29, 2014

8
2

] Abstract
I
N
Blind rendezvous is a fundamental problem in cognitive radio networks. The problem involves a
. collectionofagents(radios)thatwishtodiscovereachother(i.e.,rendezvous)intheblindsettingwhere
s
c there is no shared infrastructure and they initially have no knowledge of each other. Time is divided
[ into discrete slots and spectrum is divided into discrete channels, [n]= 1,2,...,n . Each agent may

{ }
1 access(orhopon)asinglechannelinasingletimeslotandtwoagentsrendezvouswhentheyhoponthe
v samechannelinthesametimeslot. Thegoalistodesigndeterministicchannelhoppingschedulesfor
3 eachagentsoastoguaranteerendezvousbetweenanypairofagentswithaccesstooverlappingsetsof
1
channels.
3
Theproblemhasthreecomplicatingconsiderations:first,theagentsareasymmetric,i.e.,eachagent
7
. AionlyhasaccesstoaparticularsubsetSi [n]ofthechannelsanddifferentagentsmayhaveaccessto
1 ⊂
differentsubsetsofchannels(clearly,twoagentscanrendezvousonlyiftheirchannelsubsetsoverlap);
0
second, the agents are asynchronous, i.e., they do not possess a common sense of absolute time, so
4
1 differentagentsmaycommencetheirchannelschedulesatdifferenttimes(theydohaveacommonsense
: ofslotduration);lastly,agentsareanonymousi.e.,theydonotpossessanidentity,andhencetheschedule
v
i forAimustdependonlyonSi.
X
Whetherguaranteedblindrendezvousintheasynchronousmodelwasevenachievablewasanopen
r problem.Inarecentbreakthrough,twoindependentsetsofauthors,Shinetal.[CommunicationsLetters
a
2010] and Lin et al. [INFOCOM 2011], gave the first constructions guaranteeing asynchronousblind
rendezvousinO(n2)andO(n3)time,respectively.Wepresentasubstantiallyimprovedandconceptually
simpler construction guaranteeing that any two agents, A, A , will rendezvous in O(S S loglogn)
i j i j
| || |
time. Ourresultsarethefirstthatachievenontrivialdependenceon S ,thesizeofthesetofavailable
i
| |
channels.Thisallowsus,forexample,tosaveroughlyaquadraticfactoroverthebestpreviousresultsin
theimportantcasewhenchannelsubsetshaveconstantsize. Wealsoachievethebestpossibleboundof
O(1)rendezvoustimeforthesymmetricsituation;previousworkscoulddonobetterthanO(n). Using
the probabilistic method and Ramsey theory we provideevidencein supportof our suspicion that our
constructionisasymptoticallyoptimal(uptoconstants)forsmallsizechannelsubsets: weshowbothan
W (S S )lowerboundandanW (loglogn)lowerboundwhen S , S n/2.
i j i j
| || | | | | |≤
1 Introduction
1.1 Motivation
Given the ever-increasing demand for all things wireless, spectrum has become a scarce resource. Histor-
ically, regulators around the world have employed a command and control philosophy towards managing
spectrum [24]: Somechannels werestatically licensed toparticular users(forcertain periods andincertain
geographies) while others were kept aside for community use. Cognitive radio networks have emerged as
amodern, dynamic approach tospectrum allocation [27;1]. Exploiting recent technological developments,
cognitive agents (radios) dynamically sense incumbent users and opportunistically hop to unused channels.
While they can offer improved utilization, they introduce a fundamental rendezvous problem: the problem
ofdiscovering theexistence ofpeersinamultichannel setting.
1.2 Model and Results
We work in the blind model where a collection of agents A wish to discover each other with no dedicated
i
common control channel or other shared infrastructure. Timeis divided into discrete slots and spectrum is
divided into discrete channels, [n]= 1,2,...,n . Each agent may access (or hop on) asingle channel in a
{ }
single time slot and two agents rendezvous when they hop on the same channel in the same time slot. The
challengeistodesignachannel-hopping schedule foreachagentsothattheydiscovereachother. Asstated
thus far, the problem has the trivial solution where all agents can hop on a specific channel, say channel
1, in the very first time slot. However, reality is complicated by three additional requirements: asymmetry,
asynchrony andanonymity.
AsymmetryDifferentagentsmayhaveaccesstodifferentsubsetsofchannelsasaresultoflocalinterference
orvariations inradiocapabilities. LetS [n]bethesubsetofchannels towhichagentA hasaccess. Thus
i i
⊆
thechallengeistocreateforeachagentA achannel-hopping schedules : 0,1,... S whichguarantees
i i i
{ }→
that t,s (t)=s (t) for any two agents A,A , s.t. S S =0/. (In the symmetric setting all agents have
i j i j i j
∃ ∩ 6
accesstotheidentical subsetofchannels.)
Asynchrony Different agents may not share a common notion of time. They may commence at different
“wake-up” times inducing a relative shift in their progress through their schedules. Note that agents do
possess a common understanding of slot duration. The goal, therefore, is to ensure rendezvous between a
pair of agents in the shortest possible time once they have both woken up. (In the synchronous setting all
agentsshareacommonnotionofabsolute time.)
AnonymityInoursettinganagent’s schedule mustdependonlyonthesubsetofchannels available to,and
notonadistinctidentityof,theagenti.e.,s mustdependonlyonS. NotethatS isunknowntoA fori= j
i i j i
6
anditisallowedfortwodifferentagentstohavethesamesetofaccessible channels, i.e.,S =S fori= j.
i j
6
Now, the problem has the naive randomized solution, in which each agent, at each time step, selects a
channeluniformlyandindependentlyatrandomfromitssubset. Itisnothardtoseethatthisprovidesahigh-
probability guarantee of rendezvous for agents A,A in time O(S S logn). However, the deterministic
i j i j
| || |
settingisthegold-standard inthecognitiveradionetworkingcommunity: itmakestheweakestassumptions
aboutthedevices,whichneednothaveanavailablesourceofrandomness,andprovidesabsoluteguarantees
onrendezvous time.
Herewebrieflysummarizeofourmainresults:
Algorithms
1. WegiveanO(loglogn)timealgorithm forrendezvous forthespecialcaseofagentswith S =2.
i
| |
2. Wethenshowhowtoapplythisalgorithm toyieldalgorithms forarbitrary subsets of[n]thatguaran-
teesrendezvous timeO(S S loglogn)forallpairsofsetsS andS .
i j i j
| || |
1
3. We show that a minor adaptation of this algorithm can furthermore guarantee O(1) time rendezvous
forthesymmetriccase.
4. Finally, weexplore the “one bit beacon” case, where the agents have the luxury of asingle common
random bitduring eachtimeslot. Inthis model, weshow that O(S + S +logn)timeissufficient,
i j
| | | |
withhighprobability, torendezvous.
LowerBounds
1. Weprove anW (loglogn)lowerbound ontherendezvous time,even forsynchronous agents withthe
promise that the channel sets S have constant size. This shows that some dependence on n, the size
i
ofthechannel universe, isalwaysnecessary. Inparticular, thisshowsthatthealgorithm of1aboveis
tightuptoaconstant.
2. For channel subsets of size k we prove a k2 lower bound on even the synchronous rendezvous time,
under the promise that k=O(logn/loglogn). For larger values of k, we obtain a weaker family of
results.
3. In the asynchronous time model, weprove that S S steps are necessary to rendezvous, so long as
i j
| || |
S + S n.
i j
| | | |≤
We also consider a one-round version of the problem; instead of minimizing the number of rounds we
consider the problem of maximizing the number of pairs of agents that can achieve rendezvous in a single
round. In particular for the “graphical” case where channel sets are of size 2 we show how a variant of
thecelebrated Goemans-Williamson semi-definiteprogram [7]forMAX-CUTcanbeemployedtoobtaina
0.439approximation fortheone-round maximization version. Thisresultispresented intheAppendix.
1.3 Related work
Rendezvousproblemshavealonghistoryinmathematicsandcomputerscience—anearlyexampleisRado’s
famous “Lion and Man” problem [3]. Over time a variety of problems and solutions have evolved in both
adversarial [12] and cooperative settings [2]. Rendezvous in networks has been extensively studied in the
computersciencecommunity[18]. Thoughthestudyofrendezvousincognitiveradionetworksisrelatively
recent there already exists a comprehensive survey [17] that contains a detailed taxonomy of the different
models including the specific one relevant to this work. The problem of guaranteed blind rendezvous in
theasymmetric,asynchronous andanonymouscasewasfirstconsidered in[4]andsubsequently in[20;16].
The use of prime numbers and modular algorithms was initiated in [22]. However, the general case of the
problemwithstoodattackuntil[21;15]. Thecurrentstateoftheartis[9]whichachievesanO(n2)algorithm
fortheasymmetriccaseandO(n)forthesymmetriccase. Acrucialdifference betweentheseconstructions
and ours is that we explicitly exploit the fact that the schedule s can depend arbitrarily on S, whereas the
i i
earlier constructions [21; 15; 9] derive the schedule for a channel subset by (essentially) projecting onto
the desired subset from a single uniformly generated schedule for the full set of channels. Our work is
notableforprovidingaconceptually cleanandsignificantly moreefficientO(S S loglogn)algorithm for
i j
| || |
the general asymmetric setting. Real-world cognitive networks [25] create a pooled hyperspace occupied
by signals with dimensions offrequency, time, space, angle ofarrival, etc., created by advances inantenna
design, and comprising spectrum ranging from radio frequencies and TV-band white spaces to lasers. In
these networks the total number of channels, n is large while the channel subsets accessible to any given
device aresmall. Asimilarsituation prevails inmilitary situations wheredifferent membersofa(dynamic)
coalition operate in a small portion of the available spectrum that guarantees overlap with allies. In such
situationsourschemeachievesanear-quadratic factorgainoverthepreviousresults. Andforthesymmetric
2
setting our construction achieves O(1) rendezvous time which clearly cannot be bettered. Table 1 presents
asummaryofourupperbounds inthecontextofpriorwork.
Table1: Upperboundsfordeteministic rendezvous
Paper Asymmetric Symmetric
Shin-Yang-Kim [21] O(n2) O(n2)
Lin-Liu-Chu-Leun [15] O(n3) O(n)
Gu-Hua-Wang-Lau[9] O(n2) O(n)
Ourresults O(S S loglogn) O(1)
i j
| || |
We are also the first to provide nontrivial lower bounds employing tools from Ramsey theory and the
probabilistic method. [6] is a closely related work; its globally synchronous and locally synchronous mod-
els correspond to our asynchronous and synchronous models respectively. However, while [6] explicitly
requires that exactly one node transmits on a single fixed channel for a successful broadcast, we implicitly
assumethatonceasetachieves rendezvous (onanyoneofseveralchannels) thentheyemploythestandard
“chirpandlisten”technique [26]toensuremutualidentification ofthesetmembers.
2 Definitions and notation
LetSbeacollectionofsubsetsof[n]. AnS-scheduleisafamilyofscheduless :N S,oneforeachS S.
S
→ ∈
Infact,wefocussolelyontwospecial cases:
Ann-schedule isa2[n]-schedule, onethatsupplies aschedule foreverysubsetof[n].
•
An(n,k)-schedule isaS-schedule, whereSconsists ofallsubsetsof[n]ofsizek.
•
We will typically reserve the notation S =(s A)A S to denote an S-schedule; departing from the notation
usedintheintroduction, theschedule associated w∈iththesetAissimplydenoted s .
A
Lets :N Aands :N BbetwoschedulesforoverlappingsubsetsAandBof[n]. Wesaythats
A B A
→ →
and s rendezvous synchronously intime T ifthere is atimet T so that s (t)=s (t); this corresponds
B A B
≤
torendezvous inthesynchronous model discussed intheintroduction. Recallthat theasynchronous model
introducesarbitrary“wake-up”timest andt intoeachofthetwoschedules, afterwhichtheyproceedwith
A B
their schedules. Ofcourse, inthiscase theycannot possibly rendezvous before timemax(t ,t ),whenthey
A B
arefinallyboth“awake.” Thus,wesaythatthesetwoschedulesrendezvous asynchronously intimeT if,for
allt ,t 0,thereisatimemax(t ,t ) t max(t ,t )+T sothats (t t )=s (t t ).
A B A B A B A A B B
≥ ≤ ≤ − −
Forafixed(n,k)-schedule S ,wedefineR (S )tobetheminimumT forwhichs ands synchronously
s A B
rendezvous in timeT for allA,B S. Welikewise define R (S )for asynchronous rendezvous. Finally, we
a
∈
define:
R (n,k),minR (S ) and R (n,k),minR (S ),
s s a a
S S
where these are minimized over all (n,k)-schedules S . Of course, R (n,k) R (n,k), and the simple ran-
s a
≤
domizedalgorithm described intheintroduction suggeststhatperhaps
R (n,k) k2.
a
≈
Finally, we remark that even a precise understanding of R (n,k) does not necessarily yield n-schedules
a
that guarantee satisfactory bounds on pairwise rendezvous because it is not, in general, clear how to stitch
together (n,k)-schedules fordifferentvaluesofktoprovideguarantees forpairsofsetsofdifferent sizes.
3
Notation Weuse[n]= 1,...,n andinventtheshorthandnotationlog♯n, log n . Wheneveravariable,
2
{ } ⌈ ⌉
x, represents a natural number, we use x to denote the canonical base-two encoding of x, zero-padded on
2
theleftouttolengthlog♯m,wheremisthemaximumvaluethatxmighttake.
3 Schedules for efficient rendezvous
Sets of size two We begin with a construction of a family of schedules for channel sets of size 2 that
achievesrendezvousintimeO(loglogn);thesewillbeusedasasubroutineforthegeneralconstruction. We
shall seeinSection 4that theseschedules arewithin aconstant ofoptimal. Thus, thegoalofthissection is
toprovethefollowingtheorem.
Theorem 1. For all n>0, R (n,2)=O(loglogn). Specifically, for any n>0, there is an (n,2)-schedule
a
so that for any two sets A and B of size two, s and s rendezvous asynchronously in time no more than
A B
O(loglogn).
Thesize2construction isbasedontheremarkablefactthatthereisanedgecoloring ofthelinearposet,
using only log♯n colors, for which no path of length two is monochromatic. Specifically, consider the
directed graphL =(V ,E ),withvertexsetV =[n]anddirected edgesE = (a,b) a<b . A2-Ramsey
n n n n n
{ | }
edgecoloringofL isamappingc :E Pwiththepropertythatc (a,b)=c (b,c)foranypairofdirected
n n
→ 6
edges(a,b)and(b,c)thatformadirected pathoflength2.
Lemma2. ThegraphL hasa2-Ramseyedgecoloringwithapaletteofsizelog♯n.
n
Proof. Withhindsight, associate witheachvertexk V theset
n
∈
X = i theithbitofk isa1 1,...,log♯n .
k 2
{ | }⊂{ }
Observe that if a<b, there is an element in X X . In this case, we may safely color the edge (a,b) with
b a
\
any element of X X , as it follows immediately that any pair of edges forming a directed path must have
b a
\
distinctcolors. Theschemeusesnomorethanlog♯ncolors.
ProofofTheorem1. Webegin with aconstruction for the simpler synchronous model, and then show how
toreducetheasynchronous modeltothiscase.
Thesynchronousmodel. In the synchronous model, we will simplify the presentation by discussing finite
length schedules with the understanding that rendezvous is guaranteed by the time the schedule has been
exhausted. ConsidernowasubsetoftwochannelsA= a ,a ,wherea <a . Wewilltreatsuchsize-two
0 1 0 1
{ }
subsets as directed edges of the linear poset (directed from the smaller element to the larger element). In
thissize-twocase,wemayexpressascheduleasabinarystrings s s ... 0,1 withtheconvention that
0 1 2 ∗
∈{ }
attimet,theschedule callsfora : thus,whens =0theschedule callsforthesmallerofthetwochannels;
st t
whens =1,theschedule callsforthelargerofthetwochannels.
t
Consider now a pair of overlapping subsets A= a ,a and B= b ,b with a <a and b <b .
0 1 0 1 0 1 0 1
{ } { }
When these two edges form a directed path (so that their common element is the larger of one set and the
smaller of the other), a sufficient condition for two schedules r r ...r and s s ...s to rendezvous is
0 1 ℓ 1 0 1 ℓ 1
− −
thateachofthetwotuples (0,1),(1,0) canberealizedas(r ,s )forsomet,whichistosaythat
t t
{ }
(0,1),(1,0) (r ,s ) 0 t <ℓ . (1)
t t
{ }⊂{ | ≤ }
Wereservethenotationr♦ stodenotethestatementthatthestringsrandssatisfycondition (1). Likewise,
1
when a ,a and b ,b do not form a path of length two (that is, share a common largest or smallest
0 1 0 1
{ } { }
element), asufficientcondition forrendezvous isthat
(0,0),(1,1) (r ,s ) 0 t <ℓ . (2)
t t
{ }⊂{ | ≤ }
4
Wereservethenotation r♦ stodenotethestatementthatrandssatisfy(2).
0
Intheremainder oftheproofweidentify amapx C(x)withtheproperty that
7→
x=y C(x)♦ C(y), (3)
0
⇒
x=y C(x)♦ C(y). (4)
1
6 ⇒
Withsuchamapinhand,weadoptthescheduleC(c (a ,b ) )fortheset a ,b ,wherec istheedgecoloring
2
ofLemma2. ObservethatifA= a ,a andB= b ,b formapath{oflen}gthtwo, c (a ,a )=c (b ,b )
0 1 0 1 0 1 0 1
{ } { } 6
and this schedule guarantees rendezvous by dint of property (4). Otherwise, these schedules guarantee
rendezvous bydintofproperty (3).
We return to the problem of constructing the map C(). By adopting the convention that all schedules
·
startwiththeprefix01,wecanimmediatelyguaranteeproperty(3): (0,0)and(1,1)appearin (r ,s ) 0
t t
{ | ≤
t<ℓ . Itiseasytocheckthatthemapx 01 x x,where denotesconcatenationandxthecoordinatewise
} 7→ ◦ ◦ ◦
negationofx,hasthedesired properties.
Aleanermappingcanbeobtainedbytherule
C(x),01 x wt(x) ,
2
◦ ◦
where wt(x) denotes the weight (number of 1s) of the string x. To see that this encoding has property (4),
observe thatwhenwt(x)=wt(y),both(0,1) and(1,0)mustappear intheset (x,y) 1 i x (where
i i
{ | ≤ ≤| |}
x is the ith bit of x) as x=y and they have common weight. When wt(x)<wt(y), it follows immediately
i
6
that (0,1) (x,y) 1 i x ; as for the tuple (1,0), this must be realized by one of the coordinates
i i
∈{ | ≤ ≤| |}
of wt(x) and wt(y) as the canonical encoding of integers in binary ensures that when n<m, there is a
2 2
coordinate inwhichn contains a0andm contains a1. Thecasewhenwt(x)>wt(y)ishandled similarly.
2 2
Finally, we remark that when x has length ℓ, C(x) has length ℓ+log♯ℓ+2. AsL can be edge colored
n
withapaletteofsizelog♯n,thisyieldsafamilyofschedules forsetsofsize2thatguarantees rendezvous in
timenomorethanlog♯log♯n+log♯log♯log♯n+2.
Theasynchronousmodel.Wereturnnowtotheasynchronous modeldescribedintheintroduction, inwhich
thetwoagents’schedulesaresubjectedtoanunknownshiftduetopotentiallydistinctstart-uptimes. Inthis
model, we are obligated to define schedules for all nonnegtive times (that is, our schedules have the form
s :N S [n]); one straightforward method for describing such schedules is to adopt cyclic schedules,
→ ⊂
whichcycliclyrepeatthesamefinitesequence ofchannels. Inparticular, ifs : 0,...,ℓ 1 S [n],we
{ − }→ ⊂
lets :N Sdenotetheschedule s :t s (t mod ℓ).
◦ ◦
→ 7→
Continuing in the spirit of the previous discussion, we observe that if r=r ...r and s=s ...s
0 ℓ 1 0 ℓ 1
− −
aretwoschedules forapairofsetsA= a ,a andB= b ,b forming apath, thecyclic schedules they
0 1 0 1
{ } { }
inducewillguarantee rendezvous (intimeℓ)if,foralliand j,
Sir♦ Sjs, (5)
1
whereSixdenotestheresultofcycliclyshiftingxforwardisymbols. Tosaveink,wedefiner(cid:7) stodenote
1
the condition (5): Sir♦ Sjs for all i and j. Likewise, we define r(cid:7) s when Sir♦ Sjs for all i and j. As
1 0 0
above,whenthesetwosetsdonotformapath,r(cid:7) sisasufficientcondition forrendezvous.
0
Thusourstrategyshallbetodefineamapx R(x)withtheproperty thatfortwostringsx,y,
7→
x=y R(x)(cid:7) R(y) and x=y R(x)(cid:7) R(y). (6)
0 1
⇒ 6 ⇒
With such a map defined, the construction follows that of the previous construction: the cyclic schedule
adopted bythepair(a ,b )isgivenbyR(c (a ,b ) )wherec isanedgecoloring ofL .
2 n
5
Anticipating theconstruction, wesetdownsometerminology. Forastringz,wedefinethe“graph”ofz
tobethefunction G : 0,..., z Zgivenby
z
{ | |}→
k
(cid:229)
G (0)=0, G (k)= (2z 1)
z z i
−
i=1
sothat G traces outthe“walk” prescribed byzinwhich each 1corresponds toastep northeast and each 0
z
corresponds to a step southeast as in Figure 1a. We say that a binary string z is balanced if wt(z)= z/2
| |
(sothat z isnecessarily even);equivalently G (z)=0,seeFigure1b. AbalancedstringzisCatalanifG
z z
| | | |
isnevernegative. IfG ispositive, which istosaythat G (i)>0for all0<i< z, wesaythat zisstrictly
z z
| |
Catalan; see Figure 2. We remark that if z is Catalan, 1 z 0 is strictly Catalan. Finally, we say that z is
◦ ◦
t-maximaliftheset i G (i)=max G (j) hassizeexactlyt;thenotiont-minimalisdefinedanalogously.
z j z
{ | }
Note that a strictly Catalan sequence z is 1-minimal and this single minimum appears at i=0. We remark
thatifthestringzist-maximal(ort-minimal),thesamecanbesaidofallshiftsofz.
(b) The graph of the balanced sequence
(a)Thegraphofthesequence11010. 110001.
Figure1: Graphsandbalancedstrings.
Our strategy is to work with an injective map R() with the property that R(x) is balanced, strictly
·
Catalan, and 2-maximal. Before describing a construction, we observe that such a map has the properties
outlined in(6)above.
Observe, firstof all, that iftwo distinct strings R(x) and R(y) are balanced, it follows immediately that
R(x)♦ R(y),indeed,thenumberofappearances of(0,1)isthesameasthenumberofappearancesof(1,0)
1
and cannot be zero because the strings are distinct. Thus, when R(x) and R(y) are balanced, the condition
that R(x) SiR(y) i [ℓ] is enough to guarantee that R(x)(cid:7) R(y). Note that if a string z is strictly
1
6∈ { | ∈ }
Catalan,nonontrivialshiftofzcanbestrictlyCatalan. Inparticular, allnontrivialshiftsofastrictlyCatalan
string are 1-minimal (as this is a property enjoyed by strictly Catalan strings) with adifferent unique point
ofminimality. Itfollowsthatx=y R(x)(cid:7) R(y),asdesired.
1
6 ⇒
Toensure that R(x)♦ R(y), when R(x) and R(y) arebalanced it suffices to exclude the possibility that
0
R(x)=R(y); similarly, the number of appearances of (0,0) is the same as the number of appearances of
(1,1), and cannot be zero unless the strings are complements. We conclude that, for two balanced strings
R(x)and R(y),the condition R(x) SiR(y) i [n] implies thatR(x)(cid:7) R(y). Observe that asstring zis
0
6∈{ | ∈ }
k-maximalifandonlyifzisk-minimal. ThusifR(x)andR(y)are1-minimal(astheymustbeiftheystrictly
Catalan),and2-maximal, thenR(x)=R(y). ThusR(x)(cid:7) R(x)forallx,asdesired.
0
6
Itremainstoshowthatwecanefficiently construct suchafunction.
Our starting point shall be the “Knuth mapping” x K(x) on all the binary strings; this is an efficient,
7→
injectivemappingwiththeproperty thatK(x)isbalanced; moreover,
K(x) x +log♯ x +(1/2)log♯log♯ x .
| |≤| | | | | |
(SeeKnuth[13]forfurther discussion.) Observethatifzisbalanced, thereisatleastoneshiftSczwhichis
Catalan. Toyieldaninvertible process, weconsider themap
U(z)=(Scz) 1ℓ/2 K(c ) 0ℓ/2,
2
◦ ◦ ◦
( )
∗
| {z }
6
(b) The graph of a shifted strictly Catalan se-
(a)ThegraphofastrictlyCatalansequencez;re- quencez. RemainingG valueslieintheshaded
z
mainingG valuesmustlieintheshadedarea. region.
z
Figure2: Catalansequences.
(a)Thegraphofasequence,showingamaximum(b) The sequence after the transformation to 2-
value. maximality.
Figure3: Thetransformation to2-maximality.
where ℓ= K(c ). Note that the string ( ) is Catalan, as K(c ) is balanced and hence has no more than
2 2
| | ∗
ℓ/2zeros. ItfollowsthatU(z)isCatalan(astheconcatenation oftwoCatalanstrings isCatalan). Sincethe
shift c is encoded into U(), the function is clearly injective. It follows that the map z 1 U(K(z)) 0 is
· 7→ ◦ ◦
invertible, andcarriesztoastrictly Catalanimage. Finally, weobserve thatinserting thestring 1010atany
maximal point inastring ztransforms itintoa2-maximal string inaninvertible fashion (and preserves the
otherpropertieswecareabout). WeletM(z)denotethistransformation; seeFigure3. Tocompletethestory,
wedefine
R(z),M(1 U(K(z)) 0)
◦ ◦
andobserve that R(z) z +4log♯ z +16. Sincezisanedgecolorwithlengthlog♯log♯n,thetheorem is
| |≤| | | |
proved.
3.1 A general n-schedule
In this section we show how to apply the previous result to yield n-schedules that provide rendezvous in
timeO(A B loglogn). Specifically, weprovethefollowingtheorem.
| || |
Theorem 3. There is an n-schedule so that for all overlapping A,B [n], the schedules s and s ren-
A B
⊆
dezvous asynchronously intimeO(A B loglogn).
| |·| |
Proof. ConsiderasetA= a ,...,a . Theschedule forAdepends onapairofprimes p,p intherange
0 k 1 ′
{ − }
[k,3k](there always exist twoprimes in this range). Wethen construct aschedule consisting ofasequence
of epochs, where the rth epoch calls for the size-two schedule of Theorem 1 involving the twochannels a
i
and a , where i r mod p and j r mod p. (If either i or j do not fall in the range 0,...,k 1 , then
j ′
≡ ≡ { − }
wechoose anarbitrary elementofAtofillitsplace.)
Inthefollowing, wewillsayapairofprimenumbers (p,q) ishelpful fortherendezvous oftwoagents
A and B if: (i.) p is one of the primes selected by the first agent as described above, (ii.) q is one of the
primes selected by the second agent as described above, and(iii.) p=q. Theconstruction above specifies
6
7
that each agent mustchoose twoprimes toensure that anytwoagents areguaranteed tohave ahelpful pair
betweenthem.
Now, suppose A B= c , and that c=a =b (so that c is the xth channel in A and the yth channel
x y
∩ { }
in B). In the synchronous model, we use the construction described in the proof of Theorem 1 to get a
schedule for (a,a )in each epoch. In this case, itsuffices to show that there is an epoch r satisfying r x
i j
≡
(mod p)andr y (mod q),where pandqareahelpfulpairasdescribedabove. AccordingtotheChinese
≡
Remainder Theorem, there exists asolution for r that isno more than pq. Therefore, inthe worst case, the
twoagentswillbothaccessthecommonchannelatonetime,nolaterthan pq(loglogn+logloglogn+2)=
O(kℓloglogn)stepsaftertheirschedules commence.
Theasynchronousmodelrequiresonlyaslightmodification. Supposethat,foragivenepoch,r,anagent
using the scheme described immediately above with subset Aexecutes schedule s r of length R (all epochs
A
havethesamelength). Then, wecanhandle asynchronous rendevous bydoubling thelength ofeachepoch
and executing s rs r. Assume the commencement time for the s is to be t and the commencement time
A A A a
fors istobet where,withoutlossofgenerality,t t . Letm denotetheclosestintegerto tb ta. Thenfor
B b a≤ b 2−R
anyr,therth epochofs willoverlapwiththe(r m )th epochofs byatleastRtimesteps. Foranyrsuch
A B
−
thatr x (mod p)and(r m ) y (mod q), wherethepair(p,q)ishelpful, thentherth epoch ofs will
A
≡ − ≡
overlap with the (r m )th epoch of s no less than R timeslots. Since Theorem 1 guarantees rendezvous
B
betweens Ar andany−cyclicshiftofs Br−m ,thisoverlapmustcontainsucharendezvous point.
Again by the Chinese Remainder Theorem, we know that there exists a epoch r such that r m is no
−
more than pq. Therefore, in the worst case, the two agents will access the same channel in time 2pqR=
O(kℓloglogn)aftert .
b
3.2 A general reduction thatguarantees fastsymmetricrendezvous
The rendezvous literature has given special attention to the symmetric case, where A=B. For a general
schedule that guarantees rendezvous for all (perhaps distinct) pairs of sets, one specifically examines the
rendezvous time in this symmetric case. In this section, we observe that any schedule that guarantees
rendezvous for all pairs of sets can be transformed into one that additionally guarantees O(1) rendezvous
time in the symmetric case, at the expense of a constant blow-up in the rendezvous time for all other pairs
ofsets.
Specifically, for afamily of schedules S =(s ) , for each A [n], wedefine anew schedule sˆ as
A A [n] A
follows: when s calls for the channel c , sˆ carrie⊂s out a short se⊂quence of accesses, consisting of the
A 1 A
channel c and the channel c =min A (the smallest element of A) in the pattern c c c c c c repeated
1 0 0 1 0 0 1 1
{ }
twice. Thesignificance ofthis pattern isthat 010011(cid:7) 010011: thus any pairofrotations ofc c c c c c ,
0 0 1 0 0 1 1
will yield simultaneous accesses to both (c ,c ) and (c ,c ). To ensure that there is sufficient overlap in
0 0 1 1
these short sequences ofaccesses, werepeat them twice: asinthe proof ofTheorem 3,this guarantees that
a full rotation of the sequence overlaps. By a similar argument, it follows that the time to rendezvous, for
anypairofsets,isnomorethanaconstantfactor(12,bythisconstruction) largerthaninS . However,when
A=B,suchapairwillrendezvous (attheirsmallestelement)inconstant time.
4 Lower bounds
Inthissectionweestablishthat
1. R (n,k)=W (loglogn)foranyk n/2. (Theorem4andCorollary5.)
s
≤
2. R (n,k) k2 forallk=O(logn/loglogn)and,ingeneral, R (n,k) a kforallk n1/2a (solongas
s s
≥ ≥ ≤
a k). (Theorem6.)
≤
8
3. R (n,k) k2 forall2 k n/2. (Theorem 7.)
a
≥ ≤ ≤
Thelowerbounds provided byitems 2and 3exhibit anenormous gapfor large k and, indeed, thebehavior
of R (n,k) and R (n,k) must diverge for k √n. In particular, R (n,k) =W (k2) while there is a simple
s a a
≈
algorithm that shows that R (n,k) n for all k: each agent hops on channel t at time t when t is in the
s
≤
channelset,andremainssilentotherwise.
Thedependenceofrendezvoustimeonn. WebeginwithtwolowerboundsthatestablishthatR (n,k)
s
¥ asn ¥ . →
→
Theorem 4. For all n 2, R (n,2)=W (loglogn). Rendezvous requires at least W (loglogn) time, even in
s
≥
thesynchronous modelwhenagentsarepromisedtohavesetsofsize2.
Proof. Consider the complete graph K , with the interpretation that each vertex represents a channel and
n
each edge represents a set of size two. In this case where agents correspond to two channels, we represent
N
schedules as binary sequences, s 0,1 , with the convention that a 0 calls for hopping on the smaller
∈{ }
channeland1callsforhoppingonthelargerchannel.
Let S be an (n,2)-schedule which guarantees rendezvous synchronously in T. In this case, we may
treat each s as a finite length string in 0,1 T, with the understanding that rendezvous is guaranteed
(i,j)
{ }
before any schedule is exhausted. Treat the schedules s 0,1 T as a coloring of the edges of K .
(i,j) n
∈ { }
AccordingtoavariantofRamsey’stheorem,anym-coloringoftheedgesofthecompletegraphmusthavea
monochromatictrianglewhenn em!. (See,e.g.,[8].) Note,however,thatamonochromatictriangleyields,
≥
inparticular,anorderedtriplei< j<kforwhichtheschedulesassociatedwith(i,j)and(j,k)areidentical;
such schedules never rendezvous. It follows that e(2T)! n and, by Sterling’s estimate x! √2p x(x/e)x
≥ ∼
thatT =W (loglogn).
Corollary 5. Foranyk n/2,R (n,k)=W (loglogn).
s
≤
Proof. Write [n] as the disjoint union of two sets A = 1,...,m and B = m+1,...,n , where B
{ } { } | | ≥
A (k 2)=m(k 2);ourstrategy willbetoextendthesetsofsizetwoinAtoafamilyofsubsetsof[n]of
| | − −
sizek insuchawaythatschedules fortheseextended setscanbe“pulled back” toschedules forthesetsof
sizetwo(forwhichthepreviouslowerboundapplies). Toproceedwiththisidea,weexpressBasadisjoint
union B=(B1∪···∪Bm)∪Brest, where each Bi has size exactly k−2. Now, we consider the |A2| sets of
theform
(cid:0) (cid:1)
X , i,j B ,
i,j i+jmodm
{ } { }∪
where i,j A. Let S be an (n,k)-schedule. Observe that a schedule s for the set X can be treated
as schedul∈e sˇ (for i,j ) by restriction, simply replacing all referenXc{ie,js} to elements o{iu,jt}side i,j with,
i,j
{ } { } { }
say, the smaller of i and j. In general, restriction of an (n,k)-schedule to an (n,ℓ)-schedule (for ℓ < k)
does not provide any guarantee on rendezvous, even when the original (n,k)-schedule does. However, the
intersection pattern of the sets X above is chosen in such a way that the (m,2)-schedule Sˇ obtained by
i,j
definingsˇ tobetherestriction{of}theschedule s willguarantee rendezvous.
i,j Xi,j
Considertwosubsets i,j and i,j ofA,eac{ho}fsizetwo. Ifthesetwosetsarenotidenticalbutshare
′ ′
{ } { }
acommonelement,itfollowsthati+ j mod m=i + j mod m. Thus,
′ ′
6
B B =0/
i+jmodm i+j modm
∩ ′ ′
and
X X = i,j i,j
i,j i,j ′ ′
{ }∩ {′ ′} { }∩{ }
9