the statistical and probabilistic problems considered in this work are motivated by questions arising from data traffic analysis in modern communication networks such as the internet . over these networks, information is sent in the form of packets , and packets are grouped into flows .( a flow corresponds to a group of packets sharing common characteristics .ideally , a flow can be thought of as a set of packets that arises in the network through a remote terminal session or a web page download . )each packet carries information about the flow it belongs to and also whether it is the first or the last packet in the flow . examining each packetthen allows all flows to be reconstructed . knowing the structure of flows ( flow level characteristics ) helps network operators and networking researchers to understand and discover characteristics of data traffic .a key difficulty with capturing each packet is that this rapidly leads to a huge amount of data to store and analyze .for instance , capturing all traffic for a few hours on a gigabit / sec .link at a medium load level yields several hundred gigabytes of data . a way to reducethe volume of data is by sampling packets .one of the simplest approaches is probabilistic sampling , where each packet , independently of the others , is captured and analyzed with a fixed probability .the basic problem , known as the _ inversion problem _ in the network traffic literature , is then to deduce the structure of the original flows from sampled packets .this problem has recently attracted much attention in the networking community , with the focus almost exclusively on inference of the original distribution of flow sizes ( total number of packets ) ; see . in this work , we take a closer look at statistical properties of the previously considered estimator of the original distribution of flow sizes and are also interested in inference of interarrival times between packets from sampled data . this distribution allows traffic burstiness to be characterized and leads to other major characteristics such as the duration of a flow .   from a mathematical standpoint ,flows and their probabilistic sampling can be described as follows .as suggested by data traffic measurements ( e.g. , ) , a flow can be modeled as a finite renewal process , where the total number of renewals is random and the interrenewal times , , are positive i.i.d .random variables , independent of .suppose that the finite renewal process is sampled probabilistically with a fixed probability to obtain a sampled finite point process .( as shown below , the resulting sampled point process is generally not a finite renewal process . )the sampled number of renewals ( which could be zero ) and sampled interrenewal times ( which are available only when the number of sampled renewals is greater than ) will be denoted by and , , respectively .we illustrate this notation in figure [ frp - ssp ] , where all circles ( both empty and full ) correspond to the finite renewal process , and full circles correspond to the sampled finite point process .given i.i.d .copies of a sampled finite point process ( some of which are empty when the number of sampled renewals is zero ) , one of our main goals is to infer the original distribution of interrenewal times .the focus is on nonparametric inference of , statistical properties of the resulting estimator and its performance in simulations .the estimator of involves a nonparametric estimator of the original distribution of the total number of renewals , which has previously been considered in the network traffic context .statistical properties of the estimator of will also be studied here for the first time .   to be able to make inference about , we first need to relate it to , where the latter indicates the conditional distribution function of given ( with ) . as part of this work ( see theorem [ t : lt_sit ] in section [ s : renewal - to - sampled ] ) , we show that where denotes the convolution . in particular , the right - hand side of ( [ e : dq - d - wknown ] ) , and hence , does not depend .we also note for later reference that the definition of the sequence involves the distribution . somewhat independently of the main objectives above , we examine ( [ e : dq - d - wknown ] ) for several underlying distributions , such as geometric and heavy - tailed , that arise in network traffic studies ( e.g. , ) .for example , in the geometric case , the distribution does not depend on , although this seems to be an exception to the general rule .we also provide a result similar to ( [ e : dq - d - wknown ] ) for a multidimensional vector .this allows ( conditional ) dependencies to be examined among sampled interrenewal times .for example , when the distribution is geometric , the sampled interrenewal times turn out to be independent , in which case the sampled finite point process is a finite renewal process .we should also note that forms of conditioning other than on are possible , such as ; these will be discussed briefly .having the relation ( [ e : dq - d - wknown ] ) , we would naturally expect that it can be inverted , in the sense that where is obtained by reversion of , that is , their respective -transformations ( or formal power series ) and satisfy .the relation ( [ e : d - dq - wknown ] ) suggests a natural estimator of , defined as ( see section [ s : times - sampled - to - renewals ] ) where is the empirical distribution function of given and is the reversion of the sequence , where the latter is defined as by replacing in its definition by the empirical distribution .note that the estimator is defined for fixed and .   under technical assumptions , we will show that is an asymptotically normal estimator of , namely , where is a limiting gaussian process and the convergence in distribution holds in a suitable space of functions ( theorem [ t : an - f_d - sampled ] ) .the approach and some techniques behind this result are closely related to the works in the so - called decompounding framework by buchmann and grbel , bgsted and pitts ( see also ) .these authors consider analogous estimators , but where the sequence in ( [ e : dq - d - wknown ] ) is known and hence so is its reversion sequence .we thus deviate from these earlier works by assuming that also needs to be estimated , which makes the analysis substantially more complex .more specifically , the limit in ( [ e : hat - f_d - normal ] ) can be written as where is a gaussian process , is a composition operation , is the `` derivative '' of and is the convolution .the second term in ( [ e : hat - f_d - normal - limit ] ) is the term found in the decompounding literature when and are supposed to be known .the first term in ( [ e : hat - f_d - normal - limit ] ) is new and accounts for the variations in here .since involves the estimator via , we also need the asymptotic normality result for .although already appears in the network traffic literature , its statistical properties have not been studied before to the best of our knowledge , and the asymptotic normality result is also derived here .we show that , under suitable assumptions , where is a gaussian process ( theorem [ t : an - estim - f - w ] and proposition [ p : an - estim - f - w - function ] ) .it is interesting to note that the result ( [ e : hat - f_w - normal ] ) is shown under technical assumptions which do not cover distributions with heavier tails ( such as heavy - tailed distributions ) .it is proved , however , that the assumptions can be thought of as sharp ( in the sense of proposition [ p : necerssary ] of section [ s : number - sampled - to - renewals ] ) .what the asymptotics of are beyond these assumptions remains an interesting open question .we would like to make several other related comments .first , it is well known from estimation of flow sizes in the network traffic context that the performance of the estimator degrades rapidly as the sampling probability decreases .for example , for , the inference becomes impractical for reasonable sample sizes .we shall discuss this fact in light of the derived asymptotic normality result for ( sections [ s : number - sampled - to - renewals ] and [ s : simulations - number ] ) by indicating two different regimes for the estimator performance in terms of the limiting asymptotic variance .the regimes are defined as : \\ [ -8pt ]   & \bullet & \mbox { explosive , if }  \sup_{w\in{\mathbb{n}}}es(\xi)_w^2 = \infty,\nonumber  \end{aligned}\ ] ] where is the limiting gaussian process in ( [ e : hat - f_w - normal ] ) and is expressed in terms of , and or .the performance of the estimator is satisfactory in the stable regime , but poor in the explosive regime . for small , typically belongs to the explosive regime and hence to the case of poor performance .analogous difficulties for smaller remain when using the estimator  .because of these practical considerations , in the network traffic context , alternative sampling schemes have been sought and considered , such as the sample - and - hold method , and the dual sampling technique .we plan to study inference of interrenewal times in one of these frameworks and postpone any real data application to future work .this work will therefore be limited to a simulation study ( section [ s : simulation ] ) .  in another direction ,the results on characterizing the sampled number of renewals and sampled interrenewal times , such as the relation ( [ e : dq - d - wknown ] ) , contribute to a substantial literature on sampling of point processes . sampling ( also known as _ thinning _ )is discussed in several manuscripts such as . to the best of our knowledge ,sampling ( thinning ) of finite renewal processes has been largely unexplored .the results of our work ( section [ s : renewal - to - sampled ] ) show that sampling of finite renewal processes does not generally lead to finite renewal processes ( in that , conditionally on the number of sampled renewals , sampled interrenewal times are generally dependent ) .the rest of the paper is organized as follows . in section[ s : preliminaries ] we collect the notation used throughout the work and include other preliminaries .section [ s : renewal - to - sampled ] concerns properties of the finite point process obtained from sampling the finite renewal process .inference of the original distributions of the total number of renewals and interrenewal times is studied in section [ s : sampled - to - renewal ] .a simulation study can be found in section [ s : simulation ] . for better readability ,most of the proofs are deferred to appendix [ s : proofs ] .the main body of the article contains the proofs of only those results which we consider key in individual sections .finally , appendix [ s : appendix - bounds - taylor ] contains a technical result used in section [ s : times - sampled - to - renewals ] , with bounds on remainder terms in taylor expansions of compositions of formal power series .here , we introduce notation and make a number of assumptions that will be used throughout the paper . as in section [s : intro ] , we consider a finite renewal process consisting of a finite but random number ( with ) of renewals and positive i.i.d .interrenewal times , , independent of .the terminology of finite renewal processes here follows that of , example 5.3(b ) , page 125 .we denote by the variable with a common distribution of .we also let denote the total duration of the finite renewal process defined by ( if , then ) .suppose now that each renewal , independently of the others , is sampled with a fixed probability to form a sampled finite point process .we let denote the total number of sampled renewals , , , the sampled interrenewal times and the total duration of the sampled finite point process .the following notation will be used many times throughout the paper . for a discrete random variable , its distribution or probability mass function ( p.m.f . )will be denoted by .for example , we shall write , , etc . for a continuous random variable  , its distribution function will be denoted by and its laplace transform ( lt ) will be denoted by , ( ) .we shall also use the conditional distribution functions and .so , for example , we shall write , , , etc . in several instances, we shall use analogous notation , but where is replaced by a multidimensional .  for a sequence , where and is the set of natural numbers , we denote its formal power series or its -transformation by .conversely , any such formal power series is associated with a sequence .when stands for a p.m.f . , we shall also write in place of .since a sequence will not necessarily be nonnegative , we shall also use the notation , defined as , and . in several instances, we shall use a multidimensional power series associated with and .we shall also use the following common operations on sequences ( or their formal power series ) and .the composition of and will be denoted and is defined by its formal power series as .the derivative of will be denoted by and is defined by .the reversion of a sequence will be defined as a sequence with its formal power series satisfying .the reversion is defined for any sequence with . as usual , the symbol will stand for convolution in either the discrete or continuous setting .with all these notions for sequences and their formal power series , we follow the nice monograph of  .in this section , we study several characteristics of the sampled finite point process in terms of the original finite renewal process .we first briefly consider the number of sampled renewals and then turn to sampled interrenewal times .the relation between the probability mass functions of the number of sampled renewals and the number of original renewals is given by \\ [ -10pt ]  & = & \sum_{w = s}^\infty { w\choose s } q^{s } ( 1-q)^{w - s } f_{w}(w),\qquad  s \geq 0.\nonumber\end{aligned}\ ] ] the function is given by & = & \sum_{w=1}^{\infty } f_{w}(w ) \bigl(zq+(1-q)\bigr)^w \\[-2pt ] & = & g_{w}\bigl(zq+(1-q)\bigr),\qquad |z|<1.\nonumber\end{aligned}\ ] ] the same relations ( [ e : pmf - s_w ] ) and ( [ e : rel - g_w : g_s_w ] ) also appear in and will be used in section [ s : number - sampled - to - renewals ] to obtain inversion results .given that renewals have been sampled , the next result characterizes the lt of the  sampled interrenewal time in terms of the lt of the original interrenewal times .[ t : lt_sit ] the lt of given can be expressed as where and denotes the probability that the number of original renewals between the and sampled renewals is equal to given ,   for and , we have \\ [ -10pt ]      & = &  \sum_{w = s}^\infty f_w ( w ) p(w_{q}=s|w = w ) p(d_{q , i } \leq t |w = w , w_{q}=s).\nonumber  \end{aligned}\ ] ] let denote the number of original renewals not sampled between the and sampled renewals plus 1 . if the number of original renewals is and the number of sampled renewals is , then can take the values , and by considering the location of the sampled renewal in the total number of renewals along with the possible distinct locations of the sampled renewals before the and after the sampled renewals , we obtain that \\ [ -8pt ]   & = & { w - m\choose s-1}\bigg/{w\choose s},\nonumber\end{aligned}\ ] ] where is the location of the sampled renewal and the last equality follows from the identity by using ( [ p : lt_sit_aux1])([p : lt_sit_aux3 ] ) , we deduce that where is given by ( [ eq : lt_sit_coef ] ) .the relation ( [ e : lt_sit ] ) follows by taking the lt in ( [ e : df_sit ] ) .theorem [ t : lt_sit ] implies that , when conditioning on , the distribution of the  sampled interrenewal time depends , in general , only on and not on , and hence that the times between consecutive sampled renewals are identically distributed conditionally on .in contrast to the original finite renewal process which assumes independence of and , the sampled quantities and are dependent .other forms of conditioning on the number of sampled renewals are possible .for example , from ( [ e : lt_sit ] ) , the lt of given can be expressed as where and is the probability that the number of original renewals between the and sampled renewals is equal to given , \\ [ -8pt ]  & = & \frac{q ( 1-q)^{m-1}}{p(w_q \geq s ) } \sum_{w = m+s-1}^\infty f_w(w )  \biggl ( 1- \sum_{s^{\prime}=0}^{s-2 }  { w - m \choose s^{\prime } } q^{s^{\prime } } ( 1-q)^{w - m - s^{\prime } } \biggr).\nonumber\end{aligned}\ ] ] in the rest of the paper , we shall focus only on the conditioning but similar results can be derived for other forms of conditioning such as .the next result gives the joint lt for a finite number of sampled interrenewal times .we shall use this general result below to investigate dependence between sampled interrenewal times ( see section [ s : examples ] ) ; see appendix [ s : proofs ] for the proof .[ t : lt_joint_sit ] the joint lt of with given , , can be expressed as where with and denotes the probability that the number of renewals between the and sampled renewals is equal to , , given : with .  by theorem [ t : lt_joint_sit ] , the joint distribution of a subset of sampled interrenewal times given depends only on and not on the indices of the sampled times considered .another relation of interest is between the duration of the sampled finite point process and the original interrenewal times ; see appendix [ s : proofs ] for the proof .[ p : lt_s_v ] the lt of given that can be expressed as where and is the probability that the number of renewals between the first and last sampled renewals is equal to , given :   since the lt of the duration of the original finite renewal process , given , can be expressed as we expect that where is the reversion of ( defined in section [ s : preliminaries ] ) . by plugging into ( [ e : lt_s_v ] ) , we can obtain an expression for the duration of the sampled finite point process in terms of the duration of the finite renewal process .  also , note that , equivalently, the relations ( [ e : lt_sit ] ) , ( [ e : lt_joint_sit-1 ] ) and ( [ e : lt_s_v ] ) can be expressed in terms of the distributions functions ( see ( [ e : df_sit ] ) , and relations ( [ e : df_joint_sit ] ) and ( [ e : df_s_v ] ) in appendix [ s : proofs ] , respectively ) or via the characteristic functions .   in the following ,we examine characteristics of the sampled finite point process for particular distributions of .suppose that is a geometric random variable with parameter , that is , , , .substituting this into ( [ eq : lt_sit_coef ] ) , we obtain , after straightforward calculations , that therefore , is the p.m.f . of a geometric distribution with parameter which does not depend on . from ( [ e : lt_sit ] ) and using the generating function of the geometric distribution , we have now , substituting into ( [ e : lt_joint_sit_coef ] ) , we conclude , after some algebra , that and hence , from ( [ e : lt_joint_sit-1 ] ) , the latter expression shows that when is geometrically distributed , the sampled interrenewal times are independent and identically distributed ( i.e. , the sampled finite point process is a finite renewal process ) .the lt of in ( [ e : lt_s_v ] ) , where is now the p.m.f . of a geometric distribution with parameter , simplifies to    suppose now that is a discrete pareto distribution , that is , , , and is the riemann zeta function . unlike in the previous example , closed forms for lt s of the sampled interrenewal times are not available .nevertheless , we can show that for a particular example , the sampled interrenewal times are not conditionally independent . taking , and , the random variables and are independent given if and only if for all . since is continuous in , it takes all the values in .then , ( [ ind : pareto ] ) implies that for all and hence for all .the coefficients of the term on the right- and left - hand sides of ( [ ind : pareto ] ) are , respectively , and where , , is the polylogarithm function . since , the random variables and are not conditionally independent .this is also why we use the term `` sampled finite point process '' throughout the paper instead of `` sampled finite renewal process '' .consider the case of a heavy - tailed distribution for in the sense that as , where and .such distributions are common models for long flow sizes in network traffic studies ( e.g. , ) . as in the example with pareto distribution above ,closed forms for lt s of the sampled interrenewal times are not available under ( [ e : ht ] ) .we can nevertheless provide a number of interesting qualitative results concerning the case ( [ e : ht ] ) .the relation ( [ e : ht ] ) implies that as .as indicated in , page 70 , by the results of , page 333 , this implies that that is , is also heavy - tailed with the same parameter .( note that ( [ e : ht2-q ] ) does not , in general , imply that . ) also , note that in the case ( [ e : ht2 ] ) , the total duration is expected to be heavy - tailed , even for light - tailed interrenewal times . indeed , by robert and segers , theorem 3.2 , if ( [ e : ht2 ] ) holds and then that is , the tail of total duration is dominated by that of .  on the sampling side ,the total duration is also characterized by heavy tails .indeed , by proposition [ p : lt_s_v ] above and , theorem 3.2 , where a random variable is such that , where is defined in ( [ e : lt_sit_coef ] ) , as long as ( [ e : ht - cond ] ) holds and to show ( [ e : ht - c ] ) , observe that for large , using ( [ e : ht ] ) ,   note also that the argument ( [ e : ht - c - proof ] ) above does not apply to the sampled interrenewal times .for example , for the first sampled interrenewal time , the analogous argument would show that , for coefficients in ( [ eq : lt_sit_coef ] ) , as , which is not heavy - tailed , in contrast to ( [ e : ht - c ] ) .in this section , we study inference of the original distributions of the number of renewals and interrenewal times .we are interested here in estimating the p.m.f . from i.i.d.observations of .we revisit a nonparametric estimator of introduced in and clarify several issues surrounding its use and properties .a number of open questions will also be raised .estimation of is based on a theoretical inversion of the relation ( [ e : pmf - s_w ] ) which will be discussed first .the relation ( [ e : rel - g_w : g_s_w ] ) can be written as which has the same form as the original relation ( [ e : rel - g_w : g_s_w ] ) when and are exchanged , and is replaced by . in view of ( [ e : pmf - s_w ] ) , we would then expect that \\ [ -8pt ]   & = & \sum_{s = w}^\infty { s\choose w } \frac{(-1)^{s - w}}{q^s } ( 1-q)^{s - w }   f_{w_q}(s),\qquad w\geq 1.\nonumber\end{aligned}\ ] ] hohn and veitch claim that ( [ e : f_wq->f_w ] ) holds when ( with this choice , note that in ( [ e : f_wq->f_w ] ) ) . as the following elementary result shows , this is not a necessary condition .the relation ( [ e : f_wq->f_w ] ) also holds for  ] , on the other hand , a number of distributions of interest , such as heavy - tailed distributions ( section [ s : examples ] ) , do not satisfy ( [ e : f_wq->f_w-1-cond - main ] ) . in these cases , and generally for  ] is a standard brownian motion .it is then enough to show almost sure convergence of the corresponding terms for and .doing this for is elementary and thus we do so only for .note that where are i.i.d . random variables . by the three series theorem , a.s .as long as the condition ( [ e : an - estim - f - w - cond - main ] ) holds .  for the convergence in ( iii ) ,note that the probability in ( iii ) can be expressed and bounded as where .it remains to observe that as , by the assumption ( [ e : an - estim - f - w - cond - main ] ) .the proof of ( [ e : an - var - limit ] ) is now elementary .note that the condition ( [ e : an - estim - f - w - cond - main ] ) is quite strong and , for example , does not allow for heavy - tailed distributions when  ] , with fixed .justification of the bootstrap procedure is beyond the scope of this paper .here , we provide a simulation study for the inference of the distributions of the number of renewals and interrenewal times .   as mentioned in , the estimator in ( [ e : estim - f_w-1 ] )is known to perform poorly as gets smaller . here, we shall re - examine this fact and the performance of the estimator via the asymptotic variance in ( [ e : an - var - limit ] ) of the normal limit in ( [ e : an - estim - f - w ] ) . in this regard ,two regimes should be distinguished : \\ [ -8pt ]   & \bullet & \mbox { explosive }  \bigl(\sup_{w\in{\mathbb{n } } } r_{q , w } = \infty \mbox { or }  \sup_{w\in{\mathbb{n}}}es(\xi)_w^2 = \infty\bigr),\nonumber\end{aligned}\ ] ] where is defined in ( [ e : r_w-1 ] ) , that is , as will be seen below , the performance of the estimator largely depends on which of the two regimes is considered . before providing some simulations , we take a closer look at these regimes for the distributions considered in section [ s : examples ] .suppose that follows a geometric distribution with parameter as in section [ s : examples ] .we shall derive lower and upper bounds for .for the lower bound , observe that for some constant ( which depends on and ) and hence that for some constant ( which depends on and ) . for the upper bound of ,we need a bound on binomial coefficients .we shall use the standard ( but rough ) bound given by .we shall also use the following auxiliary result , proved in appendix [ s : proofs ] .[ l : bound - integral ] let . then , for some constant and all ,   using the lemma , with generic constants , where the last inequality follows from stirling s formula .hence , by arguing similarly and using ( [ e : bound - integral2 ] ) , we have \\ [ -8pt ]   & \leq & c_3  \biggl ( \frac{\mathrm{e}^3c}{q } \biggr)^w  \biggl (   1 + \biggl(\ln \frac{1}{c(1-q)}\biggr)^{-w } + \biggl(\frac{1}{2 } \ln \frac{q}{(1-q)^2 \mathrm{e}c }   \biggr)^{-2w}\nonumber \\   & & \hphantom{c_3  \biggl ( \frac{\mathrm{e}^3c}{q } \biggr)^w  \biggl(}{}+ \biggl ( \frac{1}{2}\ln \frac{q\ln ( c(1-q))^{-1}}{(1-q)^2 \mathrm{e}c }   \biggr)^{-2w}\nonumber\\   & & \hphantom{c_3  \biggl ( \frac{\mathrm{e}^3c}{q } \biggr)^w  \biggl ( } { } + \biggl(\ln \frac{1}{c(1-q)}\biggr)^{-w }   \biggl ( \frac{1}{2}\ln \frac{q\ln ( c(1-q))^{-1}}{(1-q)^2 \mathrm{e}c } \biggr)^{-2w } \biggr ) \nonumber\\   & \leq & c_4 \biggl ( \frac{\mathrm{e}^3c}{q } \biggr)^w,\qquad \mbox{if }   \frac{\mathrm{e}^3c}{q } \leq 1,\nonumber\end{aligned}\ ] ] where , for the last inequality , we used the fact that all three log terms ( before raising to the powers and ) are bigger than 1 under .  the bounds ( [ e : r - lower - bound - geom ] ) and ( [ e : r - upper - bound - geom ] ) show that , for the geometric distribution , \\ [ -8pt ]   \frac{c}{q } > 1 \quad & \rightarrow & \quad \mbox{explosive regime}.\nonumber\end{aligned}\ ] ] what happens in the range  ] . on the other hand , observe that since , it is clear that the absolute value of the element of ( [ expansion - aux1 ] ) is less than or equal to the element of ( [ expansion - aux2 ] ) .this completes the proof of ( [ e : first - order ] ) .the proof of ( [ e : second - order ] ) is similar , but involves the observation that ^{\mathrm{t } } df_m(y_1,\ldots , y_n )  \bigr ) \\ & & \quad= \frac{1}{2}\sum_{m=1}^n [ { \varepsilon}_1 \cdots { \varepsilon}_n]^{\mathrm{t } } d^2 f_m(d_{m,1},\ldots , d_{m , n } ) [ { \varepsilon}_1 \cdots { \varepsilon}_n ] \nonumber \\ & & \quad= \frac{1}{2}\sum_{m=1}^n x_m m(m-1 )  \biggl(\sum_{k=1}^n d_{m , k } z^k \biggr)^{m-2 }  \biggl(\sum_{k=1}^n { \varepsilon}_k z^k  \biggr)^2,\end{aligned}\ ] ] where denotes the hessian of and with $ ] .the authors would like to thank the anonymous referees for useful comments and suggestions .the second author was supported in part by nsf grant dms-06 - 08669 .estan , c. and varghese , g. ( 2002 ) .new directions in traffic measurement and accounting . in _ sigcomm02 .conference on applications , technologies , architectures , and protocols for computer communications _ 323336 .

Abstract: consider a finite renewal process in the sense that interrenewal times are positive i.i.d . variables and the total number of renewals is a random variable , independent of interrenewal times . a finite point process can be obtained by probabilistic sampling of the finite renewal process , where each renewal is sampled with a fixed probability and independently of other renewals . the problem addressed in this work concerns statistical inference of the original distributions of the total number of renewals and interrenewal times from a sample of i.i.d . finite point processes obtained by sampling finite renewal processes . this problem is motivated by traffic measurements in the internet in order to characterize flows of packets ( which can be seen as finite renewal processes ) and where the use of packet sampling is becoming prevalent due to increasing link speeds and limited storage and processing capacities .


the multiarmed bandit problem is a problem based on an analogy with playing a slot machine with more than one arm or lever .each arm has a reward distribution and the objective of a gambler is to maximize the collected sum of rewards by choosing an arm to pull for each round .there is a dilemma between exploration and exploitation , namely the gambler can not tell whether an arm is optimal unless he pulls it many times , but it is also a loss to pull an inferior ( i.e. non - optimal ) arm many times .we consider an infinite - horizon -armed bandit problem .there are arms and arms are pulled infinite number of times . has a probability distribution with the expected value and the player receives a reward according to independently in each round . if the expected values are known , it is optimal to always pull the arm with the maximum expected value .a policy is an algorithm to choose the next arm to pull based on the results of past rounds .this problem is first considered by robbins . since then , many studies have been conducted for the problem . there are also many extensions for the problem .for example , auer et al . removed the assumption that rewards are stochastic , and for the stochastic setting , the case of non - stationary distributions , or the case of infinite ( possibly uncountable ) arms have been considered .  in our setting ,lai and robbins established a theoretical framework for determining optimal policies , and burnetas and katehakis extended their result to multiparameter or non - parametric models . consider a model , a generic family of distributions .the player knows and that is an element of .let denote the number of times that has been pulled over the first rounds .a policy is _ consistent _ on model if ={\mathrm{o}}(n^a) ] .it is larger than the model used in , which assumes a specific finite support .we also demonstrate simulation results of med policy comparable to ucb policies .our med policy is motivated by the observation of .when a policy achieving is used , an inferior arm waits roughly rounds to be pulled after the -th play of .then , it can be expected that a policy pulling with probability will achieve .med policy is obtained by plugging into in , where is the empirical distribution of rewards from and is the current best sample mean .med policy requires a computation of at each round whereas upper confidence bound requires the computation of  and are quantity dual to each other but the former has two advantages in practical implementation .first , is smooth in which converges to . therefore the value in the previous round can be used as a good approximation of for the current round . on the other handcontinues to increase according to and it has to be computed many times .second , as shown in theorem [ thm8 ] below , can be expressed as a _ univariate _ convex optimization problem for our model .although is also a convex optimization problem , the nonlinear constraint is harder to handle .med policy is categorized as a probability matching method ( see , e.g. for classification of policies ) . in this methodeach arm is pulled according to the probability reflecting how likely the arm is to be optimal . for example , wyatt proposed probability matching policies for boolean and gaussian models by bayesian approach with prior / posterior distributions . in our approachthe probability assigned to each arm is determined by ( normalized ) maximum likelihood instead of posterior probability .this paper is organized as follows . in section[ section - preliminary ] , we give definitions used throughout this paper and show the asymptotic bound by , which is satisfied by any consistent policy . in section [ section - proposed - policy ] , we propose med policy and prove that it is asymptotically optimal for finite support models .we also discuss practical implementation issues of minimization problem involved in med . in section [ section - experiments] , some simulation results are shown .we conclude the paper with some remarks in section [ section - remarks ] .in this section we introduce notation of this paper and present the asymptotic bound for a generic model , which is established by .let be a generic family of probability distributions on and let be the distribution of , .  ] denotes the probability and the expectation under , respectively .when we write e.g. \;\,(a\subset\mathbb{r}) ] , denotes a random variable with distribution .we define  ] .  a set of probability distributions for arms is denoted by .the joint probability and the expected value under are denoted by  ] , respectively .the expected value of is denoted by .we denote the optimal expected value by .let be the arm chosen in the -th round. then , { \nonumber}\end{aligned}\ ] ] where  ] . for notational simplicitywe often write , e.g. ,  ] .finally we define an index for and  where kullback - leibler divergence is given by &\frac{\mathrm{d}f}{\mathrm{d}g } \mbox { exists,}\\ + \infty&\mbox{otherwise . }\end{cases}{\nonumber}\end{aligned}\ ] ] represents how distinguishable is from distributions having expectations larger than .if is empty , we define .we adopt lvy distance for distance between two distributions .we use only the fact that the convergence of the lvy distance is equivalent to the weak convergence of to distribution and we write in this sense .  lai and robbins gave a lower bound for  ] be the family of distributions with a _finite _ bounded support , where is the support of distribution and are constants known to the player .we assume without loss of generality .we write and where is an arbitrary subset of  ] .however our proof of the optimality in this paper is for the above .the advantage of assuming the finiteness is that we can employ the method of types in the large deviation technique .this enables us to consider all empirical distributions obtained from each arm .  in this modelit is convenient to use instead of .properties of the minimizer of the right - hand side will be discussed in section [ section - dinf ] .[ dmindinf ] holds for all and .we will prove in lemma [ conti ] that is continuous in . follows easily from the continuity .we now introduce our med policy . in med an armis chosen randomly in the following way :  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ * [ minimum empirical divergence policy ] *  * initialization .*  pull each arm once .* for the -th round ,  1 .for each compute .2 .  choose arm according to the probability _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _  note that for any currently best since .as a result , it holds for all that   intuitively , for a currently not best arm corresponds to the maximum likelihood that is actually the best arm . therefore in med an arm is pulled with the probability proportional to this likelihood .  notethat our policy is a randomized policy .therefore probability statements below on med also involve this randomization .however for notational simplicity we omit denoting this randomization .now we present the main theorem of this paper .[ optmd ] fix satisfying and for all . under med policy , forany and it holds that \le \frac{1+{\epsilon}}{{d_{\mathrm{min}}}(f_i,\mu^*,{\mathcal{a}})}\log n + { \mathrm{o}}(1).{\nonumber}\end{aligned}\ ] ]  note that we obtain }{\log n } \le \frac{1}{{d_{\mathrm{min}}}(f_i,\mu^*,{\mathcal{a}})},\ ] ] by dividing both sides by , letting and finally letting . in view of ( [ eq : liminf - etj ] )we see that med policy is asymptotically optimal .we give a proof of theorem [ optmd ] in section [ section - proofmd ] .the following corollary shows that the optimality of med policy given in theorem [ optmd ] is a generalization of the optimality in .let  ] such that .fix satisfying and for all . under med policy , for any and it holds that \le \frac{1+{\epsilon}}{{d_{\mathrm{min}}}(f_i,\mu,{{\mathcal{a}}_{{\mathcal{x}}}})}\log n + { \mathrm{o}}(1).\label{coro}\end{aligned}\ ] ]  we prove in lemma [ suppp ] that .on the other hand , holds from .then we obtain from theorem [ optmd ] .note that is achieved also by the policy used in the if is fixed and assumed to be known .our result establishes the same bound without this assumption .   for implementing med policyit is essential to efficiently compute the minimum empirical divergence for each round . in this subsection, we clarify the nature of the convex optimization involved in and show how the minimization can be computed efficiently .in addition , for proofs of lemma [ dmindinf ] and theorem [ optmd ] , we need to clarify the behavior of as a function of .first we prove that it is sufficient to consider for the computation of :  [ suppp ] holds for any .take an arbitrary such that and .define as since and , we obtain the converse inequality is obvious from .  in view of this lemma , we simply write instead of when the third argument is obvious from the context .let and denote the finite symbols in by , i.e.  .we assume and for without loss of generality and write .  nowthe computation of is formulated as the following convex optimization problem for from lemma [ suppp ] : where we define , , and .it is obvious that is the optimal solution with the optimal value when .also , the unit point mass at 0 , is the unique feasible solution if .for the problem is infeasible .since these cases are trivial , we consider the case in the following .define and its first and second order derivatives as = \sum_{i=1}^m f_i\log(1-(x_i-\mu)\nu ) , \label{h0}\\ h'(\nu)&\equiv&{\frac{\partial } { \partial \nu}}h(\nu)= -\sum_{i=1}^{m}\frac{f_i ( x_i-\mu)}{1-(x_i-\mu)\nu } , \label{h1}\\ h''(\nu)&\equiv&{\frac{\partial ^2}{\partial \nu^2}}h(\nu)= -\sum_{i=1}^{m}\frac{f_i(x_i-\mu)^2}{(1-(x_i-\mu)\nu)^2 } .\label{h2}\end{aligned}\ ] ] now we show in theorem [ thm8 ] that the computation of is expressed as maximization of . since is concave , it is a univariate convex optimization problem .therefore can be computed easily by iterative methods such as newton s method ( see , e.g. , for general methods of convex programming ) .[ thm8 ] define =\infty ] , and is simply written as   on the other hand for the case \ge 1 ] .then we obtain {\nonumber\\}&=&{\mathrm{e}}_f[\log(1-(x-\mu)\nu_0)]\label{levy_conv}\\ & \ge&{d_{\mathrm{min}}}(f,\mu)-{\epsilon}.{\nonumber}\end{aligned}\ ] ] note that is continuous and bounded in  ] as an expectation and a probability under and the randomization in the policy .now we define events as follows : where is a constant satisfying which is set sufficiently small in the evaluation on .note that and each  ] is bounded from above by &\le &  { \mathbb{i}}[j_n(i)\cap a_n ] + { \mathbb{i}}[j_n(i)\cap c_n ] + { \mathbb{i}}[j_n(i)\cap a^c_n \cap b_n ] + { \mathbb{i}}[j_n(i){\,\cap\,}d_n].\quad \label{eq : five - terms}\end{aligned}\ ] ]  in the following lemmas [ lema]-[lemd ] we bound the expected values of sums of the four terms on the right - hand side of ( [ eq : five - terms ] ) in this order and they are sufficient to prove theorem [ optmd ] .[ lema ] fix an arbitrary .then it holds that \right ] \le \frac{1+{\epsilon}}{{d_{\mathrm{min}}}(f_i,\mu^*)}\log n + { \mathrm{o}}(1).{\nonumber}\end{aligned}\ ] ]  [ lemc ] \right]={\mathrm{o}}(1).{\nonumber}\end{aligned}\ ] ]  [ lemnab ] \right]={\mathrm{o}}(1).{\nonumber}\end{aligned}\ ] ]  [ lemd ] \right]={\mathrm{o}}(1).{\nonumber}\end{aligned}\ ] ]  before proving these lemmas , we give intuitive interpretations for these terms . represents the event that the estimator of is already close to and is pulled with a small probability . after sufficiently many rounds holds with probability close to and the term  ] according to the number of occurrences ] , we obtain \le \frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^*)}}+ \sum_{n=1}^n{ \mathbb{i}}\left[{j_n}(i){\,\cap\,}a_n { \,\cap\,}t'_i(n ) > \frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^ * ) } } \right].{\nonumber}\end{aligned}\ ] ] taking the expected value we have \right ] } \quad{\nonumber\\}&\le & \frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^*)}}+ \sum_{n=1}^n p\left[{j_n}(i){\,\cap\,}a_n { \,\cap\,}t'(n)>\frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^ * ) } } \right].{\nonumber\\}&\le & \frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^*)}}+  \sum_{n=1}^n  p\left[j_n(i ) \,\bigg|\ , a_n{\,\cap\,}t_i'(n)>\frac{(1+\epsilon)\log  n}{{{d_{\mathrm{min}}}(f_{i},\mu^*)}}\right ] { \nonumber\\}&\le & \frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^ * ) } } + n\exp\left(-\frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^ * ) } } \frac{{{d_{\mathrm{min}}}(f_{i},\mu^*)}}{1+{\epsilon}/2}\right ) \qquad(\mbox{by \eqref{bound_pj}}){\nonumber\\}&=&\frac{(1+\epsilon)\log n}{{{d_{\mathrm{min}}}(f_{i},\mu^ * ) } } +  n^{-\frac{1+{\epsilon}}{1+{\epsilon}/2}+1}{\nonumber}\end{aligned}\ ] ] the lemma is proved since .first we have & \le & \sum_{n=1}^n { \mathbb{i}}[{j_n}\in l{\,\cap\,}c_n]{\nonumber\\}&\le & \sum_{t=1}^n\sum_{n=1}^{\infty}{\mathbb{i}}[{j_n}\in l{\,\cap\,}t_1'(n)=t{\,\cap\,}c_n].\label{7 - 0}\end{aligned}\ ] ] from the technique of type ( * ? ? ?* lemma 2.1.9 ) , it holds for any type that \le \exp(-td(q||f_1 ) ) \le\exp(-t{d_{\mathrm{min}}}(q,\mu_1)).\label{7 - 19}\end{aligned}\ ] ]  let be the smallest integers in . is well defined on the event  \hat{\mu}^*(r_l)<\mu_1-{\delta} ] by \le \sum_{n=1}^\infty{\mathbb{i}}[d_n].\ ] ] since it holds that & \le & \sum_{k\in l}\sum_{n=1}^\infty { \mathbb{i}}[{{\hat{\mu}}_{k}(n)}=\hat{\mu}^*(n)>\mu_1-{\delta}]{\nonumber\\}&= & \sum_{k\in l}\sum_{t=1}^{\infty}\sum_{n=1}^{\infty } { \mathbb{i}}[{{\hat{\mu}}_{k , t}}=\hat{\mu}^*(n)>\mu_1-{\delta}{\,\cap\,}t'_k(n)=t].\label{dn1}\end{aligned}\ ] ]  now we use a reasoning similar to .let be the smallest integers in . is well defined on the event  ] . in this case , it is easily checked that  , and satisfy kuhn - tucker conditions since and for . therefore is obtained .follows from and the concavity of .now we consider the second case \ge 1 ] and we consider the case \ge 1 ] , is convex in  ] and for =1 ] . from (* corollary 3.4.3 ) , is differentiable in with therefore we obtain for > 1 ] .in this section , we present some simulation results on our med and ucb policies in .first we give an algorithm for computing and with parameters , which we denote by . here is a repetition number and is an initial value of for the optimization in theorem [ thm8 ] .recall that are defined in , and .+ * [ computation of *   , ; ; ; ; ; ; ; ; ;  in this algorithm , a lower and an upper bound of are given by and , respectively .in each step , the next point is determined based on newton s method by .when does not improve the bounds , the next point is determined by bisection method , .the complexity of the algorithm is given by .the complexity is not very small when is large .especially it requires computations when it is adopted for a continuous support model since . on the other hand, is differentiable in ( with slope ) and the argument converges to after sufficiently many rounds .therefore it is reasonable to approximate by past value of until the variation of is small . in this point of view, we implemented our med policy for our simulations in the following way :  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ * [ an implementation of med policy ] *  * parameter : * integer and real .  * initialization : *  1 .pull each arm once .2 .  set and for each .* loop : * for the -th round ,  1 .update variables for each : * if and then . *otherwise and .2 .  choose arm according to the probability _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _  now we describe the setting of our experiments .we used med , ucb - tuned and ucb2 .each plot is an average over 1,000 different runs .the parameter for ucb2 is set to , the choice of which is not very important for the performance ( see ) .first we check the effect of the choice of the parameters and . then med and ucb policies are compared .  in the following simulations , we use the model where the support is included in  ] for computational convenience. then , all rewards are passed to computation after is subtracted from them in med .table [ diss ] gives the list of distributions used in the experiments .they cover various situations on the computation of and how distinguishable the optimal arm is .distributions 1 - 4 are examples of 2-armed bandit problems . in distribution 1 , in theorem [ thm8 ] always holds with equality since .therefore the exact solution can be obtained by regardless of .also in distribution 2 , does not require the repetition after sufficiently many rounds since <1 ] .distribution 4 is an example of a difficult problem where the optimal arm is hard to distinguish since the inferior arm appears to be optimal at first with high probability .distribution 5 and 6 are examples of more general problems where the numbers of arms and the support sizes are large . in distribution 6 denotes beta distribution which has the density function { \nonumber}\end{aligned}\ ] ] where is beta function .note that beta distributions have continuous support and are not included in and therefore the performance of med is not assured theoretically .however , med is still formally applicable since the supports are bounded .the labels of each figure are as follows .`` regret '' denotes , which is the loss due to choosing suboptimal arms .`` % best arm played '' is the percentage that the best arm is chosen , that is , in these problems .`` dmin '' stands for the asymptotic bound for a consistent policy , .the asymptotic slope of the regret ( in the semi - logarithmic plot ) of a consistent policy is more than or equal to that of `` dmin '' .figure [ ex1 ] shows an experiment on the choice of the parameters and of med for distribution 3 .our implementation of med approaches the ideal med as and .however , we see from the figure that the performance is not sensitive to the choice of . this may be understood as follows : ( 1 ) the linear approximation for the case is accurate , ( 2 ) the initial value in seems to be a good approximation of and the repetition number does not have to be large .we use and in the remaining experiments based on this result .now we summarize the remaining experiments on the comparison of the policies ( figure 27 ) .* med always seems to be achieving the asymptotic bound even for continuous support distributions , since the asymptotic slope of the regret is close to that of `` dmin '' .* med performs best except for distribution 1 where med performs worst .however , the consistency of ucb - tuned is not proved unlike med and ucb2 .it appears that ucb - tuned might not be consistent , because the asymptotic slope of seems to be smaller than that of `` dmin '' .note that the theoretical logarithmic term of the regret is very near between med and ucb2 for distribution 1 ( and , respectively ) .therefore this result can be interpreted as follows : med achieves the asymptotic bound but needs some improvement in the constant term of the regret compared to ucb2 .is computed by repetitions ) .,width=245 ]   is computed by repetitions ) .,width=245 ]   is computed by repetitions ) .,width=245 ]   is computed by repetitions ) .,width=245 ]   is computed by repetitions ) .,width=245 ]   is computed by repetitions ) .,width=245 ]we proposed a policy , med , and proved that our policy achieves the asymptotic bound for finite support models .we also showed that our policy can be implemented efficiently by a convex optimization technique .  in the theoretical analysis of this paper, we assumed the finiteness of the support although med worked nicely also for distributions with continuous bounded support in the simulation .we conjecture that the optimality of med holds also for the continuous bounded support model . in addition, there are many models that can be computed explicitly , such as normal distribution model with unknown mean and variance .we expect that our med can be extended to these models .furthermore , our med is a randomized policy and the theoretical evaluation of the expectation includes randomization in the policy .we may be able to construct a deterministic version of med .in addition to the above theoretical analyses , it is also important to consider the finite horizon case. then it is necessary to derive a finite - time bound of med for this case .especially , med policy itself should be improved when the number of rounds is given in advance . in thissetting , the value of `` exploration '' becomes smaller and a current best arm is to be pulled more often as the number of remaining rounds becomes smaller .

Abstract: multiarmed bandit problem is an example of a dilemma between exploration and exploitation in reinforcement learning . this problem is expressed as a model of a gambler playing a slot machine with multiple arms . a policy chooses an arm so as to minimize the number of times that arms with inferior expectations are pulled . we propose minimum empirical divergence ( med ) policy and prove asymptotic optimality of the policy for the case of finite support models . in a setting similar to ours , burnetas and katehakis have already proposed an asymptotically optimal policy . however we do not assume knowledge of the specific support except for the upper and lower bounds of the support . furthermore , the criterion for choosing an arm , minimum empirical divergence , can be computed easily by a convex optimization technique . we confirm by simulations that med policy demonstrates good performance in finite time in comparison to other currently popular policies .


rendezvous and treasure hunt are two basic tasks performed by mobile agents in networks . in rendezvous , two agents , initially located at distinct nodes of the network , traverse network edges in synchronous rounds and have to meet at some node . in treasure hunt , a single agent has to find a stationary target ( called treasure ) situated at an unknown node of the network .the network might model a labyrinth or a system of corridors in a cave , in which case the agents might be mobile robots .the meeting of such robots might be motivated by the need to exchange previously collected samples , or to agree how to share a future cleaning or decontamination task .treasure hunt might mean searching a cave for a resource or for a missing person after an accident .in other applications we can consider a computer network , in which the mobile entities are software agents .the meeting of such agents might be necessary to exchange data or share a future task of checking the functionality of network components .treasure hunt in this case might mean looking for valuable data residing at some node of the network , or for a virus implanted at some site .the network is modeled as a simple undirected connected graph whose nodes have distinct identities .ports at a node of degree are numbered .the agents are anonymous , i.e. , do not have identifiers .agents execute a deterministic algorithm , such that , at each step , they choose a port at the current node .when an agent enters a node , it learns the entry port number , the label of the node and its degree .the cost of a rendezvous algorithm is the total worst - case number of edge traversals performed by both agents until meeting .the cost of a treasure hunt algorithm is the worst - case number of edge traversals performed by the agent until the treasure is found .if the agents have no information about the network , the cost of both rendezvous and treasure hunt can be as large as for networks with edges .this is clear for treasure hunt , as all edges ( except one ) need to be traversed by the agent to find the treasure in the worst case .the same lower bound for rendezvous follows from proposition [ eq ] in the present paper . on the other hand ,if is the distance between the initial positions of the agents , or from the initial position of the agent to the treasure , a lower bound on the cost of rendezvous and of treasure hunt is .  in this paper , we study tradeoffs between the amount of information available _ a priori _ to the agents and the cost of rendezvous and treasure hunt . following the paradigm of algorithms with advice , this information is provided to the agents at the start of their navigation by an oracle that knows the network , the starting positions of the agents and , in the case of treasure hunt , the node where the treasure is hidden .the oracle assists the agents by providing them with a binary string called _ advice _ , which can be used by the agent during the algorithm execution . in the case of rendezvous, the advice given to each agent can be different .the length of the string given to the agent in treasure hunt and the sum of the lengths of strings given to both agents in rendezvous is called the _ size of advice_.   using the framework of advice permits us to quantify the amount of information needed for an efficient solution of a given network problem ( in our case , rendezvous and treasure hunt ) regardless of the type of information that is provided .our goal is to find the smallest size of advice which enables the agents to solve rendezvous and treasure hunt at a given cost in a network with edges .this size turns out to depend on the initial distance ( between the agents in rendezvous , and between the agent and the treasure in treasure hunt ) and on the ratio , which is the _ relative cost gain _ due to advice . for arbitrary graphs , we give upper and lower bounds of and , respectively , on the optimal size of advice .hence our bounds leave only a logarithmic gap in the general case . for the class of trees ,we give nearly tight upper and lower bounds of and , respectively .our upper bounds are obtained by constructing an algorithm for all graphs ( respectively , for all trees ) that works at the given cost and with advice of the given size , while the lower bounds are proved by exhibiting networks for which it is impossible to achieve the given cost with smaller advice .treasure hunt , network exploration and rendezvous in networks are interrelated problems that have received much attention in recent literature .treasure hunt has been investigated in the line , in the plane and in other terrains .treasure hunt in anonymous networks ( without any information about the network ) has been studied in with the goal of minimizing cost .the related problem of graph exploration by mobile agents ( often called robots ) has been intensely studied as well .the goal of this task is to visit all of the nodes and/or traverse all of the edges of a graph .a lot of research considered the case of a single agent exploring a labeled graph . in agent explores strongly - connected directed graphs . in a directed graph , an agent can move only in the direction from tail to head of an edge , not vice - versa .in particular , investigated the minimum time of exploration of directed graphs , and gave improved algorithms for this problem in terms of the deficiency of the graph ( i.e. , the minimum number of edges that must be added to make the graph eulerian ) .many papers , e.g. , studied the scenario where the graph to be explored is labeled and undirected , and the agent can traverse edges in both directions . in , it was shown that a graph with nodes and edges can be explored in time . in some papers , additional restrictions on the moves of the agent were imposed , e.g. , it was assumed that the agent is tethered , i.e. , attached to the base by a rope or cable of restricted length . in , a log - space construction of a deterministic exploration for all graphs with a given bound on size was shown .the problem of rendezvous has been studied both under randomized and deterministic scenarios . in the framework of networks, it is usually assumed that the nodes do not have distinct identities .an extensive survey of randomized rendezvous in various models can be found in , cf .also . deterministic rendezvous in networks has been surveyed in .several authors considered geometric scenarios ( rendezvous in an interval of the real line , e.g. , , or in the plane , e.g. , ) .gathering more than two agents was studied , e.g. , in .for the deterministic setting , many authors studied the feasibility and time complexity of rendezvous of synchronous agents , i.e. , agents that move in rounds . in authors studied tradeoffs between the time of rendezvous and the number of edge traversals by both agents . in , the authors presented a rendezvous algorithm whose running time is polynomial in the size of the graph , the length of the shorter label and the delay between the starting times of the agents . in ,rendezvous time is polynomial in the first two of these parameters and independent of the delay .the amount of memory required by the agents to achieve deterministic rendezvous was studied in for general graphs .the amount of memory needed for randomized rendezvous in the ring was discussed , e.g. , in  .several authors investigated asynchronous rendezvous in the plane and in network environments .providing nodes or agents with information of arbitrary type that can be used to perform network tasks more efficiently has been proposed in .this approach was referred to as algorithms with _advice_. the advice is given either to nodes of the network or to mobile agents performing some network task .several of the authors cited above studied the minimum size of advice required to solve the respective network problem in an efficient way .  in , given a distributed representation of a solution for a problem , the authors investigated the number of bits of communication needed to verify the legality of the represented solution . in , the authors compared the minimum size of advice required to solve two information dissemination problems using a linear number of messages . in , it was shown that a constant amount of advice enables the nodes to carry out the distributed construction of a minimum spanning tree in logarithmic time . in ,the advice paradigm was used for online problems . in ,the authors established lower bounds on the size of advice needed to beat time for 3-coloring a cycle and to achieve time for 3-coloring unoriented trees . in the case of ,the issue was not efficiency but feasibility : it was shown that is the minimum size of advice required to perform monotone connected graph clearing . in ,the authors studied radio networks for which it is possible to perform centralized broadcasting with advice in constant time .they proved that bits of advice allow to obtain constant time in such networks , while bits are not enough . in , the authors studied the problem of topology recognition with advice given to nodes . in , the authors considered the task of drawing an isomorphic map by an agent in a graph , and their goal was to determine the minimum amount of advice that has to be given to the agent for the task to be feasible .  among the papers using the paradigm of advice , are closest to the present work .both concerned the task of graph exploration by an agent . in ,the authors investigated the minimum size of advice that has to be given to unlabeled nodes ( and not to the agent ) to permit graph exploration by an agent modeled as a -state automaton . in , the authors established the size of advice that has to be given to an agent completing exploration of trees , in order to break competitive ratio 2 . in , the authors studied the minimum size of advice that must be provided to labeled agents , in order to achieve rendezvous at minimum possible cost , i.e. , at cost , where is the initial distance between the agents .they showed that this optimal size of advice for rendezvous in -node networks is , where the labels of agents are drawn from the set .this paper differs from the present one in two important aspects .first , as opposed to the present paper , in , agents get identical advice , and nodes of the network are unlabeled .second , instead of looking at tradeoffs between cost and the size of advice , as we do in the present paper , the focus of was on the size of advice sufficient to achieve the lowest possible cost .in this section we show that , in the context of advice , treasure hunt and rendezvous are essentially equivalent .more precisely , the following proposition shows that the minimum advice sufficient to solve both problems at a given cost in the class of graphs with edges and with the initial distance is the same , up to constant factors . throughout the papera _ graph _ means a simple connected undirected graph .the number of nodes in the graph is denoted by , and the number of edges is denoted by .all logarithms are to base 2 .  [ eq ]let be positive integers .  1 .if there exists an algorithm th that solves treasure hunt at cost with advice of size in all graphs with edges and with initial distance between the agent and the treasure , then there exists an algorithm rv that solves rendezvous at cost with advice of size in all graphs with edges and with initial distance between the agents .if there exists an algorithm rv solving rendezvous at cost with advice of size less than in all graphs with edges and with initial distance between the agents , then there exists an algorithm th that solves treasure hunt at cost at most with advice of size at most in all graphs with edges and with initial distance between the agent and the treasure .  __ consider a graph with edges , and two agents , and , that have to meet .suppose that and start at nodes and in graph , and that is the distance between and .let be the advice string of size that enables an agent starting at to find the treasure located at at cost using algorithm .give advice string to agent and advice string to agent .the sum of the lengths of these strings is .the rendezvous algorithm rv is the following . with advice string inert ; with advice string execute algorithm using advice . by the correctness of th ,this rendezvous algorithm is correct and its cost is .  _consider a graph with edges and with initial distance between the agent ( initially located at ) and the treasure ( initially located at ) .we construct the following graph .it consists of two disjoint copies of with the respective nodes in each copy joined by an additional edge .the graph has edges .label nodes of the graph as follows .if some node of has label , then the corresponding node in has label and the corresponding node in has label .place two agents in , each at the node of a different copy of graph .hence , the initial positions of the agents are at distance in .let and be the advice strings ( whose lengths sum to less than ) that are provided to the agents starting in and , respectively , in the execution of rv in . in this execution ,at least one of the agents has to traverse edge , and , hence , it has to reach the node in its copy of .therefore it travels from to in with an advice string of size less than , at cost at most .algorithm th for treasure hunt in is given the advice string with the single bit appended .the algorithm consists of the solo execution of rv where the agent transforms the label of each visited node to .  in view of proposition [ eq ] ,in the rest of the paper we can restrict attention to the problem of treasure hunt .all of our results , both the upper and the lower bounds , also apply to the rendezvous problem ( with the provision that , if treasure hunt can be solved at cost with no advice , then rendezvous can be solved at cost with constant advice ) .notice that the equivalence of rendezvous and treasure hunt depends on the fact that , in rendezvous , the oracle can give different pieces of advice to the two agents .if the oracle was forced to give the same advice to both agents , then symmetry could not be broken in all cases since agents are anonymous , and rendezvous would be impossible in some networks .in this section , we proceed to prove upper and lower bounds on the advice needed to solve treasure hunt in arbitrary graphs .these bounds are expressed in terms of , which is the distance between the treasure and the initial position of the agent , and in terms of the ratio , where is the number of edges in the graph and is an upper bound on the cost of the algorithm .this ratio is the relative cost gain due to advice .we first provide an algorithm that solves treasure hunt using bits of advice , and then prove that any deterministic algorithm for this task uses at least bits of advice .consider an -node graph and a node of , which is the initial position of the agent .let be a shortest path from to the treasure , where is the node at distance from along path .let .intuitively , is an upper bound on the total number of bits needed to fully describe the sequence of ports leading from to the treasure . for any fixed integer , we describe a binary advice string of length and an algorithm that uses this advice when searching for the treasure .we do not consider values of greater than since we will show that , when , our algorithm has optimal cost .  to construct the advice ,the idea is to use bits to produce advice substrings to guide the agent along path . in particular , the first bits of advice consist of binary substrings . for each , the substring is created by considering the node on path that is at distance from in .the length of is dictated by the ratio of the number of bits needed to describe the degree of to the total number of bits needed to describe the degrees of all nodes on path .the set of ports at is partitioned into numbered _ sectors _ ( i.e. , subintervals ) of size at most .in fact , at most one of the sectors can have size smaller than this value .the substring is taken to be the binary representation of the number of the sector containing the port that leads to the next node on path towards the treasure .below , we provide pseudocode that describes how the advice is created .first , algorithm [ createadvice ] finds a shortest path from to the treasure .the path consists of node / port pairs for each , where and , for each , port leads from node to node .the sum is calculated and stored in .for ease of notation , we define .each pair is passed to the subroutine described in algorithm [ encodesector ] , along with .this subroutine uses and the degree of to determine the appropriate number of advice bits via the formula , then divides the set of ports at into numbered sectors , determines to which sector port belongs , and outputs the binary representation of this sector number as a -bit string .the resulting sequence of substrings , along with the binary string representing the value of , is encoded into a single advice string to pass to the algorithm .more specifically , these strings are encoded by doubling each digit in each substring and putting 01 between substrings .this permits the agent to unambiguously decode the original sequence , to calculate the value of by looking at the number of separators 01 , and to calculate the value of by looking at the lengths of the first advice substrings .denote by this encoding and let be the inverse ( decoding ) function , i.e. . as an example , .note that the encoding increases the total number of advice bits by a constant factor .the advice string , calculated by algorithm [ createadvice ] using the strings supplied by algorithm [ encodesector ] , is .the advice string is given to the agent .  find a shortest path in from node to the node containing the treasure .  port number leading from to node on path at distance from  ` encodesectornumber` binary representation of output   [ createadvice ]    //_ port _ is contained in the range return the -bit binary representation of   [ encodesector ]  [ advicesize ] the advice string has size .each of the strings has length , and the sum of these lengths is at most .the string is the binary encoding of the sum .this sum is maximized when all have the same degree , hence it is .it follows that the length of is .therefore , the length of is in .next , we describe the algorithm findtreasure , which is the agent s algorithm given an advice string . for the purpose of description only , we define the _ trail _ of the agent , which is a stack of edges that it has previously traversed .the stack gets popped when the agent backtracks .the agent performs a walk in starting at node . in each step of the algorithm, the agent chooses an edge to add to the trail , or it backtracks along the trail edge that it added most recently .the number of edges in the agent s trail will be used to measure the agent s progress . in particular , when the agent is located at a node and there are edges in the agent s trail , we will say that the agent is at _ progress level _ .the agent keeps track of its current progress level by maintaining a counter that is incremented when it adds a trail edge and decremented when it backtracks .the agent maintains a table containing the labels of the nodes that it has visited , and , for each node label , the smallest progress level at which the agent visited the node so far .when the agent arrives at a node from a lower progress level and does not find the treasure , it checks if its current progress level is lower than the progress level stored in the table for node . if this is not the case , or if , then the agent backtracks by going back along the edge it just arrived onalso , the agent backtracks immediately if it sees that the degree of does not `` match '' the size of in the following sense : using , the value of that is encoded in , and the degree of , the agent checks if is equal to the number of bits that the oracle would have provided if was indeed on the path from to the treasure , i.e. , if .otherwise , if the agent has determined that it should not backtrack immediately , then it uses the advice substring in the following way : it divides the set of port numbers at into sectors ( i.e. , intervals of port numbers ) of size , gives numbers to the sectors , and then interprets as the binary representation of an integer that specifies one of these sectors . for each port number in the specified sector ,the agent takes the port and arrives at some neighbour of .the agent terminates if it finds the treasure at node , or , otherwise , repeats the above at node .if , after trying all ports at node in the specified sector , the treasure has not been found , the agent backtracks .note that the advice was created with the goal of ` steering ' the agent in the right direction , i.e. , along path , but we can only guarantee that this will happen when the agent is located at nodes on path . in fact , an even stronger condition must hold : for any node on path at distance from , we can only guarantee that the advice will be helpful if the agent is located at node at progress level , since this is when the agent reads the advice substring .in other words , it is possible that the agent visits a node on at the ` wrong ' progress level , in the sense that it wo nt use the advice that was created specifically for .this is why it is not sufficient to simply have the agent backtrack whenever it arrives at a previously - visited node , since during its previous visit , it may have used the wrong advice .moreover , we must ensure that the algorithm gracefully deals with the situation where the agent is at a node at progress level , but the advice substring specifies ports that do not exist at . in our algorithm , the agent ignores any port numbers that are greater than or equal to the current node s degree .to summarize , in our algorithm , the agent searches for the treasure in a depth - first manner , but it can not perform dfs ( even only to distance ) because the cost would be too large . instead , the agent takes only a fraction of ports at each node , but may possibly have to pay for it by traversing the same edge several times ( while in dfs every edge is traversed at most twice ) . as our analysis will show , this gives an overall decrease of the total cost , especially when the advice is large .the pseudocode of the search conducted by algorithm findtreasure is described by algorithm [ takestep ] .it shows how the agent takes a step in the graph , i.e. , for each , how it uses to move from a node at progress level to a node at progress level . in order to initiate the search ,this algorithm is called at node with progress level 0 ( and ) .algorithm [ decodesector ] , used as a subroutine in algorithm [ takestep ] , shows how the agent decodes substring to obtain a range of port numbers .we assume that we have two functions related to the agent - maintained table of visited nodes : ` updatetable` that writes into the entry for node as the smallest progress level at which the agent has ever visited node , and ` currentmin` that reads the entry of the table for node .each table entry is initialized to .stop  [ line : ifline ]  ` updatetable` [ line : updatetable ]    integer value encoded in binary string  [ line : sizematch ] [ line : forloop ] take port [ line : takeport ] the node reached after taking port call   return to node [ line : backtrack ]  [ takestep ]   number of bits in _ sectornumberencoding _ integer value of _ sectornumberencoding _ return   [ decodesector ]   in what follows , let be the path from to the treasure that is used to create the advice string .suppose that consists of the nodes , where , for each , is at distance from , and the treasure is located at node .also , for each , let be the port at node that leads to node .  to prove the correctness of the algorithm, we first consider an arbitrary node on path and suppose that the agent is at progress level .clearly , this occurs at least once during the execution of ` findtreasure ` since the agent is initially located at at progress level 0 .one of the ports at that are specified by the advice substring leads to node , but the agent may try some other of these ports first .we show that either the agent finds the treasure by recursively calling ` takestep ` after taking one of these other ports , or , the agent eventually takes the port that leads to node  .[ makesprogress ] for any , consider the first time that the agent is located at node at progress level . during the execution of ` takestep` , for some node , either :  1 .  the agent moves to node at progress level , or , 2 .there is a node such that the agent moves to node at progress level , calls ` takestep` , and , during its execution , the treasure is found by the agent .since we are considering the agent s first visit to node at progress level , and it is not possible for the agent to visit at a progress level less than , it follows that .so , the * if * condition on line [ line : ifline ] evaluates to true .further , since node was used in the creation of the advice substring , it follows that , so the * if * condition on line [ line : sizematch ] evaluates to true .suppose that the treasure is not found during any execution of ` takestep` with . by the choice of , port located in the range of port numbers returned by ` getsector ` . since taking port at node leads to node , there exists an iteration of the loop in ` takestep ` such that the agent moves to node and increments its progress level to .using induction , we extend lemma [ makesprogress ] to show that the agent eventually reaches node .[ eventuallyfinds ] for any , consider the first time that the agent is located at node at progress level . during the execution of ` takestep` , for some node , the agent finds the treasure .the proof proceeds by induction on . in the base case , , andthe agent finds the treasure when it is first located at node at progress level since the treasure is located at . as induction hypothesis ,assume that , for some , when the agent is first located at node at progress level , the agent finds the treasure during the execution of ` takestep ` .now , consider the first time that the agent is located at node at progress level .note that , by the induction hypothesis , the agent was not previously located at node at progress level , since otherwise , during the execution of ` takestep ` at the first such visit , the agent would have found the treasure and terminated .  by lemma [ makesprogress ] , when the agent is first located at node at progress level , either :  1 .the agent moves to node at progress level , or , 2 .there is a node such that the agent moves to node at progress level , calls ` takestep` , and , during its execution , the treasure is found by the agent .  in the first case, the induction hypothesis implies that the agent finds the treasure . in the second case, the treasure is found by the agent , so we are done .  by lemma [ eventuallyfinds ] with ,the agent finds the treasure during the first execution of takestep , hence is correct .next , we consider the cost of algorithm ` findtreasure ` .our analysis considers the cases and separately . we proceed to find upper bounds on the cost of algorithm ` findtreasure ` in terms of a fixed upper bound on the amount of advice provided . to prove the upper bounds , we first give upper bounds on the size of the sector returned by ` getsector ` .in the first case , we show that when ( i.e. ) the cost of algorithm ` findtreasure ` is optimal .[ sectorupperone ] suppose that .for all , if the agent is located at node at progress level , then the size of the sector returned by is exactly 1 .  by the advice construction , .since , it follows that .hence , in the execution of , the value of is a positive integer , as required .[ findtreasurecostone ] suppose that . when provided with advice , the algorithm has cost .  by lemma [ sectorupperone ] , for each , when the agent is located at node at progress level , the execution of returns exactly 1 port number leading to node .since the agent starts at node at progress level 0 , it follows that the agent takes exactly steps to find the treasure .therefore , when , algorithm has cost exactly .in the second case , we assume that ( i.e. ) .[ sectorupper ] suppose that . for all ,the size of the sector returned by is at most .note that , when is executed , it must be the case that line [ line : sizematch ] evaluated to true , i.e. , that . in the execution of ,the variable is assigned the value .note that since , it follows that and , so .therefore , we have shown that , which implies that , as required .we are now ready to calculate an upper bound on the cost of algorithm ` findtreasure ` .we denote by an index such that .[ findtreasurecost ] suppose that . when provided with advice , the algorithm has cost at most .it suffices to count the total number of times that line [ line : takeport ] of ` takestep ` is called and multiply this value by 2 .this is because the cost incurred by backtracking ( i.e. , line [ line : backtrack ] of ` takestep ` ) is at most 1 for each execution of ` takestep ` , which amounts to an overall multiplicative factor of at most 2 .so , we consider the number of times that line [ line : takeport ] of ` takestep ` is called at an arbitrary node . the number of times that the * for * loop at line [ line : forloop ] is iterated is at most when is visited at progress level , since , by lemma [ sectorupper ] , this is an upper bound on the size of the range returned by ` getsector ` .since line [ line : updatetable ] ensures that the condition on line [ line : ifline ] is true at most once at each progress level , it follows that the total number of times that line [ line : takeport ] is executed is bounded above by . taking the sum over all nodes ,the total number of calls to ` takestep ` is bounded above by next , since , it follows that   since , it follows that   finally , we fix an upper bound on the cost of ` findtreasure ` and re - state lemmas [ findtreasurecostone ] and [ findtreasurecost ] to obtain an upper bound on the amount of advice needed to solve treasure hunt at cost .[ ub ] let be any graph with edges , and let be the distance from the initial position of the agent to the treasure .let be any integer such that .the amount of advice needed to solve treasure hunt at cost at most is at most bits .first , consider the case where . in this case , ( by lemma [ findtreasurecostone ] ) and .next , consider the case where . by lemma [ findtreasurecost ] ,algorithm ` findtreasure ` solves treasure hunt with cost .it follows that , so .since for each , it follows that .therefore , regardless of the value of , we have shown that . by lemma [ advicesize ] ,the size of advice is .the following lower bound follows immediately from theorem [ tree - lb ] , which is proven by constructing a tree for which treasure hunt requires bits of advice .this theorem will be proven in section 4 .[ lb ] let .there exists a graph with edges , and a position of the treasure at distance from the initial position of the agent , such that treasure hunt at cost requires bits of advice .the gap between the upper bound given by theorem [ ub ] and the lower bound given by theorem [ lb ] is at most a factor logarithmic in .moreover , it should be noted that our bounds differ only by an additive term whenever is polynomial in the gain .we now proceed to prove upper and lower bounds on the advice needed to solve treasure hunt in trees . unlike in the case of arbitrary graphs , where our upper and lower bounds may differ by a logarithmic factor , for trees our bounds differ only by an additive term . again , our bounds will be expressed in terms of , which is the distance between the treasure and the initial position of the agent , and in terms of the ratio , where is the number of edges in the tree , is the number of nodes , and is an upper bound on the cost of the algorithm .also , for any two nodes , we will denote by the distance between and in the tree , i.e. , the number of edges in the simple path between them .   to obtain our upper bound , we will use algorithm ` findtreasure ` that was defined and proven correct in section [ findtreasure ] for arbitrary graphs . in this section ,we provide an analysis of the algorithm specifically for the case of trees , which gives a strictly better upper bound .we start with the following technical lemma , which shows that , if we take the agent s initial position as the root of the tree , the agent s progress level and the agent s current depth in the tree ( i.e. , its current distance from the root ) do not differ . essentially , this is because there is only one simple path from the agent s initial position to each node , and the algorithm ensures that the agent s trail does not contain the same edge multiple times .[ progresslevel ] consider algorithm findtreasure executed in any tree .suppose that , for some neighbouring nodes and , is executed at node .if line [ line : ifline ] evaluates to true , then progress level .we proceed by induction on the agent s progress level . in the base case , consider progress level . since the first call to ` takestep ` has , and every subsequent call increments the current progress level , the agent must be located at node .next , assume that , for some progress level and any neighbouring nodes and , in the execution of ` takestep` , if line [ line : ifline ] evaluates to true , then .now , for some neighbouring nodes and , consider the execution of ` takestep` .` takestep ` was executed at node at progress level and line [ line : ifline ] of this execution evaluated to true . by the induction hypothesis , it follows that .next , consider the value of . in a tree, there is only one simple path from to and one simple path from to . since and are neighbours , either is on the path from to ( in which case ) or is on the path from to ( in which case ) .if line [ line : ifline ] of the execution of ` takestep` evaluates to true , then , i.e. , was not previously visited at a progress level less than .it follows that is not located on the path from to .therefore , it must be the case that , so .next , we proceed to find an upper bound on the cost of algorithm ` findtreasure ` in trees in terms of a fixed upper bound on the amount of advice provided .the proof is analogous to the proof of lemma [ findtreasurecost ] , the main difference being that we do not need to multiply by a factor of in order to account for the different paths that the agent could use to reach a given node .as before , we denote by an index such that .[ findtreasurecostintrees ] suppose that .when provided with advice , the algorithm has cost at most .  as in lemma[ findtreasurecost ] , it suffices to count the total number of times that line [ line : takeport ] of ` takestep ` is called and multiply this value by 2 .so , we consider the number of times that line [ line : takeport ] of ` takestep ` is called at an arbitrary node . since line [ line : takeport ] is only executed if line [ line : ifline ] evaluates to true , then , by lemma [ progresslevel ] , it follows that at line [ line : takeport ] . by lemma [ sectorupper ] ,the * for * loop at line [ line : forloop ] is iterated at most times . taking the sum over all nodes ,the total number of calls to ` takestep ` is bounded above by   next , since , it follows that since and , it follows that   finally , we fix an upper bound on the cost of ` findtreasure ` and re - state lemmas [ findtreasurecostone ] and [ findtreasurecostintrees ] as an upper bound on the amount of advice needed to solve treasure hunt in trees at cost .[ tree - ub ] let .the amount of advice needed to solve treasure hunt on trees of size with cost at most is at most bits .first , consider the case where .in this case , ( by lemma [ findtreasurecostone ] ) and .next , consider the case where . by lemma [ findtreasurecostintrees ] ,algorithm ` findtreasure ` solves treasure hunt with cost .it follows that , so . since for each , it follows that .therefore , regardless of the value of , we have shown that . by lemma [ advicesize ] ,the size of advice is .we now set out to prove a lower bound on the amount of advice needed to solve treasure hunt at cost at most .we consider a collection of _ caterpillar trees _, each constructed as follows .take a path graph consisting of nodes , where and are adjacent , for every .place the treasure at node . for each , add nodes to the graph such that each of them has degree 1 and is adjacent only to node .the resulting graph is a tree on nodes . for each node in this tree, the ports at are labeled with the integers so that , for each , the port numbers at both ends of the edge are equal .finally we fix node labels as follows . for each ,node has label , and each leaf adjacent to has label , where the port number at leading to it is .notice that all labels are distinct .  for each ,let be the port number at corresponding to the edge .the trees in are in one - to - one correspondence with the sequences because the label of each leaf is determined by the port number ( at the adjacent node ) leading to it .it follows that the number of distinct caterpillar trees in ( taking into consideration the placement of the treasure ) is .figure [ caterpillardiagram ] gives a diagram of a caterpillar tree in and shows how nodes are labeled .   with ports on path labeled .( b ) the labels of the added leaves adjacent to are shown .node is labeled . ]consider any fixed caterpillar tree .we set the starting node of the agent to be . to find the treasure, the agent must traverse the edges of path .suppose that , for some , the agent is located at node .if the agent takes port , it will arrive at node , and we say that this edge traversal is _successful_. we may assume that the agent does not return to node , i.e. , away from the treasure , because such a move would only increase the cost of the algorithm .further , the agent can detect when it has found the treasure and terminate immediately .when an agent s step is not successful ( that is , when located at node , it chooses a port other than ) it arrives at a leaf adjacent to . in this case , we say that the agent _ misses_. after a miss , the agent s next step is to return to node .let be the number of times that the agent takes a port other than when located at node in .the _ cost at node _ , denoted by , is , since there are two edge traversals for each miss and one successful edge traversal .this implies the following fact .[ cost ] for any , the total cost of any treasure hunt algorithm in is .we now prove a lower bound on the size of advice needed to solve treasure hunt for the class of caterpillar trees .[ tree - lb ]let .there exists a tree of size , and a position of the treasure at distance from the initial position of the agent , such that treasure hunt at cost requires bits of advice .consider any algorithm that solves treasure hunt at cost at most using bits of advice .let .let be a set of maximum size consisting of trees from such that , for all trees in , the agent is given the same advice string . by the pigeonhole principle , it follows that .we proceed to find an upper bound on the size of such a set .consider any two different trees such that the agent is given the same advice string for both of them .let be the smallest index such that the port at leading to is different in and .then the behaviour of the agent prior to visiting for the first time is the same in and in .hence , . by fact[ cost ] , we know that , so . therefore , the number of trees in is bounded above by the number of distinct integer - valued -tuples of non - negative terms whose sum is at most .( these tuples correspond to sequences ) . )  if , then there is only one such -tuple , i.e. , the tuple with all entries equal to 0 .it follows that .recall that was chosen as a set of maximum size such that , for all trees in , the same advice is given to the agent .it follows that , for each tree in , the agent is given a different advice string .therefore , the number of different advice strings is , so the size of advice is at least .since , and implies that , it follows that , as required .so , we proceed with the assumption that .the following claim will be used to obtain an upper bound on the number of distinct integer - valued -tuples of non - negative terms whose sum is at most . in the sequel , -tuples with integer coordinates will be called _ integer points_.[ tuples ] fix any .let be the set of integer - valued -tuples of non - negative terms whose sum is at most .then , .  to prove the claim, we note that is the number of integer points in the simplex .let denote the simplex .since is a translation of the points in by in every coordinate , it follows that is also the number of integer points in the simplex . for each integer point in , we construct a small -dimensional _ box _ centered at .more specifically , for each such that , we construct . note that , for any two distinct integer points , the boxes and are disjoint .further , the volume of each such is .finally , we wish to find an upper bound on the volume of the union of all boxes where is an integer point in . to this end , we define a simplex ( a scaled version of ) such that , for each integer point , the box is completely contained in . in particular , we define .it follows that is bounded above by the volume of . from ,the volume of is equal to , which implies that .this completes the proof of the claim .  by claim [ tuples ] with , the number of trees in is bounded above by .combined with our earlier lower bound on the number of trees in , we have , which implies that  {d!}}{3(c - d ) } \geq \frac{k \cdot \sqrt[d]{d!}}{3c}.\ ] ] so , {d!}}{3c } \right).\ ] ]  by stirling s formula we have , for sufficiently large . hence {d ! } \geqd^{1/(2d ) } \cdot ( d / e) ] .hence , the above bound on implies . since , it follows that , so the size of advice is in , as required .we established upper and lower bounds on the minimum size of advice sufficient to solve the problems of rendezvous and of treasure hunt at a given cost . for the class of trees our bounds are almost tight , up to constant factors and a summand of . for the class of arbitrary graphs , our bounds leave a gap of a logarithmic factor . closing these gapsis a natural open problem .it should be noted , however , that , even for arbitrary graphs , our bounds are asymptotically tight whenever is and is polynomial in the gain .this is the case , for example , when we want to accomplish treasure hunt or rendezvous at cost in an -node graph .there are only two special situations when our gap for arbitrary graphs remains non - constant .one of them is if is very large with respect to the gain , e.g. , for an -node graph with edges in which the treasure is located at distance at cost ; our ( multiplicative ) gap is in this case .the other situation is when both and are very small with respect to , e.g. , when the treasure in an -node graph is located at distance and we want to do treasure hunt at cost . in this case we have an additive gap of .it should also be noted that , in the context of advice , treasure hunt is not only equivalent to rendezvous of two agents , as shown in proposition [ eq ] , but also to rendezvous of many agents , which is often called _ gathering_. this task consists in gathering several agents at the same node in the same round . in this case , the cost should be defined as the maximum number of edge traversals per agent , and the advice size as the maximum number of bits per agent .the reduction given by the first part of proposition [ eq ] should be modified as follows .one of the agents , starting at some node , is given advice string indicating that it should be inert .each other agent is given the advice string , where is the advice enabling agent to find a treasure located at .this work was partially supported by nserc discovery grant 8136 2013 and by the research chair in distributed computing at the universit du qubec en outaouais .e. bampas , j. czyzowicz , l. gasieniec , d. ilcinkas , a. labourel , almost optimal asynchronous rendezvous in infinite multidimensional grids , proc .24th international symposium on distributed computing ( disc 2010 ) , 297 - 311 .p. fraigniaud , d. ilcinkas , a. pelc , communication algorithms with advice , journal of computer and system sciences 76 ( 2010 ) , 222232 .p. fraigniaud , d. ilcinkas , a. pelc , tree exploration with advice , information and computation 206 ( 2008 ) , 12761287 .p. fraigniaud , a. korman , e. lebhar , local mst computation with short advice , theory of computing systems 47 ( 2010 ) , 920933 .    c. gavoille , d. peleg , s. prennes , r. raz .distance labeling in graphs , journal of algorithms 53 ( 2004 ) , 85 - 112 .hipke , c. icking , r. klein , e. langetepe , how to nd a point on a line within a xed distance . disc .93 ( 1999 ) , 6773 .a. miller , a. pelc , fast rendezvous with advice , proc .10th int . symp . on algorithms and experiments for sensor systems , wireless networks and distributed robotics ( algosensors 2014 ) .full version at arxiv:1407.1428v1 [ cs.ds ]

Abstract: in rendezvous , two agents traverse network edges in synchronous rounds and have to meet at some node . in treasure hunt , a single agent has to find a stationary target situated at an unknown node of the network . we study tradeoffs between the amount of information ( _ advice _ ) available _ a priori _ to the agents and the cost ( number of edge traversals ) of rendezvous and treasure hunt . our goal is to find the smallest size of advice which enables the agents to solve these tasks at some cost in a network with edges . this size turns out to depend on the initial distance and on the ratio , which is the _ relative cost gain _ due to advice . for arbitrary graphs , we give upper and lower bounds of and , respectively , on the optimal size of advice . for the class of trees , we give nearly tight upper and lower bounds of and , respectively .   * keywords : * rendezvous , treasure hunt , advice , deterministic algorithm , mobile agent , cost .  0.2 in 0.1 in 0.0 in