Continuous Patrolling Games
CContinuous Patrolling Games
Steve Alpern ∗ Thomas Lidbetter † Katerina Papadaki ‡ August 18, 2020
Abstract
The continuous patrolling game studied here was first proposed in Alpern et al. (2011),which studied a discrete time game where facilities to be protected were modeled asthe nodes of a graph. Here we consider protecting roads or pipelines, modeled as thearcs of a continuous network Q . The Attacker chooses a point of Q to attack during achosen time interval of fixed duration (the attack time , α ). The Patroller chooses a unitspeed path on Q and intercepts the attack (and wins) if she visits the attacked pointduring the attack time interval. Solutions to the game have previously been given incertain special cases. Here, we analyze the game on arbitrary networks. Our resultsinclude the following: (i) a solution to the game for any network Q , as long as α issufficiently short, generalizing the known solutions for circle or Eulerian networks andthe network with two nodes joined by three arcs; (ii) a solution to the game for all treenetworks that satisfy a condition on their extremities. We present a conjecture on thesolution of the game for arbitrary trees and establish it in certain cases. ∗ Warwick Business School, University of Warwick, Coventry CV4 7AL, UK, [email protected] † Rutgers Business School, Newark, NJ 07102, USA, [email protected] ‡ Department of Mathematics, London School of Economics, London WC2A 2AE, UK,[email protected] a r X i v : . [ c s . D M ] A ug Introduction
Patrolling games were introduced at the end of Alpern et al. (2011) to model the operationalproblem of how to optimally schedule patrols to intercept a terrorist attack, theft or infil-tration. That paper, contrasting with earlier adversarial patrolling (Stackelberg) versions,modeled the problem as a zero-sum game between an Attacker and a Patroller, who wish torespectively maximize and minimize the probability of a successful attack. The domain onwhich the game was played out was taken to be a graph, with attacks restricted to the nodesand taking a given integer number of periods. A patrol is a walk on the graph, and interceptsthe attack if it visits the attacked node during the attack period. This could model a guardin an art museum who enters a room while a thief is in the midst of removing a valuablepainting from the wall. That paper was able to make some key observations about theirgame, giving bounds on the value, but was unable to find the value precisely or give optimalstrategies except in some very limited cases. Papadaki et al. (2016) solved the game for linegraphs, but the solution was very complicated even for this apparently simple graph. In theConclusion section of the original paper Alpern et al. (2011), an extension of the problem tocontinuous space and time was suggested:“It may be natural to consider a continuous-time formulation of this problem.An attack takes place at any point of the network (not necessarily a node) on acontinuous time interval of fixed length. The Patroller uses a unit speed path andwins if she is at the attacked point at some time during the attack interval. Thiswould model, for example, the defense of a pipeline system, and would resembleto a greater extent the classical search game problem.” [p. 1256]The purpose of this paper is to carry out the suggestion in the quote above for anarbitrary network. We allow attacks that have a prescribed duration α to occur at any pointof a continuous network Q . A unit speed patrol on Q is said to intercept the attack (andwin for the Patroller) if it arrives at the attacked point at some time during the attack. Thevalue of the game is the probability of interception, with best play on both sides. We findthat optimal play for the Attacker typically involves mixing pure attacks that take place atdifferent times.After this type of continuous game was first proposed in 2011, it has been solved for somespecial networks. The circle network (or any Eulerian network) is easy to solve: a periodictraversal of the Eulerian tour, starting at a random point, is optimal for the Patroller;2ttacking starting at a fixed time at a uniformly random location is optimal for the Attacker(see Alpern et al. (2016) and Garrec (2019)). The line segment network was solved in Alpern et al. (2016). In Garrec (2019) a solution for some values of α is given for the network withtwo nodes connected by three unit length arcs, and a complete formulation of the generalgame is given, including a proof of the existence of the value. The present paper extends tosome extent all three of these prior results to general classes of networks: Eulerian networksto networks without leaf arcs; the line segment network to trees; the three-arc network tonetworks with large girth - for small attack times.Our main results and chapter organization are as follows. Section 3 presents several(mixed) strategies for the players that can be used or adapted to obtain solutions of thegame for various classes of networks in later sections. We note that Eulerian networks haveno leaves, and Section 4 generalizes the solution of the former to networks without leaves.In particular, as long as the attack time is sufficiently short, we show that the attack strat-egy that chooses a point uniformly at random is still optimal; an optimal strategy for thePatroller is to follow a double cover tour of the network which never traverses an arc con-secutively in opposite directions (as described in Theorem 4). We also give a new algorithmfor constructing such a tour in Theorem 3. In Section 5 we allow the network to have leaves,and modify the optimal strategies of the previous section to generate optimal strategies forarbitrary networks, as long as the attack time is sufficiently short (see Theorem 6).Section 6 considers trees and in particular those that satisfy a condition we call theLeaf Condition. We give a precise definition of the condition, which requires some delicacy(Definition 8). In fact, any tree satisfies the Leaf Condition as long as the attack time issufficiently short. Star networks (trees with only leaf arcs) also satisfy the Leaf Condition forsufficiently large attack times, and the only stars that do not satisfy the Leaf Condition arethose that have an arc that is longer than half the total length of the network. In Theorem 8we solve the game for all trees in the case that the Leaf Condition holds, giving a simpleexpression for the value of the game in terms of the length of the network, the attack timeand another parameter. In Section 7 Conjecture 1 states that this expression is always equalto the value of the game on trees. We establish the conjecture for some stars that do notsatisfy the Leaf Condition. 3 Literature Review
In addition to the papers discussed in the Introduction, which were the most relevant tocontinuous patrolling, there is a more extensive literature on adversarial patrolling. Theproblem of patrolling a perimeter has been analyzed by Zoroa et al. (2012) (where theattack location can move to adjacent locations) and Lin (2019), the latter in a continuoustime context. Extensions of Alpern et al. (2016) where the costs of successful attacks aretime and node dependent have been studied by Lin et al. (2013) (for random attack times),Lin et al. (2014) (with imperfect detection) and Yolmeh and Baykal-G¨ursoy (2019) (whichincludes an application to an urban rail network).Stackelberg approaches, with the Patroller as first mover, have been pioneered in anartificial intelligence context by Basilico et al. (2012) (which includes an algorithm for largecases) and Basilico et al. (2017) (where the optimal strategy in certain cases is for thePatroller to stay in place until the sensor reveals an attack an unknown location).More applied approaches to patrolling are of practical importance. Applications toscheduling randomized security checks and canine patrols at Los Angeles Airport have beendeveloped and deployed in Pita et al. (2008). The United States Coast Guard also uses agame-theoretic system to schedule patrols in the Port of Boston (An et al. et al. (2016) (where the novel algorithm PAWS was introduced) and Xu et al. (2019)(where the success of deploying PAWS in the field is described). Patrolling to detect radi-ation and consequently nuclear threats was modeled in the novel paper of Hochbaum et al. (2014).The possibility that the Attacker could know when the Patroller is nearby (perhaps atthe same node), raised in Alpern et al. (2011), has recently been studied in Alpern andKatsikas (2019) and Lin (2019) in different contexts. In the former this knowledge helpedthe Attacker, in the latter, it did not. Multiple patrollers have been considered in the roboticsand computer science literatures, where an important paper with a similar network structureto ours is Czyzowicz et al. (2017). A connection between patrols and inspection games ismade in Baston and Bostock (1991) and between patrols and hide-seek games in Garrec(2019). Restricting the Patroller to periodic paths creates difficulties analyzed in Alpern etal. (2018). 4
Formal Definitions for Network and Game
In this section we define the continuous patrolling game and present definitions related tothe connected network Q on which it is played. For Q , standard graph theoretic definitionsmust be modified for a network which is considered as a metric space and a measure space,not simply a combinatorial object.To define Q , we begin with the usual combinatorial structure of arcs and nodes, withthe addition of a length λ ( a ) assigned to each arc. We can then identify an arc a withan open interval of length λ ( a ) , endowed with Lebesgue measure and Euclidean distance d , and consider λ as a measure on Q , called length . The total length of Q is denoted by µ = λ ( Q ) . The topology on the interval arcs gives a topology on their union Q . A path in Q is a continuous function from a closed interval to Q . We take the metric d ( x, y ) on Q as the minimum length of a path between x and y . A point x of Q is called a regular point if it has a neighborhood homeomorphic to an open interval. Regular pointsare the interior points of arcs. The remaining points of Q are the nodes, which are theboundary points of the arcs. The degree of a node y can be defined either combinatoriallyor topologically. Combinatorially, it is the number of adjacent nodes; topologically, it is thenumber of components of a neighborhood of y from which y has been deleted. Note thatthis means that strictly speaking, nodes of degree two are not permitted. A node of degree1 is called a leaf node , and its adjacent arc is called a leaf arc . To ensure that every leaf archas a single leaf node in its closure, we exclude the line segment network from consideration.In any case the continuous patrolling game has been solved for the line segment in Alpern et al. (2016).A circuit in Q is a closed path (that is, with the same startpoint and endpoint) consistingof distinct adjacent arcs. A tour of Q is a closed path visiting all points of Q , and a tour ofminimum length is called a Chinese Postman Tour (CPT) . The length of this path is denoted¯ µ . It was shown by Edmonds and Johnson (1973) that a CPT can be found in polynomialtime, with respect to the number of nodes. A closed path which is a circuit and a tour iscalled an Eulerian tour. As is well known, a connected network has an Eulerian tour if andonly if it is Eulerian, defined as having nodes all of even degree. If we double every arc of anetwork Q , the resulting network is Eulerian with length 2 µ , so Q has a tour of length 2 µ and hence ¯ µ ≤ µ .The continuous patrolling game is played on Q as follows. The Attacker chooses a point x in Q to attack, and a closed time interval J of given length α during which to attack it.5ince α is fixed, the attack interval J = [ τ, τ + α ] is determined by its starting time τ. Thegame and its value are determined by the pair ( Q, α ) . The Patroller chooses a path S ( t ) ,where t ≥
0, which we call a patrol , satisfying d ( S ( t ) , S ( t ′ )) ≤ ∣ t − t ′ ∣ , for all t, t ′ ≥ . (1)For simplicity, we shall call a path satisfying the 1 − Lipshitz condition (1) a unit speed path .We don’t specify an upper bound on the starting time of the attack, but in every case wehave studied there is an optimal mixed attack strategy in which all its (pure strategy) attacksare over by time 4 µ . A patrol is said to intercept an attack if it visits the attacked pointwhile it is being attacked. The game is very simply defined: the maximizing Patroller wins(payoff P =
1) if her patrol intercepts the attack. Otherwise, the Attacker wins (payoff P = S and the attack is at point x during the interval J = [ τ, τ + α ] , then the payoff P to the maximizing Patroller is given by P ( S, ( x, J )) = ⎧⎪⎪⎨⎪⎪⎩ x ∈ S ( J ) , V , is the interception probability,with best play on both sides.Garrec (2019) used the fact that P is lower semicontinuous to establish the existence ofa value V for this infinite game. We note that if α = Q (according to λ ) at a fixed time; if α ≥ ¯ µ , the Patrollercan ensure a win by adopting a Chinese Postman Tour, starting anywhere at time 0 andrepeating the tour with period ¯ µ . So to avoid the trivial cases where one of the player canalways win, we assume 0 < α < ¯ µ .Throughout the paper the complement Q − Y of a set Y is denoted by Y c . Some networks, as we shall see in later sections, require Attacker strategies specifically suitedto their structure, such as attacks on leaf nodes when the network is a tree. But there arealso some general strategies that are available on any network. Here we define two of theseand present the general bounds on the value that they give.6 efinition 1 (Uniform attack strategy) A uniform attack strategy is a mixture ofpure attacks that have a common attack time interval J = [ M, M + α ] . The attacked point ischosen uniformly at random. That is, the probability that the attacked point lies in a set Y is given by λ ( Y ) / µ . We restate a lemma from Alpern et al. (2016) for completeness (the proof is in the OnlineAppendix).
Lemma 1
Against any patrol S , a uniform attack strategy is intercepted with probability notmore than α / µ . Consequently V ≤ α / µ for any network. We now define independence for sets and strategies.
Definition 2 (Independent set)
A subset I of Q is called independent if the distancebetween any two of its points is at least α . For any subset Y of Q , the set W ≡ W ( Y ) is thesubset of Q consisting of all points at distance at most α / from Y . Definition 3 (Independent attack strategy)
Given an independent set I of cardinality l and the set W ≡ W ( I ) , the independent attack strategy is as follows for p = lαλ ( W c )+ lα .1. With probability p attack at an element of I chosen equiprobably at a start time chosenuniformly at random in J = [ , α ] .2. With probability − p attack uniformly on W c at start time α / . The independent attack strategy randomizes over both time and space, unlike the strategyof the same name defined in Alpern et al. (2011) for the discrete patrolling game, whichrandomizes only over space. The following result gives an upper bound on the strategy’sinterception probability.
Theorem 1
Let I be an independent subset of Q of cardinality l . Then V ≤ αλ ( W c ) + lα , which the Attacker can ensure by adopting the independent attack strategy. If λ ( W c ) = wehave V ≤ / l . Furthermore, if I is the set of leaf nodes, and leaf arcs have lengths exceeding α / , then V ≤ αµ + lα / . roof: Let S denote any patrol and suppose the independent attack strategy is adopted.If S remains in W during J , it intercepts the attack with probability at most p / l , where l isthe cardinality of I . Similarly, since S has unit speed, if it remains in W c during time J , itintercepts an attack with probability at most ( − p ) ( α / λ ( W c )) . The chosen value of p isthe one that makes these probabilities both equal to α / ( λ ( W c ) + lα ) .Finally, suppose the patrol S starts in W c at time 0, reaches a point x ∈ I at some time t , α ≤ t ≤ α , early enough to intercept some attacks on I and late enough to intercept someattacks on W c . Since the latest such a patrol can leave W c is at time t − α /
2, it can cover a setof length at most ( t − α / ) − ( α / ) = t − α in W c after the attacks at time α /
2, intercepting afraction ( t − α ) / λ ( W c ) of the attacks there. In addition, the patrol can intercept the attacksat x starting between t − α and α , so a fraction ( α − t ) / α of the attacks at x , or ( α − t ) / lα of the attacks on I . Thus the maximum probability that a patrol arriving at I at time t canintercept an attack is given by ( − p ) t − αλ ( W c ) + p α − tlα = αλ ( W c ) + lα. By time symmetry, the same bound holds if the patrol starts at a point of I and ends up in W c .If λ ( W c ) = V ≤ / l trivially.To prove the last assertion note that if I is the set of leaf nodes, and leaf arcs have lengthsexceeding α /
2, then leaf nodes form an independent set I and λ ( W ) = lα / ◻ Some patrol strategies come from finding closed paths on the network with specific proper-ties, and then have the Patroller go around them periodically starting at a random point.Normally the closed path will be a tour, but we give a more general definition in case it isnot.
Definition 4 (Randomized periodic extension) If S ∶ [ , L ] → Q is a closed unit speedpath, we can extend it to various patrols S ∆ ∶ [ , ∞) → Q of period L by the definition S ∆ ( t ) = S ( ( t + ∆ ) mod L ) , for all t ≥ . Thus S ∆ is a periodic patrol that starts at the point S ( ∆ ) at time . The randomizedperiodic extension ˜ S of S is defined as the random mixture of the pure patrols S ∆ , with ∆ chosen uniformly in the interval (or circle) [ , L ] . In the special case that S is a ChinesePostman Tour, with L = ¯ µ , we call ˜ S a Chinese Postman Tour strategy. .3 A Theorem on k − covering Tours If a network Q has an Eulerian tour, its randomized periodic extension makes an effectivepatrolling strategy, because it visits all regular points equally often (once), so the Attacker isindifferent as to where to attack. If there is no Eulerian tour (the general case), we can stilluse this idea, if there is a tour which visits all regular points equally often. In Theorem 3and Lemma 6, we will show that there is indeed such a tour which visits all regular pointstwice (a 2 − cover), with some additional properties. This idea is formalized in the following. Theorem 2
Let S ∶ [ , L ] → Q be a closed unit speed tour that visits every point of Q at k times which are separated by at least α (mod L ) . Let ˜ S be the randomized periodic extensionof S (from Definition 4). Then we have (i) ˜ S intercepts any attack with probability at least kα / L . (ii) If L = kµ , then the randomized periodic extension ˜ S (for the Patroller) and a uniformattack strategy (for the Attacker) are optimal and the value of the game is given by α / µ . Proof:
For part (i), suppose the attack takes place at a point x in Q starting at sometime τ . Let t i , i = , . . . , k be times, separated by at least α , such that S ( t i ) = x . Theattack will be intercepted by S ∆ if ∆ is in the set Y = ∪ i [ t i − τ − α, t i − τ ] (modulo L ), sincein this case the Patroller will visit x = S ( t i ) at some time in [ τ, τ + α ] . The separationassumption ensures that these intervals are disjoint, and since they all have length α , thelength (Lebesgue measure) of Y is given by ∣ Y ∣ = kα . By the definition of ˜ S , the probabilitythat ∆ ∈ Y is equal to ∣ Y ∣ / L = kα / L , as claimed in (i), so we have V ≥ kα / L = kα / kµ = α / µ under the assumption of part (ii). By Lemma 1, we also have that V ≤ α / µ , so the twoinequalities give V = α / µ , with ˜ S and the uniform attack strategy optimal. ◻ As suggested above in the introductory remarks of this subsection, taking k = et al. (2016) andGarrec (2019). Corollary 1 If Q is Eulerian, with Eulerian tour S , then for α ≤ µ we have V = α / µ .( V = if α ≥ µ .) In this case the randomized periodic extension ˜ S and the uniform attackstrategy are optimal for the Patroller and Attacker, respectively. Furthermore, for a ChinesePostman Tour S of any network Q , taking k = and L = ¯ µ gives V ≥ α / ¯ µ .
9t is useful to note for applications to patrolling by m robots, that if in Theorem 2we require that S visits every point at k times separated by time intervals mα , then m Patrollers can intercept any attack with probability at least mkα / L (or 1, if this is larger).To see this, pick ∆ as above and let the path of the i ’th Patroller (robot) be defined by S i ( t ) = S ( ∆ + i ( L / m ) + t ) . The arrival times at any point of Q are then separated by atleast α . This shows that in our later lower bounds for V , these can be multiplied by thenumber of Patrollers, with an upper bound of 1. Q Helps the Patroller
We conclude this section with an observation on the effect of identifying points of Q on thevalue. Alpern et al. (2011) considered the effect of identifying two nodes of a graph. Here,we identify two points of the network Q , using the well known quotient topology. In Figure 1we identify the arc midpoints C and D of the network Q to produce a new network Q ′ .Figure 1: Identifying points C , D of Q to obtain Q ′ .We may first look at two cases which have already been solved, the line segment Q line =[ , ] and the circle Q circle = [ , ] mod 1 (which is obtained from the line segment byidentifying the endpoints), with say α = /
2. From Alpern et al. (2016), we have V ( Q line ) = α / ( µ + α ) = /
3. However as the circle is Eulerian, we have V ( Q circle ) = α / µ = /
2, which islarger. It is easy to show that identifying points cannot decrease the value. Of course if wefurther identify points on the circle, we get new points of degree 4, so the resulting Euleriannetwork retains the value of 1 / Lemma 2
Suppose Q ′ , d ′ is the metric space obtained from Q, d by replacing the metric d with a smaller metric d ′ , that is, with ≤ d ′ ( x, y ) ≤ d ( x, y ) for all x, y ∈ Q = Q ′ . Then V ( Q ′ , d ′ ) ≥ V ( Q, d ) . Furthermore, if Q ′ is obtained from Q by decreasing the length of anarc or simply identifying two points x and y , the same result holds. The proof of Lemma 2 is given in the Online Appendix. An application of it is given atthe end of Section 4. 10
Networks Without Leaves
To extend Corollary 1 to general networks, we first note that Eulerian networks have no leafarcs, so we attempt to find such a tour S satisfying the hypothesis of Theorem 2 for networkswithout leaf arcs. It turns out that taking k = α is sufficiently small with respect to the girth g of Q , definedfor networks as the minimum length of a circuit in Q , and if Q has no circuits then g = ∞ .(For networks with unit length arcs, this coincides with the usual integer definition of thegirth of a graph.) Our first result is the following. Theorem 3
For any network Q there is a tour S which covers every arc twice and forwhich no arc is traversed consecutively in opposite directions, except for leaf arcs. The way we will prove Theorem 3 is to double every arc of Q to create an network ˆ Q .Then ˆ Q is Eulerian and has an Eulerian tour. We note that in Euler’s Theorem (finding anEulerian tour in graphs of even degree), we can control to some extent the construction ofthe tour. The following refinement of Euler’s Theorem (Lemma 3) is based on some simplemodifications of the traditional proof and shows that we can control the pairing of enteredand exited passages of the tour at every node. Formally, a passage at a node x is a pair ( x, a ) , where a is an arc incident to x . So a node of degree d has d passages and every arcis part of two passages. Lemma 3
Let Q be a connected Eulerian network such that at every node the passages areidentified in pairs (they are “paired”). Then there is an Eulerian tour S of Q satisfying S never enters and leaves a node via paired passages. (*) Proof:
This proof mimics the usual proof of Euler’s Theorem. We first construct a circuit C satisfying condition (*), which we call a ∗ -circuit, using the following rules:1. Start at any node x and leave by any passage P (we let P ′ be the paired passage of P ).2. Always leave a node by an untraversed passage not paired with the arriving passage.3. If, after arriving at a node, there are three untraversed passages with exactly two ofthem paired, leave by one of this pair. 11. If, after arriving at node x , there are two untraversed passages, leave by passage P ′ , ifit is untraversed.5. If there are no remaining untraversed passages after arriving at a node, stop.To simply obtain a circuit (not necessarily satisfying (*)) starting and ending at x , wewould follow the usual method of simply leaving a node by any untraversed passage , a simplerform of Rule 2. The existence of an untraversed passage (at any node other than the startingnode x ) follows from the fact the after arriving at a node an odd number of passages willhave been traversed, so an odd number (hence not 0) are untraversed. We show that the fullform of Rule 2 along with the other rules ensure that we can always leave a node in a waythat satisfies (*) whether the node is the initial node x or another node y .We first check that after arriving at a node y other than the starting node x , there cannotbe only one remaining untraversed passage which is paired with the arriving passage. Sinceevery node has even degree and degree two nodes are not permitted, the node y must havebeen previously arrived at. After this previous arrival at y , there must have been threeuntraversed passages with exactly two of them paired. But Rule 3 ensures the circuit leftby one of those two passages, so after arriving by the other one on the final visit, the lastuntraversed passage must have a different label.To check that the final arriving passage at the initial node x is not P ′ , note that if P ′ hadnot been traversed before the penultimate visit to x , Rule 4 ensures that it will be traversedon that visit, and it will not be the final arriving passage.If C is a tour (contains all the arcs), we are done. Otherwise, since Q is connected, thereis a node z with some passages in C and some not in C (see Figure 2). Suppose that C leaves z beginning via passage a and ends at z via passage b . We create a new ∗ -circuit startingat z , called C ′ , using the same rules and using only passages not in C . Suppose C ′ beginswith a passage called d (which we can choose) and ends with a passage called e (which wecannot control). The combined circuit CC ′ which starts at z and traverses C and then C ′ will satisfy (*) except possibly for the transitions b, d and e, a between the two circuits, sowe need d ≠ b ′ and e ≠ a ′ (this means d is not paired with b and e is not paired with a ) . Thearc d is chosen as follows. 12igure 2: How to join two ∗ -circuits at node z .1. If a ′ is not in C , take d = a ′ . This ensures that d = a ′ ≠ b ′ since a ≠ b . Also e ≠ d = a ′ .2. If a ′ is in C , take d ≠ b ′ . We know that also e ≠ a ′ because a ′ is in C .If the circuit CC ′ is not a tour, we iteratively continue to add new circuits until we endup with a tour, noting that the process is guaranteed to end since every new circuit containsat least one new arc and there are a finite number of arcs. ◻ Now we are ready to prove Theorem 3.
Proof of Theorem 3.
Let ˆ Q be the Eulerian network obtained from Q by doubling everyarc. (This has the effect of replacing leaf arcs with loops of double the length, since degreetwo nodes are not permitted.) At every node of ˆ Q we pair passages that correspond to thesame passage of Q . Now apply Lemma 3 to ˆ Q to obtain an Eulerian circuit ˆ S of ˆ Q satisfyingcondition (*). This corresponds to S , a double cover of Q (a tour of Q where every arc istraversed twice), in which consecutive arcs are distinct, except for leaf arcs. For loops, anarc may be repeated consecutively, but always in the same direction both times. ◻ Theorem 3 is not new; it was proved by Sabidussi (1977). See also Klavzar and Rus(2013) and Eggleton and Skilton (1984). Our proof based on the new result, Lemma 3, iselementary.The proof of Lemma 3 gives rise to an algorithm for constructing an Eulerian tour ofˆ Q satisfying condition (*), and hence a tour of Q of the form described in the statementof Theorem 3 (named S ). Indeed, by following the rules listed in the proof of Lemma 3,we obtain a circuit C in ˆ Q satisfying (*); by recursively applying the rules to the connectedcomponents of ˆ Q − C and appending these circuits to C at appropriate points, we can obtainan Eulerian tour of ˆ Q satisfying (*).We illustrate the creation of the ∗ -circuit described above for the network K depictedin Figure 3. Doubling each arc, we give the extra arc the same label as the original arc13ut with a prime. Applying the rules of the proof of Lemma 3, starting at the bottom leftnode, we obtain a circuit: a, b, c, d, e, c ′ , a ′ , f, d ′ . Removing this circuit leaves the networkconsisting of arcs b ′ , e ′ and f ′ , which is already a circuit. Adding this circuit at the firstpossible opportunity, we obtain the Eulerian tour a, b ′ , e ′ , f ′ , b, c, d, e, c ′ , a ′ , f, d ′ . 𝑏𝑑𝑐𝑎 𝑑 𝑒 𝑓 Figure 3: The network K . Theorem 4
Let Q be a network without leaf arcs. Then for α ≤ g , where g is the girth, wehave the following:1. The value of the game is V = α / µ .2. For the Attacker, any uniform attack strategy is optimal.3. For the Patroller, the randomized periodic extension ˜ S is optimal, for any tour S given by Theorem 3. Proof:
Let S be a tour of Q given by Theorem 3. Note that it has length L = µ . Sincethere are no leaf arcs, any two consecutive arcs of S are distinct. Suppose some point x of Q is reached by S at consecutive times t and s with t < s . Let Z denote the restriction of S to the interval [ t, s ] . Then Z is a circuit of length s − t and hence s − t ≥ g , by the definitionof girth. Hence V = α / µ , by Theorem 2(ii) with k = α ≤ g . ◻ For the network K depicted in Figure 3, assuming all arcs have length 1, the girth g is 3.So for α ≤
3, the uniform attack strategy is optimal and the Patroller strategy S is optimal,where S is the tour a, b ′ , e ′ , f ′ , b, c, d, e, c ′ , a ′ , f, d ′ .As a further example, consider Q to be a network with two nodes A and B connected bythree arcs of lengths a ≤ b ≤ c . Then g = a + b and µ = a + b + c , so we have by Theorem 4 thatthe value is V ( α ) = α / ( a + b + c ) for α ≤ a + b . This network, with a = b = c = g = V ( α ) = α / α ≤ V ( α ) ≤ f ( α ) ≡ − ( / )( − α / ) for α ∈ [ , / ] . Since f ( α ) < α / α ∈ ( , / ] ( f ( α ) = α / a = f ′ ( α ) = ( − α )/ < / α > α / α in this interval, so the bound α ≤ g = α ≤ g .Suppose we have a network Q with two nodes connected by five arcs labeled as 1 , , , , i having length i . The girth is given by g = g ( Q ) = + =
3. However, suppose weobtain a double cover (with k = S of Q described by the sequence [ , ′ , , ′ , , ′ , , ′ , , ′ ] ,where unprimed arcs go from, say, node A to node B and primed arcs go from node B tonode A . The shortest return time to a regular point is for a point x near node B on thearc of length 5. After leaving x , going to nearby B , the patrol traverses arcs of lengths1 + + + =
10 before going back to x from B . Note that S returns to A after gaps of3 , , , B is visited twice separated by a gap of 14. So for the network Q we have V = α / µ for α ≤
10 rather than just for α ≤
3. This observation leads to combinatorial questions aboutthe maximum shortest circuit in a k -cover of a network Q . As noted above based on Garrec’sanalysis of the three arc network, in certain cases V = α / µ fails for all α > g . We note thatour example Q generalizes easily to the following. Theorem 5
Let Q be a network with two nodes connected by an odd number of arcs. Then V = α / µ for α ≤ µ − L , where L is the length of the longest arc. If Q is a network with two nodes connected by an even number of arcs, then Q is Eulerianand thus V = α / µ for all α .We conclude this section with an application of our earlier result on identifying points.Consider the two networks Q and Q ′ drawn in Figure 1, with α =
3. We would like to showthat V ( Q ′ ) = α / µ = / = /
2. We know from Lemma 2 that V ( Q ′ ) ≤ α / µ = /
2. So weonly need 1 / V ( Q ′ ) . However we cannot apply Theorem 4 becauseit is not true that α is less than or equal to the girth of Q ′ , which is 2. However we knoweither from Garrec (2019) or from Theorem 4 (which applies because 3 = α < g =
4) that V ( Q ) = α / µ = /
2. So by viewing Q ′ as coming from Q by identifying points C and D ,Lemma 2 gives V ( Q ′ ) ≥ V ( Q ) = / We now extend Theorem 4 to networks with leaves. We begin with a modified Patrollerstrategy based on the tour S of Theorem 3.15 efinition 5 Let S be a tour given by Theorem 3. We denote by S α the tour that followsthe same trajectory as S but stops for time α whenever it reaches a leaf node. Lemma 4
Let Q be a network with l leaf nodes and girth g . Then V ≥ αµ + lα / , for α ≤ g. Proof:
Tour S α takes total time 2 µ + lα . Note that every point of Q is visited by S α attwo times differing by at least α . So by Theorem 2 part (i) with k = L = µ + lα , wehave V ≥ α / ( µ + lα ) . (We observe that instead of stopping for time α , the tour S α coulddo anything in this time interval, such as going away from the node a distance α / ◻ Definition 6 (Generalized girth)
We define the generalized girth g ∗ of a network Q by considering a leaf arc of length L to be a circuit of length L . So g ∗ is the minimum ofcircuit lengths of Q and twice the length of any leaf arc. In particular g ∗ ≤ g , with equality if there are no leaf arcs or if all leaf arcs have lengthgreater than g /
2. Note that if α ≤ g ∗ we know in particular that all leaf arcs have length atleast α / Lemma 5
Let Q be a network with l leaf nodes and generalized girth g ∗ . Then by adoptingthe independent attack strategy on the set I of leaf nodes, the Attacker can ensure that theinterception probability is less than αµ + lα / for α ≤ g ∗ . Hence, V ≤ αµ + lα / , for α ≤ g ∗ . Proof:
As noted above, the assumption on α ensures that all leaf arcs have length at least α /
2, so the result follows from Theorem 1. ◻ Since g ∗ ≤ g , Lemmas 4 and 5 apply when α ≤ g ∗ and hence we have the followingextension of Theorem 4 to networks with leaf arcs. Theorem 6
Let Q be a network with l leaf nodes and generalized girth g ∗ . Then V = αµ + lα / , for α ≤ g ∗ . For the Patroller, an optimal strategy is S α as defined above. For the Attacker, an optimalstrategy is the independent attack strategy, taking I to be the independent set of leaf nodes.
16t is useful for later comparisons to specialize this result to trees.
Corollary 2 If Q is a tree with l leaf arcs, then(i) V ≥ αµ + lα / ,(ii) with equality if all leaf arcs have length at least α / . Proof:
To establish (ii), note that trees have no circuits, so the generalized girth g ∗ istwice the length of its smallest leaf arc, so by assumption, α ≤ g ∗ . The result now followsfrom Theorem 6. For (i), consider the patrol S α . Note that between any two visits by S α toa point of Q , a leaf node is visited. Hence the return times exceed the time α that S α stopsat that node, and the result follows from Theorem 2(i) with k = L = µ + lα . ◻ For example, consider the tree Q depicted in Figure 4. The number of leaf arcs is l = g ∗ = µ =
9, so by Theorem 6, the value of thegame is α /( + α / ) for α ≤
2. Figure 4: The tree Q . In Corollary 2 we gave some preliminary results for trees. Lemma 4 gave a lower bound onthe value of the game based on the Patroller strategy S α . Furthermore, for α ≤ g ∗ , where g ∗ is the generalized girth, we showed in Theorem 6 that the independent attack strategyensures that this lower bound is tight. Note that for a tree, g ∗ is twice the length of theshortest leaf arc. In this section, we extend these results and give optimal Patroller andAttacker strategies for some values of α which are greater than g ∗ . We start by defining the extremity set E , a subset of Q that is essential in describing optimal Patroller and Attackerstrategies. 17 .1 The Extremity Set E The relationship between the network Q and the duration α of the attack interval determinesthe type of optimal player strategies. In this section we define the extremity set E that helpsus explore this relationship for trees.If B is a set of points then we denote by ¯ B the topological closure of B . If Q is a treenetwork, then its minimum tour time is 2 µ , as every arc must be traversed twice. If x is aregular point of tree network Q , then Q − { x } has two connected components Q = Q ( x ) and Q = Q ( x ) , whose lengths satisfy λ ( Q ) + λ ( Q ) = λ ( Q ) = µ . We introduce a subset E ( Q ) of Q . Definition 7 (The extremity set E ) Let Q be a tree. The extremity set E ≡ E ( Q, α ) isdefined as the set of all regular points x ∈ Q such that min i = , λ ( Q i ( x )) < α / . (2)Note that min i = , λ ( Q i ) ≤ µ / µ < α then (2) holds for all regularpoints, which implies that ¯ E = Q . The extremity set E consists of regular points whoseminimum return time during a CPT is less than the attack duration α . It can be partitionedinto maximal connected sets that we call components of E and we denote by E j . Example 1
We illustrate the extremity set E on the tree network of Figure 4 that has µ = .Figure 5 shows how E changes for increasing values of α on this network. As α increasesthe components grow starting from points near the five leaf nodes of the tree. Initially thereare five components (cases α = , , , ); but eventually points near non-leaf nodes becomemembers of E and the number of components increase to seven (cases α = , , , ). Notethat in case α = the closure of ¯ E of E is equal to the whole network. The results from theprevious sections (Theorem 6, Corollary 2) solve the game for cases α ≤ g ∗ = , but in thissection we extend the results to cover all cases of α ≤ . E ( Q, α ) , shown in thick red lines, for the tree Q of Figure 4and α = , . . . , Example 2
Figure 6 depicts a star network. We assume that the extremity set E is as itappears on the figure in red thick lines; we make no assumptions on the value of α or thelength of the arcs; from the shape of E we draw some conclusions. Here, E decomposes intofour components: ( A, B ) , ( A, C ) , ( D, F ) , ( E, G ) . We claim that λ ( DF ) = λ ( EG ) = α / ;this is because on leaf arc AD (similarly for AE) if λ ( DF ) < α / there would be a point Xon the right of F whose distance from D would be < α / , implying λ ( DX ) < α / and thuscontradicting X ∉ E . Similarly, if λ ( DF ) > α / there would be a point X on the left of Fwhere λ ( DX ) > α / contradicting X ∈ E . Thus, components E j that are strict subsets of aleaf arc and whose closure contains the leaf node will have length α / . However, components E j whose closure is the entire leaf arc (like AB and AC) must have length ≤ α / ; if theyhad length > α / then there would be point X on the component AB near node A where λ ( BX ) > α / contradicting x ∈ E . Figure 6: A tree, with its extremity set E in thick red. E -patrolling Strategy S E for Trees We will see that for some trees, the uniform CPT strategy is still optimal for the Patroller,but its optimality depends on the size of the attack duration, α . As mentioned earlier, for atree a CPT is simply any depth-first search which returns to its start point after completing19ts search, so that ¯ µ = µ ; every point of the tree except the leaf nodes is visited at least twiceby a CPT. This means the leaf nodes and regular points near them are left “less protected”by a uniform CPT than the other points, and for sufficiently small values of α , there willbe points in the tree whose two closest visit times (modulo ¯ µ ) are at least time α apart,meaning that they are, in a sense “twice as protected” as the leaf nodes. (In all that follows,arithmetic on time will be performed modulo the length of the tour in question).This motivates the introduction of a new Patroller strategy S E for trees that we call the E -patrolling strategy. To describe it, we use the set extremity set E ≡ E ( Q, α ) that wedefined earlier; in particular, we use the closure of ¯ E of E and its components ¯ E , . . . , ¯ E k ,each of which is a subtree of Q . We have λ ( ¯ E ) = λ ( E ) but by using the componentsof ¯ E rather than the components of E , we include the nodes and thereby unite adjacentcomponents of E into a single component of ¯ E . For example, in Figure 6 there are fourcomponents of E but only three components of ¯ E , since in ¯ E the lines AB and AC join toform a single component BAC.Let Q be a tree with ¯ E ≠ Q . We first construct a CPT S with the additional propertythat every component ¯ E j is searched in a single CPT of ¯ E j , which we call C j ; note thatsome CPTs of Q might search different subsets of ¯ E j during non-consecutive time intervals- we exclude this possibility by construction.To obtain a CPT of Q with this property, we begin at any regular point not in ¯ E and goin either direction. When arriving at any node, we leave by a passage not already traversed,if there is such a passage. (This is the usual depth-first construction and ensures we obtaina CPT.) Furthermore, if the node belongs to some component ¯ E j and there are untraversedpassages staying in that component, we take one of these. For example, in Figure 6 if westart on F A going right, and tour the leaf arc to B from A , we must then take the passageto C (staying in component BAC) rather than the other untraversed passage out of A goingto E . This ensures that the CPT say ABAEACADA (in which the component
BAC of¯ E is not traversed in a single CPT of BAC ) will not be constructed, but rather one like [ ABACA ] EADA , where the bracketed expression is a CPT of the component
BAC .Then we make two types of additions at every component. If λ ( ¯ E j ) ≥ α /
2, we followthe CPT C j of ¯ E j in S by another identical one, before continuing with S . Note that thislocal CPT takes time ≥ α , so the time between the first and second CPT of ¯ E j reaching any(regular) point is at least α .If λ ( ¯ E j ) < α / E j twice in succession because points of¯ E j will not be visited at least two times that are at least time α apart. Instead we wait20ntil S returns to ¯ E j after the first occurrence of C j in S , and then insert another C j . Wemust check that these two times that S visits ¯ E j are separated by time at least α . Let [ t , t ] be the time interval during which S tours ¯ E j so that that S ( t ) = S ( t ) = x , say, and t − t = λ ( ¯ E j ) . We claim that Q − ¯ E j has at least two components. If not, then x , which ison the boundary of ¯ E j , must be in the interior of an arc. Let x ′ = S ( t − ε ) = S ( t + ε ) ∉ E ,where d ( x, x ′ ) < α / − λ ( ¯ E j ) . Then S (( t − ε, t + ε )) is a component of Q − x ′ with length λ ( ¯ E j ) + ε < α /
2, so x ′ ∈ E , a contradiction.Thus, Q − ¯ E j has at least two components and they must all have length greater than α /
2, since any component with length at most α / E , and could notbe disjoint from ¯ E j . So the next time after t that S arrives at x is t ≥ t + α , and the nexttime after t that S arrives at x is at least t + α . Then S is updated by adding another tourof C j at time t .Observe that each additional local CPT of ¯ E j takes time 2 λ ( ¯ E j ) , so the total length ofthe resulting tour S E is 2 µ + (∑ j λ ( ¯ E j )) = ( µ + λ ( ¯ E )) and by construction it reaches everypoint of Q at two times separated by at least α (modulo the length of the tour). Note thatif ¯ E = Q , we simply take S E = S . The optimal periodic strategy is thus S E . For the network ofFigure 6, taking S as ABACADAEA we could have S E = ABACAF D [ F DF ][ ABACA ] GE [ GEG ] A ,where the brackets indicate the three inserted local CPT’s of the components of ¯ E . Note thattwo of these are inserted right after their first occurrence, but the third one [ABACA] is in-serted nonconsecutively. Our construction would not work directly on the CPT ABAEACADA .Thus we have established the following result by explicit construction.
Lemma 6
Let Q be a tree. Then there is a tour S E , an E -patrolling strategy, of length ( µ + λ ( E )) such that every point x of Q is visited at least twice at times that differ by atleast α . We conjecture that E -patrolling strategies are always optimal for trees, and we laterconfirm the conjecture in some special cases. For now we give a general bound on the valueof the game obtained by using an E -patrolling strategy. Let v ∗ ≡ α /( µ + λ ( E )) . Theorem 7
Let Q be a tree. Any E -patrolling strategy intercepts any attack with probabilityat least v ∗ . Proof:
Follows from Lemma 6 and Theorem 2 part (i) with k = S = S E , and L = ( µ + λ ( E )) . ◻ α ≤ g ∗ we have λ ( E ) = lα /
2, and the result of Theorem 7 becomes thesame as the result of Corollary 2. In that case, the patrolling strategy S α gives the samelower bound as an E -patrolling strategy. E -attack Strategy In the previous section we showed that on a tree, any E -patrolling strategy intercepts anyattack with probability at least v ∗ . Here, we define the E -attack strategy , whose attacksare intercepted with probability at most v ∗ on some trees. The condition that allows thisstrategy to be defined and to be optimal is given in Definition 8. It is useful to note thatwhile for patrolling strategies we looked at the components of the closure ¯ E of E , for theattack strategy given here we look at the components of E itself. Definition 8 (Leaf Condition)
Let Q be a tree. We say that ( Q, α ) satisfies the LeafCondition if E consists of all points on every leaf arc within distance α / of its leaf node. For example, in Figure 5 the cases that satisfy the Leaf Condition are the first four( α = , , , E consist of five components; all of these five components are subsetsof leaf arcs and they are within α / E j of E corresponds to a leaf node; this is easy to check inFigure 5. Cases α = , , , ( Q, α ) does not satisfy theLeaf Condition. Definition 9 ( E -attack strategy) Suppose ( Q, α ) satisfies the Leaf Condition, where Q is a tree. Let x j denote the leaf node contained in the closure of the component E j of E , andlet e j = λ ( E j ) and let M = max j λ ( E j ) be the maximum length of a component of E . Wedefine the E -attack strategy as follows:1. With probability λ ( E c )/( µ + λ ( E )) , attack a uniformly random point of E c at time M .2. With probability e j /( µ + λ ( E )) , attack at leaf node x j at a start time chosen uniformlyin the interval [ M − e j , M + e j ] . Note that the Leaf Condition implies that ∑ j e j = λ ( E ) , therefore the sum of the probabilitiesfrom 1. and 2. above sum to 1. Also, unlike the uniform attack strategy, the E -attackstrategy is not synchronous. That is, the attack does not start at a fixed, deterministic time.22 xample 3 We revisit Figure 6, where the leaf arcs have lengths , , , and α = . Weillustrate the E -attack strategy on this star network in Figure 7. Here µ = ; the extremityset E is shown in thick red lines. E consists of four components that are subsets of leaf arcsand whose points are within α / from the leaf node, thus the Leaf Condition is satisfied. Also,note that λ ( E ) = and µ + λ ( E ) = . The E -attack strategy then attacks as follows: withequal probabilities / it attacks at nodes D and E with a starting time chosen uniformlyon [ , ] ; with probabilities / , / it attacks leaf nodes B , C with a starting time chosenuniformly on [ , ] , [ , ] respectively; with probability / it attacks uniformly on set E c attime M = . Figure 7: The E -attack strategy on an asymmetric star with arcs lengths 2,1,6,6 with α = E is shown in thick red lines.We next prove that for trees Q , the E -attack strategy is optimal if ( Q, α ) satisfies theLeaf Condition. Lemma 7
Let Q be a tree and suppose ( Q, α ) satisfies the Leaf Condition. Then the E -attack strategy is intercepted by any patrol with probability at most v ∗ = α /( µ + λ ( E )) . The proof of Lemma 7 is in the Online Appendix. If we combine the results of Theorem 7and Lemma 7 on patrolling and attack strategies for trees, we obtain the following exact resultfor the value of the game.
Theorem 8
Let Q be a tree and suppose ( Q, α ) satisfies the Leaf Condition. Then any E -patrolling strategy is optimal, the E -attack strategy is optimal, and the value of the gameis V = v ∗ . Example 4
We revisit the network Q from Figure 7 with α = and µ = . We first considerpatrolling strategies. The S α patrolling strategy is ADDABBACCAEEA , where repeating anode means it stays there for duration α ; this tour has length µ + ( ) = . From Corollary 2 e have V ≥ αµ + lα / = / . An E -patrolling strategy is ADF DABACABACAEGEA withlength µ + λ ( E ) = ; from Theorem 7 we have V ≥ v ∗ = αµ + λ ( E ) = / . As we can see,an E -patrolling strategy, which is defined only for trees offers an improvement over the S α patrolling strategy, which is a more general strategy.Now, we consider attacker strategies. Let I be the set of leaf nodes. The sets E and W ≡ W ( I ) are shown in Figure 8 with solid thick red and dashed thick green lines respectively.Note that ( Q, α ) satisfies the Leaf Condition. The E -attack strategy is demonstrated inFigure 7 and it gives a lower bound, v ∗ = αµ + λ ( E ) = / , from Theorem 8, which is optimal.The bound given by Theorem 1 αλ ( W c )+ lα = αµ + lα / = / does not hold in this case because I is not an independent set or, equivalently, leaf arcs do not have lengths exceeding α / . Figure 8: Star with arc lengths 6,6,2,1 and α =
6. The solid thick red line is the set E andthe thick dashed green line is the set W ≡ W ( I ) , where I is the set of leaf nodes; note thathere I is not an independent set.A star is a network consisting entirely of leaf arcs. We call a star balanced if no arccomprises more than half of its total length; otherwise we say that it is skewed . It is easy tocheck that balanced stars satisfy the Leaf Condition. All symmetric stars (whose arcs are allthe same length) are balanced. An example of a skewed star is a star with arc lengths 1 , , µ = µ .It is also easy to see that if Q is a star (which may be balanced or skewed) whose longestarc has length at most α /
2, then ¯ E = Q and hence Q satisfies the Leaf Condition. SoTheorem 8 gives the following. Corollary 3
Let Q be a star. Then the E -attack strategy and any E -patrolling strategy areoptimal and the value of the game is V = v ∗ = α /( µ + λ ( E )) if either(i) Q is balanced or(ii) α is at least twice the length of the longest arc of Q .
24e study the skewed star of Figure 9 in Section 7.Note that if Q is the line segment network, then by adding an artificial node in the center,we can apply Corollary 3, part (i), recovering the result for the value of this game, givenpreviously in Alpern et al. (2016) (though the optimal strategies given here are different). In the last section we considered patrolling games on trees. We showed (Theorem 7) thatthe E -patrolling strategy intercepts any attack with probability at least v ∗ = α / ( µ + λ ( E )) and that (Lemma 7) for trees satisfying the Leaf Condition, the E -attack strategy avoidsinterception with probability at least v ∗ . Thus for trees we have V = v ∗ if the Leaf Conditionis satisfied, but what happens when it is not satisfied? In this section we present some trees Q and attack durations α for which the Leaf Condition fails but nevertheless V = v ∗ . We dothis by specifying particular attack strategies which are optimal on these trees.We conjecture the following on trees: Conjecture 1
Let Q be a tree network, then for any α the E -patrolling strategy is optimaland the value of the game is V = v ∗ = α /( µ + λ ( E )) . In Section 7.1 we find such strategies for a range of values of α on a skewed star withlengths 6 , , µ = α ≤ α ≥
12. In what follows we introduce attack strategies for thisstar that guarantee v ∗ for the attacker for 4 ≤ α ≤
8, and thus verify the conjecture. We referto this skewed star as the 6 − − v ∗ for the attacker and thus verifies theconjecture.Figure 9: The 6 − − E is shown in thick red lines and E c in grey.25 .1 Optimal Attack Strategies on the − − Star, ≤ α ≤ We define an attack strategy that we will show is optimal for the 6 − − ≤ α ≤ θ = ( µ + λ ( E )) ; this gives v ∗ = α / θ . Further, we note that for the6 − − ≤ α ≤ λ ( E ) = α . We denote the left and rightboundary points of E c with E by u and v respectively; since 4 ≤ α ≤
8, both of these pointsare on the long arc or on its boundary.We note that the Leaf Condition for this star holds for α = < α ≤
8, thusthe E -attack strategy is not defined for the latter set of values. Thus, we define a new attackstrategy. For α = Definition 10 (6-1-1-attack)
The strategy is defined as follows:
Left attacks:
With probability α / θ , attack equiprobably at nodes and , startinguniformly at times in [ , + α ] . Middle attacks:
With probability λ ( E c )/ θ , attack at a uniformly random point of E c (set of points between u and v ), starting equiprobably at times α / or α / + . Right attacks:
With probability α / θ , attack at node at starting times in [ , α + ] described as follows: conditional on the attack taking place here, the starting time isgiven by the following cumulative probability distribution function, f ( y ) = ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩ y α if ≤ y ≤ , α + y − α if ≤ y ≤ α, α − α + y − α α if α ≤ y ≤ α + . (This is uniform on [ , α ] with conditional probability ( α − )/ α and uniform on [ , ] ∪[ α, α + ] , with conditional probability / α ) .Note that the total probability of attack is αθ + λ ( E c ) θ + αθ = λ ( E ) + ( µ − λ ( E )) θ = ( µ + λ ( E )) θ = Proposition 1
For the − − star, shown in Figure 9, with ≤ α ≤ , we have V = v ∗ = α / θ . The − − -attack strategy avoids interception with probability v ∗ = α /( µ + λ ( E )) . et al. (2016). The conjecture was that for fortrees, if α is at least the diameter of the network, the value of the game is α / ¯ µ = α /( µ ) . Forthe 6-1-1 star, the diameter is 7, and by Proposition 1, for 7 ≤ α <
8, the value is α /( µ + λ ( E )) .This is not equal to α /( µ ) , since λ ( E ) < µ in that range of α , disproving the conjecture.The 6 − − v ∗ for the attacker for α >
8. Notethat for α > α −
1, then arrives at node 1 at time α + α + α +
10. This patrol will intercept the left attackswith conditional probability of 1. The attacks at node 9 end at time 2 α + α +
10 if and only only α ≥
8. Thus there will be positive probability p that the leftattacks are intercepted. Thus, the total interception probability will be αθ + αθ p > αθ = v ∗ . ¯ E = Q satisfying Conjecture 1. We now consider the tree depicted in Figure 10 with unit length arcs and α =
6. This gives¯ E = Q and thus λ ( E ) = µ . Here µ = v ∗ = α / µ = / µ = α = • At each leaf node 1 and 2 attack with probability 6 /
24 at a start time chosen uniformlyin the interval [ , ] (total attack probability 12 / • At leaf node 6 attack with attack start time uniformly: in the interval [ , ] withprobability 2 /
24, in the interval [ , ] with probability 4 /
24, in the interval [ , ] withprobability 2 /
24 (total attack probability 8 / • At leaf node 7 attack takes place with probability 4 /
24 at a start time chosen uniformlyin the interval [ , ] (total attack probability 4 / v ∗ = /
2, thus verifying the conjecture; the proof is along the same lines as that of Section7.1.
This paper is the first to analyze continuous patrolling games on arbitrary networks. Thesegames model the problem of defending transportation networks, pipelines or other networkswhich can be attacked anywhere along their length. In Section 3 we developed a number ofvery general techniques which can be used by the players to estimate the efficacy of varioustypes of strategies on arbitrary networks. In Section 4 we solved the patrolling game forall networks without leaves, using a periodic patrolling path which covers every arc exactlytwice (a double-cover), for attack lengths α not exceeding the girth g of the network. Thatresult (Theorem 4) was then extended in Section 5 to general networks (allowing leaves)by considering a notion of generalized girth g ∗ (Theorem 6). For trees, g ∗ is simply twicethe length of the shortest leaf arc, so Theorem 6 may not be very useful for trees with ashort leaf arc. To remedy this, Section 6 developed a concept of the extremity set E of atree and strategies for both players which are defined in terms of E . We then defined theLeaf Condition for a tree, which required (among other things) that the extremity set E is a subset of the leaf arcs. Conjecture 1 says that the value of the patrolling game on atree is given by v ∗ = α /( µ + λ ( E )) and that the strategies mentioned above based on E areoptimal. Our main result for trees is that Conjecture 1 holds for all trees satisfying the LeafCondition (Theorem 8). We then showed that the Leaf Condition (and hence Conjecture 1)holds for all stars where the length of any leaf arc is not more than half the total length ofthe star. Some stars without this property, as well as a certain non-star tree, are shown tosatisfy Conjecture 1 (but not the Leaf Condition) by explicit construction of strategies inSection 7. Acknowledgements
This material is based upon work supported by the National Science Foundation under GrantNo. CMMI-1935826. 28 eferences
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