Lower and Upper Bounds

A useful hiding distribution for obtaining lower bounds on the expected search time is the uniform distributionh_µ,i.e., “completely random” hiding strategy onQ.

More precisely:

Definition 5 The uniform distribution h_µis a random choice of the target location H such that for all (measurable) sets B∈ Q,

Pr(H∈ B|h_µ)=µ(B)/µ.

(Note thatµ≡µ(Q).)

The next theorem presents a useful lower bound forv.

Theorem 6 If the target location is chosen according to the uniform distribution h_µ then, for any search strategy, the expected time spent in searching is at least µ/2.

Proof.It is easy to see that the probability that the target has been found by time t satisfies Since the hiding distributionh_µguaranteesµ/2 against any starting point of the searcher, we also have the following:

Corollary 7 If there exists a search strategy with valuev=µ/2,then allowing the searcher to choose his starting point does not reduce the expected search time below µ/2.

In studying search trajectories inQ, we shall often use the notion “closed trajec-tory” defined as follows.

Definition 8 A trajectory S(t)defined for0≤t ≤τ is called “closed” if S(0)=S(τ).

(Note that a closed trajectory may cut itself and may even go through some of the arcs more than once.) If a closed trajectory visits all the points of Q , then it is called a

“tour”.

Eulerian tours, defined as follows, will have an important role in our discussion.

Definition 9 A graph Q is called “Euler i an”if there exists a tour L with lengthµ, in which case the tour L will be called an “Euler i an tour”.

A trajectory S(t),0≤t ≤µ,which covers all the points of Q in timeµ,will be called an “Eulerian path”. (Such a path may not be closed.)

It is well known thatQis Eulerian if and only if the degree of every node is even, and that it has an Eulerian path if and only if at most two nodes have odd degree. In

the case thatQhas two odd degree nodes, then every Eulerian path has to start at one of them and end at the other. (See [9].)

Chinese postman tours will play an important role in our discussion:

Definition 10 A closed trajectory which visits all the points of Q and has minimum length will be called a “minimal tour” (or a “Chinese postman tour”) and is usually denoted by L.Its length will be denoted byµ.¯

Finding a minimal tour for a given graph is called theC hi nese post manproblem.

This problem can be reformulated for any given graph Q as follows. Find a set of arcs, of minimum total length, such that when these arcs are duplicated (traversed twice in the tour) the degree of each node becomes even. This problem was solved by Edmonds [6] and Edmonds and Johnson [7] using a matching algorithm which uses O(n³) computational steps, wherenis the number of nodes inQ. This algorithm can be described as follows. First compute the shortest paths between all pairs of odd-degree nodes ofQ. Then partition the odd-degree nodes into pairs so that the sum of lengths of the shortest paths joining the pairs is minimum. This can be done by solving a minimum weighted matching problem. The arcs ofQin the paths identified with arcs of the matching are the arcs which should be duplicated (i.e., traversed twice). The algorithm is also described by Christofides [5] and Lawler [21]. (An updated survey on the Chinese postman problem is presented by Eiselt et al. [8].)

Once an Eulerian graph is given, one can use the following simple algorithm for finding an Eulerian tour (see [4]). Begin at any node Aand take any arc not yet used as long as removing this arc from the set of unused arcs does not disconnect the graph consisting of the unused arcs and incident nodes to them. Some algorithms which are more efficient than this simple algorithm were presented by Edmonds and Johnson [7]. Actually, it is possible to slightly modify Edmond’s algorithm in order to obtain a trajectory (not necessarily closed) which visits all the points of Qand has minimum length. This trajectory is a minimax search trajectory (pure search strategy).

Example 11 Consider the graph in Figure1 (having the same structure as Euler’s K¨oninsberg bridge problem). The duplicated arcs in the minimal tour can be either {a1,b1,c1}(based on the partition{A B,OC}) or{b1,d}(based on{O B,AC}) or {a1,e}(based on{O A,BC}). The corresponding sum of the lengths of the arcs is either5or4or5.Thus, the minimal tour duplicates the arcs b1 and d.The minimal tour can be traversed by the following trajectory S1

S1=Ob1Bb1Ob2BeCd Aa1Oa2AdCc1O (1) with lengthµ¯ =20.

A minimax search trajectory is based on duplicating arcs (having minimum total length) in order to make all the node degrees even except for the starting point O plus

b1 B C

e d

b2

−1−

−3−

a1

a2

−1−

−2−

−3− −4−

A

O c1

−2−

Figure 1.

another node. It can be easily seen that the arcs that have to be duplicated are a1and b1leading to the following minimax trajectory

Ob1Bb1Ob2BeCd Aa1Oa1Aa2Oc1C with length18.

The length of a Chinese postman tour is bounded by twice the size of the search space as is stated by the following:

Lemma 12 Any minimal tour satisfiesµ¯ ≤2µ.Equality holds only for trees.

This result can easily be obtained by considering a graph ¯Qobtained fromQas follows. To any arcbinQ, add another arc ¯bwhich connects the same nodes and has the same length asb. It is easy to see that ¯Qis Eulerian with a tour length 2µ.If we now map the graph ¯Qinto the original graphQsuch that both arcsband ¯bof ¯Qare mapped into the single arcbofQ, then the tour ¯L is mapped into a tourL ofQwith the same length 2µ.IfQ is not a tree, it contains a circuitC.If we remove all new arcs ¯bin ¯Qcorresponding to this circuit, then the resulting graph is still Eulerian and containsQ,but has total length less than 2µ.

Using the length ¯µ of the minimal tour, we now derive an upper bound for the value of the search game in a graph in terms of the length of its minimal tour.

Lemma 13 For any graph Q,the valuevsatisfiesv ≤µ/2,¯ whereµ¯ is the length of a minimal tour L of Q.

Proof.Assume that the searcher uses a strategy ¯swhich encircles L equiprobably in each direction. It can be easily seen that, for any target location, the sum of these two possible discovery times is at most ¯µ.Consequently, the expected discovery time does

not exceed ¯µ/2.

Definition 14 The search strategy s used for proving Lemma 13 will be called¯ the“random Chinese postman tour”.

A random Chinese tour of Figure 1 is either to randomly follow eitherS1(see (1)) or the same path in the opposite direction:

Oc1Cd Aa2Oa1AdCe Bb2Ob1Bb1O.

Combining Theorem 6, Lemma 12 and Lemma 13 we obtain the following result.

Theorem 15 For any graph Q, the search valuevsatisfies

µ/2≤v≤µ/2¯ ≤µ. (2)

In the next section we will show that the upper bound, µ,is attained if and only ifQis a tree.

The lower bound,µ/2,is attained if and only if Q is Eulerian because if Q is Eulerian then ¯µ=µso that (2) implies:

Theorem 16 If Q is Eulerian then the Random Chinese Postman tour is an optimal search strategy yielding a value ofµ/2(half the total length of Q).

Actually the Eulerian graphs are the only graphs with a search value ofµ/2 : Theorem 17 If the search value of a graph Q isµ/2then Q is Eulerian

The proof is given in [12] and in [1].

Corollary 18 For an Eulerian graph, the optimal strategies and the value of the search problem remain the same if we remove the usual restriction S(0)=O and instead allow the searcher an arbitrary starting point.

The claim of the corollary is an immediate consequence of Corollary 7.

Remark 19 Corollary 18 fails to hold for non-Eulerian graphs because (unlike the Eulerian case) the least favorable hiding distribution usually depends on the starting point of the searcher.

In general, if we allow the searcher an arbitrary starting point, thenv=µ/2if and only if there exists an Eulerian path (not necessarily closed) in Q.(If there exists such a path then the searcher can keep the expected capture time≤µ/2by an analogous strategy tos of Lemma 13. If there exists no Eulerian path in Q,¯ then the uniform hiding distribution yields the expected search time exceedingµ/2.)