We consider the symmetric rendezvous search game on a complete graph of n locations. In 1990, Anderson and Weber proposed a strategy in which, over successive blocks of n-1 steps, the players independently choose either to stay at their initial location or to tour the other n-1 locations, with probabilities p and 1-p, respectively. Their strategy has been proved optimal for n=2 with p=1/2, and for n=3 with p=1/3. The proof for n=3 is very complicated and it has been difficult to guess what might be true for n>3. Anderson and Weber suspected that their strategy might not be optimal for n>3, but they had no particular reason to believe this and no one has been able to find anything better. This paper describes a strategy that is better than Anderson--Weber for n=4. However, it is better by only a tiny fraction of a percent.
In the symmetric rendezvous search game on K n (the completely connected graph on n vertices) two players are initially placed at two distinct vertices (called locations). The game is played in discrete steps and at each step each player can either stay where he is or move to a different location. The players share no common labelling of the locations. Our aim is to find a (randomizing) strategy such that if both players independently follow this strategy then they minimize the expected number of steps until they first meet. Rendezvous search games of this type were first proposed by Steve Alpern in 1976. They are simple to describe, and have received considerable attention in the popular press as they model problems that are familiar in real life. They are notoriously difficult to analyse.
The Anderson-Weber strategy is a mixed strategy that proceeds in blocks of n -1 steps. Players begin at distinct locations, called their home locations. In each successive block a player either stays at his home location, with probability p, or makes a randomly chosen tour of his n -1 non-home locations, doing this with probability 1 -p. The motivation for the strategy comes from the wait-for-mommy strategy that is optimal in an asymmetric version of the problem. With probability 2p(1 -p) the players play the wait-for-mommy strategy over the first n -1 steps and so rendezvous in expected time (n + 1)/2.
Anderson and Weber (1990) proved that the above strategy is optimal for the game on K 2 , with p = 1/2, and conjectured that it should be optimal for K 3 , with p = 1/3. This was finally proved by Weber (2006), who established a strong AW property (SAW) that AW minimizes E[min{T, k}] for all k. Anderson and Weber suspected that their strategy might not be optimal for n > 3, but they had no particular reason to believe this and no one has been able to find any strategy that is better. Indeed, AW has been shown optimal amongst 2-Markov strategies. Fan (2009) showed that AW minimizes P (T > 2) and E[min{T, 2}]. He also found that AW is not optimal on K 4 if players have the extra information that the location can be viewed as being arranged on a circle and the players are given a common notion of clockwise. However, the question as to whether or not AW is optimal has remained open for the case in which there is no such special extra information. Fan writes, ‘The author believes that SAW still holds on K 4 , and so AW strategy is still optimal’. We were inclined to agree, but now find that AW can be bettered. For more background to the problem see Weber (2006).
Let us begin by reprising the AW strategy for the symmetric rendezvous game on K 4 . We assume that there is no special knowledge (such as a common notion of clockwise on a circle). The AW strategy is a 3-Markov strategy that repeats in blocks of 3 steps. In each successive block of 3 steps, each player, independently, remains at his home location with probability p, or does a random chosen tour of his 3 non-home locations, with probability 1 -p. This leads to rendezvous in an expected number steps ET , where
This is explained as follows.
If both stay home they do not meet.
If one stays home, while the other tours, then they meet in expected time 2. We now explain how the AW strategy can be bettered. Suppose player I has location 1 as his home, and player II has location 2 as home. We might imagine that each player labels his non-home locations as a, b, c, and so a tour of his non-home locations is one of six possible tours: abc, acb, bac, bca, cab, cba. In the case that player I has (a, b, c) = (2, 3, 4) and player II has (a, b, c) = (1, 3, 4) we can compute the matrix
where we have ordered the rows and columns to correspond to abc, acb, bac, bca, cab, cba. A number entry indicates the step at which players meet when they meet, and X indicates that they do not meet. There are 36 such matrices, over which we must average, for each possible pair of assignments by players I and II, of (2, 3, 4) and (1, 3, 4), respectively, to (a, b, c).
Let us begin by noting that if a player stays home for three steps and meeting does not occur, then the other player must also have been staying home. Similarly, if a player tours for three steps and meeting does not occur, then the other player must also have been touring (and their tours not meeting). Thus after any 3k steps (a multiple of 3) each player knows exactly how many times both have been touring.
Whenever a player makes a tour in the AW strategy he chooses his tour at random (independently of previous tours). We show how to improve AW introducing some dependence between tours. Let us adopt a notation in which the first tour a player makes is labelled A. The second distinct tour a player makes is labelled B, and so on. So, for example, AAB means that on his first three tours, a player (i) first makes a random tour, (ii) second makes the same tour as his first tour, (iii) and third makes a tour chosen randomly fro
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