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215 The Devil's Dartboard Trevor Lipscombe and Arturo Sangalli Abstract To produce the most-dicult dartboard possible, we devise a polynomial-time algorithm. The problem, and the algorithm, are shown to be closely related to the Travelling Salesman Problem. AMS Subject Classication: 68Q25 (Analysis of algorithms and problem complexity), 05A99 (Classical combinatorial problems, and 00A08 (Recreational mathematics). Introduction Students of mathematics can often be found playing darts in bars. The reason, no doubt, is their love of the subject. For example, recent analyses such as [1] and [2] have shown that the mathematically adept, yet physically inept, darts players should aim at the bull's eye in order to maximize their potential score. This raises the question: what could dartboard manufacturers do to overcome this strategy? The answer could be to renumber the board completely. One mathematician to consider the benets and strategies of possible renumbering was Ivars Peterson, in a popular article [3]. The numbers in a dartboard are (sort of) arranged so that there is a penalty for error. For instance, the number 20 at the top of the board is sandwiched between the far smaller numbers 1 and 5. Aim at the 20 with three darts and, if you are not very good, you risk a score of 20 + 5 + 1 = 26. Likewise, 19 is anked by 3 and 7 (for a possible score of 29), and the 17 has a 2 and 3 on either side (= 22). There are more than 1:2 1017 circular arrangements of the numbers 1, 2, : : : , 20 | the exact number being 19! (that is, factorial 19). The challenge is to nd the one arrangement among them that gives the least reward for throwing error. More mathematically, what is the arrangement for which the standard deviation of the three-sums (that is, the sums of three consecutive numbers) is the smallest? We call this arrangement \The Devil's Dartboard." We shall consider n{number dartboards (n > 3) and ask: Is there a polynomial-time[4] algorithm that will generate the Devil's Dartboard for each n? A Devil's Dartboard Problem (DDP) consists of nding such an algorithm or proving that it does not exist. For the record, the actual dartboard, found in drinking establishments across the globe, is the permutation: 20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12, 5. Copyright c 2000 Canadian Mathematical Society 216 A Decent Algorithm Here is a polynomial-time algorithm that, if it does not always produce a Devil's Dartboard, seems to come pretty close, especially for large n. The permutation p is dened by induction. Choose p(k + 1) such that the 3{sum p(k ; 1) + p(k) + p(k + 1) is as close as possible to the mean m = 3(n + 1)=2 of the 3{sums. In the case of a tie, select p(k + 1) so that consecutive 3{sums (dierent from m) fall on opposite sides of m, beginning with a 3{sum greater than m. To get the induction started, set p(1) = n and p(2) = 1. For example, if n = 6, we have m = 3(6 + 1)=2 = 10:5. The algorithm then produces the Devil's Dartboard 6, 1, 4, 5, 2, 3. The sequence of 3{sums is 11, 10, 11, 10, 11, 10, and the standard deviation = 0:5, clearly the smallest one possible. It is not, however, the only Devil's Dartboard for 6 numbers. Every Devil's Dartboard has its `opposite'; that is, the permutation obtained by reading the numbers right to left, which is another Devil's Dartboard. These two permutations correspond to the two ways of writing the numbers on a circular board: clockwise or counterclockwise. For n = 20, the algorithm generates the permutation 20, 1, 11, 19, 2, 10, 18, 4, 9, 17, 6, 8, 16, 7, 12, 13, 5, 14, 15, 3, having = 2:54. It is not a Devil's Dartboard, though, because the permutation that ends : : : , 3, 15, 14, 5 | and coincides with the previous permutation elsewhere | has = 2:52. Devil's Dartboards and Travelling Salesmen The Devil's Dartboard Problem, which concerns permutations and minimization, is reminiscent of a famous optimization puzzle, the so-called Travelling Salesman Problem (TSP) [5]: Given a set f1, 2, : : : , ng of n \cities" and the set of distances d(i;j ) between cities i and j , nd an ordering of cities | that is, a tour | of minimum length (the length of a tour is the sum of the distances between consecutive cities.) The \cities" may be arbitrary objects, and the \distances" d(i; j ) any non-negative real-valued function. The so-called NP problems (for Nondeterministic Polynomial Time Veriable) [6] are, roughly speaking, those problems whose solutions might be dicult to nd but are easy to check. For example, it is straightforward to verify whether a given tour of the cities in a TSP has length less than, say, 1; 000. The TSP is therefore an NP problem, but it is actually more than that: it has the property that any other NP problem may be transformed into an instance of the TSP in polynomial time. In other words, the TSP is \as hard as" any other NP problem. This implies that any polynomial-time algorithm for the TSP could be used to solve every other NP problem also in polynomial time. Problems that have this property are known as NP complete. In practical terms, it is considered highly unlikely that NP-complete problems could 217 be \eciently" solved | that is, within polynomial time | although strictly speaking the question is still open. Strategies to nd near-optimal tours rather than exact solutions are known as heuristics. The nearest-neighbour heuristic for the TSP is the common-sense rule of travelling to the nearest city not yet called upon, with the initial city being chosen at random. In other words, a tour p is constructed step by step such that each new city p(k + 1) adds the minimal length to the partial tour p(1), p(2), : : : , p(k). Consider a generalized travelling salesman problem (GTSP): There is a (reasonably simple) function L which assigns to every permutation p of the n cities and to each k (= 1, 2, : : : , n), a non-negative real number Lp (k). Find the permutation p of the cities for which the sum Lp (1)+ Lp (2) + : : : + Lp (n) is a minimum. The standard TSP is obtained by taking Lp (k) = d(p(k);p(k + 1)); the Devil's Dartboard Problem results from taking Lp (k) = (p(k) + p(k + 1) + p(k + 2) ; 3(n + 1)=2)2. Moreover, in this general setting, our algorithm is the equivalent of the nearest-neighbour heuristic. Mathematically sophisticated salespersons, therefore, will be able to spend more time in the bar, and less on the road, and be far better at darts than their colleagues. References 1. 2. 3. 4. David Percy, Mathematics Today 35, p. 54. Robert Matthews, `Go For It' New Scientist 17, April 1999, p. 16. Ivars Peterson, Ivars Peterson's MathLand, May 19, 1997. Arturo Sangalli, The Importance of Being Fuzzy, Princeton University Press, Princeton, 1998. 5. E.L. Lawler et al. (eds), The Travelling Salesman Problem, John Wiley & Sons, New York, 1985. 6. M.R. Garey and D.S. Johnson, Computers and Intractability: A Guide to the Theory of NP Completeness, W.H. Freeman, San Francisco, 1979. Trevor Lipscombe Princeton University Press 41 William Street Princeton NJ 08540, USA. Arturo Sangalli Department of Mathematics Champlain Regional College Lennoxville Campus PO Box 5003 Lennoxville QC, Canada J1M 2A1.