另类称球趣题：验证砝码所标克数的正确性

Baron Mu¨nchhausen Redeems Himself: Bounds for a Coin-Weighing Puzzle Tanya Khovanova MIT Cambridge, MA 02139 tanyakh@yahoo.com Joel Brewster Lewis MIT Cambridge, MA 02139 jblewis@math.mit.edu Submitted: Jun 18, 2010; Accepted: Dec 26, 2010; Published: Feb 14, 2011 Mathematics Subject Classification: 05D99, 00A08, 11B75 Abstract We investigate a coin-weighing puzzle that appeared in the 1991 Moscow Math Olympiad. We generalize the puzzle by varying the number of participating coins, and deduce an upper bound on the number of weighings needed to solve the puzzle that is noticeably better than the trivial upper bound. In particular, we show that logarithmically-many weighings on a balance suffice. 1 Introduction 1.1 Background Baron Mu¨nchhausen is famous for telling the truth, only the truth and nothing but the truth [6]. Unfortunately, no one believes him. Alexander Shapovalov gave him an unusual chance to redeem himself by inventing a problem that appeared in the Regional round of the All-Russian Math Olympiad in 2000 [8]. Eight coins weighing 1, 2, . . . , 8 grams are given, but which weighs how much is unknown. Baron Mu¨nchhausen claims he knows which coin is which; and offers to prove himself right by conducting one weighing on a balance scale, so as to unequivocally demonstrate the weight of at least one of the coins. Is this possible, or is he exaggerating? In [4], T. Khovanova, K. Knop and A. Radul considered a natural generalization of this problem. They defined the following sequence, which they called Baron Mu¨nchhausen’s sequence (sequence A174541 in [7]): the electronic journal of combinatorics 18 (2011), #P37 1 Let n coins weighing 1, 2, . . . , n grams be given. Suppose Baron Mu¨nchhausen knows which coin weighs how much, but his audience does not. Then b(n) is the minimum number of weighings the Baron must conduct on a balance scale, so as to unequivocally demonstrate the weight of at least one of the coins. They completely described the sequence. Namely, they proved that b(n) ≤ 2, and provided the list of n for which b(n) = 1. A similar coin-weighing puzzle, due to Sergey Tokarev [5], appeared in the last round of the Moscow Math Olympiad in 1991: You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same, except for their labels. The number {1, 2, 3, 4, 5, 6} on the top of each coin should correspond to its weight. How can you determine whether all the numbers are correct, using the balance scale only twice? Most people are surprised to discover that only in two weighings the weight of the each coin can be established. We invite the reader to try this puzzle out before the enjoyment is spoiled on page 3. 1.2 The Sequence We generalize the preceding puzzle to n coins that weigh 1, 2, . . ., n grams. We are interested in the minimum number of weighings a(n) on a balance scale that are needed in order to convince the audience about the weight of all coins. In this paper, we demonstrate that we can do this in not more than order of log n weighings. Because the sequence a(n) relates to the task of identifying all coins (while the sequence b(n) relates to the task of identifying some coin) we will call it the Baron’s omni-sequence. We also calculate bounds for how many weighings are needed to prove the weight for a given particular coin. 1.3 The Roadmap In Section 2 we give a precise definition of the Baron’s omni-sequence and calculate its first few terms. In Section 3 we prove natural lower and upper bounds for the sequence, and in Section 4 we present the values of all known terms of the sequence. Section 5 is devoted to useful notations and terminology. In Section 6 we describe the idea behind the main proof of a tighter upper bound. We put this idea into practice in the subsequent three sections: in Section 7, we show that it is possible to determine the weights of several special coins in ⌈log2 n⌉ weighings, and in Section 8 we show how to use ⌈log2 n⌉ additional weighings to prove the weights of the rest of the coins. We thus establish that a(n) does not exceed 2⌈log2 n⌉. In Section 9, we give a refined version of the argument which results in a modestly improved bound. In Section 10 we consider the related task of proving the weight of a particular (e.g., adversarially-chosen) coin and prove that it can be done in not more than seven weighings. the electronic journal of combinatorics 18 (2011), #P37 2 In Section 11 we discuss two topics. First, we discuss the question of the monotonic- ity of the Baron’s omni-sequence. We do not come to a conclusion, but just provide considerations. Second, we show how Konstantin Knop and his collaborators used the rearrangement inequality to find optimal sets of weighings for a number of different values of n. Finally, in Section 12 we offer some further comments, questions and ideas for future research. 2 The Sequence The sequence a(n) is defined as follows: Let n coins weighing 1, 2, . . . , n grams be given. Suppose Baron Mu¨nchhausen knows which coin weighs how much, but the audience does not. Then a(n) is the minimum number of weighings he must conduct on a balance scale, so as to unequivocally demonstrate the weight of all the coins. The original Olympiad puzzle is asking for a proof that a(6) = 2. 2.1 Examples Let us see what happens for small indices. For n = 1, the Baron does not need to prove anything, as there is just one coin weighing 1 gram. For n = 2, one weighing is enough. The Baron places one coin on the left pan of the scale and one on the right, after which everybody knows that the lighter coin weighs 1 gram and the heavier coin weighs 2 grams. For n = 3, by exhaustive search we can see that the Baron can not prove all the coins in one weighing, but can in two. For the first weighing, he compares the 1-gram and 2-gram coins, and for the second weighing the 2-gram and the 3-gram coins. Thus he establishes the order of the weights. For n = 4, two weighings are enough. First, the Baron places the 1-gram and 2-gram coins on the left pan and the 4-gram coin on the right pan. The only way for one coin to be strictly heavier than the combination of two others is for it to be the 4-gram coin. The 3-gram is also uniquely identified by the method of elimination. In the second weighing, the Baron differentiates the 1-gram and the 2-gram coins. For n = 5, two weighings are enough. The Baron places the 1-gram and 2-gram coins on the left pan and the 4-gram coin on the right pan. For the second weighing he places the 1-gram and the 4-gram coins on the left pan and the 5-gram coin on the right pan. It is left to the reader to check that these two weighings identify each coin. For n = 6, two weighings are enough. The first weighing is 1+2+3 = 6. This identifies the 6-gram coin and divides the other coins into two groups: {1, 2, 3} and {4, 5}. The second weighing is 1 + 6 < 3 + 5. the electronic journal of combinatorics 18 (2011), #P37 3 Another essentially different solution for n = 6 was suggested by Max Alekseyev in a private email: 1 + 2 + 5 < 3 + 6 and 1 + 3 < 5. So the sequence begins, 0, 1, 2, 2, 2, 2. Because it is something of a mouthful to always refer to the good Baron Mu¨nchhausen, we suppress further mention of him. Instead, “we” will perform the weighings, or they will take place in the passive voice. 3 Natural Bounds For all n, we have that a(n) ≤ n − 1 (see [3]): for each k < n, in the k-th weighing we compare the k-gram and (k + 1)-gram coins. Getting the expected result every time confirms the weights of all coins. On the other hand, we have that a(n) ≥ log3(n). Indeed, suppose we conduct several weighings; then to every coin we can assign a sequence of three letters L, R, O, corre- sponding to where the coin was placed during each weighing – on the left pan, on the right pan or in the out-pile (i.e., on neither pan). If two coins are assigned the same letters for every weighing, then our weighings do not distinguish between them. That is, if we switched the weights of these two coins, the results of all the weighings will be the same. If the number of weighings were less than log3(n), we are guaranteed to have such a pair of coins. Thus, at least log3(n) weighings are needed. 4 More Terms Several other terms of the sequence are known. In the cases n = 10 and n = 11, Alexey Radul found sets of three weighings that demonstrate the identity of every coin [3]. As this matches the lower bound, we conclude that a(10) = a(11) = 3. Max Alekseyev wrote a program to exhaustively search through all possible combinations of weighings, with the result that a(7) = a(8) = a(9) = 3. The program also confirmed the values for n = 10 and n = 11, but larger values of n were beyond its limits. After that Konstantin Knop calculated more terms of the sequence by finding weigh- ings that match the lower bound. In particular, he stated that he found weighings to demonstrate that a(12) = . . . = a(17) = 3 and a(53) = 4 (see comments in [3]). When we were writing this paper we asked Konstantin Knop if he would share his weighings with us. He sent them to us, explaining that they were calculated together with Ilya Bogdanov. With his permission we include some of his weighings in this paper. Here we show how to demonstrate the identities of all coins for n = 15. The technique is similar to the one used in cases for n = 4 and n = 6, and we will use a related technique in Section 8 to prove our upper bound. The first weighing is 1 + · · ·+ 7 < 14 + 15. The only way a collection of seven coins can be lighter than two coins is if the seven coins are the lightest coins from the set and the two coins are the heaviest. Thus, this weighing the electronic journal of combinatorics 18 (2011), #P37 4 divides all coins into three groups C1 = {1, 2, 3, 4, 5, 6, 7}, C2 = {8, 9, 10, 11, 12, 13} and C3 = {14, 15}. In the second weighing, the audience sees three coins from C1, one coin from C2 and both coins from C3 go on the left pan, while three coins from C1 and two coins from C2 go on the right pan: (1 + 2 + 3) + 8 + (14 + 15) = (5 + 6 + 7) + (12 + 13). Observing that the weighing balances, the audience is forced to conclude that the left pan holds the lightest coins from each group and the right pan holds the heaviest. Thus, the coins are split into the following groups: {1, 2, 3}, {4}, {5, 6, 7}, {8}, {9, 10, 11}, {12, 13} and {14, 15}. Similarly, we take the third weighing 1 + 5 + 8 + 9 + 12 + 14 = 3 + 7 + 11 + 13 + 15, and this can balance only if the lightest coins from each group are on the left pan and the heaviest are on the right. Thus, in the end all coins are identified. The other weighings that Konstantin Knop sent to us use a different technique which is not related to our proof of the upper bound for a(n), so we delay presenting it until Section 11.2. Maxim Kalenkov used the same technique and the help of a computer to find two more terms, namely a(18) = a(19) = 3. So the sequence begins, 0, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3. No sets of three weighings that identify all coins are known for 20 ≤ n ≤ 27. However, Maxim Kalenkov found solutions in four weighings for a range of numbers from n = 20 to n = 58 inclusive. 5 Notation and Terminology For integers x ≤ y, we denote by [x . . . y] the set of consecutive integers between x and y, inclusive. For x = 1, instead of using [x . . . y] we will just use [y], which is the standard notation for the range anyway. We will use the number i to denote the i-gram coin on a pan. Thus, [x . . . y] represents the set of coins of weights no smaller than x and no larger than y, and we will occasionally construct weighings using this set notation. All arithmetic operations other than addition are understood to take place on the weight of a single coin; thus 3·22−1 represents the 11- gram coin. Addition operates normally when appearing inside brackets and parentheses and operates as union when appearing outside square brackets, so 1+2 means the 1-gram and the 2-gram coins taken together, while (1 + 2) + [5 . . . 7] is the set {3, 5, 6, 7} of four coins. Equalities and inequalities represent the outcomes of particular weighings; thus [3] + [5 . . . 7] > [11 . . . 12] represents the weighing with the coins 1, 2, 3, 5, 6, and 7 on the left pan and the coins 11 and 12 on the right pan, in which the left pan had larger total weight. In particular, when representing a weighing as an equality/inequality we will refer to the the electronic journal of combinatorics 18 (2011), #P37 5 left and right sides of the equality/inequality as the left and right pans of the weighing, respectively. If A denotes a set of coins, then wt(A) denotes the total weight of those coins. We denote by #A the cardinality of the set A. Define the small half of a (finite, totally ordered) set A to be the set consisting of the⌊ #A 2 ⌋ smallest elements of A. For example, the small half of {1, 3, 4, 5, 7} is {1, 3}. A subset B of a set A is said to be upwards-closed if for every x ∈ B and y ∈ A with x < y we have y ∈ B. Thus, the set {1, 3, 4, 5, 7} has six upwards-closed subsets, three of which are {4, 5, 7}, the entire set, and the empty set. The notion of a downwards-closed subset is defined analogously. 6 An Idea Before we proceed with the main section of the proof, we present an idea that actually does not work, but that we will use as a starting point. Given a set of coins [1 . . . n], suppose we can find numbers k < m that satisfy wt([1 . . . k]) = wt([m+ 1 . . . n]). In this case the weighing [1 . . . k] = [m+ 1 . . . n] will balance. This fact demonstrates that the coins in question really are the coins we claim: the sum of k coins is at least the weight of the left pan, while the sum of n −m coins is at most the weight of the right pan. This gives us our first division into three parts: [1 . . . k], [k + 1 . . .m] and [m+ 1 . . . n]. If we have in particular that k = n/2, then the division into the three parts above supplies us with the division of the range [n] into two halves. Suppose for the second weighing we can balance [1 . . . n/4]+ [n/2+1 . . .3n/4] against an appropriately-chosen combination of upwards-closed subsets of [n/4 + 1 . . . n/2] and [n/2 + 1 . . . n]. In this way we divide each half from the previous division into halves again. For the third division, we place the small half of each of the four groups into which we have divided the coins on one pan, and we choose an upwards-closed subset of the heavier halves on the other pan so that the pans balance. This again divides each of our four subsets into two halves. Continuing such binary division we can identify all coins in log2(n) weighings. Unfortunately, this strategy fails in a very simple way: it is impossible to carry out in general. In particular, the very first step is quite often impossible. Consider, for example, 12 coins. We want to find an upward-closed subset to balance the lightest six coins. But 12 < 1 + 2 + 3 + 4 + 5 + 6 < 12 + 11. However, this problem can be overcome if we are first able to prove the identities of a small number of helper coins; these coins could then be used to make up the difference between the small half of the coins and the corresponding upward-closed set. For example, if we start with 12 coins and somehow can prove the identities of the 1-gram and the 2-gram coins, then we can balance out the small half of the leftover set: the electronic journal of combinatorics 18 (2011), #P37 6 3 + 4 + 5 + 6 + 7 = 12 + 11 + 2. This suggests that we should start by looking for easily-identifiable sets of “helper coins.” 7 Helper Coins We now give a simple procedure to identify a set of helper coins. This set of helper coins does not require many weighings to identify. In addition, it is versatile and produces many sums. Let the binary expansion of n − 1 be n − 1 = 2a1 + 2a2 + . . . with a1 > a2 > . . . ≥ 0 and a1 = ⌈log2 n⌉ − 1. We perform the weighings 1 < 2, 1 + 2 < 4, 1 + 2 + 4 < 8, . . . , 1 + 2 + 4 + . . . + 2a1−1 < 2a1 and 2a1 + 2a2 + . . . < n. From the first weighing, we learn that the coin we claim has weight 2 grams has weight at least this large. Similarly, from the second weighing we learn that the coin we claim has weight 4 grams weighs at least that much, and so on. Thus, the last weighing demonstrates that the coin we claim has weight n has weight at least n. However, all our coins weigh at most n grams, whence the coin we claim has weight n must actually have that weight. Moreover, this also shows that the coins 1, 2, . . . , 2a1 are the coins we claim. Denote by H(n) the set of coins identified by these ⌈log2 n⌉ weighings. The following useful property of H(n) is clear. Proposition 1. Using only the elements of H(n), we can construct a pile of coins whose weight is i for any i ∈ [n]. That is, [n] ⊆ {wt(T ) : T ⊆ H(n)}. We now use this set H(n) to give an effective version of the algorithm described in Section 6. 8 The Upper Bound Theorem 2. We can identify all coins in [n] in at most 2⌈log2 n⌉ weighings. That is, a(n) ≤ 2⌈log2 n⌉. Proof. Section 2.1 demonstrates the result for n = 1, 2, 3. For n ≥ 4, use the construction of Section 7 to identify the coins in the set H(n) in ⌈log2 n⌉ weighings. Set C = [n]rH(n). We perform binary search on C as follows: suppose that at some stage, we have successfully demonstrated a division of C = C1∪C2∪· · ·∪Cm into several disjoint ranges so that for every non-helper coin we know to which range it belongs, and that the ranges are numbered in order: for any i < j and any x ∈ Ci, y ∈ Cj we have x < y. (Initially, this is the case with m = 1 and C = C1.) If Ci consists of one element, then the identity of the coin in Ci is already proven, so we may set it aside. For all i for which #Ci > 1, we split Ci and place its small half on the left pan of the balance. The other non-proven elements of C have larger total weight and each is of weight at most n. Thus, we may begin adding unused non-proven elements of C to the right pan, starting with the largest, until the right pan weighs at least as much the electronic journal of combinatorics 18 (2011), #P37 7 as the left pan. As soon as the right pan reaches the weight of the left pan the difference between the weights of the two pans is at most n. Then (by Proposition 1) we may add elements from H(n) to the left pan as needed in order to make the pans balance. This weighing identifies the small half of Ci for all i, and so divides each Ci into two almost-equal-sized parts. Repeating this log2(#C) ≤ ⌈log2 n⌉ times results in a total ordering of the elements of C, and so the identification of all elements of [n]. Thus, at most 2⌈log2 n⌉ weighings are required. 9 The Refined Upper Bound The way we divide coins into piles in the previous theorem is not optimal. In particular, it leaves room for two improvements. First, when we remove the small half of each pile (to place on the left pan), we ignore the information we get from the fact that the remaining coins are divided into two parts (some on the right pan, some in the out pile). Thus, we can do better by keeping track of all three parts of the division. Second, the most profitable way of dividing coins into three piles would be to have each pile of the same, or almost the same, size. In our approach it is not possible for the set of the lightest coins to be the same cardinality as the set of the heaviest coins of the same total weight. However, it is possible to do better than in Section 8 by choosing a division in which the part of largest cardinality has less than half of the coins. Suppose we have the set of coins [n]. For some k,m, we divide the coins between the two pans (with some left out, i.e., not on either pan) by placing the lightest k coins on the left pan and the heaviest m coins on the right pan so that the right pan holds more total weight and the weight differe