Combinatorial Games. Jeffrey Kwan. October 2, 2017

Transcription

1 Combinatorial Games Jeffrey Kwan October 2, 2017 Don t worry, it s just a game... 1 A Brief Introduction Almost all of the games that we will discuss will involve two players with a fixed set of rules (such games are called combinatorial games). There will be no luck involved. Since there isn t much theory to talk about, we ll just start with a warm-up problem. Example 1.1 (Shortlist 2009/C1) Consider 2009 cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. (a) Does the game necessarily end? (b) Does there exist a winning strategy for the starting player? Solution. Intuitively, it seems like the game should end. Every time you flip over a block of 50 cards, it seems like the potential for more flips decreases: that is, the leftmost cards have more potential than the rightmost cards since the leftmost card of the block is flipped from gold to black. So how do we rigorize this idea? We do the usual trick of associating a 2009-digit binary number to the state of the game gold is a 1 and black is a 0. Notice that every move the binary number decreases since 2 x > 2 x x x 50, so eventually the game will end. For part (b), we consider the set of cards C = {10, 60, 110,..., 1960}. Why? Because exactly one card in C is flipped every move. After the starting player s turn, there are always an odd number of gold cards in C, so the second player will always be able to make a move. Thus the first player can never win. 2 Specific Techniques Now I will present two specific techniques that may be useful in solving combinatorial games. Most of the time it s best to just try playing the game a few times to get a feel for the problem, but if you get stuck, here are some things you can try. 1

2 2 Specific Techniques 2.1 Finding an Invariant After trying to play the game a few times, it is usually fairly obvious who has a winning strategy. Once this is determined, one of the best approaches is to find an invariant or some structure that the winning player can preserve that prevents the other player from winning. In fact, we employed this strategy in the warm-up when we proved that the binary number decreases. 1 Let s see an example of an invariant that is much less obvious. Example 2.1 (Shortlist 2009/C5) Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighbouring buckets, empties them to the river and puts them back. Then the next round begins. The Stepmother s goal is to make one of these buckets overflow. Cinderella s goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow? Solution. After playing the game a few times, one can see that it is very difficult for the stepmother to win. However, finding an algorithm that guarantees Cinderella a win is surprisingly hard. Let s try to find some kind of invariant which prevents the stepmother from winning. At the beginning of each round, there are 2 neighboring buckets which are empty. Notice that the only way for the stepmother to win is if at the beginning of the round, 2 non-adjacent buckets have a total of more than 1 liter of water. a 3 a 2 = 0 a 2 1 Case 1 a 3+a 5 1 a 3+a 5 1 a 1 = 0 a 3 a 4 1 a 5 a 1+a 3 1 a 1 a 2 = 0 a 4 = 0 a 3 = 0 a 5 = 0 a 1+a 4 1 Case 2 a 1+a 4 1 a 4 a5 1 a 1 WLOG assume that A 4 and A 5 are empty. Cinderella can then force the following structure: a 1 + a 3 1 and a 2 1. That is, out of the three buckets that Cinderella did 1 Technically we found a monovariant instead of an invariant, but they re pretty much the same thing anyway. 2

3 2 Specific Techniques not empty, the two non-adjacent ones have sum less than 1 and the last one has less than 1 liter of water. Clearly the stepmother cannot enforce an overflow in this scenario. Now we show that Cinderella can actually enforce this structure after every turn. Clearly this holds at the beginning of the first round. Assume that we have the a 4 = a 5 = 0, a 1 + a 3 1 and a 2 1. Then we have two cases: Case (1). If a 5 + a 3 1 after the stepmother fills the buckets, then empty buckets A 1 and A 2. Case (2). We cannot have both a 5 + a 3 > 1 and a 4 + a 1 > 1 since the stepmother only takes 1 liter of water from the river and a 1 + a 3 1 at the beginning of the round. Hence we must have a 4 + a 1 1, so we can empty A 2 and A 3. In both of these cases, the structure is preserved. Thus Cinderella can always prevent the stepmother from causing an overflow. 2.2 Induction Our second technique is induction, a classical technique which can be used in almost all areas of combinatorics. It should be no surprise that it also applies to combinatorial games. First, we showcase an easy Russian problem. Example 2.2 (Russia 2000) Two pirates divide their loot, consisting of two sacks of coins and one diamond. They decide to use the following rules. On each turn, one pirate chooses a sack and takes 2m coins from it, keeping m for himself and putting the rest into the other sack. The pirates alternate taking turns until no more moves are possible; the first pirate unable to make a move loses the diamond, and the other pirate takes it. For what initial numbers of coins can the first pirate guarantee that he will obtain the diamond? Solution. Let s start by trying to play the game a few times. Specifically, let s try a few scenarios and see if the first pirate can win. Let x, y be the number of coins in the two sacks. Then we can make a table of the starting positions and whether the first pirate can win. x y First Pirate 2 0 W 2 1 L 2 2 L 3 0 W 3 1 W 3 2 L 3 3 L 4 0 W 4 1 W 4 2 W 4 3 L 4 4 L 3

4 3 Miscellaneous Examples The pattern should now be clear. The first pirate can win if and only if x y 2. At this point the problem is morally solved ; the rest is really just a formality with induction. The main idea is that in a winning position, next pirate to move can force a losing position, and in a losing position, any move will lead to a winning position for the other pirate. We include the details for the sake of completeness. We induct on x+y, the total number of coins. WLOG x > y, so suppose that x y 2 and player P 1 is next to move. Then he should take 2 x y 2 from the first sack, keeping half for himself and moving the other half to the smaller sack. One can easily see that the new difference between the sacks is at most 1. Thus by the inductive hypothesis the second player is in a losing position. If x y 1, then no matter which move the first player makes, the difference between the sacks will be at least 2. Thus the second player can win, completing the proof. Induction is only a tool to prove conjectures that you ve made from trying small cases of the game. Usually, induction isn t the key idea in solving any problem; the reduction from one state to a smaller state of the game is usually where the difficulty lies. 3 Miscellaneous Examples Sometimes 2 there is no general strategy to solve a combinatorial game. Usually, the best way to attack a problem is to try playing the game and see what happens. We will now present some more examples that don t use the above techniques. 3.1 Writing Numbers Our first example is problem C5 from the 2004 IMO Shortlist, but it seems to be a bit easy for its placement. Example 3.1 (Shortlist 2004/C5) Given an integer N, players A and B play a game where they alternate writing down numbers. First, A writes down 1. In each subsequent turn, if the last number written is n, then the next player can write either n + 1 or 2n, but this number cannot be bigger than N. The player who writes N wins. For which values of N does B win? Solution. As always, we try some small cases first. Out of the first ten possibilities for N, B wins when N = 2, 8, 10. The first observation to make is that A always wins when N is odd. Why? Because A always chooses the n + 1 option, forcing B to write an even number every turn. So if N is odd, B can t win. This game is now all about parity. For even N, the player who writes an even k with 2k > N wins since all subsequent moves are adding 1. Claim. For even N, a player has a winning strategy for N if and only if he has a winning strategy for n = N 4. Proof. This is fairly obvious given what we know already: if the player can write n, then his opponent will write either n + 1 or 2n. In either case, the player can double and write an even number k such that 2k > N. This proves the claim. 2 Actually, most of the time. 4

5 3 Miscellaneous Examples Now we can see that since B wins for N = 2, B also wins for N = 8, 10 as 8 4 = 10 4 = 2. We can continue in this manner to show that B can win whenever N is the sum of odd powers of 2. The proof of this is some trivial induction which is omitted. 3.2 Vectors in F 2 Now we will turn to a much more challenging problem which was derived from properties of vectors in F 2. Example 3.2 (Shortlist 2014/C8) A card deck consists of 1024 cards. On each card, a set of distinct decimal digits is written in such a way that no two of these sets coincide (thus, one of the cards is empty). Two players alternately take cards from the deck, one card per turn. After the deck is empty, each player checks if he can throw out one of his cards so that each of the ten digits occurs on an even number of his remaining cards. If one player can do this but the other one cannot, the one who can is the winner; otherwise a draw is declared. Determine all possible first moves of the first player after which he has a winning strategy. Solution. For convenience, we make some definitions first. For some cards C 1, C 2,..., C k, define the sum of C 1, C 2,..., C k, denoted C 1 C 2 C k, to be the set of digits that are on an odd number of cards. 3 Obviously we have (X Y ) Y = X and the sum of all the cards is. The first player wins if he has the sum of his cards he can discard this card and the sum of the rest is! The main idea is to choose some card B and pair up all the cards (X, X B). Claim. If we select 512 cards, one from each pair, their sum will be either or B. Proof. It turns out that this is actually very easy to prove by considering the number of times a given element is counted. Call the sum of the 512 cards S and consider an element b of B. Notice that b is on exactly one card out of each pair. In the ith pair, let X i be the card containing b, and let Y i be the other card in the pair. Notice that the Y i are exactly all the cards not containing b. For any element a not in B and any pair (X i, Y i ), either a is on both X i and Y i or a is on neither. Since a is in exactly 256 of the pairs, a will not be in S. Now for any element b b in B, b will be in exactly 256 of the Y i as these are the sets not containing b. Thus the sum of the Y i is. If we exchange any of the 512 cards to an X i, the sum will change by B since every element of B is in exactly one of X i, Y i. Therefore the sum will either be or B. Strangely enough, this was the hardest part of the problem! The rest is just using the claim in a suitable way for the first player. Case (1). Suppose the first player takes A on the first move and the second player takes B. We present a winning strategy for the first player. The first player takes A B. Again, pair the cards (X, X B). Then the deck has many complete pair and one extra element (at this point, is the extra element since its partner B is already taken). If the second player takes a card from a complete pair, then 3 If we think of the cards as vector in F 2, then this is just normal addition. 5

6 3 Miscellaneous Examples the first player takes its partner. If the second player takes the extra element, then the first player takes an arbitrary card Y, and Y B becomes the new extra element. Let s see why this works. Say for a moment that the first player had instead of A. Then he would have exactly one card from every pair, so the sum of his cards would be either or B. Thus the sum of his real cards is either A or A B, but he has both of these cards! If the second player takes, choose an arbitrary card for B and continue as before. Case (2). We now present a winning strategy for the second player if the first player chooses on the first turn. After the first player chooses, the second player chooses an arbitrary card A. Then the first player selects some card B. As in the previous strategy, the second player takes A B, and in every subsequent move, the second player takes the partner of the card the first player just took. Then the second player wins for the same reasons as before. Therefore the first player has a winning strategy after taking any nonempty card on the first move. 3.3 Grand Finalé: Hunters and Rabbits Our final example is the statistically hardest problem ever at the IMO, the famed hunter-rabbit game. Example 3.3 (IMO 2017/3) A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit s starting point, A 0, and the hunter s starting point, B 0 are the same. After n 1 rounds of the game, the rabbit is at point A n 1 and the hunter is at point B n 1. In the n th round of the game, three things occur in order: (i) The rabbit moves invisibly to a point A n such that the distance between A n 1 and A n is exactly 1. (ii) A tracking device reports a point P n to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between P n and A n is at most 1. (iii) The hunter moves visibly to a point B n such that the distance between B n 1 and B n is exactly 1. Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after 10 9 rounds, she can ensure that the distance between her and the rabbit is at most 100? Solution. Before we do anything else, we should probably figure out what the answer is. The obvious strategy for the hunter is to follow the last ping. But this fails (convince yourself that this is true)! Now we can say with reasonable confidence that the rabbit wins. We have two possible approaches: 1. We can provide a strategy for the rabbit in every situation. That is, control the location of the pings to keep increasing the distance. 6

7 4 Practice Problems 2. We can find a way to increase the distance between the rabbit and the hunter from d to d + ε in f(d) moves. Hopefully f(d) will be small enough to achieve the bound of It turns out that both approaches lead to a solution, but the second approach is slightly cleaner. Claim. Suppose at some point the rabbit is exactly d 1 units away from hunter. Then it can increase it s distance from d to at least d in at most 4d steps, regardless of what the hunter knows about the rabbit beforehand. Proof. The main idea is that the rabbit can keep pinging on one line l while actually traveling on a different line. However, the hunter will not be able to tell which side of l the rabbit is on, so the hunter is forced to stay on l. In this way, the rabbit can increase it s distance from the hunter. Call the initial position of the hunter H and the rabbit R. Then line HR is l. The rabbit chooses two points X and Y which are reflections about l such that RX = RY = 4d and XY = 2. X H R H M Y The rabbit chooses either X or Y, say X, and moves on the line RX while always pinging on l. The hunter is forced to stay on l because she has no idea whether the rabbit is on line RX or RY (she is essentially minimizing max HX, HY ). Thus after 4d moves, the rabbit will be at X and the hunter will be at point H l such that HH = 4d. Now we just compute proving the claim. H X 2 = 1 + H M 2 = 1 + (RM RH ) 2 = 1 + ( 16d 2 1 (4d d)) (4d 1 4d 4d)2 = 1 + (d 1 4d )2 d , Now we re pretty much done. At the start, the rabbit moves to A 1 and pings back at A 0, so after the first move the hunter can only guarantee a distance of at most 2 between them. Then we just apply the claim a bunch of times and by some trivial inequalities we can conclude that the answer is no. The hunter cannot guarantee that the distance between her and the rabbit is less than 100 after 10 9 rounds. 4 Practice Problems Most of these games are just general problems that I found on AoPS. Some are quite enjoyable to play in real life, especially 4.2, 4.8, and

8 4 Practice Problems Problem 4.1 (Baltic Way 2013). A positive integer is written on a blackboard. Players A and B play the following game: in each move one has to choose a proper divisor m of the number n written on the blackboard (1 < m < n) and replaces n with n m. Player A makes the first move, then players move alternately. The player who can t make a move loses the game. For which starting numbers is there a winning strategy for player B? Problem 4.2 (Tournament of Towns 2005). John and James wish to divide 25 coins, of denominations 1, 2, 3,..., 25 kopeks. In each move, one of them chooses a coin, and the other player decides who must take this coin. John makes the initial choice of a coin, and in subsequent moves, the choice is made by the player having more kopeks at the time. In the event that there is a tie, the choice is made by the same player in the preceding move. After all the coins have been taken, the player with more kopeks wins. Which player has a winning strategy? Problem 4.3 (Tournament of Towns 2003). In a game, Boris has 1000 cards numbered 2, 4,..., 2000, while Anna has 1001 cards numbered 1, 3,..., The game lasts 1000 rounds. In an odd-numbered round, Boris plays any card of his. Anna sees it and plays a card of hers. The player whose card has the larger number wins the round, and both cards and discarded. An even-numbered round is played in the same manner, except that Anna plays first. At the end of the game, Anna discards her unused card. What is the maximal number of rounds each player can guarantee to win, regardless of how the opponent plays? Problem 4.4 (USAMO 2014/4). Let k be a positive integer. Two players A and B play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with A moving first. In his move, A may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, B may choose any counter on the board and remove it. If at any time there are k consecutive grid cells in a line all of which contain a counter, A wins. Find the minimum value of k for which A cannot win in a finite number of moves, or prove that no such minimum value exists. Problem 4.5 (RMM 2015/2). For an integer n 5, two players play the following game on a regular n-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the n-gon without jumping over another counter. A move is legal if the area of the triangle formed by the counters is strictly greater after the move than before. The players take turns to make legal moves, and if a player cannot make a legal move, that player loses. For which values of n does the player making the first move have a winning strategy? Problem 4.6 (Shortlist 2012/C4). Players A and B play a game with N 2012 coins and 2012 boxes arranged around a circle. Initially A distributes the coins among the boxes so that there is at least 1 coin in each box. Then the two of them make moves in the order B, A, B, A,... by the following rules: On every move of his B passes 1 coin from every box to an adjacent box. On every move of hers A chooses several coins that were not involved in B s previous move and are in different boxes. She passes every coin to an adjacent box. Player A s goal is to ensure at least 1 coin in each box after every move of hers, regardless of how B plays and how many moves are made. Find the least N that enables her to succeed. 8

9 4 Practice Problems Problem 4.7 (Shortlist 2015/C4). Let n be a positive integer. Two players A and B play a game in which they take turns choosing positive integers k n. The rules of the game are: (i) A player cannot choose a number that has been chosen by either player on any previous turn. (ii) A player cannot choose a number consecutive to any of those the player has already chosen on any previous turn. (iii) The game is a draw if all numbers have been chosen; otherwise the player who cannot choose a number anymore loses the game. The player A takes the first turn. Determine the outcome of the game, assuming that both players use optimal strategies. Problem 4.8 (USAMO 2016/6). Integers n and k are given, with n k 2. You play the following game against an evil wizard. The wizard has 2n cards; for each i = 1,..., n, there are two cards labeled i. Initially, the wizard places all cards face down in a row, in unknown order. You may repeatedly make moves of the following form: you point to any k of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the k chosen cards and then turns them back face-down. Then, it is your turn again. We say this game is winnable if there exist some positive integer m and some strategy that is guaranteed to win in at most m moves, no matter how the wizard responds. For which values of n and k is the game winnable? Problem 4.9 (Shortlist 2014/C6). We are given an infinite deck of cards, each with a real number on it. For every real number x, there is exactly one card in the deck that has x written on it. Now two players draw disjoint sets A and B of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions: 1. The winner only depends on the relative order of the 200 cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner. 2. If we write the elements of both sets in increasing order as A = {a 1, a 2,..., a 100 } and B = {b 1, b 2,..., b 100 }, and a i > b i for all i, then A beats B. 3. If three players draw three disjoint sets A, B, C from the deck, A beats B and B beats C then A also beats C. How many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets A and B such that A beats B according to one rule, but B beats A according to the other. Problem 4.10 (Shortlist 2013/C8). Players A and B play a paintful game on the real line. Player A has a pot of paint with four units of black ink. A quantity p of this ink suffices to blacken a (closed) real interval of length p. In every round, player A picks some positive integer m and provides 1/2 m units of ink from the pot. Player B then picks an integer k and blackens the interval from k/2 m to (k + 1)/2 m (some parts of this interval may have been blackened before). The goal of player A is to reach a situation where the pot is empty and the interval [0, 1] is not completely blackened. Decide whether there exists a strategy for player A to win in a finite number of moves. 9

10 References Problem 4.11 (IMO 2012/3). Let k and n be fixed positive integers. In the liar s guessing game, Amy chooses integers x and N with 1 x N. She tells Ben what N is, but not what x is. Ben may then repeatedly ask Amy whether x S for arbitrary sets S of integers. Amy will always answer with yes or no, but she might lie. The only restriction is that she can lie at most k times in a row. After he has asked as many questions as he wants, Ben must specify a set of at most n positive integers. If x is in this set he wins; otherwise, he loses. Prove that: a) If n 2 k then Ben can always win. b) For sufficiently large k there exist n 1.99 k such that Ben cannot guarantee a win. References [1] Mathematical Induction by Titu Andreescu and Vlad Crişan for some of the induction problems. [2] for the rest of the problems. [3] for official wordings and statistics. 10