Semiinvariants IMOTC 2013 Simple semiinvariants


 Magnus Montgomery
 2 years ago
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1 Seiinvariants IMOTC 2013 Siple seiinvariants These are soe notes (written by Tejaswi Navilarekallu) used at the International Matheatical Olypiad Training Cap (IMOTC) 2013 held in Mubai during AprilMay, Introduction It is quite often that we see probles in Olypaids which involve aking oves to change a given configuration. A typical proble would have a given initial configuration, rules that every ove should follow, and a final desired configuration. The oves can be copletely deterinistic, or can depend on certain choices (like in gaes). The task ight be, for exaple, to show that a final configuration cannot be achieved, or to count the nuber of oves required to get to a certain configuration. In this note, we shall explore a plethora of such probles and discuss a particular technique of associating (sei)invariant quantities to the configurations. 2 Siple seiinvariants As we shall see below, a key concept in probles with oves is to associate a quantity, an invariant or a seiinvariant, to each possible configuration such that it changes uniforly with every ove (or reains unchanged for that atter). Typical unifority we will see is the (strict) onotonicity. If, for exaple, the seiinvariant is a positive integer that strictly decreases with every ove, then by induction on this seiinvariant it follows that we can ake only finitely any oves. This is a standard way to attack any probles and we shall see this being used repeatedly below. We shall soetie hide the seiinvariant and only consider the induction, but the ideas behind each of the will be siilar. Let us start with the following (very) siple exaple. Exaple 2.1. Let n be a positive integer. Written on a board are n natural nubers x 1, x 2,..., x n. A ove consists of replacing two nubers on the board by their su (until there is only one nuber on the board). Then the nuber on the board at the end is independent of the choices of oves. It is clear fro the description of a ove that the nuber of integers on the board reduces with every ove. This is our first exaple of a seiinvariant! For a configuration C of a bunch of nubers on the board, we define I(C) to be the nuber of integers in this configuration. If a ove takes C to C then I(C ) = I(C) 1 < I(C). Therefore we will end up with one nuber after exactly n 1 oves. This sounds like a coplicated way of aking a siple observation, but we shall consider less trivial seiinvariants soon. Consider a ove that replaces two nubers x and y on the board by x+y. As entioned above, we want to associate each configuration with a quantity that changes uniforly. Given that we are replacing the nubers by their sus, it is natural that the su of the nubers on the board changes uniforly. In fact, the su does not change at all. More precisely, for any configuration C, let J(C) denote the su of all the nubers on the board in that configuration. If a ove takes C to C then we have J(C) = J(C ). Therefore J is an invariant. Hence it follows that the nuber on the board at the end is J(C 0 ) = x 1 + x x n (with C 0 the initial configuration), which is independent of the choices of oves. That was a lengthy explanation for a very siple exaple. Let us ake it a bit ore interesting by changing the exaple slightly. 1
2 Seiinvariants IMOTC 2013 Siple seiinvariants Exaple 2.2. Written on a board are the nubers 1, 2, 3,..., In a ove, Alice can choose two nubers x and y on the board and replace the with x y. Can Alice ake 2012 oves in such a way that the final nuber on the board is 1000? This exaple differs fro the previous one only by a sign, i.e., we have x y instead of x + y as part of the rule. Since we already know soe nice invariant for the previous exaple, let us instead look for the siilarity between the two. As a ore general observation, the (only) siilarity between x y and x+y is their parity. That is, either both of the are odd, or both are even. This hints that if we consider the sae quantity J(C) as in the previous exaple, then even though it ay not reain invariant under a ove, its parity will. To be precise, for a configuration C of nubers on the board, let J(C) denote the su of all those nubers. Then J(C) (od 2) is invariant under a ove. Since the su of all the nubers on the board initially is odd and 1000 is even, it follows that the desired final configuration cannot be achieved. Proble 2.3. Written on a blackboard are n nonnegative integers whose greatest coon divisor is 1. A ove consists of erasing two nubers x and y, where x y, on the blackboard and replacing the with the nubers x y and 2y. Deterine for which original ntuples of nubers on the blackboard is it possible to reach a point, after soe nuber of oves, where n 1 of the nubers on the blackboard are zeros. Discussion. Let us start exaining the seiinvariants. A ove replaces two nubers x and y with x y and 2y. What is evident at first sight is that the su of the nubers reains the sae after any ove. There is another siple observation we can ake. Note that we eventually want any zeros on the board. A ove that erases two nubers x and y, with y = 0, will replace those nubers back, and hence it has no effect. Therefore any zero that appears on the board will reain forever. Thus the nuber of zeros on the board can never decrease, so the nuber of zeros on the board is a seiinvariant. This seiinvariant is a nonnegative integer, increases onotonically with every ove and is bounded above by n. Unfortunately, it is not a strictly increasing function, so we need soething else. Working out exaples is in our opinion one of the best ways to get an insight. This is one of the few aspects that we shall repeat every now and then! When working out exaples, it is ost often the best to start with the siplest possible ones. In this proble, we can start with n = 2 and with nubers 1, a on the board. As we proceed with the first few values of a we can guess that the desired end point is reached if and only if a = 2 k 1 for soe k 0. We can try to work out soe ore exaples with a, b > 1 on the board to start with and we see in all the cases that a + b is a power of two. If we now have ore than two nubers on the board in the beginning, then to get to the desired end configuration we have to pass a stage in which there are exactly two nonzero nubers on the board. Fro what we have guessed so far, the su of these two nubers should be a power of two. Since the su of all the nubers on the board is an invariant our guess should be that a 1 + a a n = 2 k for soe k 0, where a 1, a 2,..., a n are the nubers on the board. Let us go back to the case n = 2. If a and b are the nubers on the board in the beginning, then for their su to be a positive power of 2 we need both of the to be of the sae parity. They cannot both be even since they should be coprie to each other (by the given condition of the proble), and hence they are both odd. After the first ove, we are left with two even nubers. It is evident that henceforth all the nubers that appear on the board will be even. We can therefore reove a factor of 2, or rather a factor of a power of 2, and reduce the case back to odd nubers. 2
3 Seiinvariants IMOTC 2013 Siple seiinvariants Thus the power of two dividing the gcd of the two nubers increases (strictly) at every stage. This is an excellent seiinvariant! In fact, the sae seiinvariant works in the general case. As we have observed that the gcd is what we need to consider, we note that either (x y, 2y) = (x, y) or (x y, 2y) = 2(x, y). This will be the key in the solution below. In the solution below, we shall not explicitly consider the seiinvariant, but it is there in a hidden for. Solution. We shall show that we can reach the desired configuration if and only if the su of all the nubers on the board is a power of two. We ake two observations. Firstly, the su of all the nubers on the board is invariant under a ove and secondly, (x y, 2y) = (x, y) or (x y, 2y) = 2(x, y). Since the gcd of the nubers to start with is 1 it follows that the gcd of the nubers on the board is a power of 2 after any nuber of oves. If we can reach a point where n 1 of the nubers are zero, then the nonzero nuber at this stage is a power of two. But, by the first observation, this is the su of all the nubers at the beginining. This proves the only if iplication. Our third observation is that if we have the nubers a 1, a 2,..., a n on the board, then we can reach the desired configuration by aking oves if and only if we can do the sae with the nubers a 1 /d, a 2 /d,..., a n /d on the board, where d = (a 1, a 2,..., a n ). Now suppose that a 1 a 2... a n are n nonnegative integers on the board such that a 1 + a a n = 2 k, with k 0. We shall show by induction on k that we can reach the desired configuration. The stateent is obvious if k = 0. Suppose that k > 0. Then the nuber of odd nubers aong a 1, a 2,..., a n is even. Pair the up in any order and for each pair (x, y) with x y, replace the with x y and 2y (i.e., ake a ove with these nubers). Note that both x y and 2y are even, and therefore we will end up with all even nubers, say b 1, b 2,..., b n. Since b 1 /2 + b 2 /2 + + b n /2 = 2 k 1 it follows by the induction hypothesis that we can reach the desired configuration starting with the nubers b 1 /2, b 2 /2,..., b n /2. By the third observation, we can therefore reach the desired configuration starting with the nubers b 1, b 2,..., b n. This copletes the induction step and concludes the solution. Proble 2.4. Let M be a set of six distinct positive integers whose su is 60. These nubers are written on the faces of a cube, one nuber to each face. A ove consists of choosing three faces of the cube that share a coon vertex and adding 1 to the nubers on those faces. Deterine the nuber of sets for which it is possible, after a finite nuber of oves, to produce a cube all of whose sides have the sae nuber. Discussion. Note that the desired configuration in this proble is not an end configuration since we can get different configurations if we ake ore oves thereafter. So we are not really interested in a seiinvariant, but rather we are looking for soe invariant. Let us ake the notation a bit ore precise, and we shall see iediately what we need to look for. We let i, j = 1, 2, 3 in this entire paragraph. Let A i and B i denote the opposite faces of the cube and let a i and b i denote the nubers on the respectively. Then a ove consists of choosing one face fro each of {A i, B i } and increasing the nuber on it by 1. So for any i, the su a i + b i will increase by 1. Therefore the difference of two such sus is an invariant. Thus to get the desired configuration it is necessary to have these sus equal. For the converse, we have to ake the right choices of the oves to ake all the nubers equal. This is not very difficult we can ake the nubers on each of the opposite sides equal. 3
4 Seiinvariants IMOTC 2013 Siple seiinvariants Solution. For i = 1, 2, 3, let A i and B i denote the opposite faces of the cube and let a i and b i denote the nubers on the respectively. Then for any 1 i, j 3, the quantity a i + b i a j b j is invariant under a ove. Therefore, if after finite nuber of oves, all the sides have the sae nuber then a i + b i = 20 for all i = 1, 2, 3. Thus M = {x, y, z, 20 x, 20 y, 20 z} where x, y, z are integers with 1 x < y < z 9, and there are ( 9 3) = 84 such possible sets. Conversely, suppose that M is one of these 84 sets. We ay assue that a i b i for i = 1, 2, 3 and that b 1 a 1 b 2 a 2 b 3 a 3. We add (b 1 a 1 )/2 + (b 3 a 3 )/2 to the faces containing a 1, a 2 and a 3, (b 2 a 2 )/2 (b 1 a 1 )/2 to the faces containing b 1, a 2 and a 3, and (b 3 a 3 )/2 (b 2 a 2 )/2 to the faces containing b 1, b 2 and a 1. It is easy to see that all the nubers added are integers. With these oves we will have all the nubers equal to b 3. Proble 2.5 (USAMO 2003, Proble 6). At the vertices of a regular hexagon are written six nonnegative integers whose su is n. One is allowed to ake oves of the following for: (s)he ay pick a vertex and replace the nuber written there by the absolute value of the difference between the nubers written at the two neighbouring vertices. Prove that if n is odd then one can ake a sequence of oves, after which the nuber 0 appears at all six vertices. Discussion. There are two seiinvariants that we can think of at first sight: the nuber of zeros and the axiu of the six nubers. The forer is onotonic only if we choose our oves carefully (otherwise this nuber could both increase and decrease). The latter however is a onotonically decreasing quantity. The only instance in which this latter quantity cannot be reduced further is when we have the nubers a, a, 0, a, a, 0 at the vertices (in that order), a deadlock situation. So if we soehow ake sure that this configuration is never reached then we can get to all zeros. We cannot blindly keep reducing one of the largest nubers, because the configuration a+1, a, 0, a, a, 0 will then result in a deadlock. So we have to be bit ore careful. The given condition that n is odd is quite crucial (note that the deadlock situation has the su even). One can therefore try to keep the su of all nubers odd (unless all the nubers becoe zero). This ay not be possible after every ove, but we can group the oves together such that this is the case. Such grouping was also part of the solution to Proble 2.3, albeit in a hidden way. To say a little bit ore about such groupings, note that we have either one, three or five odd nubers to start with. It is a coon idea to reduce any such possibilities to one which is ost suited to the situation. In this case, there is no such configuration that is ost suited. Soe configurations ight be ore intuitive to one and soe to the others. Here we give one reason why the solution below is chosen aong any possible ones. Reeber that we want to reduce the axiu value. If this axiu value is even and is adjacent to vertices with odd nubers, then we can reduce the value without changing the configuration type. For this purpose, it is good to have even and odd nubers at every alternate vertices. We therefore consider this as a special configuration and try to reduce every other configuration to this. Solution. By an odd configuration C we ean six nonnegative nubers C = (a 1, a 2,..., a 6 ) such that 6 i=1 a i is odd. Let t k denote the operation at vertex k, i.e., replacing a k with a k 1 a k+1, with k written odulo 6. If M(C) denotes the axiu of the six nubers, then note that application of t k will not increase the value of M. We call an odd configuration C very odd if the six nubers are alternatively odd and even. The following twostep algorith terinates since after every application of Step 2 to a configuration C we get a configuration C with M(C ) < M(C) or C = (0, 0, 0, 0, 0, 0). 4
5 Seiinvariants IMOTC 2013 Moving on with oves Step A. Suppose that C is odd but not very odd. Suppose first that we have five odd nubers. Then we ay assue that C (1, 1, 1, 1, 1, 0) (od 2). Then apply t 2 and t 4 to get a very odd configuration. Suppose that we have three odd nubers, but they are not on the alternative vertices. Then there are two possibilities (after syetry): C (1, 1, 1, 0, 0, 0) or C (1, 1, 0, 1, 0, 0) odulo 2. In the forer case apply t 4 and t 6 to reduce to the case of five odd nubers, and in the latter case apply t 6 followed by t 1 to get a very odd configuration. Finally suppose that we have one odd nuber. Then we ay assue that C (1, 0, 0, 0, 0, 0) (od 2). Then apply t 2 and t 6 to reduce to the previous case. Step B. Suppose that C is very odd, so C (1, 0, 1, 0, 1, 0) (od 2). Applying t 2, t 4 and t 6 if necessary, we ay assue that M(C) is odd. If M(C) a 1 then apply t 3 and t 5 to get an odd configuration C with M(C ) < M(C), and then go to Step A. We take siilar actions if M(C) a 3 or M(C) a 5. If M(C) = a 1 = a 3 = a 5, the apply t 2, t 4, t 6 followed by t 1, t 3, t 5 to get (0, 0, 0, 0, 0, 0) and terinate the algorith. 3 Moving on with oves We have seen a few probles with siple seiinvariants. We now ove on to slightly ore involved quantities. Soe of these are not very easy to find, but we shall give soe intuitive arguents towards obtaining the. Proble 3.1 (IMO 1986, Proble 3). To each vertex of a regular pentagon an integer is assigned so that the su of all five nubers is positive. If three consecutive vertices are assigned the nubers x, y, z respectively, and y < 0, then the following operation is allowed: x, y, z are replaced by x+y, y, z +y respectively. Such an operation is perfored repeatedly as long as at least one of the five nubers is negative. Deterine whether this procedure necessarily coes to an end after a finite nuber of steps. Discussion. Considering a few exaples with sall nubers, one should quickly guess that the procedure necessarily coes to an end after a finite nuber of steps. It is then a question of assigning a seiinvariant that changes onotonically with every operation. There are any possible choices for this. First let us look at soe quantities that fail to serve the purpose. Before anything else, note that the su of all the nubers is an invariant. A natural quantity to look at in such cases is the su of all the squares. A siple calculation then tells us that this is not onotonic with respect to the operations. The next (wrong) guess could be the iniu of the five nubers. If x, y, z are all negative, then we have bigger negative nubers than before. This leads to the intuition that if a negative nuber is next to a positive nuber then the value of the seiinvariant should be better. In other words, the quantity should take into account the neighbours of every vertex. We can therefore consider a i + a i+1 or (a i + a i+1 ) 2 where a i s are the nubers at the vertices and the index i is taken odulo 5. After an operation, these quantities will have ters of the for a i (or a 2 i ), and of the for a i +a i+1 +a i+2 or (a i +a i+1 +a i+2 ) 2. For the first quantity, to copensate for these extra 5
6 Seiinvariants IMOTC 2013 Moving on with oves ters we can consider ( a i + a i +a i+1 + a i +a i+1 +a i+2 + a i +a i+1 +a i+2 +a i+3 ). This is then a seiinvariant. For the second quantity, it is even sipler: we just consider a 2 i + (a i + a i+1 ) 2. We give both these solutions below. 1 Solution. Let a 1, a 2,..., a 5 be the nubers at the vertices, in that order, and let s > 0 denote the su of these five nubers. Let I = I(a 1, a 2,..., a 5 ) = ( ) a i + a i + a i+1 + a i + a i+1 + a i+2 + a i + a i+1 + a i+2 + a i+3, where all the sus are taken over 1 i 5 and the indices are written odulo 5. Suppose that we perfor an operation with x = a 1, y = a 2 and z = a 3. Then we get I = I(a 1 + a 2, a 2, a 2 + a 3, a 4, a 5 ) = I s a 2 + s + a 2. Since s > 0 and a 2 < 0 it follows that I < I. Thus I is a positive quantity that reduces with every operation. Therefore the procedure necessarily coes to an end. Alternate solution. With the notation as in the above solution, let I = I(a 1, a 2,..., a 5 ) = a 2 i + (a i + a i+1 ) 2. Then I = I(a 1 + a 2, a 2, a 2 + a 3, a 4, a 5 ) = I + 2a 2 (s a 2 ) < I. Thus the quantity I is a positive integer that reduces with every operation and hence it follows that the procedure necessarily coes to an end. Proble 3.2 (USAMO 2011, Proble 2). An integer is assigned to each vertex of a regular pentagon so that the su of the five integers is A turn of a solitaire gae consists of subtracting an integer fro each of the integers at two neighboring vertices and adding 2 to the opposite vertex, which is not adjacent to either of the first two vertices. (The aount and the vertices chosen can vary fro turn to turn.) The gae is won at a certain vertex if, after soe nuber of turns, that vertex has the nuber 2011 and the other four vertices have the nuber 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the gae can be won. Discussion. This is another proble where the su of all the nubers is an invariant. The ai of this proble is a bit different in coparison with the ones we have discussed so far. In this case, we are looking for an invariant that associates a vertex to every configuration. Since there are five vertices, this invariant can be thought of as an integer odulo 5, each corresponding to a vertex. In particular, the vertices are to be nubered 0, 1, 2, 3, 4. Note that any change in the nubering of these vertices should also change the invariant in a siilar fashion. So the invariant should be a weighted su of the nubers. Let us look at the trivial exaple where the nubers are 1, 0, 0, 0, 0. Then if the vertex to which 1 is assigned is nubered i, then the invariant should be i. Does that give any ideas? On the other hand, consider a turn in which we add 2 to the integer at vertex v. The operation is syetric with respect to this vertex. If we are looking for a weighted su that is invariant under every turn, then the weight at v should equal to the su of the weights at the other two vertices (at which subtraction takes place). Suppose the weight at v is a then the su of the weights 1 A third solution can be given by consider the seiinvariant (a i a i+2) 2. 6
7 Seiinvariants IMOTC 2013 Moving on with oves at the other two vetices should be 2a. Intuitively, this leads to assigning weights a 2 and a + 2 at these vertices. These ideas lead to the following solution. One ore point to note is that there are two parts to the given proble: to show that there is at ost one vertex at which the gae can be won; and to show that the gae can in fact be won. The latter part can be solved by trying to ake the integers zero this has less to do with invariants, and ore to do with anipulations of nubers. Solution. Let a 0, a 1, a 2, a 3, a 4 be the nubers at the vertices in that order. Let 0 I 4 be such that I a 1 + 2a 2 + 3a 3 + 4a 4 (od 5). We clai that I is invariant under any turn. This is because, any turn consists of changing I by k (k + 1) + 2(k + 3) 0 (od 5). If we end up with 2011 at one vertex and zero at every other vertex then 2011 will have to be at the vertex I. It now reains to show that we can reach this point. We first show that (0, 0, 0, 0, 0) can be reached fro (0, 0, 5, 5, 0). This can be acheived with the following turns: (0, 0, 5, 5, 0) (2, 0, 1, 5, 2) (0, 2, 1, 1, 2) ( 2, 2, 2, 0, 2) (0, 2, 2, 0, 4) (0, 0, 0, 0, 0). Now suppose that a 0 + a 1 + a 2 + a 3 + a 4 = 2011, and assue without loss of generality that I = 0. Play a turn if necessary, and assue that a 0 = By a turn (, k) we ean the turn where we add 2 to a k and subtract fro a k 2 and a k+2 (with indices odulo 5). Play the following turns: (a 1, 3) and ( a 1, 4). Our configuration then looks like (2011, 0, b 2, b 3, b 4 ). Play the turns ( b 4, 4) and ( b 4, 2) and (b 4, 3) to get a configuration (2011, 0, c 3, c 4, 0). Since the su of the five nubers is always 2011 it follows that c 3 = c 4. Moreover, since I = 0 is an invariant it follows that c 3 is divisible by 5. We have show that (0, 0, 5, 5, 0) can be brought to (0, 0, 0, 0, 0), so it follows that we can reach (2011, 0, 0, 0, 0) fro (2011, 0, c 3, c 4, 0). Proble 3.3. There are 60 girls sitting around in a circle. One of the is initially given M chocolates. In each ove, every girl who has at least 3 chocolates gives one chocolate each to both her neighbours. This procedure stops either if all the girls have a chocolate or if no girl has ore than 2 chocolates. Find the iniu value of M for which every girl gets a chocolate. Discussion. One of the first things to notice is that there are only finitely any configurations. In such cases it is worthwhile to analyse if there is a loop. Obviously, if we there is a seiinvariant associated with each configuration such that it (strictly) decreases (or increases) with every ove then there cannot be any loop. Suppose that there cannot be a loop, then the procedure will stop. In that case, if M 119 then it is not hard to see (by PigeonHolePrinciple) that the procedure cannot stop because of the second condition. Note that if a girl gets a chocolate at any tie, then she will have at least one chocolate thereafter. Since there is syetry in the given proble it is therefore enough to that girl opposite to G 0, the girl to who the chocolates were given initially, gets a chocolate. Also, to show that there is no loop one can consider 59 girls sitting in a line (instead of a circle). Another iportant thing to observe here is that the when a girl gives chocolates to her neighbours the chocolates ove away fro her. This hints that the (su of all) distances of chocolates ight be a good choice for a seiinvariant. When talking about distances, there is a choice we need to ake the choice of the origin, or the choice of the girl fro who the distances are to be easured. In this case, the natural choice would be G 0 (or the girl opposite to her). With this 7
8 Seiinvariants IMOTC 2013 Moving on with oves idea, if we naively construct the seiinvariant I(C) = i a i where a i is the distance fro G 0 of the ith chocolate, then we realise that I(C) is onotonically (but not strictly) increasing. That is because, if a girl gives a chocolate each to her neighbour then the seiinvariant does not increase unless that girl is G 0. It is good to keep in ind that when we talk about distances it is often useful to consider the squares of the distances rather than their absolute value. In this situation as well, if J(C) = i a2 i then it follows easily that J(C) increases strictly with every ove. This is enough to conclude that M 119. Solution. We shall only give here half the solution. That is, we shall only prove that if M 119 then every girl gets a chocolate. The other part of the solution (which is not directly related to any seiinvariant calculation) is left as an exercise to the reader. We shal denote by G 0 the girl to who the chocolates were given initially. It is clear that if a girl gets a chocolate after soe oves, then she will have a chocolate with her thereafter. Also it is clear that if the girl G 0 sitting diagonally opposite to G 0 gets a chocolate then everyone gets a chocolate. Nuber the chocolates 1, 2,..., M. By a configuration, we ean a distribution of chocolates aong the girls. So in the initial configuration C 0 one girl, say G 0, has 60 chocolates, and the rest have none. In a configuration C, we denote by a i (C) the distance of the ith chocolate fro G 0. Here the distance between two girls is taken to be 1. Define J(C) = M a i (C) 2. i=1 If there is a ove that takes C to C, then by definition it follows that G 0 does not have any chocolate in C. Consider a girl who had ore than two chocolates in C. Let her distance fro G 0 be d. Then the change in J(C) because of her action of distribution of chocolates is 2d 2 +(d 1) 2 +(d+1) 2 = 1. This proves that J(C ) > J(C). Suppose that M 119. If G 0 does not have a chocolate with her, then b PigeonHolePrinciple it follows that at least one of the other girls has at least three chocolates, and therefore the procedure continues (and thus increasing the seiinvariant). But the seiinvariant is bounded above by M 29 2, and hence it follows that all the girls get a chocolate. Proble 3.4. There are n 1 bins B 1, B 2,..., B n and there are a total of n balls in the. A ove consists of one of the following: (a) if B 1 has at least two balls, then we can ove one ball to B 2 ; (b) if B n has at least two balls, then we can ove one ball to B n 1 ; or (c) if B i has at least two balls for soe 2 i n 1, then we can ove one ball to B i 1 and one to B i+1. Show that starting fro any initial configurations any sequence of oves will lead to the configuration in which all the bins have exactly one ball each. Discussion. Let us first fix soe notation. For 1 i n, let a i denote the nuber of balls in bin B i, so we have a 1 + a a n = n. We can blindly try the kind of seiinvariants we have encountered so far, for exaple, a 2 i. It is easy to see that this quantity does not satisfy any sort of onotonicity. Let us look at the seiinvariant that we want fro a slightly different perspective. We want it to easure how far we are fro the desired configuration. In other words, we want to easure how far the ith ball is fro the ith bin. Obviously, this easure is trivial in case of the desired 8
9 Seiinvariants IMOTC 2013 Moving on with oves configuration. Now let us see how we can calculate this easure. Since there are a 1 balls in the first bin, for i = 1, 2,..., a 1, the distance of ith ball fro the ith bin is (i 1) 2. We have taken the square so as to avoid the difficulty in dealing with the absolute values. For i = a 1 + 1,..., a 1 + a 2, the easure equals (i 2) 2 and so on. We will leave it to the reader to check that we eventually get that the su of all these easures is I(a 1, a 2,..., a n ) = n (s k k) 2, where s k = a 1 + a a k. This seiinvariant is onotonically decreasing, but it fails to decrease strictly in one particular case. This failure can be set right if we assign a weight. After the solution based on the above seiinvariant idea, we shall also give an alternate solution purely based on induction. We give this solution for two reasons. First of all, the seiinvariant used in the solution below looks bit tricky! And secondly, the induction also provides a bit ore flexibility, in the sense that the solution is sae for a siple generalization stated at the end. We want to stress that induction is ore powerful than what we think, but we ay have to deal with it carefully at ties. This proble is a nice exaple where a careful application of nested induction leads to an alternate solution. There are three quantities, naely r, c r and s, on which we apply induction, but every step is quite siple. What akes the solution a bit coplicated is the nested property. Solution. For 1 k n, let a k denote the nuber of balls in bin B k and let s k = a 1 +a 2 + +a k. Let I(a 1, a 2,..., a n ) = n k=1 (s k θk) 2, where 0 < θ < 1 is a quantity we shall choose later. We then have k=1 I(a 1, a 2,..., a n ) I(a 1 1, a 2 + 1,..., a n ) = 2(a 1 θ) 1 > 0, I(a 1, a 2,..., a n ) I(a 1,..., a k 1 + 1, a k 2, a k+1 + 1,..., a n ) = 2a k 2 2θ > 0, I(a 1, a 2,..., a n ) I(a 1,..., a n 1 + 1, a n 1) = 2θ(n 1) (2n 3), where the last quantity is positive provided θ > (2n 3)/(2n 2). Thus choosing such a θ we get that the value of I decreases with any ove. Since there are only finitely any possible configurations it follows that the process should terinate. It reains to show that the iniu value of I is achieved at the desired configuration. This is true because (s k θk) 2 (k θk) 2 = (s k k)(s k k+2k(1 θ)) 0 for all k = 1, 2,..., n, and the equality holds if and only if s k = k. Alternate solution. Let t k, for 1 k n, denote the ove in (a) if k = 1, ove in (b) if k = n or the ove in (c) with k = i. Suppose that we do not have the desired configuration. Let r be the sallest index such that B r has at least two balls. We shall induct on n r. The base case is when r = n. Let c n denote the nuber of balls in B n. We induct again on c n. The base case now is when c n = 2, and in this case let s < n be the unique index such that B s does not contain any ball. We induct (for the third tie!) on n 1 s. If s = n 1 then applying t n will give us the desired configuration. If s < n 1 then apply t n, t n 1,..., t s+1 to get a configuration in which c n = 2 and B s+1 does not contain any ball in it, and hence we are done by induction on n 1 s. This copletes the base case c n = 2. We now consider the (base!) case r = n and c n > 3. Let s be the largest index such that B s does not contain any ball in it. We again induct on n 1 s. As before, if s = n 1 then applying 9
10 Seiinvariants IMOTC 2013 Moving on with oves t n will give us a configuration in which c n is saller and we are then done by induction on c n. If s < n 1 then applying t n, t n 1,..., t s+1 will give us a configuration in which B s+1 does not contain any ball, and thus we are done by induction on n 1 s. This copletes the base case r = n. Now suppose that every configuration with r > R can be brought into the desired configuration. Consider a configuration with r = R and let c r 2 denote the nuber of balls in B r. We apply induction on c r. Consider the base case c r = 2. If there is a bin B i, with i < R, that does not contain any ball in it then let s denote the largest of such indices. Otherwise let s = 0. Applying t R, t R 1,..., t s+1 will then result in either the desired configuration or a configuration with r > R. This copletes the proof of the base case c r = 2. Now assue that c r > 2. Again let s be as in the case c r = 2. In this case, applying t R, t R 1,..., t s+1 will reduce c r and hence we are done by induction. The solution is now coplete. And here is a siple generalization of the previous proble. Proble 3.5. Let n, a 1, a 2,..., a n be positive integers and let M = a 1 + a a n. Consider n bins B 1, B 2,..., B n which contain a total of M balls in the. A ove consists of one of the following: (a) if B 1 has at least a balls, then we can ove one ball to B 2 ; (b) if B n has at least a n + 1 balls, then we can ove one ball to B n 1 ; or (c) if B i has at least a i + 1 balls for soe 2 i n 1, then we can ove one ball to B i 1 and one to B i+1. Show that starting fro any initial configurations we can ake appropriate oves to get a configuration in which, for all 1 i n, B i contains exactly a i balls. Proble 3.6 (IMO Shortlist 1996). A finite nuber of coins are placed on an infinite row of squares. A sequence of oves is perfored as follows: at each stage a square containing ore than one coin is chosen. Two coins are taken fro this square; one of the is placed on the square iediately to the left while the other is placed on the square iediately to the right of the chosen square. The sequence terinates if at soe point there is at ost one coin on each square. Given soe initial configuration, show that any legal sequence of oves will terinate. Reark. In fact, it is also true that any legal sequence of oves will terinate after sae nuber of oves and in the sae configuration. Discussion. This proble is a variation of Proble 3.4. There are two differences: we now have an infinite row of squares instead of finitely any bins; and, we also have to show that the nuber of steps is sae. Since there are only finite nuber of coins, we should be able to restrict ourselves to a finite set of rows. For this we have to show that the coins cannot go beyond certain squares. Intuitively, if S k is the leftost square with a coin in it, then the squares to the left of it can never have all the coins in the. We shall use this in the solution below and show that only finitely any squares need to be considered. We notice the siilarity of this proble with Proble 3.2. A ove at square S k is syetric at the square. With a k denoting the nuber of coins in square S k, we see that a k will be an invariant, and oreover the weighted su ka k will also be an invariant (exactly like in the solution to Proble 3.2). We are however looking for a seiinvariant. An iediate thought would be to assign different weights (that alost balance each other out). Since (k 1) 2 +(k+1) 2 2k 2 = 2, the seiinvariant k 2 a k should be a good choice. 10
11 Seiinvariants IMOTC 2013 In a different direction Solution. Suppose that there are n coins. Let S i denote the squares with i ranging over the set of all integers and let a i denote the nuber of coins in S i. A ove eans choosing an integer i with a i 2 and replacing a i 1, a i, a i+1 with a i 1 + 1, a i 2, a i respectively. Suppose that S k is the leftost square that has a coin in the initial configuration. Then we notice that a k +a k at any point of tie. This is quite evident. We now prove by induction that for any i 0 we have a k i + a k i+1 + i + 1. The base case i = 0 is what we just noticed. Now consider the general case of i > 0 (assuing the stateent for i 1). If the inequality does not hold at soe point of tie, then consider the ove because of which this fails. This ove has to happen at the square S k i which will then have to have at least two coins. We then get that a k i+1 + a k i+2 + < i, a contradiction to the induction hypothesis. We have thus proved that the coins cannot ove to the left of S k n+1. Siilarly, we can give a bound on the righthand side as well, so we only have to consider finitely any squares. Let S 1, S 2,..., S N be the squares to be considered. Let I(a 1, a 2,..., a N ) = N k=1 k2 a k. Then I(a 1, a 2,..., a N ) I(a 1,..., a k 1 + 1, a k 2, a k+1 + 1,..., a N ) = 2a k 2 > 0. This proves that the any sequence of oves will terinate. 2 4 In a different direction The last proble we shall consider here is of a different nature than the ones discussed above. We have seen probles where our goal is to get to a final configuration. The proble below is to prove just the opposite that there is no final configuration, but instead there is a repetition. Proble 4.1 (IMO 1993, Proble 6). Let n > 1 be an integer. In a circular arrangeent of n laps L 0, L 1,..., L n 1, each of which can be either ON or OFF, we start with the situation where all laps are ON, and then carry out a sequence of steps Step 0, Step 1,.... If L j 1 (with j taken odulo n) is ON then Step j changes the state of L j (it goes fro ON to OFF, or vice versa), but does not change the state of any other lap. If L j 1 is OFF then Step j does not change anything at all. Show that: (a) There is a positive integer M(n) such that after M(n) steps all laps are ON again. (b) If n = 2 k for soe positive integer k then all the laps are ON after n 2 1 steps. (c) If n = 2 k + 1 for soe positive integer k then all the laps are ON after n 2 n + 1 steps. Discussion. The key idea in such probles where we have to look for repetition is to consider a backward ove. If there is a unique backward ove, then it follows by the finiteness of the nuber of configurations that they will have to repeat after a while. In ters of graph theory, this is sae as saying that if the degree of every vertex is exactly two (one for the forward ove and one for the backward ove) then it coprises of disjoint cycles. It is tie to again ention two things: exaples and induction! For the second part, it is quite evident how the configuration changes if we work out couple of exaples. It is then natural to 2 We note that N k=1 s2 k, with s k = a 1 + a a k, is another strictly increasing seiinvariant. 11
12 Seiinvariants IMOTC 2013 In a different direction consider a solution using induction. Working out exaples will also help us see the siilarities between the configurations in the second and third part. We shall also briefly ention two alternative solutions using polynoials and generating functions. These ideas differ very uch fro our current topic, so we shall keep the arguents to bare inial. Solution. To ake soe of the notations sipler, we shall assue that we begin with Step 1 rather than Step 0. Note that this does not change any of the stateents of the proble since we can always renuber the laps and get the exact stateent as in the proble. (a) Write (j, (a 0, a 1,..., a n 1 )) for a configuration of the lights, with each a i being 0 if L i is off or 1 otherwise, and j indicating that we have just finished Step j, so the initial configuration is ( 1, (1, 1,..., 1)). Obviously, j is written odulo n. Clearly, there are only finitely any such configurations. This proves that the configurations repeat after a while. We note that (j 1, (a 0, a 1,..., a j 1, a j +a j 1, a j+1,..., a n 1 )) is the unique predecessor of (j, (a 1,..., a n )), and hence it follows that the initial configuration repeats. (b) Suppose that n = 2 k. We shall show the following by induction: L n 1 is OFF after Step n 1 and reains off until Step n 2 1; L 0 is ON, and all the other laps are OFF after n 2 n steps; all the laps are ON after n 2 1 steps. The stateent is obvious for the base case k = 1. Suppose that k > 1 and that the stateent holds if we replace k with k 1. Let = 2 k 1. The configuration after n 1 steps is ( 1, }{{} 1010 }{{ 10} ). Thanks to the first stateent in the induction, it follows that after 2( 2 ) steps the configuration is (0, 1000 }{{ 00} 1000 }{{ 00} ), and thus after steps we will have ( 1, }{{} 0000 }{{ 00} ). Another n steps will result in ( 1, }{{} 0000 }{{ 00} ). Again fro the first stateent in the induction, it follows that after n 2 n steps we will have (0, }{{} 0000 }{{ 00} ), and hence all the lights will be ON after n 2 1 steps. 12
13 Seiinvariants IMOTC 2013 Conclusion (c) The configuration after two steps is (2, ) which is nothing but (0, ) but shifted by 2. We shall consider this rotated configuration henceforth (and add 2 in the end). After another 2 k 1 steps we get (2 k 1, }{{} 2 k 0). Fro the previous part it follows the last two laps reain off until we reach the configuration (0, ). We further note that it takes (2 k 1)(2 k + 1) = n 2 2n steps to reach this. After another n 1 steps we will have all the lights ON, and counting the first two steps, we have thus perfored exactly n 2 n + 1 steps. Alternate solution. There is a clever, but less intuitive solution. Consider the polynoial P (x) = a 0 + a 1 x + a 2 x a n 1 x n 1. Then the corresponding polynoial after Step 0 and an index shift by 1 is Q(x) = a n 1 + (a 0 + a n 1 )x + a 1 x a n 2 x n 1. We note that Q(x) xp (x) (od x n + x n 1 + 1) (where we consider all the polynoails with coefficients odulo 2). The corresponding polynoial after k steps (and with the index shifted by k) is therefore x k P (x) (od x n + x n 1 + 1). Hence it is enough to show: (a) x M(n) 1 (od x n + x n 1 + 1) for soe integer M(n). (b) If n = 2 k then x n2 1 1 (od x n + x n 1 + 1). (c) If n = 2 k + 1 then x n2 n+1 1 (od x n + x n 1 + 1). For part (a) we note that there are only finitely any residue classes odulo x n + x n and hence there are integers k > l such that x k x l (od x n + x n 1 + 1). This proves that x k l 1 (od x n + x n 1 + 1). For part (b) we note that x n2 (x n 1 + 1) n x n2 n + 1 (od x n + x n 1 + 1) since ( n k) is even for all 1 k n 1. On the other hand x n2 = x n2 n x n x n2 1 + x n2 n (od x n + x n 1 + 1). This proves the required equivalence. Proof of part (c) is siilar to that of part (b). In this case we have x n2 1 x n2 n + x n 1 (od x n + x n 1 + 1) and then the required equivalence follows. Alternate solution. The above solution can also be seen in a slightly different way using generating function. Let a i+rn equal 0 if the lap L i is OFF after it was operated for r ties (i.e., after i + (r 1)n + 1 steps) and equal 1 otherwise. Then we have a j+n = a j + a j+n 1 for all j 0 and a 0 = a 1 = = a n 1 = 1. If we let G(x) = a j x j it then follows that G(x) = 1/(1 x x n ) (assuing that all the coefficients are considered odulo 2). This leads to a solution. 5 Conclusion We have seen a few different kind of seiinvariants in this section. One should always look for invariants, which ight be at ties copletely useless for the solution, but will give an idea as to what seiinvariant we should look for. It is good to keep in ind soe of the standard 13
14 Seiinvariants IMOTC 2013 Extra probles seiinvariants one can look for: ak, a 2 k, ka k, k 2 a k, (a k 1 + a k ) 2, s k..., and also 1/ak, 1/a 2 k,..., or a cobination of these. Below are a few ore probles. The last two probles, which can possibly be solved using seiinvariants, are of different nature. 6 Extra probles Proble 6.1. The nubers 1, 2,..., n are written on a blackboard. It is peritted to erase any two nubers a and b and write the new nuber ab + a + b. Find with proof, the nuber that can be on the blackboard after n 1 such operations? Proble 6.2. A circle is divided into six sectors. Then the nubers 1, 0, 1, 0, 0, 0 are written into the sectors. You ay increase two neighboring nubers by 1. Is it possible to equalize all nubers by a sequence of such steps? Proble 6.3. Let n be a positive integer and a 1, a 2,..., a n positive real nubers. We put these nubers into groups G 1, G 2,... G r, with r 1, with each group being nonepty. For every a i we let its relative rating to be the ratio of a i to the su of all the eleents in the group containing a i. A ove consists of oving an eleent fro one group to the other provided its relative rating increases. Show that we can only ake finitely any oves. Proble 6.4. On each square of a chessboard is a light bulb which has two states on and off. A ove consists of choosing a square and changing the state of the bulbs in that square and in the neighbouring squares (which share a side). Show that starting fro any configuration we can ake finitely any oves to reach a point where all the bulbs are switched off. Proble 6.5. Let a 1, a 2,..., a 100 be an ordered set of nubers. At each ove it is allowed to choose any two nubers a n, a and change the to a2 n a n ( a2 an a ) and a2 an n ( a2 n a a n ) respectively. Deterine if it is possible, starting with the set with a i = 1 5 for i = 20, 40, 60, 80, 100 and a i = 1 otherwise, to obtain a set consisting of integers only. Proble 6.6. We have n(n+1) 2 stones and they are divided into a few groups. In each ove we take one stone fro each group and for a new group with taken stones. Prove that after a few oves we will get groups with 1, 2, 3,..., n stones (and after that position doesn t change). Proble 6.7. Several boxes are arranged in a circle. Each box ay be epty or ay contain one or several chips. A ove consists of taking all the chips fro soe box and distributing the one by one into subsequent boxes clockwise starting fro the next box in the clockwise direction. (a) Suppose that on each ove (except for the first one) one ust take the chips fro the box where the last chip was placed on the previous ove. Prove that after several oves the initial distribution of the chips aong the boxes will reappear. (b) Now, suppose that in each ove one can take the chips fro any box. Is it true that for every initial distribution of the chips you can get any possible distribution? Proble 6.8 (IMO Shortlist 2009). 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 player, standing 14
15 Seiinvariants IMOTC 2013 Extra probles by the sae long side of the table, play a gae with alternating oves. Each ove consists of choosing a block of 50 consecutive cards, the leftost of which is showing gold, and turning the all over, so those which showed gold now show black and vice versa. The last player who can ake a legal ove wins. (a) Does the gae necessarily end? (b) Does there exist a winning strategy for the starting player? Proble 6.9 (Korean National Matheatical Olypiad 2009, Proble 3). There are 2008 white stones and one black stone in a row. A ove eans the following: select one black stone and change the colour of its neighbouring stone(s). Find all possible initial position of the black stone for which all the stones can be ade black after finitely any oves. Proble There is (2n 1) (2n 1) square and for every sall square there is one arrow, up or down or right or left. A bug is on one of the sall squares. It travels to a next square following the arrow. If bug leaves the square, the arrow of the square turn π 2 in counter clock wise. Prove that this bug will be out of this big square before 2 3n 1 (n 1)! 3 oves. 15
IV Approximation of Rational Functions 1. IV.C Bounding (Rational) Functions on Intervals... 4
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