The Magnus-Derek Game

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1 The Magnus-Derek Game Z. Nedev S. Muthukrishnan Abstract We introduce a new combinatorial game between two layers: Magnus and Derek. Initially, a token is laced at osition 0 on a round table with n ositions. In each round of the game Magnus chooses the number of ositions for the token to move, and Derek decides in which direction, + (clockwise) or (counterclockwise), the token will be moved. Magnus aims to maximize the total number of ositions visited during the course of the game, while Derek aims to minimize this quantity. We define f (n) to be the eventual size of of the set of visited ositions when both layers lay otimally. We rove a closed form exression for f (n) in terms of the rime factorization of n, and rovide algorithmic strategies for Magnus and Derek to meet this bound. We note the relevance of the game for a mobile agent exloring a ring network with faulty sense of direction, and we ose variants of the game for future study. 1 Introduction We will introduce a new class of combinatorial games that we call the Magnus-Derek games, and we will study in detail the basic version of the game. The game is layed on a circular table with n ositions labeled consecutively from 0 to n 1, and with a token initially laced at osition 0. The two layers are called Magnus (from magnitude) and Derek (from direction). Suose the token s current osition is i. A round consists of Magnus calling a magnitude l with 0 < l n followed by Derek calling a direction + (clockwise) or (counterclockwise). The token is then moved to osition i + l mod n or i l mod n according to Derek s choice of direction. Thus, Magnus must choose a set N = {i + l, i l} of two ossibilities for the next osition, and Derek chooses which of these ositions is realized. (In our exlanations of the algorithms, rather than stating the token moves we will often say the next osition will be or even say Magnus moves identifying Magnus with the token.) Magnus s goal is to maximize the cardinality of the set of all ositions visited by the token in the course of the game. Derek s goal is to minimize the cardinality of this set. Note that revisiting a osition does not affect the cardinality. The current token osition, the size of the table (n), and the set of already visited ositions are known to both layers at all times. The algorithmic strategies we resent for both layers are adative: that is, a layer s move deends both on the set of already visited ositions and on the current token osition. We will assume that both layers know the factorization of n at the beginning of the game. We define f (n) to be the eventual cardinality of the set of ositions visited when both layers lay otimally. We will show that f (n) is well defined: Magnus has a strategy to ensure that at least f (n) ositions are visited, and Derek has a strategy to ensure that at most f (n) ositions are visited, irresective of each others strategies. Thus we examine three roblems: Univ. of Victoria, Part of the work was done while visiting DIMACS. DIMACS Permanent Member Rutgers Univ., 1

2 1. What is f (n) for a given n? Clearly f (n) n.. What algorithm for Magnus will ensure that at least f (n) ositions are visited? We refer an algorithm that minimizes the number of moves required to achieve this goal. Less imortantly, we want to minimize the time required for Magnus to comute a magnitude in each round. 3. What algorithm for Derek will ensure that no more than f (n) ositions are visited? Again, we want to minimize the time require for Derek to comute a direction in each round. We are also interested in a lower bound on the number of moves until f (n) ositions are visited. The Main Result Theorem 1 If n = k for some k, then f (n) = n. In this case Magnus has a strategy for visiting all ositions in exactly n 1 moves; the strategy can be formulated as a sequence of n 1 magnitudes at the start of the game, and calculating the sequence requires O(n) time. If n = m where is the smallest odd rime factor of n, then f (n) = ( 1)n and there exists a strategy for Derek to ensure at most f (n) ositions are visited. Derek s calculations require O(1) time in each round. Furthermore there exists a strategy for Magnus to visit at least f (n) ositions in g(n) moves where: If m = 1, g(n) = O(n ) moves. If m = k, g(n) = O( + n) moves. Otherwise, if m = k m 1 where m 1 is bigger than two and odd, g(n) = O( n ) moves. Magnus s comutations require O(1) time in each round. As n increases from 1 to, f (n) is 1,,, 4, 4, 4, 6, 8, 6, 8, 10, 8, 1, 1, 10, 16,.... This sequence has not reviously aeared in the Encycloedia of Integer Sequences [3]. 3 Motivation A mobile agent (MA) is an autonomous comuter rogram that can move itself from host to host in an MAenabled network. Mobile agents have been roosed as efficient solutions for many distributed comuting and network maintenance tasks; one examle is intrusion detection in comuter networks [7]. A common task for an MA is to visit as many as ossible of the hosts or nodes in a network. The notion of a network with a sense of direction is defined in [4, 6]. In [5], the following informal definition is given for a sense of direction in a network with a ring architecture: Sense of direction refers to the caability of a rocessor to distinguish between its adjacent communication lines, according to some globally consistent scheme. In a ring network this roerty is usually referred to as orientation, which exresses the rocessor s ability to distinguish between left and right, where left means the same to all rocessors. A network may be without a sense of direction due to a hardware or software fault, or due to malicious activity by an intruder. We consider a ring network of n hosts in which there is inconsistent global sense of direction. That is, if H 0, H 1,..., H n 1 are our n hosts, H i is connected to (and can communicate only with)

3 H i+1( mod n) and H i 1( mod n). Furthermore, although each host can distinguish between its two adjacent neighbours, the labelling left and right is not consistent across hosts and may not be consistent over time. The game is designed to model an MA attemting to exlore or visit all hosts in such a network. In our model we assume that the each time the MA moves it has no control over the initial direction ( left or right ) of its movement. However, the MA may redetermine how many hosts to ass through before stoing (the magnitude of its movement), and is able to kee moving in the same direction on the ring until it stos. The MA can only exlore hosts at which it stos; it is unable to exlore the hosts through which it asses while moving. In this aer we give an algorithmic strategy for an MA exloring such a network. The strategy ensures that the MA exlores as many hosts as ossible, even when its directions of movement are maliciously chosen to force the MA to visit a minimal number of hosts. Secondary goals of the MA are to (1) minimize the number of moves necessary, and () efficiently comute the magnitude of each move. We formulate the roblem as an adversarial game between two layers: Magnus, reresenting the MA, attemts to visit as many hosts as ossible. Derek, reresenting a malicious adversary, attemts to revent Magnus from achieving this goal. 4 When n = k In this section, we consider the case when n = k. We use C(l, d, s) to denote {s + i d 0 i < l}; that is, the set of l ositions beginning at s that are successively d aart. Lemma If n = k, there is a strategy for Magnus forcing the token to visit all ositions, and Magnus does not need to adat as the game rogresses. Rather, Magnus s strategy (a sequence of n 1 magnitudes) can be determined at the start of the game. Proof. The strategy (and the roof) can be described inductively on the ower k. Let k = 1 so that n =. Magnus calls the magnitude 1 and irresective of Derek s chosen direction, the token moves to the unvisited osition. Hence f () = and only a single move is required. Now let k > 1. We assume that f ( k 1 ) = k 1 and that there exists a sequence d 1, d,..., d k 1 1 of magnitudes to achieve visiting all ositions irresective of Derek s choice of directions. Let n = k. We artition the set C(n, 1, 0) into C( n,, 0) and C( n,, 1). Magnus first uses d 1, d,..., d n to visit all ositions in C( n,, 0). Then, regardless of the resultant osition, Magnus calls 1 forcing the token to move to some osition j in C( n,, 1). Thereafter, Magnus s choices would be d 1, d,..., d n 1 to visit all ositions in C( n,, 1). Therefore, all n ositions are visited and the total number of moves is n 1. The roof is constructive and Magnus s strategy is easy to generate in O(n) time at the start of the game. If n = k for some k, then Magnus can declare the following sequence of moves: ( n 1 ; n ; n 1 ); n 3 ; ( n 1 ; n ; n 1 ); n 4 ; (:::);...; n k ; (:::); Here (:::) means a reetition of all the moves starting from the beginning of the sequence until, but not including, the move before the current (:::). Notice that the sequence as a whole is a alindrome, that its first and second halves are each alindromes, and so forth. 1 Examle. For n = 16, Magnus s choices are (in order and with brackets for clarifications): ((8; 4; 8; ); (8; 4; 8; ))1; ((8; 4; 8; ); (8; 4; 8; )) Henceforth, we will consider values of n that are not exact owers of two. 3

4 5 Derek s Strategy and Proof of f (n) ( 1)n This is the easier art of Theorem 1. Lemma 3 Let n = m where is the smallest odd rime factor of n ( > ). Then f (n) ( 1)n and there exists a strategy for Derek with O(1) running time in each round ensuring that at most f (n) ositions are visited. Proof. Consider the following strategy for Derek. At the beginning of the game, he chooses a osition k where 1 k 1. We claim that Derek can revent any osition congruent to k mod from being visited. The token starts at osition i = 0, which is not congruent to k mod. In each round Derek must choose the next osition from the set {i + l, i l} where i denotes the current osition (not congruent to k mod ) and l denotes the magnitude chosen by Magnus. At least one of i + l and i l is not congruent to k mod, for, otherwise, both i + l = k mod and i l = k mod and by adding the two equations we obtain the contradiction i = k mod, or i = k mod. Derek s algorithm insures that the token never visits any osition in C( n,, k), the set of ositions congruent to k mod. This set has cardinality n and this immediately imlies that f (n) (n n ( 1)n ) = Derek s strategy does not deend on the entire set of reviously visited ositions, but only on the currently visited osition and Magnus s choice of l. In each turn, the calculations for Derek s choice run in O(1) time. Thus, we have roved the Derek s art in Theorem 1. In fact, we have roved a stronger result. Derek can secify in advance a set of ositions (C( n,, k) for a suitable k) and ensure that the token never visits those ositions. It may be of further interest to find a strategy for Derek that not only achieves at most f (n) ositions visited but also revents the token from visiting f (n) ositions before a certain minimum number of rounds are layed in the game. 6 Magnus s Strategy and Proof of f (n) ( 1)n For greater insight into the roblem, we will show two quite different algorithmic strategies for Magnus to visit at least f (n) different ositions. The two strategies share a common idea: although in many situations Magnus cannot reach a given redetermined osition, he can often reach at least one of two redetermined ositions. 6.1 First Strategy for Magnus To rove the general result for n k, we will first state the above idea as a lemma, and then use it to derive an algorithm for Magnus. Lemma 4 Let n k, and let the current osition be K, and let I and J be two distinct ositions such that J > I and gcd(j I, n) = f for some non-negative integer f (gcd is the greatest common divisor). Then Magnus can find a set of magnitudes with cardinality O(n) such that the token can be moved from K to one of I and J regardless of Derek s choices. These magnitudes can be found in time O(n ). 4

5 Proof. With resect to the air of distinct ositions I and J, we artition all ositions 0,..., n 1 into classes as follows. Class C 0 contains I and J. Class C 1 contains every osition C 0 for which there exists a magnitude d such that both +d mod n C 0 and d mod n C 0. That is, if the current osition is in C 1, Magnus can call a magnitude such that the token ends in a osition in C 0 irresective of Derek s choice. More generally, class C m contains every osition C 0 C 1 C m 1 such that there exists a magnitude d for which both +d mod n C 0 C 1 C m 1 and d mod n C 0 C 1 C m 1. We define C m = C 0 C 1 C m. We will rove that if C m+1 = 0 for any m, then C m = {0,..., n 1}. In other words, as we let i increase and construct C i, if we cannot find at least one osition for C m, then we have exhausted all ositions. We rove this by contradiction. Suose C m+1 = 0 for some m and that there exists a osition α C m. Since C 1 contains at least 1 osition (see construction, below), (m + 1) and we can always find at least three ositions in C m. Consider all ositions in C m sorted in increasing order α 1 < α <... < α t. In the next eleven lines, the index arithmetic is by mod t; for examle, index t (by mod t). If α i α i 1 is even for some i, then there are an odd number of ositions between these ositions. The osition equidistant from both, namely α i 1 + α i+α i 1, belongs to C m+1, giving the contradiction. Suose α i α i 1 is odd for all i. Consider any three consecutive ositions α l 1, α l, α l+1 from Cm. Then, α l+1 α l 1 must be even. If the midoint α l 1 + α l+1+α l 1 is not in Cm, it must belong to C m+1, giving the contradiction. Otherwise, the midoint equals α l. Reeating this argument for each consecutive trile from Cm, we obtain either a contradiction, or that α 1, α,... are arranged as vertices of a regular olygon of less than n vertices (since α Cm) with sides α i α i 1 = L for all i, where L 3 and L is odd. Therefore, n = al for some ositive integer a. Furthermore, since I, J {α 1, α,...} they must be vertices of the olygon and, consequently, J I = bl for some ositive integer b. This imlies that gcd(j I, n) = cl for some ositive integer c. Then gcd(j I, n) f, contradicting the assumtion in the lemma. We have now roven that Cl contains all ositions 0, 1,..., n 1 for some l n 1. We now show how to construct the classes by induction and find an algorithm for Magnus to reach I or J starting from K. For efficiency, we consider each class to be stored as a list. Trivially, C 0 = {I, J}. For convenience, we call the elements of {0, 1, 1, 1 + 1,..., n, n 1 1, n 1, n 1 } oints on the table. In other words, the set of all oints on the the table is the union of the set of all ositions and the set of all (non-integer) midoints between them. To construct C 1, there are only four candidates to consider: the two oints diametrically oosite I and J, and the two midoints between I and J (namely, the two oints equidistant from I and J). From these four ossible oints, we take only those that are integers, and therefore valid ositions. If n is odd, C 1 contains either I + J I or I + J I+n mod n, deending on which is an integer. If n is even, C 1 contains the two ositions oosite to I and J, resectively I + n/ mod n and J + n/ mod n. When n is even, the two midoints are either both integers or both non-integers; if they are integers, we also include them in C 1. Notice that it is not ossible for I and J to be diametrically oosite to each other. If they were, gcd((j I), n) = gcd( n, n) = n and cannot be f since n k. Therefore, C1 3 regardless of whether n is even or odd. Now suose we have already constructed C 0, C 1,, C m, C m 1 as lists. Notice that this defines an order for each element in Cm 1 (and, eventually, will define an order for {0, 1,..., n 1}): first come the elements from C 0, in their order, then the elements from C 1, in their order, and so forth. For each C m 1, we consider its diametrically oosite oint, + n mod n, and all midoints between and any q Cm 1 with q occuring before in the order. We add to C m whichever of these are integers and do not already belong to Cm 1 (or C m). The construction requires O(j) time for a table 5

6 osition in the jth lace in the order; therefore, the entire rocess takes O(1) + O() O(n) = O(n ) time. Each osition i = 0,..., n 1 belongs to some class C h(i). The construction also shows how to determine for each initial osition i a magnitude d i such that Magnus can force the token to a osition in class C h(i) 1, irresective of Derek s choice. Now, suose K belongs to some class C h(k). Magnus can choose d K to move to a osition in C h(k) 1. Iterating this rocess, the token finally moves to a osition in C 0, roving the lemma. Using lemma 4, we can now rove that f (n) ( 1) n. Lemma 5 Let n = m k where m, k are non-negative integers and is the smallest odd rime factor of n. Then there exists an O(n 3 ) time algorithm for Magnus that roduces a sequence of O(n ) moves such that f (n) ( 1) n. Proof. From any current osition K, if there exist two unvisited ositions I and J, J > I, such that gcd(j I, n) = f for some integer f, Magnus emloys the algorithm in lemma 4 to reach either I or J. After Magnus reaches either I or J, Magnus reeats this rocedure until no air of unvisited ositions I and J satisfy gcd(j I, n) = f. We claim that at most n ositions remain unvisited. Suose otherwise. We artition all n ositions into n consecutive grous of ositions as follows: (0, 1,,..., 1), (, +1, +,..., 1),..., (n, n +1,..., n 1). By the igeonhole rincile, there will be at least one grou containing two unvisited ositions, say I and J. We have J I 1 since they both belong to the same grou. Since is the smallest odd rime factor of n, gcd(j I, n) must be a ower of (including 0 = 1), which is a contradiction. The above strategy for Magnus roves the main theorem and yields insight into the structure of the roblem. We will now give an alternative, more efficient strategy for Magnus. 6. Second Strategy for Magnus First, we will rovide a strategy for the simle case when n is rime and then use it as a tool for the general case n = m. Although the techniques of this strategy differ from the revious one, it is built on the same sub-strategy: in many situations Magnus can reach at least one osition from a set of two. Lemma 6 If n > is rime, then f (n) n 1 and there exists a corresonding strategy for Magnus that requires O(n ) moves. At each move Magnus s comutations run in O(1) time. Proof. First, we rovide a O(n) stes rocedure for Magnus to reach at least one osition from a set of two different ositions α and β, β > α. Let d = β α mod n and say Magnus is at osition 0 currently. Then we can reresent o as [ 0 = α + ( 0 α)1 = α + ( 0 α)d 1] 0 d = α + µ 0 0 d where d 1 is the inverse of d and µ 0 = [ ( 0 α)d 1]. Let i be the token s osition at the end of ste i for i = 0, 1,..., n 1. A ste does not always corresond to a round, so it is ossible i = i 1, in which case only the formula for the current osition changes. Consider the following strategy for Magnus. At each ste i, if µ i 1 is even, let µ i µ i 1 / and no move is erformed; only the current osition changes its reresentation to i 1 = i = α+µ i i d mod n. If µ i 1 is odd, Magnus calls l = i 1 d mod n and moves to i = [ ] α + µ i 1 i 1 d ± i 1 d = α + [µ i 1 ± 1] i 1 d = α + 6 [ µi 1 ± 1 ] i d

7 deending on whether Derek s resonse is + or. Setting µ i if Derek s resonse is + or µ i = µ i 1 1 otherwise, the osition at the end of ste i can be reresented again as i = α + µ i i d mod n = µ i 1+1 After n 1 iterations, µ n 1 = 0 or 1 since µ 0 n 1 and µ i < µ i 1 unless µ i 1 was 0 or 1. Hence, n 1 = α + 0 n 1 d = α or n 1 = α + 1 n 1 d mod n. By Fermat s little theorem [], we know that n 1 = 1 mod n for rime n. Hence, Magnus is in either α or α + d = β after O(n) stes. Therefore, the rocedure takes O(n) moves (since one ste corresonds to zero or one round), and each ste takes O(1) time to calculate µ i. Now starting with any osition, Magnus has a strategy to achieve f (n) n 1. Whenever there are two or more unvisited ositions, he alies the above substrategy of O(n) moves for a total of O(n ) moves, giving the lemma. The lemma holds regardless of the starting osition. Also, the easiest way to calculate the new i d is to multily the old i 1 d by mod n. Combining lemma 6 with lemma 3 gives the value for f (n) when n is rime. As an aside, the study of the length of the µ i sequence with adversarial sequence of + and resonses from Derek may be of indeendent interest. Now we turn to the general case of n. Lemma 7 Let n = k 1... s where 1,,..., s are rime numbers, s and s 1 is an integer. Then there exists a strategy for Magnus such that f (n) ( 1)n and the number of moves and the running time for calculating the strategy is bounded as follows. If s = 1, that is, n = k 1, so (n/ 1 ) = k, then the uer bound is O( 1 + n). Otherwise, the uer bound is O(n / 1 ). Proof. For convenience let = 1, that is, we denote with the smallest odd rime factor of n. We artition the set of all ositions, C(n, 1, 0), into sets, C(n/ = m,, i), for i = 0, 1,..., 1. As a reliminary, we show that with O() moves Magnus can visit a osition from any two sets C(n/,, j 1 ) and C(n/,, j ), where j 1 j. Magnus uses an adatation of the technique from lemma 6. He calculates the magnitudes to call as if the table size were ; if his current osition is k, he assumes that the current osition is k mod ; and he lays as if he wanted to visit one of the two ositions j 1 or j on that imaginary table. The ositions he visits will not be the same as if he were laying on a table with size, but since the table size n = m, they will be congruent by (mod ) to the ositions he would have visited if the table size were. After O() moves, he will be in a real osition that is either congruent to j 1 or j, and, therefore, the osition will belong to C(n/,, j 1 ) or C(n/,, j ). Magnus roceeds in two hases and inductively uses the same two hase strategy on smaller roblem instances. Phase 1. Phase 1 consists of the reetition of two main stes: A and B. Magnus alies ste A after the token comes for the first time on a osition from any set C(n/ = m,, i). In ste A, Magnus lays as if the table size were n/: if n/ is a ower of, he alies the strategy from Section 4. On each round, he calculates an intermediate magnitude l, laying as if only the ositions in C(n/,, i) existed, but then he calls l. if n/ is not a ower of, he alies recursively this two hase algorithm. On each round, he calculates an intermediate magnitude l, laying as if only the ositions in C(n/,, i) existed, but then he calls l. 7

8 When he has visited at least f (n/) ositions from C(n/,, i), we say that Magnus has exlored the set C(n/,, i) and then he erforms ste B. There may be some unvisited ositions left from C(n/,, i). In ste B, Magnus chooses any two sets C(n/,, j 1 ) and C(n/,, j ) never exlored before and using the adatation of the technique from lemma 6, Magnus comes to one of these sets in O() moves. Then he alies ste A again. Phase 1 ends after 1 of the C(n/,, i) sets have been thus exlored and in each navigated set at least f (n/) ositions have been visited. Let at the end of this hase, the unexlored set be C(n/,, γ). There may also exist ositions not in C(n/,, γ) that Magnus has not visited; if no such ositions exist, Phase is not needed. Phase. Consider any osition α / C(n/,, γ) Then α C(n/,, β), with β γ. Magnus can use the following strategy to get with O() moves into either α or a osition from C(n/,, γ). First, using the adatation of the technique from lemma 6 described above, in O() moves, Magnus comes into a osition in either C(n/,, γ) or C(n/,, (β + γ)/) where the division / is erformed on rime field []. Suose Magnus gets onto a osition δ in C(n/,, (β + γ)/). Then Magnus s next move is α δ. If Derek resonds +, Magnus reaches δ + (α δ) which is α. Else, if Derek resonds, then Magnus reaches δ + (δ α) = δ α which is a osition in C(n/,, γ). This is because we have δ = (β + γ)/ + l 1 and α = β + l for some non-negative integers l 1 and l. Thus δ α = γ + (l 1 l ). Using the above strategy reeatedly with unvisited ositions outside of the set C(n/,, γ) Magnus can either visit them all in which case the hases are comlete or come onto a osition in C(n/,, γ). In this second case, Magnus exlores C(n/,, γ) by using this two hase algorithm recursively as in ste A of Phase 1. Thus, Magnus visits f (n/) ositions in C(n/,, γ) and the hases are comlete. We can now count the number of visited ositions. There are two ossibilities: either all ositions in each set C(n/,, i), where i γ are visited, or all C(n/,, i) s are visited and exlored using the two hase strategy. Hence, we have f (n) min{( 1)n/, f (n/)} Observe that inductively we can set f (n/) ( 1)(n/) since either n/ is exact ower of two in which case f (n/) = n/ or otherwise because is no larger than the smallest odd rime factor of n/. That roves f (n) ( 1)n. We can uer bound the number of moves, say g(n). Case 1. If s = 1, that is, n = k so (n/) = k. Then once Magnus reaches for the first time a set C(n/,, i γ), he can use the algorithm from Section 4 to visit all ositions in it for (n/) 1 moves. So g(n) + ( 1)( k 1) + n. Case. If n has at least two odd rime factors, that is, s and n = k 1... s where 1,,..., s are the odd rime factors of n sorted in increasing order. Then Magnus needs at most 1 moves in total to get for the first time into each of the sets C(n/ 1, 1, i γ) s. Together for Phase 1 and Phase, the total number of moves for navigating within the various C(n/ 1, 1, i) s is uer bounded by ( 1 1)g(n/ 1 ) + g(n/ 1 ) + ( 1 1)(n/ 1 )(1/ 1 )( 1 ) where the first summand is for navigating within the 1 1 sets C(n/ 1, 1, i γ) s in Phase 1, and the second summand is for navigating C(n/ 1, 1, γ) if Magnus reaches ositions in it in Phase. The third summand is for visiting all unvisited nodes left after Phase 1 in the 1 1 number of sets C(n/ 1, 1, i γ). To visit each of these unvisited nodes takes at most 1 moves and in each such set there are at most (1/ 1 )(n/ 1 ) nodes left unvisited after Phase 1. Hence, g(n) g(n/ 1 ) + n. (1) 8

9 Using the above inequality (1), we get that for j = 1,,..., s we have From Case 1 it follows that n g( ) n n j j+1 g( ) () 1... j 1... j 1... j+1 n g( ) n s + (3) 1... s s 1 Alying recursively the inequalities from () and from (3) to the right hand side of inequality (1), we get g(n) n g(n/ 1 ) n n + 1 g( n 1 ) 4n n + n n 1 3 g( 1 3 )... n(s 1) s 1 g( 1... s 1 ) n(s 1) s 1 ( n s s 1 ) n(s 1) + n s + n n s + n(s 1) (4) Since obviously (s 1) 1... s 1, we have the simler g(n) n s +n / s n / 1 = O(n / 1 ). Because, at each move, Magnuss comutations in lemma 6 run in O(1) time, here, in the general case of n, Magnuss comutations run also in O(1) time. Both the number of moves and the running time of the algorithm for Magnus are non-monotonic functions over n; similarly, f (n) is also non-monotonic with increasing n. These three factors deend on the factorization of n, and esecially on its smallest odd rime factor. 7 Concluding Remarks We have initiated the study of two erson games we call the Magnus-Derek games. We have roved tight existential bounds and shown matching algorithmic strategies on the basic version of the roblem. An immediate oen roblem is if we can imrove the number of moves (or the running time) of the algorithms. The following variants for the game may be of further interest. Dual roblems. Say the game stos after no more than m rounds, and let f m(n) be the size of visited set if both layers lay otimally. What is f m(n) and what are associated strategies? Off line strategy construction. Given a sequence d 1, d,..., d k of Magnus s moves, design corresonding + s or s for the Derek so that f (n) is minimized. Non-adative strategies. Limit Magnus and/or Derek to use only non-adative strategies, that is, Magnus (res. Derek) resecifies a sequence of moves without knowing Derek s (res. Magnus s) resonses, and evaluate the sequence when Derek (res. Magnus) later secifies resonses in order to otimize their goal. What is f (n) and what are associated non-adative strategies? Limit revisits. Suose Magnus may not visit a osition more than say k times. Then, what is f k (n) and what are associated strategies? Random strategies. Suose Magnus or Derek used random strategies. The roblem is to estimate the exected f (n). This is akin to random walks with random stes, but differs in having either one of the layer being deterministic or adversarial. Limiting information. Suose Magnus and/or Derek do not know n or knows n but does not have a way to factorize it. This is similar to grah exloration roblems studied in theoretical comuter science with limited number of tokens or memory. [1]. 9

10 Making each layer nag as well as move. The two layers have asymmetrical roles. For each move, first, Magnus chooses its magnitude, then Derek determines its direction. While Magnus wants to maximize the cardinality of the set of visited ositions, Derek wants to maximize the cardinality of the set of unvisited ositions. In other two-erson games too, this tye of asymmetry is seen (for examle, the Pusher-Chooser games of [8]). We roose a modified game in which Magnus and Derek alternate their roles each round. Positions visited by Magnus are colored red, and those by Derek are colored blue. Derek is restricted to non-red ositions and Magnus to non-blue. What are suitable strategies and how many ositions are visited by each layer? 8 Acknowledgements We are grateful to the following eole and institutions for their advice, discussion, and suort: Jeffrey O. Shallit, Noam Sturmwind for editing assistance, Valerie King, Uriel Feige, Nicola Santoro, Graham Cormode, Richard Guy, Susan Hoe and Peter Winkler; the School of Comuter Science at University of Waterloo and the DIMACS Center.. References [1] M. A. Bender, A. Fernandez, D. Ron, A. Sahai, and S. Vadhan. The Power of a Pebble: Exloring and Maing Directed Grahs. Proc. of the 30th Annual ACM Symosium on Theory of Comuting (STOC), ages 69-78, [] I. Niven and H. Zuckerman. An introduction to the theory of numbers. John Wiley and Sons Inc, [3] N. Sloane. On line Encycloedia of Sequences. htt:// njas/sequences/ [4] N. Santoro. Sense of Direction, Toological Awareness and Communication Comlexity. SIGACT News, 16:, 1984, [5] H. Attiya, J. van Leeuwen, N. Santoro, S. Zaks. Efficient Elections in Chordal Ring Networks. Algorithmica (1989) 4: [6] P. Flocchini, B. Mans, N. Santoro. Sense of Direction: Definitions, Proerties, and Classes. Networks 3 (1998), no. 3, [7] Wayne A. Jansen. Intrusion detection with mobile agents. Comuter Communications, Volume 5, Issue 15, 15 Setember 00: [8] J. Sencer and P. Winkler. Three thresholds for a liar. Combinatorics, Probability and Comuting, 1:1 (199),

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