Beyond Knights and Knaves

Transcription

1 Beyond Knights and Knaves Christine T. Cheng 1, Andrew McConvey 2,, Drew Onderko 1,, Nathaniel Shar 3,, and Charles Tomlinson 4 1 Department of Computer Science, University of Wisconsin-Milwaukee, Milwaukee, WI 53211, USA {ccheng,donderko}@uwm.edu 2 Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA mcconve2@illinois.edu 3 Department of Mathematics, Rutgers University, Piscataway, NJ 08854, USA nshar@math.rutgers.edu 4 Department of Mathematics, University of Nebraska-Lincoln, Lincoln, NE 68588, USA ctomlinson2@math.unl.edu Abstract. In the classic knights and knaves problem, there are n people in a room each of whom is a knight or a knave. Knights always tell the truth while knaves always lie. Everyone in the room knows each other s identity. You are allowed to ask questions of the form Person i, is person j a knight? and you are told that there are more knights than knaves. What is the fewest number of questions you can ask to determine a knight? How about to determine everyone s identity? In this paper, we consider the knights and no-men problem, where a noman is a person who always answers no. Assuming there are at least k knights, we show that ( ) n 1 2 (k 2)(n 1) 2 questions are necessary and 2(k 1) sufficient in the worst case to identify a knight. We also show that n 2 questions suffice to identify a no-man, and ( ) n 1 2 (k 2)(n 1) 2 +n 2 2(k 1) questions suffice to identify everyone in the room. We then consider a generalization of the knights and knaves problem that captures most of the variants of the knights and knaves problem in the literature. In the agent labeling problem, we wish to identify everyone s type; in the agent identification problem, we wish to identify an agent having a particular type. We present results with regards to the fewest number of questions needed in the worst case to solve both the agent labeling and agent identification problems. 1 Introduction In the classic knights and knaves problem, there are n people in a room labeled 1 to n. Each person is either a knight or a knave. Knights always tell the truth Supported in part by the National Science Foundation grant DMS EMSW21-MCTP: Research Experience for Graduate Students.

2 while knaves always lie. Everyone in the room knows each other s identity. You areallowedto askquestions ofthe form Personi, is personj aknight? and you are told that there are more knights than knaves. What is the fewest number of questions you can ask to determine a knight? How about to determine everyone s identity? Popular variations to this problem introduce new types of people such as spies, who lie or tell the truth arbitrarily, and yes-men, who answer yes all the time. Several equivalent formulations of the knights and knaves problem have been studied by other authors including the problem of classifying computer chips as faulty or reliable [2], the problem of identifying a coin of the majority weight by performing a series of weight comparisons [10], the problem of identifying a ball of the most common color by doing a series of color comparisons [1], [3], [5], [7], [8], etc. Saks and Werman [8] studied a version of the knights and knaves problem where knights are in the majority if such a majority exists. Their goal was to identify a knight or declare the nonexistence of such a majority. They showed that n B(n) questions are necessary and sufficient in the worst case, where B(n) is the number of ones in the binary representation of n. Simpler proofs were given later by Alonso et al. [3] and Wiener [10]. If the number of knights is at least k, and k > n/2, Aigner [1] showed that n 1 p questions are necessary in the worst case, where p is the highest power of two dividing ( n 1 k 1), and that 2(n k) B(n k) questions are necessary and sufficient in the worst case when it is known that a majority exists. When we want to identify everyone in a room of knights and knaves, Aigner [1] observed that n 1 questions are necessary in the worst case whenever n > 2. Aigner also showed that when the number of knights is at least k, k > n/2, questions are necessary in the worst case, up to a possible error of 1. The knights and spies problem is just like the knights and knaves problem except that agents which are not knights are spies. Alonso et al. [2] showed that, surprisingly, any strategy that can identify a knight in the knights and knaves setting can also be used to identify a knight in the knights and spies setting. Therefore, the two problems require the same number of questions to identify a knight in the worst case. Aigner [1] also proved that the same result is true when the number of knights is bounded below by k, k > n/2. If we wish to identify everyone in a room of knights and spies and there are k > n/2 knights, Blecher [4] and Wildon [11] showed independently that 2n k 1 questions are necessary and sufficient in the worst case. Thus, it follows that 3n/2 1 questions are necessary and sufficient in the worst case if it is only known that knights are in the majority. Most recently, Hanajiri [6] introduced the concept of yes-men and studied the knights and yes-men problem. Suppose there are n people in the room, at most p ofwhomareknights and atmost q ofwhom areyes-men.hanajirishowed that n+ log 2 (p+q n) questions suffice for identifying everyone in the room. Additionally, he conjectured that the same number of questions are necessary (n k)n+(n k) n k+1

3 in the worst case when p,q / {0,n}. He was able to verify his conjecture when p+q {n,n+1,2n 1}. Our Results. Inspired by Hanajiri s work, we introduce the notion of a noman, a person who always answers no. Assume that there are n people each of whom is a knight or a no-man, and that there are at least k knights. We show that ( ) n 1 2 (k 2)(n 1) 2 2(k 1) questions are necessary and sufficient in the worst case to identify a knight. We also show that n 2 questions suffice to identify a no-man, and ( ) n 1 2 (k 2)(n 1) 2 2(k 1) +n 2 questions suffice to identify everyone in the room. Next, we present a generalization of the knights and knaves problem that allows us to consider new types of people and/or put multiple types (not just two types) of people in a room. An instance consists of n people, the m types that each person can have, and a 0-1 matrix Q of order m that describes how the m types of people interact with each other. We use D Q to denote the direct graphon[m] whoseadjacencymatrixisq. In the agent labeling problem, wewish to determine the type of every individual in the room; in the agent identification problem, we wish to find an agent having a particular type. Itturnsoutthatevenwhenwehaveaskedtheagentsallthepossiblequestions we can ask, we may not be able to solve the agent labeling or agent identification problems. When this happens, we call the instance ill-formed; otherwise, it is well-formed. We provide a characterization of the well-formed instances of both the agent labeling and agent identification problems in terms of D Q. For the agent labeling problem, we prove that when an instance s D Q has what we call an uninformative partition, identifying everyone s type requires Ω(n 2 /m 4 ) questions in the worst case. On the other hand, when the instance s D Q hasnouninformativepartitions,everyone stypecanbedeterminedino(mn) questions. This implies that when m = o(n 1/5 ) there are hard and easy variants of the knights and knaves problem. In particular, when m is a constant, the hard variants need Ω(n 2 ) questions to solve the agent labeling problem in the worst case. The easy variants, however, can solve the agent labeling problem using O(n) questions. For the agent identification problem, we prove that when an instance s D Q has uninformative partitions and s is an informative type with respect to one of the uninformativepartitions, then finding anagent oftype s requiresω(n 2 /m 4 ) questions in the worst case. On the other hand, when the instance s D Q has no uninformative partition, finding an agent of any type can be solved using O(mn) questions. Thus, setting aside the case when s is an uninformative type with respect to every uninformative partition of D Q, our results again imply that when m = o(n 1/5 ), there are easy and hard variants of the knights and knaves problem when it comes to solving the agent identification problem.

4 2 The Knights and No-Men Problem Let A = {a 1,a 2,...,a n } be a set of agents each of whom is a knight or a noman. We are allowed to ask questions of the form a i, is a j a knight? for i j. A knight always answers truthfully while a no-man always answers no. Additionally, we are told that there are at least k 2 knights. Let us consider the problem of finding a knight. Suppose we asked the agents a set of questions S. Let G S be the undirected graphwherev(g S ) = Aand E(G S ) = {{a i,a j } : a i, is a j a knight? or a j, is a i a knight? is in S}. We shall say that a labeling of the agents f : A {knight, no-man} is consistent with respect to S if based on the agents answers it is plausible that each a i has type f(a i ). More specifically, f is a consistent labeling with respect to S if for each question a i, is a j a knight? in S, the answer given by a i is the same as the answer an agent of type f(a i ) would give if asked about an agent of type f(a j ). Lemma 1. After asking the questions in S, we can conclude that some particular agent is a knight if and only if at least one of these conditions hold: (1) some question s answer is yes, or (2) the answers to all the questions are no and there is some a i that is in every independent set of size k in G S. Proof. First, we argue that the conditions are sufficient. We get a response of yes if and only if one knight is asked about another knight. Thus, if condition (1) holds and the answer to the question associated with {a i,a j } E(G S ) is a yes, we can conclude that both a i and a j are knights. So suppose the answers to the questions in S are all no. It must be the case that no knight has been asked about another knight. The set of knights form an independent set of size at least k in G S. If condition (2) holds, some a i is part of every independent set of size k in G S, including the independent sets containing k knights. We can then conclude that a i is a knight. Next, let us show that the conditions are necessary. Suppose neither condition holds. Since condition (1) does not hold, the responses to the questions in S are all no. Once again the set of knights form an independent set I of size at least k in G S. Since condition (2) does not hold, for every a i A, there exists an independent set I ai in G S of size k such that a i I ai. Now, for each a i A, let f i be a labeling of the agents that assigns the agents in I ai as knights and all other agents as no-men. Clearly, f i is consistent with respect to S. Since it is possible that each a i is not a knight, we cannot conclude that any particular agent in A is a knight. The Turán graph T(n,k) is the graph whose n vertices are partitioned into k parts of as equal size as possible, and two vertices are adjacent if and only if they belong to different parts. Turán s theorem [9] states that among n-vertex graphs without a clique of size k +1, the graph with the most number of edges is T(n,k). In particular, T(n,k) has (k 1)n2 2k edges. The next lemma shows the connection of Lemma 1, condition (2) to the complement of T(n 1,k 1).

5 Lemma 2. Let G(n,k) contain all graphs G on n vertices such that G has a vertex that is part of every independent set of size k. Let G G(n,k) so that among all the graphs in G(n,k) it has the fewest number of edges. Then G is the complement of T(n 1,k 1) unioned with an isolated vertex. Proof. Let v be a vertex of G that is part of every independent set of size k. Notice that deg(v) = 0. Otherwise, any edge incident to v can be removed and v will still part of every independent set of size k, contradicting our assumption about the minimality of E(G ). Next, consider G v. By our assumptions about G and v, it must be the case that G v has no independent sets of size k, and among such graphs with n 1 vertices, it has the fewest number of edges. Consequently, its complement has the property that it has no cliques of size k, and among such graphs with n 1 vertices, it has the most number of edges. According to Turán s theorem [9], this complement is T(n 1,k 1). Our lemma follows. Here is our main result. Theorem 1. Let I(n,k) be the fewest number of questions needed in the worst case to identify a knight in the knights and no-men problem when there are n agents at least k 2 of whom are knights. Then I(n,k) = ( ) n 1 2 (k 2)(n 1) 2 2(k 1), the number of edges in the complement of T(n 1,k 1). Proof. Let t (n 1,k 1) denote the number of edges in the complement of T(n 1,k 1). First, we describe a strategy for finding a knight. Start by setting aside an agent. Partition the remaining n 1 agents into k 1 parts of as equal size as possible. For each part, ask every pair {a i,a j } of agents the question a i, is a j a knight?. (The order of a i and a j doesn t matter here.) If the response is yes, stop and conclude that that a i and a j are knights. If none of the agents answered yes, conclude that (1) there are exactly k knights, (2) each part has exactly one of them, and (3) the agent we had set aside is a knight. Using the fact that there are at least k knights and at most k 1 parts, it is easy to verify that our strategy s conclusions are correct. It also asks the most number of questions when the answers to all its questions are no. Now, we have chosen the questions so that they form a graph that is isomorphic to G so the number of questions our strategy will ask in the worst case is t (n 1,k 1); i.e., I(n,k) t (n 1,k 1). Next, consider an arbitrary strategy. Suppose there are situations where the strategy will conclude that some agent is a knight even though all the answers to its questions are no. Let G be the graph formed by these questions. According to Lemma 1, G must have the property described in condition (2). Thus, G G(n,k). According to Lemma 2, G must have at least t (n 1,k 1) edges. Now, suppose the strategy will conclude that some agent is a knight only when it receives a yes response. Assume that in the worst case it will ask r questions. This means that in every situation where the strategy has already asked r 1 questions and the answers were all no, the rth question must generate a yes answer and the two agents that are part of the question are knights. Thus, if

6 the strategy knows about the worst case bound r, it can avoid asking the rth question since after receiving the no answer to the (r 1)st question, it can already identify an agent that is a knight. Applying the reasoning we used in the previous case, we have that r 1 t (n 1,k 1) so r > t (n 1,k 1). We have shown that every strategy for finding a knight where there are n agents and at least k 2 knights will need to ask at least t (n 1,k 1) questions. Interestingly, it is much easier to find a no-man than a knight as shown in this next theorem. Theorem 2. Let Î(n,k) be the fewest number of questions needed in the worst case to identify a no-man in the knights and no-men problem when there are n agents, at least k 2 knights, and at least 2 no-men. Then Î(n,k) n 2. Proof. Choose an arbitrary agent a and ask a about n k agents. If all the answers are no then a has to be a no-man because at least one of the n k agents we asked a about is a knight. But suppose at least one answer is a yes, then a is a knight. If there is also a no answer, then the agent we asked a about must be a no-man. But suppose all the answers are yes. This means that a and the n k agents are all knights. Then we just ask a about each of the remaining k 1 agents we have not asked a about until we get a no response. This leads us to a no-man using at most n 2 questions. Next, consider the problem of identifying all n agents as either a knight or a no-man. Theorem 3. Let L(n,k) be the fewest number of questions needed in the worst case to identify all agents as a knight or no-man when there are n agents at least k 2 of whom are knights. Then I(n,k) L(n,k) I(n,k)+n 2. Proof. Since identifying all agents involves identifying at least one knight, A(n, k) I(n,k). To prove the upper bound, we describe a strategy for identifying all agents. Start by using the strategy described in the proof of Theorem 1 to identify a knight. Suppose a yes answer was obtained. We can immediately determine that two agents are knights. We then ask one of them about the remaining n 2 agents. On the other hand, suppose the answers to the initial strategy for identifying a knight are all no. Then we know that the agent that was set aside is a knight, and there are exactly k knights. We then ask the knight about n 2 of the remaining unidentified n 1 agents. If k knights have been identified altogether, the last unidentified agent has to be a no-man; otherwise, he is a knight. In both cases, the number of questions we have asked is at most I(n,k)+n 2. We suspect that the upper bound given in Theorem 3 is the correct value for A(n,k).

7 3 Generalizing the Knights and Knaves Problem We now consider a generalization of the knights and knaves problem that encompasses most of the variants we discussed in the introduction. It has the following set-up: There are n agents in a room each of whom is one of m types. The agents have been together long enough so they know each other s types. Outsiders are allowed to communicate with the agents by asking only one kind of question. This question is directed it is addressed to some agent a i and is about another agent a j and has a yes or no answer. Agent a i s response is a deterministic function of his and a j s types. 5 You happen to come across this room and are told how agents respond to the directed questions as a function of their types. Your goal is to determine the types of all the agents or identify an agent of a particular type using as few questions as possible. Notice that in our set-up, the actual question itself is not a factor since it is the only question an outsider can ask. What matters instead is the direction of the question and the response to it. Assume the types are from the set [m] = {1,2,...,m}. We shall use a (0,1)-matrix Q of order m to encode the responses of the m types of agents to the directed question. That is, Q st is equal to 0 if the answer to the question addressed to an agent of type s about an agent of type t is no and is equal to 1 otherwise. Throughout the paper, we shall represent the structure of agent types in terms of D Q = ([m],q), the directed graph whose vertex set is [m] and whose adjacency matrix is Q. Formally, our model has a set of n agents A = {a 1,a 2,...,a n }, where t(a i ) denotes the type of agent a i. Every t(a i ) [m]. As an outsider, we do not know the agents types. However, we are allowed to ask one agent about another agent using some standard format. Let q(a i,a j ) denote the answer to the question addressed to a i about a j, where a i a j. 6 This answer, given by a i, is Q t(ai),t(a j). We are interested in the following problems: The agent labeling problem. Given A, Q, and the ability to query the agents, determine the types of all the agents in A. The agent identification problem. Given A, Q, s [m], and the ability to query the agents, find an agent in A of type s. Our goal is to solve both problems using as few questions as possible. Some examples. Consider an instance where the agents can have three types: knights, knaves, or yes-men. The table on the left shows how the three types of agents respond to the question Agent a i, is agent a j a knight?, the matrix in the middle encodes these responses, while the directed graph on the right has the matrix as its adjacency matrix. 5 Thus, none of the agents are spies since spies lie arbitrarily. 6 We emphasize that a i and a j must be distinct. Otherwise, in the knights and knaves problem, etc., we can just ask each agent if he is a knight and the agent labeling and agent identification problems become trivial.

8 Knight Knave Yes-man (v 1 ) Knight Yes No No (v 2 ) Knave No Yes Yes (v 3 ) Yes-man Yes Yes Yes v 2 v 1 v 3 In the coin partitioning problem, there are n coins which have to be partitioned into m distinct weight classes by a series of weight comparisons. That is, there are n agents with m different types. We are allowed to ask the question Coin a i, is coin a j s less than or equal to your weight? The matrix below encodes the responses, and the associated directed graph is shown on the right v1 v2 v3... vm Assumptions. Finally, we make the following assumptions about our model: First, for each s [m], there are at least two agents with type s. This is a technicality we need because we cannot ask an agent about himself. For example, if there is only one agent of a certain type, we will not be able to tell if this agent s type has a loop or not in D Q. Second,foranytwotypessands,theyeitherhaveadifferentin-neighborhood or a different out-neighborhood in D Q. Otherwise, we have two versions of the same type, and it would be impossible for us to distinguish between these two versions. Third, we have infinite computational power. That is, we can develop strategies and process the agents responses using algorithms that may not be efficient. This will allow us to focus solely on analyzing the number of questions we need to ask to solve the agent labeling and agent identification problems. 3.1 Similar Agents and Consistent Labelings In the next section, we will consider the scenario where we asked the agents all the possible questions we can ask. We would like to group them according to how they answered as well as how others answered when asked about them. Let agent b a i,a j. We say that a i and a j are similar with respect to b, denoted as a i b a j, if q(a i,b) = q(a j,b) and q(b,a i ) = q(b,a j ). Additionally, we say that a i and a j are similar, denoted as a i a j, if they are similar with respect to every agent b a i,a j. Lemma 3. For any two agents a i and a j, a i a j if and only if t(a i ) = t(a j ). Hence, the relation is an equivalence relation over A and has m equivalence classes.

9 Proof. Let b a i,a j. When t(a i ) = t(a j ), q(a i,b) = q(a j,b) and q(b,a i ) = q(b,a j ) because these answers are simply based on the type of b and the types of a i and a j. Thus, a i a j. Let us consider the other direction. Suppose a i a j but t(a i ) t(a j ). By our second assumption, t(a i ) and t(a j ) must differ in their in-neighborhoods or their out-neighborhoods in D Q. Assume the former is true. There there is some type s so that Q s,t(ai) Q s,t(aj). By our first assumption, there is an agent b a i,a j such that t(b) = s. Then q(b,a i ) q(b,a j ), contradicting the assumption that a i a j. The same argument holds when t(a i ) and t(a j ) have different out-neighborhoods in D Q. It follows that t(a i ) = t(a j ). We have shown that for any two agents a i and a j, a i a j if and only if t(a i ) = t(a j ). This fact immediately implies that the relation is reflexive, symmetric and transitive i.e., is an equivalence relation. Since there are m types of agents and by our first assumption there is at least two agents of each type, has m equivalence classes. Another issue we will need to resolve is how we determine the agents types. For this, we extend the notion of consistent labelings described from the previous section to our general setting. Let S be a set of questions we have asked the agents. Based on their answers, we assign each agent a label from [m]. We say that such a labeling of the agents f : A [m] is consistent with respect to S if for each question addressed to a i about a j in S, the answer given by a i is the same as the answer an agent of type f(a i ) would give if asked about an agent of type f(a j ). When S contains all the questions we can ask the agents of A, we simply say that f is a consistent labeling. One such example is the labeling f that assigns each agent their type (i.e., f (a) = t(a) for each a A). Proposition 1. A labeling f : A [m] is consistent if for any ordered pair (a i,a j ) of agents, Q f(ai),f(a j) = Q t(ai),t(a j). That is, (f(a i ),f(a j )) is an edge in D Q if and only if (t(a i ),t(a j )) is an edge in D Q. The next lemma describes the relationships between similar agents and consistent labelings. Lemma 4. For any two agents a i and a j, every consistent labeling f assigns a i and a j the same labels if and only if t(a i ) = t(a j ). Proof. Let f be a consistent labeling. Suppose f(a i ) = f(a j ) but t(a i ) t(a j ). Since t(a i ) t(a j ), there must be some type s so that Q s,t(ai) Q s,t(aj) or Q t(ai),s Q t(aj),s according to our second assumption. By the first assumption, there is some agent b a i,a j such that t(b) = s. Then Q f(b),f(ai) = Q s,t(ai) and Q f(b),f(aj) = Q s,t(aj). Since f(a i ) = f(a j ), Q s,t(ai) = Q s,t(aj). By a similar argument, Q t(ai),s = Q t(aj),s. We have arrived at a contradiction. Hence, t(a i ) = t(a j ). Thus, if agents have the same labels under f, they also must have the same type. For each s [m], let b s denote an agent of type s. According to our previous argument, the m agents in {b s : s [m]} are assigned to different labels by f.

10 Now suppose f assigned an agent of type s, b s, to a label different from f(b s). Then there must be two agents of different types assigned to the same label. Since this is not possible, f must assign all agents of type s to f(b s ). 4 Ill-Formed Instances and Consistent Labelings When we tradequestions and answerswith the agents,we arein effect narrowing down the set of consistent labelings of the agents. In the agent labeling problem, our goal is to get to a point where there is only one consistent labeling left so that we are certain about each agent s type. In the agent identification problem where we want to identify an agent of type s, our goal is to reach a point where there is an agent a i, so that in all of the labelings consistent with the set of questions asked so far, a i is assigned the label s. We call (A,Q) an ill-formed instance of the agent labeling problem if there is more than one consistent labeling of the agents. That is, the agent labeling problem cannot be solved even after we have asked all the questions. Similarly, (A,Q,s ) is an ill-formed instance of the agent identification problem if there is no agent a i so that a i is assigned the label s in every consistent labeling of the agents. For both problems, we call an instance well-formed if it is not ill-formed. As an example, consider an instance (A,Q) of the classic knights and knaves problem. Assume we have asked all the possible questions. Let a be one of the agents. Partition the set of agents into two groups: the first group contains a and all the agents a identified as knights (i.e., a answered yes when asked about these agents); the second group contains the remaining agents (i.e., a answered no when asked about these agents). Notice that if a is a knight, the first group is composed entirely of knights and the second group entirely of knaves. On the other hand, if a is a knave, the opposite is true. Thus, there are two consistent labelings: f 1 assigns the first group as knights and the second group as knaves; f 2 does the opposite. In other words, after we have asked all the questions, we can partition the agents into a group of knights and a group of knaves but we cannot tell them apart. This is because knights and knaves behave in the same way they both say yes toagentswhohavetheirtype and no to others.this is captured in D Q where both the knight-type and the knave-type are singleton vertices with a loop. Hence, (A,Q) is an ill-formed instance of the agent labeling problem. Similarly, (A,Q,s ) where s is the type corresponding to a knight (or knave) is an ill-formed instance of the agent identification problem. The astute reader would note though that we are able to identify a knight in the classic knights and knaves problem. The reason is quite subtle in the problem s description, we are told that there are more knights than knaves. Thus, the group with more agents is the knights group. In what follows, we provide a graph-based characterization of the ill-formed instances (and equivalently the well-formed instances) of the agent labeling and agent identification problems. Lemma 5. Let α be an automorphism of D Q. Then f α such that f α (a) = α(t(a)) for each a A is a consistent labeling of A.

11 Proof. Foranyorderedpairofagents(a i,a j )ofa,q f(ai),f(a j) = Q α(t(ai)),α(t(a j)). Butthe latterisequaltoq t(ai),t(a j) becauseαisanautomorphismofd Q.Hence, f is a consistent labeling of A. Lemma 6. Suppose f is a consistent labeling of A. For each s [m], let b s be an agent of type s. Then α f such that α f (s) = f(b s ) for each s [m] is an automorphism of D Q. Proof. Let B = {b s,s [m]}. We note that by Lemma 4, no two agents in B are mapped by f to the same label. Hence, α f is a bijection. Consider an ordered pair of distinct types (s,s ). By the definition of α f, (α f (s),α f (s )) is an edge in D Q if and only if (f(b s ),f(b s )) is an edge in D Q. But the latter is true if and only if (t(b s ),t(b s )) = (s,s ) is an edge in D Q. Thus, α f preserves edges between distinct types in D Q. Now, consider the ordered pair (s,s). Let b s be another agent with label s. According to Lemma 4, f(b s ) = f(b s). Hence, (α f (s),α f (s)) is an edge in D Q if and only if (f(b s ),f(b s )) is an edge in D Q. But the latter is true if and only if (t(b s ),t(b s )) = (s,s) is an edge in D Q. It follows that α f also preserves edges between the same types in D Q. Hence, α f is an automorphism of D Q. Theorem 4. An instance (A, Q) of the agent labeling problem is ill-formed if and only if D Q has more than one automorphism. Proof. Suppose D Q has an automorphism α that is different from the identity map. From Lemma 5, f α is a consistent lebeling of A. Since α is not the identity map, f f so (A,Q) is ill-formed. Suppose f f is another consistent labeling of A because (A,Q) is illformed. By Lemma 4, there is some s [m] so that for every agent a of label s, f(a) s. By Lemma 6, α f is an automorphism of D Q. Since f(b s ) s, α f is not the identity map. Therefore, D Q has more than one automorphism. Theorem 5. An instance (A,Q,s ) of the agent identification problem is illformed if and only if some automorphism of D Q does not fix s (i.e., the automorphism does not map s to itself). Proof. Suppose α is an automorphism of D Q that does not fix s. By Lemma 5, f α is a consistent labeling of A. Since f assigns all agents of type s to a label different from s, (A,Q,s ) is ill-formed. Suppose (A,Q,s ) is an ill-formed instance. By Lemma 4, there must be a consistent labeling f that assigns all agents of type s to a label s so that s s. According to Lemma 6, α f is an automorphism of D Q. Since f(b s ) = s, α f (s ) = s s. That is, α f does not fix s. 4.1 Finding a Consistent Labeling Let us onceagainassumethat we haveaskedthe agentsallthe possible questions we can ask. How do we go about constructing a consistent labeling of the agents?

12 Let E 1,E 2,...,E m denote the equivalence classes of. For each E i, let b i and b i be two agents in E i. Denote by D the directed graph whose vertex set is {E 1,E 2,...,E m } and whose edge set is {(E i,e j ) : q(b i,b j ) = 1} {(E i,e i ) : q(b i,b i ) = 1}. Since the equivalence classes are in one-to-one correspondence with the m types of agents, D is in fact isomorphic to D Q. We now describe a procedure for labeling the agents of A: Label(A,Q,{q(a i,a j ) : a i,a j A}) Partition the agents according to and form the equivalence classes of. Construct the directed graph D. Find an isomorphism α from D to D Q. For each agent a in equivalence class E, set f(a) = α(e). Return f. Theorem 6. The labeling f returned by Label is consistent. Consequently, (1) when (A,Q) is a well-formed instance of the agent labeling problem, f = f. (2) when (A,Q,s ) is a well-formed instance of the agent identification problem, f(a) = s if and only if t(a) = s for each a A. Proof. First, consider any ordered pair of agents (a,b) where a E i, b E j, a b but E i may be the same as E j. We know that (α(e i ),α(e j )) is an edge in D Q if and only if (t(a),t(b)) is an edge in D Q. Thus, Q f(a),f(b) = Q α(ei),α(e j) = Q t(a),t(b). That is, f is a consistent labeling. When (A, Q) is a well-formed instance of the agent labeling problem, there is only one consistent labeling of the agents so f must be f. When (A,Q,s ) is a well-formed instance of the agent identification problem, there is at least one agent a of type s so that every consistent labeling assigns a the label s. But according to Lemma 4 every consistent labeling must assign all agents of type s the label s. Furthermore, agents of type s s will not be assigned the label s. Since f is a consistent labeling, f(a) = s if and only if t(a) = s for each a A. Theorem 6 shows that when we have asked the agents all the questions we can ask and the instances are well-formed, Label will solve the agent labeling and agent identification problems. Starting with the next section onwards, we will assume that the inputs to the agent labeling or agent identification problems are well-formed. 5 Uninformative Partitions Two of our objectives for studying the generalized knights and knaves problem is to understand (1) why solving the agent labeling problem requires significantly more questions in the worst case in the knights and no-men setting than in the knights and yes-men setting, and (2) why finding a knight requires significantly

13 more questions in the worst case than finding a no-man in the knights and nomen setting. In this section, we begin to address these issues by introducing what we call uninformative partitions. Any two (not necessarily distinct) types s and s have four possible relationships basedon the questionsthat can be askedbetween agentsoftype s andwith those of type s : (1) Q s,s = Q s,s = 1; (2) Q s,s = 1, Q s,s = 0; (3) Q s,s = 0, Q s,s = 1; (4) Q s,s = Q s,s = 0. Keeping this in mind, let P = {P 1,P 2,...,P r } be a partition of [m] with r < m. Type s is uninformative with respect to P if, for i = 1,...,r, the relationship of s with each type in P i is the same. That is, we cannot use questions addressed to or about agents of type s to distinguish between agents whose types belong to the same part in P. Otherwise, s is informative with respect to P. We call P an uninformative partition of D Q = ([m],q) if each of its parts contains an uninformative type with respect to P. For example, in the knights and yes-men problem, the only possible candidate for an uninformative partition is P = {{knight, yes-man}} since we can only consider partitions with just one part. A knight is informative with respect to P because two knights have a relationship that is described by (1) whereas a knight and a yes-man have a relationship that is described by (3). Similarly, a yes-man is also informative with respect to P because the relationship between a yes-man and a knight is described by (2) whereas that between a yes-man and another yes-man is described by (1). Thus, the knights and yes-men problem has no uninformative partitions. On the other hand, in the knights and no-men problem, we leave it up to the reader to verify that the no-man is uninformative with respect to P = {{knight, no-man}} while the knight a informative with respect to the same partition. Hence, the knights and no-man problem has an uninformative partition. Proposition 2. Let P = {P 1,P 2,...,P r } be an uninformative partition of D Q. Each part of P contains exactly one uninformative type with respect to P. Proof. For each P i, let s i be an uninformative type with respect to P. Without loss of generality, suppose P 1 contains two uninformative types: s 1 and s 1. We will argue that this implies that s 1 and s 1 have exactly the same in- and outneighborhoods in D Q, contradicting the second assumption we made about our model. Suppose s P i is in the in-neighborhood of s 1 in D Q. We know that Q si,s 1 = Q si,s since s i uninformative. Thus, if s = s 1 i, s is also in the inneighborhood of s 1. But suppose s s i. Then Q si,s 1 = Q s,s1 and Q si,s = Q 1 s,s 1 since both s 1 and s 1 are uninformative. Using the first fact we cited, Q s,s 1 = Q s,s 1 so again s is in the in-neighborhood of s 1. It follows that s 1 and s 1 have the same in-neighborhood. A similar argument shows that they have the same out-neighborhood. Thus, no part of P can have two or more uninformative types with respect to P. Fromhereon,assumethatP = {P 1,P 2,...,P r }isanuninformativepartition of D Q with s i as the uninformative type in P i. An agent a A is uninformative agent with respect to P if t(a) is an uninformative type with respect to P; otherwise, a is informative with respect to P. How difficult is it to find an agent

14 who is informative with respect to P? We note that the agent labeling problem is at least as hard as this problem since knowing each agent s type allows us to immediately identify an informative agent. The agent identification problem where s is an informative type with respect to P is also at least as hard as this problem since every agent of type s is informative. Let A P = {A 1,A 2,...,A r } denote the partition of A such that a A i if and only if t(a) P i. We shall now show that even when we know A P, finding an informative agent requires Ω(n 2 /m 4 ) questions in the worst case. We will prove this result in the same way that we showed that finding a knight in the knights and no-men problem requires Ω(n 2 ) questions in the worst case. Let a A i and b A j. Suppose we asked a about b. We say that the question is unhelpful when q(a,b) = Q si,s j because we did not gain any definite information about a or b s type from q(a,b). On the other hand, when q(a,b) Q si,s j, we say that the question is helpful because we can conclude that both a and b are informative agents. In the knights and no-man problem, questions with no responses were unhelpful while those with yes responses were helpful. Let G = (V,E) be a graph and V = {V 1,V 2,...,V r } be a partition of V. Let k i beanon-negativeintegersuchthatk i V i fori = 1,...,r. A(k 1,k 2,...,k r )- independent set I of G is an independent set of G such that I V i = k i for i = 1,...,r. We are now ready to present the counterparts to Lemmas 1, 2, and Theorem 1. Lemma 7. After asking the questions in S, we can conclude that some agent is informative if and only if one of these conditions hold: (1) some question in S is helpful, or (2) every question in S is unhelpful and there is some agent a that is in every (k 1,k 2,...,k r )-independent set in G S where k i = 2( P i 1) for i = 1,...,r. Proof. First, we show that the conditions are sufficient. By definition, a question is helpful if and only if the two involved agents are informative. Thus, if condition (1) holds, we can immediately identify two informative agents. So suppose all questions in S are unhelpful. The set of informative agents must form a (k 1,k 2,...,k r)-independent set in G S where k i 2( P i 1) for i = 1,...,r since each informative label in P i has at least two agents in A i. If condition (2) holds, a must be part of this independent set so we can conclude that a is an informative agent. Next, let us show that the conditions are necessary. Suppose neither condition holds. This means that all questions in S are unhelpful so once again the set of informative agents form a (k 1,k 2,...,k r )-independent set in S where k i 2( P i 1) for i = 1,...,r in G S. Furthermore, for each agent a, there is some (k 1,k 2,...,k r )-independent set I a in G S that does not contain a. We shall show below that there is a labeling f a that is consistent with respect to S such that f(a) is an uninformative type. This implies that it is plausible that a is an uninformative agent. Since we chose a arbitrarily, we cannot conclude that any particular agent in A is informative. We now construct f a. For i = 1,...,k do the following: (1) for each informative type s i in P i, pick two unlabeled agents from I a A i and set their f a -label

15 to s i, and (2) assign all agents in A i I a the uninformative type s i. The labeling f a is consistent with respect to S because whenever there is an edge between two agents in G S, f a assigned at least one of them an uninformative type. Lemma 8. Let k = (k 1,k 2,...,k r ) such that k i = 2( P i 1) for i = 1,...,r. For each set A of n agents partitioned according to A P, let G(A P,k) contain all graphs G on A that has a vertex that is part of every k-independent set in G. Then there is some A such that every graph in G(A P,k) has Ω(n2 /m 4 ) edges. Proof. Let A be a set of n agents such that the size of each part of A P is either n/r or n/r. Let G G(A P,k) so that among all graphs in the set it has the fewest number of edges. Denote the parts in A P as A 1,A 2,...,A r. Assume a A i is a vertex of G that is part of every k-independent set. Notice that deg(a) = 0; otherwise any edge incident to a can be removed and a will still be part of every k-independent set, contradicting the minimality of E(G ). Hence, E(G ) = E(G a), and G a has no k-independent sets relative to the partition (A 1,A 2,...,A r ) of its vertex set, where A i = A i \ {a} and A j = A j for j i. Let us now bound the size of E(G a). Suppose that we wish to construct G a on A\{a} starting with an empty edge set. At this beginning stage, there are ( )( ) ( ) A X = 1 A 2 A... r k 1 k 2 k-independent sets. As we add each edge one by one, some k-independent sets are destroyed until all X of them are completely eliminated. If Y is the maximum number of k-independent sets that the addition of a single edge can destroy, then G a has at least X/Y edges. So let us determine Y. We consider two cases: (1) the added edge e is completely contained within A j for some j [r] or (2) e has one endpoint in A p and one endpoint in A q for some p,q [r],p q. For case (1), the maximum number of k-independent sets that can be destroyed is ( A 1 Y 1 = k 1 )( A 2 k 2 )... ( A j 1 k j 1 )( A j 2 k j 2 k r )( A j+1 k j+1 )... ( A r For case (2), the maximum number of k-independent sets that can be destroyed is Y 2 = ( A 1 ) ( A k 1... p 1 )( A p 1 )( A k p 1 k p 1 p+1 k p+1 ) ( A q 1 )( A q 1 )( A q+1 k q 1 k q 1 Since Y = max{y 1,Y 2 }, X/Y = min{x/y 1,X/Y 2 }. Now, X Y 1 = ( A j ) k j ( A j 2 k j 2 ) = A j 2 A j kj 2 k j k r ). ) ( k q+1... A r ) k r.

16 and X Y 2 = ( A p ( A p 1 k p 1 ) k q )( A q 1 k q 1 k p )( A q ) = A p A q. k p k q Both X/Y 1 and X/Y 2 are Ω(n 2 /m 4 ) because A j,a p,a q have Θ(n/r) agents and r m 1 while each k i is at most m. Thus, G a, and consequently G, has Ω(n 2 /m 4 ) edges. Theorem 7. Given P, {s i : i = 1,...,r} and A P, finding an informative agent with respect to P requires asking Ω(n 2 /m 4 ) questions in the worst case. Proof. LetA be thesetofnagentsdescribedinlemma8.consideranarbitrary strategyfor finding an informativeagent ofa. Suppose that there aresituations where the strategy will conclude that some agent is informative even though all the questions asked were unhelpful. Let G be the graph formed by these questions. According to Lemma 7, G must have the property described in its second condition. Thus, G G(A P,k). By Lemma 8, G has Ω(n2 /m 4 ) edges so this strategy will ask Ω(n 2 /m 4 ) questions in the worst case. Now, suppose the strategy will conclude that some agent is informative only when a helpful question has been asked. Assume that in the worst case it uses w questions. It follows that in situations where the strategy has already asked w 1 questions all of which are unhelpful, the wth question must be helpful and the two agents that are part of this question are informative. Thus, if the strategy is aware of the worst case bound w, it can avoid asking the wth question; the first w 1 unhelpful questions are sufficient for identifying an informative agent. Applying the reasoning we used in the previous case, w 1 Ω(n 2 /m 4 ). The theorem follows. Let L(n,Q) denote the fewest number of questions needed in the worst case to solve the agent labeling problem whose input is a well-formed instance (A, Q) with A = n. Similarly, let I(n,Q,s ) be the fewest number of questions needed in the worst case to solve the agent identification problem whose input is a wellformed instance (A,Q,s ) with A = n. In light of our earlier discussion about the difficulty of finding an informative agent in relation to that of the agent labeling and agent identification problems, we have the following result: Corollary 1. When D Q = ([m],q) has an uninformative partition, L(n,Q) Ω(n 2 /m 4 ). Additionally, when s is an informative type in some uninformative partition of D Q, I(n,Q,s ) Ω(n 2 /m 4 ). 6 Label or Detect Given (A,Q) where A = n and Q has order m, we now present an algorithm called LabelOrDetect that either outputs a consistent labeling of the agents in A or detects the presence of an uninformative partition in D Q using O(mn) questions. In the first part of the algorithm, a set of special agents are chosen.

17 All questions addressed to the special agents or about the special agents are asked during the course of the algorithm. The answers are then used to partition asubset ofthe agentsinto parts.the goalisto get toapoint whereeachpart has exactly one special agent because the algorithm intends to make a special agent represent the agents in its part. This seems valid because if special agent b i is in part A i, b i s answers are just like those of the agents in A i. In the second part of the algorithm, the answers to all the unanswered questions are then obtained by assuming that, for each part A i, the agents in it have the same type as b i. The agents in A are then partitioned according to the relation. If the number of equivalence classes is less than what is expected which is m the earlier assumption must be false. The algorithm concludes that it had encountered an uninformative partition. However, if the number of equivalence classes is m, the algorithm passes the answers to all the questions to the procedure Label, which then outputs a labeling f. Consider any two agents of A. Just like any two types, these two agents have four possible relationships based on how they answer queries about each other. Let A A, let A be a partition of the agents in A, and let b be an agent not in A. In our algorithm, we use refine(a,a,b) to denote the operation that partitions the agents in each part of A according to their relationship with b. The first part of the algorithm is done through the procedure AskAndPartition, shown in Figure 1. Initially, A and A are set to A and {A} respectively, and all agents are unmarked. The values q (a,a ) for each a,a A are set to 1 as an indication that the question addressed to a about a has not been asked. The algorithm then enters a while loop. At the beginning of each iteration, the parts in A may or may not contain special agents. For those parts A i that are missing a special agent, an agent b i A i is chosen, marked special and noted as new. All questions between b i and every agent in A b i are asked and then stored in q. Based on the answers, b i is used to refine A i b i and every other part of A later. If b i subdivided A i b i into two or more parts, b i is moved from A i to B because it is no longer obvious which part b i should belong to; otherwise, b i stays in A i. Notice that AskAndPartition is type-preserving. That is, if two agents have the same type and neither one was moved to B, they stayed in the same part of A throughout the execution of the procedure. Since there are only m types, the number of parts in A never exceeds m. Additionally, from one iteration to the next in the while loop of AskAndPartition, the number of parts in A either increases or stays the same. Keeping these observations in mind, we now prove some important lemmas about the procedure. Lemma 9. The algorithm AskAndPartition terminates. Proof. To prove the lemma, it suffices to show that the while loop in AskAndPartition ends. Let A (y) denote the set A at the beginning of the yth iteration of the while loop. Suppose that the sizes of the A (y) s are always increasing. We note that the while loop will never terminate in this case because at the beginning of each iteration there are always some parts that have no special agents. But this

18 AskAndPartition(A, Q) A A, A {A} B For every pair of agents a and a, initialize q (a,a ) to 1. Mark all agents in A as not special. while (some parts of A has no agent marked special) Denote the parts in A as A 1,A 2,...,A k. B new for each part A i in A A i {A i} if A i does not have an agent marked special choose an arbitrary agent b i A i, mark it as special, add it to B new ask all questions between b i and every other agent in A and record the answers in q if A i {b i} A i refine(a i b i,a i b i,b i) if A i contains two or more parts move b i from A i to B else A i {A i} endfor for each b i B new for each part A j that does not contain b i /* Questions between b i and A j have already been asked earlier.*/ A j refine(a j,a j,b i) endfor endfor A k i=1 A i, A k i=1 A i endwhile return( A,A,B,{q(a,a ) : a,a A}) Fig. 1. The procedure AskAndPartition.

19 will also mean that A (y) will exceed m at some point, a contradiction. So it must be the case that for some y, A (y) = A (y+1). Choose y so that it has the smallest possible value. Since A (y) had more parts than A (y 1), some parts of A (y) were assigned new special agents. But since A (y) = A (y+1), none of these special agents refined the partsof A (y). That is, none of them wereremovedfrom their part and moved to B. It follows that in the next iteration of the while loop, all parts in A (y+1) had special agents so the while loop had to terminate. Lemma 10. Let A f, A f, and B f denote the sets A, A and B at the end of the while loop in AskAndPartition. Either each part in A f consists of agents of the same type or D Q has an uninformative partition. Proof. Let A 1,A 2,...,A m be the parts in A f, and let b i be the special agent in A i for i = 1,...,m. Suppose that some part A i has two agents whose types are different. To prove the lemma, we will show that D Q has an uninformative partition. Let P = {P 1,...,P m,...,p m } such that (1) for 1 i m, P i contains the types of the agents in A i and (2) P m +1,...,P m are singleton sets containing the types not found in m j=1 P i. (When all the types can be found in m j=1 P i, we do not need P m +1,...,P m.) Since AskAndPartition is type-preserving, no two parts in P contain the same type. Additionally, because m i=1 P i = [m] and some P i contains at least two types, P is a partition of [m] whose number of parts is strictly less than m. Next, we claim that if b is a special agent in B, t(b) is an uninformative type with respect to P. Recall that this means that t(b) has the same relationship with each type in P i for i = 1,...,m. If P i containsonly one type, this is clearly true. If P i contains two or more types, this is also true because when b was first marked special and used to refine the parts of A, it had the opportunity to subdivide A i but it never did. Thus, t(b) is an uninformative type with respect to P. Consequently, the single type in P i is uninformative with respect to P for i = m +1,...,m. Let us now consider t(b i ) for i = 1,...,m. If there is some special agent b in B with t(b) = t(b i ), t(b i ) has to be uninformative with respect to P. So suppose no special agent in B has the same type as t(b i ). Then all agents of type t(b i ) are in A i. If there is another agent b i in A i such that t(b i ) t(b i), then the questions between b i and b i were answered in the same way as the questions between b i and every other agent of type t(b i ) because b i did not subdivide A i. This shows that t(b i ) has the same relationship with itself and t(b i ), and in general with every other type in P i. It also has the same relationship with every type in P j, j i, j = 1,...,m because when b i was first marked special and used to refine the parts of A, it had the opportunity to subdivide A i but again it never did. So once again t(b i ), for i = 1,...,m, is uninformative with respect to P. We have now proven that P is an uninformative partition. Figure 2 shows the algorithm LabelOrDetect. Theorem 8. The total number of questions asked in LabelOrDetect is O(mn). Furthermore, if the algorithm outputs a function f, the function is a consistent