1 The consecutive integer game

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A xel B orn

^∗

C or A.J. H urkens

^†

G erhard J. W oeginger

^†

The movie “Romanoﬀ and Juliet” (Universal International Studios, 1961) shows Peter Ustinov as the Prime Minister of the tiny, peaceful, sleepy nation of Concordia, only a few miles across somewhere in Europe. It is the time of the cold war, and Concordia is squeezed between East and West. Americans and Sovi- ets are ﬁercly determined to dominate the neutral country. Prime Minister Ustinov learns by chance from the American Ambassador that the Americans know about the latest top secret Soviet manoeuvre in this contest to control Concordia.

After a bit of reﬂection, Ustinov visits the Soviet Ambassador and tells him in a voice fraught with signiﬁcance: “They know.”

The Soviet Ambassador is not impressed: “We know they know.”

Ustinov goes back to the American Ambassador: “They know you know.”

The American Ambassador smiles: “But we know they know we know.”

Ustinov returns to the Soviet Ambassador: “They know you know they know.”

The Soviet Ambassador replies triumphantly: “We know they know we know they know!”

Once again Ustinov returns to the American Ambassador. Wearily he says:

“Well, they know you know they know you know.”

The American Ambassador repeats it after him, counting on his ﬁngers.

“What?” he suddenly cries in horror. “They know we know they know we know?”

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This is the third article — in a series of three — dedicated to the many variants and variations of the Freudenthal problem [10]. The common crux in all these problems is how to draw the right conclusions from strange conversations that mainly consist of“I-do-not-know”and“Now-I-do-know”and“I-know-that-you- know” statements. In two predecessor articles [4, 5] we have surveyed a number of knowledge propagation puzzles with short conversations of strictly bounded

∗Oberstufen-Realgymnasium Ursulinen, Leonhardstrasse 62, 8010 Graz, Austria.

†Email:{wscor|gwoegi}@win.tue.nl. Department of Mathematics and Computer Science, TU Eindhoven, P.O. Box 513, 5600 MB Eindhoven, The Netherlands.

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length. In the current article we consider knowledge propagation puzzles with longconversations, where the number of statements comes as a parameter. Typi- cally, the answer to such a problem varies with the length of the conversation. In all these problems it is crucial to recognize the diﬀerence between what is true and what a player knows to be true. Aumann [1] elaborated on this distinction by introducing the notion ofcommon knowledge, that is, facts that are true, that each player knows to be true, that each player knows that each player knows to be true, that each player knows that each player knows that each player knows to be true, that each player knows that each player knows that each player knows that each player knows to be true, and so forth.

The article is structured in the following way. We ﬁrst introduce and analyze three concrete knowledge propagation puzzles: The consecutive integer game, the arithmetic mean game, and the sum game. Then we discuss in detail a fairly general result on a fairly general game due to Conway and Paterson [8], and ﬁnally we review some of the history of the considered problems.

1 The consecutive integer game

This section is centered around the so-called consecutive integer game. The following paradoxical situation from the year 1953 (taken from page 4 of Littlewood [14]) might be one of the earliest predecessors of this game.

There is an indeﬁnite supply of cards marked 1 and 2 on opposite sides, and of cards marked 2 and 3, 3 and 4, and so on. A card is drawn at random by a referee and held between the players A and B, so that each sees one side only. Either player may veto the round, but if it is played the player seeing the higher number wins.

The point now is that every round is vetoed. If A sees a 1, the other side is 2 and he must veto. If he sees a 2, the other side is 1 or 3; if 1 thenBmust veto; if he does not thenAmust. And so on by induction.

In the year 1980, Van Emde Boas, Groenendijk, and Stokhof [9] then in- troduced and analyzed the following closely related consecutive integer game.

(Later, also David Gale discussed this game in detail, in his mathematical entertainments column [11] in the Mathematical Intelligencer.) Here is our rewording:

The teacher says to Xavier and Yvo: I have written a positive integer x on Xavier’s hat, and a positive integery on Yvo’s hat. You both see the number on the hat of the other, but you do not know the number on your own hat. Furthermore, I tell you that x andy are consecutive integers, which means {x,y} = {n,n+1} for some positive integern.

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1. Xavier and Yvo say: I don’t know my number.

. . . . . . . .

Then Xavier suddenly says: Aha. Now I know my number.

Determine x andy!

There is nothing special about the 71 rounds of conversation in this puzzle: In the more general consecutive integer game CIG(r) a corresponding conversation goes over r ≥ 0 rounds.

For r ≥ 0, we let A^r denote the set of integers n that are still alive after r rounds, that is the set of all integers n for which {x,y} = {n,n+1} is compatible withr rounds of I-don’t-know statements by Xavier and Yvo. We show by induction on r that A^r = {n : n ≥ r +1}. It is obvious that A⁰ = {n : n ≥ 1}. In the inductive step, consider the moment in time right after the (r −1)th I-don’t-know round: (a) If at that moment one of the players sees the number r written on the other hat, then he concludes from r −1 A^r−1 that he himself has r +1 on his own hat. (b) If at that moment one of the players seesr +with ≥ 1 written on the other hat, then he cannot decide for sure whether he has r+−1 orr ++1 on his own hat. Hence indeedA^r = {n : n ≥ r +1}. This completes the inductive argument, and yields the following theorem.

Theorem 1.1. (Van Emde Boas, Groenendijk, and Stokhof [9])

For any number r ≥0of rounds, the answer to problem CIG(r) is that Xavier has x = r+2written on his hat, whereas Yvo has y= r+1on his hat.

For the consecutive integer game CIG(71) displayed above we derive x = 73 and y = 72. Now let us turn to another problem, which we will call the distinct integers game. It is an easy variation on the consecutive integer game:

The teacher says to n ≥ 2 players P₁, . . . ,P_n: For k = 1, . . . ,n, I have written a positive integer a_k on the hat of player P_k. You all see the numbers on the hats of the others, but you do not know the number on your own hat. Furthermore, I tell you that thesenintegers

a1, . . . ,a_n are all pairwise distinct. Can you ﬁnd out who has the

smallest number written on his hat?

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1. Players P1, . . . ,P_n say: I don’t know whose number is smallest.

2. Players P₁, . . . ,P_n say: I don’t know whose number is smallest.

3. Players P1, . . . ,P_n say: I don’t know whose number is smallest.

. . . . . . . .

13. PlayersP1, . . . ,P_n say: I don’t know whose number is smallest.

14. PlayersP₁, . . . ,P_n say: I don’t know whose number is smallest.

Then suddenly then−1 players P2, . . . ,P_n all say: Aha. Now I know that player P₁ has the smallest number. To which player P₁ reacts:

Aha. Then I know the numbera1 on my hat.

Determine the value ofa₁!

In the general distinct integers game, DIG(r) this conversation goes over r ≥ 0 rounds instead of the 14 listed rounds. The exact number n ≥ 2 of players is irrelevant for the answer. The analysis of the consecutive integer game CIG(r) can easily be adapted and carried over to DIG(r): If the game ends after r rounds, then the player with the smallest number has r +1 written on his hat. None of the other players is able to deduce the number on his hat. For the distinct integers game DIG(14) displayed above we derive a1 = 15.

2 The arithmetic mean game

The following puzzle was posed in 1970 by David Silverman [19] in the problem corner of the Journal of Recreational Mathematics. Silverman’s original version is embedded into a sentimental story around three identical triplet brothers who all fall in love with the beautiful Wanda. Here is our sober rewording:

The teacher says to Xavier, Yvo, and Zeno: I have written a positive integer x on Xavier’s hat, a positive integer y on Yvo’s hat, and a positive integer z on Zeno’s hat. You all can see the numbers on the hats of the others, but you do not know the number on your own hat.

Furthermore, I tell you that one of the three integers is the arithmetic mean of the other two; that is, either x = ¹₂(y +z) or y = ¹2(x +z) or z = ¹₂(x+y) holds.

1. Xavier, Yvo, and Zeno say: I don’t know my number.

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. . . . . . . .

Then after several rounds Zeno suddenly says: Aha. Now I know my number.

What is the value ofz, if Zeno sees x =4 andy = 6 on the hats of his friends?

The general case of this arithmetic mean game with numbers x,y,z on the three hats is denoted Mean(x,y,z). For r ≥ 0, we let set Ar contain all possible pairs (a,b) of positive integers a and b, that are compatible with r rounds of I-don’t- know statements, if they are seen on the hats of the other two players. The pairs in Ar are said to be alive after r rounds. Furthermore for r ≥ 0 we deﬁne Dr = A^r− A^r+1 as the set of all pairs that allow one to determine the own number after preciselyr rounds of I-don’t-know statements.

Under which conditions is a pair (a,b) with a < b contained in a set Ar? Well, if one sees a < bwritten on the hats of the others, then there are only three potential candidates for the own number:

• The ﬁrst candidate is 2a−b; it can be ignored a priori if 2a ≤b.

• The second candidate is ¹₂(a+b); it can be ignored a priori ifa+bis odd.

• The third candidate is 2b−a; it can never be ignored a priori.

The pair (a,b) lies inA^r, if afterr−1 rounds at least two of these three candidates are still a feasible option for the number on the own hat. And a candidate x is a feasible option, if and only if (x,a) and (x,b) are both in Ar−1: Then the player with the numberaon his hat sees (x,b) and the player withbon his hat sees (x,a), and they both have nothing to tell except I-don’t-know. Let us summarize these observations.

Lemma 2.1. A pair (a,b)with a b lies in setA^rwith r ≥1if at least two out of the following three statements are satisﬁed:

• (2a−b,a) and(2a−b,b)are both inAr−1

• (¹₂(a+b),a) and(¹₂(a+b),b) are both inA^r−1

• (2b−a,a) and(2b−a,b)are both inAr−1

Furthermore, the set A⁰ contains all pairs (a,b) of positive integers. The pairs (a,a) occur in setA⁰, but in none of the other setsAr with r ≥1.

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Any arithmetic mean game Mean(x,y,z) can be fully analyzed by repeated application of Lemma 2.1. In the rest of this section, we will provide an explicit analysis of Mean(x,y,z) that is based on the following crucial definitions. For two positive integers a and bwith a < b, we define u = u(a,b) and k = k(a,b) as the unique integers that satisfyb−a =u·2^k whereuis odd. We define

R(a,b) = k +(a−1)/u.

Fora >b, we set R(a,b) =R(b,a). Finally, we setR(a,a) = 0 for alla.

Lemma 2.2. For positive integers a and b,(a,b) ∈ Dr if and only if r =R(a,b).

Proof. The proof is done by induction on r. For r = 0, observe that (a,b) ∈ D⁰ holds, if and only ifa= bholds, or if 2a≤ bwith odda+b. But these are exactly the cases withR(a,b) =0.

Let us turn to the inductive step for r ≥ 1. For the (if)-part, consider a andb withR(a,b) = r. We distinguish three cases. In the ﬁrst case, assume a < b < 2a anda+beven. Let u = u(a,b) and k = k(a,b) be deﬁned as above. Observe the following.

R(2a−b,a) = k+(2a−b−1)/u = R(a,b)−2^k

R(2a−b,b) = (k +1)+(2a−b−1)/u = R(a,b)+1−2^k R((a+b)/2,a) = (k −1)+(a−1)/u = R(a,b)−1

R((a+b)/2,b) = (k −1)+(a+b−2)/(2u) = R(a,b)−1+2^k⁻¹ R(a,2b−a) = (k +1)+(a−1)/u = R(a,b)+1

R(b,2b−a) = k+(b−1)/u = R(a,b)+2^k

SinceR(a,2b−a) andR(b,2b−a) both are strictly greater thanr−1, the inductive assumption implies that the corresponding pairs are still alive after r −1 rounds, and hence candidate 2b−ais still alive afterrrounds. SinceR((a+b)/2,b) > r−1 andR((a+b)/2,a) =r−1, the inductive assumption yields that candidate ¹₂(a+b) can be ignored afterr−1 rounds (but it could not have been ignored at any earlier moment in time). SinceR(2a−b,a)≤ r−1, the inductive assumption yields that also the candidate 2a−bcan be ignored. Summarizing, we conclude that 2b−a is the only surviving candidate, which yields (a,b) ∈ Dr.

In the second case 2a ≤ band a+bis even. Then 2a−bcan be ignored right from the beginning and ¹₂(a+b) can be ignored after r −1 rounds. Since 2b−a is the only surviving candidate, we conclude (a,b) ∈ Dr. Finally, in the third case a < b < 2a anda+bis odd. Then k = 0, and candidate ¹₂(a+b) is ignored right from the beginning. Since R(2a−b,a) = r −1 and R(2a− b,b) = r, candidate

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2a−b will be ignored after r −1 rounds. This once again implies that 2b−a is the only surviving candidate, and that (a,b) ∈ Dr.

For the (only if)-part, consider a pair (a,b) ∈ Dr. The inductive assumption yields R(a,b) ≥ r. Suppose that R(a,b) ≥ r +1. If a+ b is even, the inductive assumption yields that the four pairs ((a + b)/2,a), ((a + b)/2,b), (a,2b − a), (b,2b−a) all are inAr−1. Hence, neither candidate ¹₂(a+b) nor candidate 2b−a can be ignored; a contradiction. If a + b is odd, then k = 0. The inductive assumption yields that the four pairs (2a−b,a), (2a−b,b), (a,2b−a), (b,2b−a) all are in Ar−1. Hence, neither candidate 2a − b nor candidate 2b − a can be ignored; another contradiction. This completes the proof of the lemma.

Note that all cases in the above inductive proof lead to the largest number 2b− a as the only surviving candidate. This yields the following summarizing theorem.

Theorem 2.3. In the arithmetic mean game Mean(x,y,z) with x ≤ y ≤ z, the player with the largest number z = 2y −x will announce his number after R(x,y) rounds.

Hence in the original puzzle by David Silverman with x = 4 and y = 6, the number on Zeno’s hat is 8, which he recognizes after R(4,6) = 4 rounds of I- don’t-know statements.

3 The sum game

In February 2004, Andy Liu [15] posed the following cute puzzle in the problem corner of the journal Math Horizons. Note that in Liu’s puzzle the conversation is round-robin, which is diﬀerent from the preceding sections where conversations consisted of simultaneous announcements.

The teacher says to Xavier, Yvo, and Zeno: I have written a positive integer xon Xavier’s hat, a positive integeryon Yvo’s hat, and a positive integer z on Zeno’s hat. You all see the numbers on the hats of the others, but you do not know the number on your own hat. Further- more, I tell you that one of the three integers is the sum of the other two.

1. Xavier says: I don’t know my number.

2. Yvo says: I don’t know my number.

3. Zeno says: I don’t know my number.

Then Xavier suddenly says: Aha. The number on my hat is x = 50.

Determiney andz!

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For r ≥ 0, let Ar denote the set of all triples (x,y,z) that are compatible with the ﬁrst r rounds of I-don’t-know statements, and letDr = Ar − A_r₊1 denote the set of all triples for which the game terminates after exactly r rounds. Clearly, A0

contains all triples (x,y,z) of positive integers, for which one coordinate equals the sum of the other two coordinates. If in statement #1 Xavier seesy z, then he cannot decide whether x = y +z or whether x = |y−z|. But if he seesy = z, then he deduces x =2y. This yields

D0 = {(2t,t,t) : tpositive integer}.

By similar reasoning, we derive that

D1 = {(t,2t,t), (2t,3t,t) : tpositive integer}

D2 = {(t,t,2t), (2t,t,3t), (t,2t,3t), (2t,3t,5t) : tpositive integer} D3 = {(3t,2t,t), (4t,3t,t), (3t,t,2t), (4t,t,3t), (5t,2t,3t),

(8t,3t,5t) : t positive integer}

The triple (50,20,30) is the unique triple in D3 with x-coordinate 50. Therefore, Xavier seesy =20 and z =30 on the other hats.

In the rest of this section, we will consider a generalization of Andy Liu’s problem to conversations with simultaneous I-don’t-know announcements:

The teacher says to Xavier, Yvo, and Zeno: I have written a positive integer x on Xavier’s hat, a positive integer y on Yvo’s hat, and a positive integer z on Zeno’s hat. You all can see the numbers on the hats of the others, but you do not know the number on your own hat.

Furthermore, I tell you that one of the three integers is the sum of the other two.

. . . . . . . .

Then after several rounds Zeno suddenly says: Aha. Now I know my number.

What is the value of z, if Zeno sees x = 21 andy = 33 on the hats of his friends?

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The general sum game with arbitrary positive integers x,y,z on the hats will be denoted Sum(x,y,z). Similarly as in the preceding sections, we let set Ar with r ≥ 0 contain all pairs (a,b) of positive integers aand bthat are compatible with r rounds of I-don’t-know statements, if one sees them written on the hats of the others. And we deﬁne Dr = Ar − Ar+1 as the set of all pairs that allow one to identify the number on the own hat after precisely r rounds of I-don’t-know statements.

Lemma 3.1. A pair (a,b) of positive integers lies in set A^r with r ≥ 1 if the following two statements are satisﬁed:

• (a+b,a)and(a+b,b) are both inA^r−1

• (|a−b|,a) and(|a−b|,b) are both inAr−1

Furthermore, the set A⁰ contains all pairs(a,b)of positive integers.

Let us deﬁne values R(a,b) for all pairs of positive integers a and b: We set R(a,a) = 0 for all a, and we let R(a,b) = R(b,a) for all a > b. For a < b, we recursively deﬁne

R(a,b) = 1+ R(b−a,a).

Since the greatest common divisor of integers a and b satisﬁes the relation gcd(a,b) = gcd(b− a,a), our function R(a,b) essentially counts the number of steps in the execution of a variant of the Euclidean algorithm. And as for all other variants of the Euclidean algorithm, there is no nice way of describing R(a,b) in closed form.

Lemma 3.2. For positive integers a and b,(a,b) ∈ D^r if and only if r = R(a,b).

Proof. Fora< b, the four pairs listed in Lemma 3.1 become (a+b,a), (a+b,b), (b −a,a), and (b −a,b). Since the first pair satisfies R(a +b,a) = 1+R(a,b), since the second pair satisfies R(a+b,b) = 1+R(a,b), and since the fourth pair satisfies R(b−a,b) = 1+R(b−a,a) = R(a,b), we see that the statement that we want to prove mainly depends on the third pair (b−a,a). The proof is completed by a straightforward induction on r.

We summarize our observations in the following theorem.

Theorem 3.3. In the sum game Sum(x,y,z) with x ≤ y ≤ z, the player with the largest number z = x +y will announce his number after R(x,y)rounds.

Finally, let us discuss our introductory example where Zeno sees x = 21 and y = 33. In evaluatingR(21,33), we are lead step by step to the ﬁve pairs

(21,33) → (12,21) → (9,12) → (3,9) → (3,6) → (3,3)

Therefore Zeno recognizes afterR(21,33) = 5 rounds of I-don’t-know statements that he is participating in the game Sum(21,33,54), and that the number on his hat is 54.

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4 The Conway-Paterson game

There are many classes of knowledge propagation puzzles that ﬁt the theme of this article. One of the most general variations on this theme is due to John Horton Conway and Mike Paterson [8].

The teacher says ton ≥ 2 players P1, . . . ,P_n: Fork = 1, . . . ,n, I have written a non-negative integera_k on the hat of player P_k. You all see the numbers on the hats of the others, but you do not know the number on your own hat. Furthermore, I have writtenmintegers s1, . . . ,s_m on the blackboard. You can all see these m numbers on the blackboard, and I tell you that one of them equalsn

k=1a_k.

1. The players P₁, . . . ,P_n all say: I don’t know my number.

. . . . . . . .

Suddenly one of the players says: Aha. Now I know my number.

Under which conditions on the n+m numbersa₁, . . . ,a_n and s₁, . . . ,s_m will such a game indeed terminate?

For r ≥ 0, letAr denote the set of alln-tuples (a1, . . . ,a_n) that are compatible with the ﬁrst r rounds of I-don’t-know statements. Then the set A⁰ contains all n-tuples with non-negative integer coordinates that add up to one of the sums s1, . . . ,s_m. We say that two n-tuples (a1, . . . ,a_n) and (b1, . . . ,b_n) are k-twins, if they agree in all coordinates with the only exception of the kth coordinate; in other words, a_i = b_i holds for all i k, whereas in the kth coordinate we have a_k b_k.

Lemma 4.1. Let r ≥ 1. An n-tuple (a₁, . . . ,a_n) ∈ Ar−1 also lies in set Ar, if and only if for every index k with 1 ≤ k ≤ n the set A^r−1 contains some k-twin of (a1, . . . ,a_n).

Proof. Consider some n-tuple (a1, . . . ,a_n) is inAr−1. If for some indexk there is no appropriate k-twin in A^r−1, then player P_k deduces that his number equals a_k and terminates the game. But if for every index k, the set A^r−1 contains a k-twin of (a1, . . . ,a_n), then no player can eliminate (a1, . . . ,a_n) as a feasible option.

Let us look at some special cases with n = 2 players and m = 3 sums. As a ﬁrst example, we will demonstrate that the two-player game with sums 4, 9, and 15 will always terminate after at most 10 rounds.

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0 1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9

x x

x x x

x x

Figure 1: Every Conway-Paterson game with (s1,s2,s3) =(4,9,15) terminates.

The setA⁰ contains all points (x,y) with non-negative integer coordinates that satisfy x +y ∈ {4,9,15}. The points with x ≥ 10 or y ≥ 10 disappear after the first round. The set A¹ is depicted in Figure 1. Note that A¹ only contains one point with x-coordinate 5 and only one point with y-coordinate 5. According to Lemma 4.1 these two points do not survive the second round; the reader should verify that all other points stay alive. The third round kills off the points with x- coordinate ory-coordinate equal 4, the fourth round kills the coordinate 0, and the fifth to tenth round kill points with coordinates 9, 6, 3, 1, 8, and 7, respectively.

Now the only surviving point inA¹⁰ is (2,2). Hence, such a game terminates after at most 10 rounds.

Next, let us consider the two-player game with sums 5, 9, and 15. The set A⁰ contains all points (x,y) with non-negative integer coordinates that satisfy x+y ∈ {5,9,15}. The ﬁrst round kills all points with a coordinate greater than 9, and the resulting set A¹ is depicted in Figure 2. For every point inA¹, the horizontal and the vertical line through that point both contain another point in A¹. Therefore Ar = A¹ for all r ≥ 2. The game either terminates right at the beginning (if one of the players sees 10, 11, 12, 13, 14, or 15 written on the other hat), or it stalls for ever.

Lasry, Morel, and Solimini [13] have characterized all triples (s1,s2,s3) for which the two-player game must terminate. It is instructive to compare the condition in the following lemma to the two cases (s₁,s₂,s₃) = (4,9,15) and (s1,s2,s3) =(5,9,15) discussed above.

Theorem 4.2. (Lasry, Morel, and Solimini [13])

The Conway-Paterson game with n = 2players and m = 3 sums must terminate,

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0 1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9

x x

x x x

x x

x x x

x x

Figure 2: A stalled situation in a Conway-Paterson game with (s1,s2,s3) = (5,9,15).

whenever the sums s1 < s2 < s3on the blackboard satisfy the following condition:

0 < gcd(s3 −s2, s2 −s1) < s3− s2 −s1

In all other cases, the game may stall for ever (if the values of a₁ and a₂ are appropriately chosen).

In the rest of this section, we discuss the main results of Conway and Paterson [8]. A non-empty, ﬁnite set A of n-tuples is called stalled, if for each of its ele- ments and for each coordinatek it contains an appropriatek-twin of the element.

Lemma 4.3. Let n ≥ 1, and let A be a non-empty, ﬁnite set of n-tuples over the non-negative integers. If A is stalled, then the n-tuples in A must have at least n+1distinct coordinate sums.

Proof. By induction onn. The case n =1 is immediate. In the inductive step, let a^∗ denote the smallest value that occurs as nth coordinate in ann-tuple in A. Let A contain all n-tuples in A with nth coordinate equal to a^∗. Let Bdenote the set of all (n−1)-tuples (b1, . . . ,b_n−1) for which (b1, . . . ,b_n−1,a^∗) is in A.

Since A is stalled, also B is stalled. By the inductive assumption, the vectors in B have at least ndistinct coordinate sums. Consequently, also the vectors (b1, . . . ,b_n−1,a^∗) in A have at least n distinct coordinate sums. Finally, consider an n-tuple in A with largest coordinate sum. Since A is stalled, A contains an n-twin of thisn-tuple. This n-twin has a coordinate sum that is strictly larger than the coordinate sums of alln-tuples inA.

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If an instance of the Conway-Paterson game fails to terminate, then it must reach a point — say in roundr— where it becomes impossible to eliminate further elements from the set A_r. By Lemma 4.1 this set A_r must be stalled, and by Lemma 4.3 it must have at least n+ 1 distinct coordinate sums. This yields the following.

Theorem 4.4. (Conway and Paterson [8])

Any Conway-Paterson game with n ≥2players and m ≤n sums on the blackboard must eventually terminate.

We remark that the bound n+1 in Lemma 4.3 is best possible and cannot be increased to any higher value, as illustrated by the (stalled) set of alln-tuples all of whose coordinates are either 0 or 1. The same example demonstrates that also the statement in Theorem 4.4 is best possible in the following sense: For any n ≥ 2, there exist Conway-Paterson games with n players and m = n +1 sums on the blackboard that do not terminate. The game where every player has 0 or 1 written on his hat and where the blackboard contains the values 0,1,2, . . . ,n will never terminate.

The above analysis can easily be extended to the variant where the n players make there announcements in a round-robin fashion: First player P1 announces that he doesn’t know, then player P2 announces that he doesn’t know, then player P₃, and so on. This round-robin variant is actually the game that was analyzed in the original article [8].

5 History lesson: Muddy faces and cheating wives

One of the oldest puzzles involving common knowledge and knowledge propagation (as discussed in this article) is the so-called Muddy Children puzzle: In a school class with n ≥ 3 very intelligent, perceptive, and truthful children, one morning k ≥ 2 of the children come to school with some mud on their forehead.

Everybody can see the mud of the others, but no one can see the mud on his own forehead. The teacher says: “We are going to play a game. This beeper is going to sound every10seconds. If you know that your forehead is dirty, then please raise your hand after the beep.” The game commences, but after several minutes nothing has happened. The teacher says: “I think I should give you a hint. At least one of you has mud on his forehead.” The game starts again, and after several rounds all muddy children raise their hands. How did this happen?

What new information was gained from the teacher’s hint? After all, every child already knew that there were some muddy foreheads. . . . Of course, the muddy children puzzle is just a simple special case of the Conway-Paterson game:

Every child has the number 0 (no mud) or 1 (mud) written on its forehead. Before

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the hint the sum of all n numbers may be any of the n+1 integers 0,1,2, . . . ,n.

After the hint, everybody knows that the sum is one of the integers 1,2, . . . ,n.

And when the beeper sounds for thekth time, then allk muddy children will raise their hands.

One variant of this puzzle goes back to the year 1832, when the German writer and translator Gottlob Regis (1791-1854) completed his French-to-German translation [18] of Gargantua and Pantagruel by François Rabelais. On page 103, Regis comments on his translation of the French “pincer sans rire” by the Ger- man“ungelacht pfetz ich dich”:

Ungelacht pfetz ich dich. Gesellschaftsspiel. Jeder zwickt seinen rechten Nachbar an Kinn oder Nase; wenn er lacht, giebt er ein Pfand.

Zwei von der Gesellschaft sind nämlich im Complot und haben einen verkohlten Korkstöpsel, woran sie sich die Finger, und mithin denen, die sie zupfen, die Gesichter schwärzen. Diese werden nun um so lächerlicher, weil jeder glaubt, man lache über den anderen.

In English, the preceding paragraph would roughly read as follows: I pinch you without laughing. Parlor game. Everybody pinches his right neighbor into chin or nose; if one laughs, one must give a pledge. Two in the round have secretely blackened their ﬁngers on a charred piece of cork, and hence will blacken the faces of their neighbors. These neighbors make a fool of themselves, since they both think that everybody is laughing about the other one.

This cruel game is also mentioned in the book [17] published in 1883 by William Wells Newell, who gathered the melodies, rules, and movements of games and songs that were popular with children of those times. On page 137, in the chapter on “Laughter games”, we ﬁnd the following paragraph:

In a Swiss game, this performance is complicated by a jest. Each child pinches his neighbor’s ear; but by agreement the players blacken their ﬁngers, keeping two of the party in ignorance. Each of the two victims imagines it to be the other who is the object of the uproarious mirth of the company.

Finally in 1935, two clear mathematical formulations of the muddy children problem popped up in the Mathematics literature [2, 6]. Werner E. Buker [6] from the Pittsburgh Public Schools posed the following variant with n = 3 persons in the journal School Science and Mathematics. He wrote: Here is a puzzle that is going the rounds. I don’t know just who the originator was, but it’s not so bad, and I thought some of the readers of School Science and Mathematics who do not know it might like to try it.

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A king blindfolded three of his subjects and touched each on the forehead. They knew that the ﬁnger touching them might or might not be covered with lampblack. When the blindfolds were removed, they were given instructions to start whistling if they saw one or more black spots. As soon as any one of them could ﬁgure out whether he himself had lampblack on his forehead, he was to stop whistling.

There were spots on all three foreheads, and presently A stopped whistling. How did he know he had lampblack on his forehead?

A solution to Buker’s problem appeared in [7]. Around the same time, Albert A.

Bennett [2] from Brown University formulated the general variant of the muddy children problem in the problem section of the American Mathematical Monthly:

A car with n > 2 passengers of diﬀerent speeds of mental reaction passes through a tunnel, and each passenger acquires unconsciously a smudge of soot upon his forehead. Suppose that each passenger

(1) laughs and continues to laugh as soon as and only so long as he sees a smudge upon the forehead of a fellow passenger;

(2) can see the foreheads of all his fellows;

(3) reasons correctly;

(4) will clean his own forehead when and only when his reasoning forces him to conclude that he has a smudge;

(5) knows that (1), (2), (3) and (4) hold for each of his fellows.

Show that each passenger will eventually wipe his own forehead. (For the case of n = 3, this has been proposed in conversation by Dr.

Church of Princeton.)

A better formulation of condition (5) might have been to state that each passenger knows that (1), (2), (3), (4), and also (5) hold for each of his fellows. Dr. Church of Princeton should of course be Alonzo Church (although we do not have formal evidence for this). Two nice and complete solutions to Bennet’s problem appeared two years later in the same journal [3].

In 1953, John Littlewood (on page 3 of [14]) refered to the n = 3 case of the muddy children problem already as awell-known puzzle that swept Europe a good many years ago and in one form or another has appeared in a number of books:

Three ladies A, B,C in a railway carriage all have dirty faces and are all laughing. It suddenly ﬂashes on A: Why doesn’t B realize C is laughing at her? — Heavens! I must be laughable. (Formally: If I, B,

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am not laughable, thenC has nothing to laugh at. Since Bdoes not so argue, I, A, must be laughable.)

This is genuine mathematical reasoning, and surely with minimum material. But further, which has not got into the books so far as I know, there is an extension, in principle, to n ladies, all dirty and all laughing. There is an induction: In the (n+1)-situation Aargues: If I am not laughable, then B,C, . . .would constitute ann-situation and B would stop laughing, but does not.

Since these times, the muddy children problem has become a frequent guest in puzzle corners and problem collections. The children sometimes turn into philoso- phers, prisoners, students, or monks, and the muddy foreheads sometimes turn into painted faces, black crosses, green stamps, red hats, or blue blobs; see for instance Stewart [20]. One particularly popular version of the muddy children puzzle deals with unfaithful wives. This version showed up in 1958 in the book [12] by George Gamow and Marvin Stern, who state in a footnote that it was com- municated to them by Dr. Victor Ambarzuminian (which might be a misspelling of the astrophysicist Viktor Amazaspovich Ambarzumian).

The great Sultan Ibn-al-Kuz was very much worried about the large number of unfaithful wives among the population of his capi- tal city. There were forty women who were openly deceiving their husbands, but, as often happens, although all these cases were a mat- ter of common knowledge, the husbands in question were ignorant of their wives’ behavior.

In order to punish the wretched women, the sultan issued a proclamation which permitted the husbands of unfaithful wives to kill them, provided, however, that they were quite sure of the inﬁdelity. The proclamation did not mention either the number or the names of the wives known to be unfaithful; it merely stated that such cases were known in the city and suggested that the husbands do something about it. However, to the great surprise of the entire legislative body and the city police, no wife killings were reported on the day of the proclamation, or on the days that followed. In fact, an entire month passed without any result, and it seemed the deceived husbands just did not care to save their honor.

“O Great Sultan,” said the vizier to Ibn-al-Kuz, “shouldn’t we announce the names of the forty unfaithful wives, if the husbands are too lazy to pursue the cases themselves?”

“No,” said the sultan. “Let us wait. My people may be lazy, but they are certainly very intelligent and wise. I am sure action will be taken very soon.”

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And, indeed, on the fortieth day after the proclamation, action suddenly broke out. That single night forty women were killed, and a quick check revealed that they were the forty who were known to have been deceiving their husbands.

“I do not understand it,” exclaimed the vizier. “Why did these forty wronged husbands wait such a long time to take action, and why did they ﬁnally take it on the same day?”

The sultan then presents the ﬁrst three steps of an inductive argument, by dis- cussing the situation of the three husbands Abdula, Hadjibaba, Faruk, and their respective wives. The vizier is deeply impressed by this.

“O Great Sultan,” exclaimed the vizier, “you have certainly opened my eyes on that problem. Of course, if there were four un- faithful wives, each of the four wronged husbands would reduce the case to that of three and not kill his wife until the fourth day. And so on, and so on, up to forty wives.”

Note that the story is somewhat fuzzy about the numbering of days and nights:

Action does not break out on the day of the proclamation (day 0), nor one day later (day 1), nor on any of the following thirty-eight days, but only on the fortieth day after the proclamation (day 40). Since the counting starts with 0, there actually should be forty-one unfaithful wives. Gamow and Stern use this to construct a nice punch line with deadly consequences for the vizier’s wife:

“I am glad,”said the sultan,“that you ﬁnally understand the situ- ation. It is nice to have a vizier whose intelligence is so much inferior to that of the average citizen. But what if I tell you that the reported number of unfaithful wives was actually forty-one?”

Moses, Dolev, and Halpern [16] use cheating husbands to illustrate the sub- tle relationship between knowledge, communication, and action in a distributed environment. They discuss a number of entertaining variants with asynchronous channels, weakly synchronous broadcast, delayed communication, and ring-based communication.

6 Instead of a conclusion

This article discussed and surveyed a number of common knowledge problems.

We will complete it with a puzzle (taken from Gamow and Stern [12]) in which it is crucial that knowledge doesnot propagate.

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During the Nazi occupation of France four passengers were riding in a compartment of a train rolling from Paris to Marseille: A young and very good-looking girl, an old lady, an officer of the German occupation forces, and a middle-aged Frenchman of indefinite profes- sion. None of them knew any of the others, and no conversation was started while the train was moving south. As the train entered a tunnel, the electric lights in the car failed, and for several minutes the four passengers were sitting in complete darkness. Suddenly there was the sound of a kiss, followed by the sound of the impact of somebody’s fist against somebody’s face.

When the train came out of the tunnel, the positions of the passengers were unchanged, but the German oﬃcer had a large bruise under his eye. The old lady thought: “Serves him right. French girls know how to defend themselves against the Boches.” The young girl thought: “This German has a strange taste. Instead of kissing me, he kissed the old hag. Well, he got what he deserved.” The German held his hand against his injured eye, and thought: “The Frenchman tried to kiss the girl, and when she struck out in the darkness, she accidentally hit me in the face.”

The problem is to ﬁnd out what the Frenchman thought, and what actually had happened.

To make things more exciting, we have encoded the solution: ecaf s’namreG eht ni ti dehsams dna tsif nwo sih dessik dah namhcnerF ehT.

References

[1] R.J. Aumann(1976). Agreeing to disagree.Annals of Statistics 4, 1236–1239.

[2] A.A. Bennett(1935). Problem No. 3734.American Mathematical Monthly 42, 256.

[3] A.A. Bennett, E.P. Starke, and G.M. Clemence (1937). Solution to problem No. 3734.American Mathematical Monthly 44, 333-334.

[4] A. Born, C.A.J. Hurkens, and G.J. Woeginger (2006). The Freudenthal problem and its ramiﬁcations (Part I).Bulletin of the EATCS 90, 175–191.

[5] A. Born, C.A.J. Hurkens, and G.J. Woeginger (2007). The Freudenthal problem and its ramiﬁcations (Part II).Bulletin of the EATCS 91, 189–204.

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[7] W.E. Buker, R. Wood,and O.B. Rose (1935). Solution to science question 686: A puzzler for the thinkers.School Science and Mathematics 35, 429.

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[11] D. Gale(1991). Mathematical Entertainments: Paradoxes and a pair of boxes. The Mathematical Intelligencer 13, 31–33.

[12] G. Gamow andM. Stern(1958).Puzzle-Math. Macmillan, London.

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[14] J.E. Littlewood(1953).A Mathematician’s Miscellany. Methuen, London.

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A case study of knowledge, action, and communication. Distributed Computing 1, 167–176.

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[18] G. Regis(1832).Meister Franz Rabelais der Arzeney Doctoren Gargantua und Pan- tagruel, aus dem Französischen verteutscht, mit Einleitung und Anmerkungen, den Varianten des zweyten Buches von 1533, auch einem noch unbekannten Gargantua, herausgegeben durch Gottlob Regis. Barth, Leipzig.

[19] D.L. Silverman (1970). Problem No. 124: The triplets. Journal of Recreational Mathematics 3, 105.

[20] I. Stewart (1998). Mathematical recreations: Monks, blobs, and common knowledge.Scientiﬁc American(August 1998), 96–97.