2.3 The Muddy Children Puzzle
2.3.1 The Original Version
The puzzle goes like this:nchildren are playing in the mud andkof them have dirty foreheads. Each child can see all the other foreheads except for his own. Their father appears at the door and tells them that at least one of them has a dirty forehead. Then he asks if any one knows that it is him that has a dirty forehead, the children think and they all, simultaneously, answer no. Then having heard all the no answers of other children, they think again and again answer no. The question is after how many rounds of no answers, each dirty child will come to know that he is dirty? We will prove in this section, that the answer isk−1, that is afterk−1rounds of no answers, all the dirty children will know that they are dirty.
We encode the puzzle in an epistemic system(M, Q,{fA}A∈A). The set of agentsAincludes the
children, that is
{C1,· · · , Cn} ⊆ A
We assume that the firstkchildren C1,· · ·, Ck for 1 ≤ k ≤ nare the dirty ones. The module M
includes all possible initial propositions about which child is dirty. So we have a proposition that says no child is dirty, and thus corresponds to the situation where no child is dirty. We denote this proposition ass∅. Similarly we have a proposition that says only child one is dirty, denoted ass{C1},
which corresponds to the situation where only child one is dirty and so on for all the children. So we have2npossible propositions, denoted ass
βwhere the subscriptβis the set of those children that have
mud on their forehead, that is
β ⊆ A
For example the propositions{C1,...,Cn}says that all the children are dirty. So among our propositions
are the two extreme proposition: s∅ ands{C1,...,Cn}, but the proposition that is true in the real case is
s{C1,...,Ck}, that stand for our assumption that children 1 tokare dirty.
We have to set the appearance maps for each child about all this proposition. That is we have to see how does each proposition appear to each child. Since the children cannot see their own foreheads (which might either be dirty or not), in each proposition, they think they might be dirty or not. In other words, in each proposition they think they might be in the set of dirty children or not. This is encoded as
fCMi(sβ) =sβ\{Ci}∨sβ∪{Ci}
wheresβ\{Ci} is where the childCi is not dirty andsβ∪{Ci} is where he is dirty. Each of our propo-
sitions correspond to an epistemic situation, but they also satisfy some facts, that is the facts that correspond to the situation described by the proposition. For example propositions∅corresponds the
fact that ‘no child is dirty’. We denote this fact byD∅and set an order between the proposition and its
corresponding fact, that is
s∅ ≤D∅
Similarly for the other propositions, we assume Di denotes the fact that the i’th child has a dirty
forehead, hence we have
sβ ≤Di for allCi ∈β
and the set of our facts will be
{D∅} ∪ {Di∈M |Ci∈ A} ⊆Stab(Q)
which is a subset of the set of all facts of our epistemic system, that is the stabilizer ofQ.
Now we start to encode the dynamic part of the puzzle, that is the communication actions that happen between the father and the children. The first of these happens when the father tells the children
that at least one child is dirty. We denote this action by q0 and assume that it is an element of the
quantale
q0 ∈Q
Since this is a public action, all the children hear it, so it appears as it is to all the children:
fCQ
i(q0) =q0 for all1≤i≤n
The kernel of this action, is the set of all propositions (epistemic or fact) that it cannot be applied to, that is the proposition that says no child is dirtyD∅, and everything that implies it:
ker(q0) =↓D∅
Note that the propositions∅is included in the kernel sinces∅ ≤D∅.
Each round of no answers of children is also a communication action between the children. Since they all announce the same proposition, that is the no answer, we denote them all equally as the action
qin our quantale
q∈Q
Since these answers are told to every one, and each child hears them, they appear as they are to each child, that is
fCQ
i(q) =q for all1≤i≤n
The kernel of this action, that is the kernel of each round of no answers, is the set of all propositions in which some child knows that he is dirty. The set contains all the propositions that say for example child one knows that he is dirty, child two knows that he is dirty and so on, that isWi=n
i=12CiDi. So the
kernel of eachqwill be
ker(q) =↓
i=n
_
i=1
2CiDi
We are done with our encoding: we have encoded the assumptions of the puzzle in the propositions and their appearances in the module, we have also encoded the dynamic assumptions in the actions of the quantale, and their appearances. The kernel of each action connects the action to the proposition that it contains, that is the communicated proposition, it can be seen as the propositional representative of the action in the module.
We now claim that afterk−1rounds of no answers of children, all the dirty children know that they are dirty. This claim is formalized using the dynamic and epistemic modalities as follows: Proposition 2.3.1 After thek−1’s rounds of answers, dirty child jfor1 ≤ j ≤ kknows that he is dirty i.e.
s{C1,···,Ck} ≤[q0(•q)
(k−1)]2
where(•q)(k−1)denotes(k−1)times sequential composition of no answersq• · · · •q.
Proof.The proof goes by induction on the number of dirty childrenk. Base Case.For the base case we prove the proposition fork= 1, that is
sC1 ≤[q0]2C1D1
By adjunction between dynamic modality and update this is equivalent to
sC1·q0 ≤2C1D1
and by adjunction between epistemic modality and appearance this is equivalent to
fC1(sC1 ·q0)≤D1
By the update inequality, to prove the above inequality it is enough to prove the following
fC1(sC1)·fC1(q0)≤D1
which is by the initial assumptions of the puzzle onfC1 of propositions and actions equal to (sC1∨s∅)·q0 ≤D1.
By distributivity of update over joins of module, we have to show two cases
sC1·q0 ≤D1 , s∅·q0 ≤D1 .
For the first case, by the initial assumption for the states satisfying their corresponding facts we have
sC1 ≤D1, now since update is order preserving, we update both sides byq0and getsC1·q0 ≤D1·q0.
ButD1is a fact and stable under any update soD1·q0 ≤D1and by transitivity we obtainsC1·q0≤D1.
For the second case, by our initial assumptions about the kernel of the actions we have thats∅ ∈
ker(q0), and thus by definition of kernels∅·q0 =⊥, which is less than any proposition, and again by
transitivity we get⊥=s∅·q0 ≤D1.
Induction Hypothesis. We assume that the above proposition holds for k− 1 dirty children and 1≤l≤k−1
s{C1,···,Ck−1} ≤[q0(•q)
(k−2)]2
ClDl.
By the dynamic adjunction, this is equivalent to
s{C1,···,Ck−1}·(q0(•q)
(k−2))≤2
and by the module equation for associativity of update over sequential composition this is equivalent to s{C1,···,Ck−1}·q0(·q) (k−2) ≤2 ClDl Now since2ClDl ≤ W
i2CiDiand by the initial kernel assumptions we have
W
i2CiDi ∈ker(q), we
get
s{C1,···,Ck−1}·q0(·q)
(k−2)∈ker(q)
We will use this in the proof of our induction step below.
Induction Step.Finally we have to show the proposition forkdirty children
s{C1,···,Ck} ≤[q0(•q)
(k−1)]2
CjDj.
We apply the dynamic and epistemic modalities and obtain the following equivalent form
fCj(s{C1,···,Ck}·(q0(•q)
(k−1)))≤2
CjDj
For which by the update and multiplication inequality and initial assumptions on the appearance of actions, it is enough to show the following
fCj(s{C1,···,Ck})·(q0(·q)
(k−1)))≤D
j
Now we use the assumptions on the appearance of propositions and have to show the following two
cases s{C1,···,Ck}·(q0(·q) (k−1)))≤D j , s{C1,···,Ck}\{Cj}·(q0(·q) (k−1)))≤D j .
Similar to the base case, the first case follows by the initial assumption on facts s{C1,···,Ck} ≤ Dj
and stability ofDj under updates. The second case follows from the induction hypothesis, where we
obtained
s{C1,···,Ck−1}·q0(·q)
(k−2)∈ker(q)
and since dirty children are the firstkones, we gets{C1,···,Ck}\Cj·q0(·q)
(k−2) ∈ker(q). By definition
of kernel
s{C1,···,Ck}\{Cj}·q0(·q)
(k−2)·q=⊥
and we are done since⊥ ≤Dj. 2