A Method to Solve the Puzzles of Knights and
Knaves ∗
L. Aszalós
†
Abstract
In his book ”What is the name of this book?” Raymond M. Smullyan
used a lot of puzzles to illustrate the background of the Gödel incompleteness
theorem. These puzzles became popular and nowadays are being published in
amusement magazines, too. In each section of the book different conditions
are met. In the best known type of puzzles we have only two types of people,
knights and knaves. Knights always tell the truth and knaves always lie.
We describe a method to formulate this kind of puzzles, then we prove
the soundness and the completeness of this logic. At the end we give Prolog
programs for solving the puzzles.
1
The logic of puzzles — Introduction
In the puzzles the inhabitants make statements about themselves and about the
others, for example ‘Mr. A is a knave.’ or ‘Mrs. B said that she was not a knave.’
Usually, to solve a puzzle we must determine the type of persons. If we want
to formulate the puzzles, we need to express statements of the following kinds:
the person x is a knight, the person x is a knave and the person x can say the
statement B. In these cases we shall write Tx , Fx , and Sx B, respectively. In his
puzzles Smullyan uses say which sounds better than can say, but we shall use the
latter. If somebody said something, then he is able to say it, so he can say it, hence
we do not loose information when we use can say instead of say. If somebody didn’t
say a sentence B, then we have no information whether he cannot say B or he can
but just for some reason he didn’t. In [2] there is no puzzle formulated such that
it contains the negation of say, but why should conform to this restriction? Of
course, we could have a problem in the formalization of puzzles with say, because if
x said that y has not said the statement B earlier (which is true), and later y says
B, then we must use tense logic to handle this situation or we must restrict our
tools and definitions in some way. But if we use can say, then the former situation
cannot occur so our life will be easier, we can formulate puzzles easily and we can
use the negation of can say in the puzzles, too.
∗ Research
was partly supported by the grant OTKA no. 354-19341
of Debrecen, 4010 Debrecen, PO Box 12, Hungary, e-mail: aszalos@math.klte.hu
† University
1
2
L. Aszalós
We could demonstrate our method better if we complicate the puzzles and if we
introduce a third type of inhabitants: normals. Normals sometimes tell the truth
and are sometimes lying. If the person x is a normal, then we shall write Nx . The set
of islanders is denoted by P. Of course we have only finitely many islanders, so the
set P is finite. Our connectives are the usual ones: ¬ (negation), ∧ (conjunction),
∨ (disjunction) and ⊃ (implication) and we are using the > (constant true) and the
⊥ (constant false). In the propositional logic the atomic statements are denoted
by propositional letters (for example by Q, R, ...). The set of atomic statements
(propositional constants) is denoted by S. The definition of the set of formulae is
the following: Let P and S be finite sets. The set of formulae is the smallest set
F such that
• > ∈ F and ⊥ ∈ F.
• S ⊂ F,
• If A ∈ F, then ¬A ∈ F.
• If A ∈ F and B ∈ F, then (A ⊃ B) ∈ F, (A ∨ B) ∈ F and (A ∧ B) ∈ F.
• If x ∈ P then Tx ∈ F, Fx ∈ F and Nx ∈ F.
• If x ∈ P and B ∈ F, then Sx B ∈ F.
The valuation ϑ assigns a truth value to every formula. If the formula A is true
in a valuation ϑ, this is denoted by ϑ |= A. The properties of the islanders can be
coded into the valuation: Let ϑS ⊂ S (ϑS is the set of true atomic statements),
ϑT : P → {t, f, n} (ϑT is the classification of islanders). ϑ = hϑS , ϑT i is a valuation
over F if:
• ϑ|= >, ϑ|6 = ⊥.
• If Q ∈ S then ϑ|= Q iff Q ∈ ϑS .
• ϑ|= ¬A iff ϑ|6 = A.
• ϑ|= A ⊃ B iff ϑ|6 = A or ϑ|= B.
• ϑ|= A ∨ B iff ϑ|= A or ϑ|= B.
• ϑ|= A ∧ B iff ϑ|= A and ϑ|= B.
• If x ∈ P then ϑ|= Tx iff ϑT (x) = t
• If x ∈ P then ϑ|= Fx iff ϑT (x) = f
• If x ∈ P then ϑ|= Nx iff ϑT (x) = n
• ϑ|= Sx B iff ϑ|= Tx and ϑ|= B or ϑ|= Fx and ϑ|6 = B or ϑ|= Nx .
A Method to Solve the Puzzles of Knights and Knaves
3
A formula A is valid if for every valuation ϑ ϑ|= A and it is satisfiable if there exists
a valuation ϑ such that ϑ|= A.
We shall use analytic tableaux [1, pp. 20-21] to check the provability of formulae.
We shall use the notation and definitions of this book, but 1) in the nodes (except
the root) we allow not only a formula but the set of formulae, 2) we need rules
about can say:
¬Sx B
Sx B
Tx
¬Tx ¬Tx
Tx ¬Tx
and
Fx ¬Fx
Fx
¬Fx
¬Fx
¬Nx
¬Nx
Nx
¬Nx ¬Nx
¬B
B
B
¬B
2
Soundness
First we prove that our proving tool, the tableaux method is correct, that is, if we
can prove a formula using it, then the formula is valid.
Theorem 1 If the tableaux for a formula F is closed, then F is unsatisfiable.
Proof. Let ϑ be a valuation that makes F true. We prove by mathematical
induction (on the size of the tree) that the tableaux for F is open. If the tableaux
has only one node (root), then the claim is true. Let us assume that Υ is a tableaux
for F , which is true in ϑ. Hence, Υ is open. Let Υ0 be an extension of Υ. Υ has an
open path, say p. If Υ0 is obtained by an application of a rule for a formula from
another path, then the path p is an open path of Υ0 , so Υ0 is an open tableaux.
Otherwise, we get Υ0 by applying a rule for the formula A in the path p of Υ.
• If A is an α formula then the new node contains formulae α1 and α2 so the
new path contains them, too. The formula α is on the true path, so it is true.
By definition of the valuation, formulae α1 and α2 are true, too. Hence the
new path is open.
• If A is a β formula then the new nodes contain formulae β1 and β2 , respectively. So the new path contains exactly one of them. β is true, so by
definition, one of β1 and β2 is true, hence one of the new paths is open.
• If A is a Sx B type formula, it is true in ϑ and by definition of ϑ |= Sx B we
know that (ϑ |= Tx and ϑ |= B) or (ϑ |= Fx and ϑ |= ¬B) or ϑ |= Nx is true. In
the first case ϑ|= ¬Fx and ϑ|= ¬Nx , in the second case ϑ| = ¬Tx and ϑ|= ¬Nx ,
and in the third case ϑ| = ¬Tx and ϑ|= ¬Fx by definition of ϑT , so one of the
new paths is open.
• If A is a ¬Sx B type formula, then by ϑ6|= Sx B we know that ϑ|= Fx and ϑ|= B
or ϑ |= Tx and ϑ |= ¬B. In the first case ϑ |= ¬Tx and ϑ |= ¬Nx , in the second
case ϑ|= ¬Fx and ϑ|= ¬Nx by definition of ϑT , so one of the new paths is open,
too.
4
L. Aszalós
Now a closed tableaux Υ cannot be true under any interpolation, so the origin of
any closed tableaux must be unsatisfiable. From this follows that every formula
provable by the tableaux method must be valid.
2
3
Completeness
In this section we want to show that any valid formula is provable, so the method
is complete. For this purpose we shall introduce a function that assigns a natural
number to every formula. This number represents a measure of complexity of the
formula. In the inductive proof below we use these numbers. The degree of a
formula is defined as follows
• The degree of >, ⊥, of all elements of S, of Tx , Fx and Nx is 0.
• The degree of ¬A and the degree of A are the same.
• If the degree of A is n and the degree of B is m, then the degrees of (A ∧ B),
(A ∨ B) and (A ⊃ B) are max(m, n) + 1.
• If the degree of B is n, then the degree of Sx B is n + 1.
The Hintikka set is a usual tool to show the completeness of a logic. Our logic is
more complicated than the propositional logic, so the following definition contains
four new items. The set of formulae U is a Hintikka set if the following holds:
H0 U does not contain a predicate letter and its negation simultaneously.
H1 If α ∈ U, then α1 ∈ U and α2 ∈ U.
H2 If β ∈ U, then β1 ∈ U or β2 ∈ U.
H3 If x ∈ P, then only one of Tx , Fx and Nx is in U.
H4 If x ∈ P then not all of ¬Tx , ¬Fx and ¬Nx are the members of U
H5 If Sx B ∈ U, then {Tx , ¬Fx , ¬Nx , A} or {¬Tx , Fx , ¬Nx , ¬A} or {¬Tx , ¬Fx , Nx , }
is the subset of the set of formulae U.
H6 If ¬Sx B ∈ U, then {¬Tx , Fx , ¬Nx , A} or {Tx , ¬Fx , ¬Nx , ¬A} is the subset of
the set of formulae U.
Lemma 2 Every Hintikka set is satisfiable.
Proof. Let U is a Hintikka set. We make a suitable ϑ, in which all the members
of U are true. Let ϑS be the set of all predicate letters which are in U.
• If Tx ∈ U or {¬Fx , ¬Nx } ⊂ U, then ϑT (x) is t.
• If Fx ∈ U, or {¬Tx , ¬Nx } ⊂ U, then ϑT (x) is f .
A Method to Solve the Puzzles of Knights and Knaves
5
• If Nx ∈ U, or {¬Tx , ¬Fx } ⊂ U, then ϑT (x) is n.
If none of the former three points hold, then
• If ¬Tx ∈
/ U, then ϑT (x) is t.
• If ¬Fx ∈
/ U, then ϑT (x) is f .
• If ¬Nx ∈
/ U, then ϑT (x) is n.
By H3 and H4, by this rules we can assign a value to every x. Now all of the
formulae of degree 0 in U are true in hϑS , ϑT i. Let A be a formula with a degree
greater than 0.
• If A is an α formula, then by H1 α1 ∈ U and α2 ∈ U and their degrees are
smaller than the degree of α; by induction hypotheses α1 and α2 are true in
ϑ and so α.
• If A is a β formula, then the proof is similar.
• If A is a Sx B formula, then by H5 {Tx , ¬Fx , ¬Nx , B} or {¬Tx , Fx , ¬Nx , ¬B} or
{¬Tx , ¬Fx , Nx } is the subset of the set of formulae U. All this formulae are
of degree smaller than the degree of Sx B, so all of them are true in ϑ and by
definition ϑ|= Sx B.
• The case of ¬Sx B is similar.
We have constructed a valuation ϑ such that every formula of U is true in ϑ, so U
is satisfiable.
2
Theorem 3 Let be F is a formula. If F is unsatisfiable then the tableaux for F is
closed.
Proof. Let F be unsatisfiable. Construct the full tableaux for F and assume that
this tableaux has an open path. By the construction, the set of formulae of this
path is a Hintikka set. By the previous lemma there exists a valuation ϑ such that
all formulae of this path are true in it, hence F is true, too. This contradicts our
hypothesis. The origin of this contradiction was our assumption, so the tableaux
has no open path.
2
The theorem above claims that our logic is complete. If we have a valid formula
A, then we can construct a tableaux for ¬A. This formula is unsatisfiable, so its
tableaux is closed.
4
4.1
Other type of puzzles
The Island of Knights and Knaves
Let us return to the original puzzles! If we have only two type of inhabitants
then the definitions are simpler than before: in the definition of the formulae we
6
L. Aszalós
do not need Nx , we have only two types of inhabitants, hence we need a different
ϑT : P → {t, f } and one item of the definition of the valuation is changed: ϑ|= Sx B
iff (ϑ|= Tx and ϑ|= B) or (ϑ|= Fx and ϑ6|= B). The rules for can say are the following:
¬Sx B
Sx B
Tx
¬Tx
Tx ¬Tx
and
Fx
Fx
¬Fx
¬Fx
B
¬B
¬B
B
The lemmas and theorems are the same as before, but the definition of the Hintikka
set is changed due the changes in preconditions. We leave to the reader to rewrite
and check the definitions and proofs.
4.2
Island of Knights, Knaves, Normals and Mutes
To get a symmetric type of puzzles, we shall introduce a fourth type of inhabitants.
This new type is the opposite to the normals. Any inhabitant of this kind can say
nothing, so we call them mutes, and we shall write Mx if the person x is a mute.
We need to revise the definition of formulae according to this new type of persons.
Of course we need a different ϑT : P → {t, f, n, m}, but the definition of ϑ |= Sx B
will be remain the same. The other definitions and proofs can be extend according
to the rules:
¬Sx B
Sx B
Tx
¬Tx ¬Tx
Tx ¬Tx ¬Tx
Fx ¬Fx
Fx ¬Fx
¬Fx
¬Fx
and
¬Nx
¬Nx
Nx
¬Nx ¬Nx ¬Nx
¬Mx ¬Mx
Mx
¬Mx
¬Mx ¬Mx
¬B
B
B
¬B
but we left it to the reader.
4.3
Island of Bahava
On the Island of Bahava there are exactly tree types of islanders. Each knight’s
spouse is a knave and conversely. Let us denote by x0 the spouse of x. Of course,
we shall identify x00 and x, so we shall write x instead of x00 without notifying it.
Our rules are extended to
Sx B
¬Sx B
Tx
¬Tx
¬Tx
Tx
¬Tx
¬Fx
Fx
¬Fx
¬Fx
Fx
¬Nx
Nx
¬Nx
¬Nx
¬Nx
and
¬Tx0
Tx0 ¬Tx0
¬Tx0
Tx0
Fx0 ¬Fx0 ¬Fx0
Fx0 ¬Fx0
¬Nx0 ¬Nx0
N x0
¬Nx0 ¬Nx0
B
¬B
¬B
B
A Method to Solve the Puzzles of Knights and Knaves
5
7
Proof of the solution of two puzzles
The puzzle 40 of [2] is the following: On the Island of Knights, Knaves and Normals
A said that B is a knight. B said that A is a knave. Let us prove that one of them
told the truth but is not a knight or that one of them told a lie but is not a knave.
To prove the statement above we need to construct the tableau-proof of the
formula Sa Tb ∧ Sb Fa ⊃ (Tb ∧ ¬Ta ) ∨ (Fa ∧ ¬Tb ) ∨ (¬Tb ∧ ¬Fa ) ∨ (¬Fa ∧ ¬Fb ). After
a few of uninteresting steps we get the proof-tree on Fig. 1. On this picture we
have boxed the subtrees instead of writing them vertically. The formulae closing
the paths are denoted by a star.
Ta , ¬Fa , ¬Na , Tb
Tb , ¬Fb , ¬Nb , Fa ∗
¬Ta , Fa , ¬Na , ¬Tb
Tb ∗ , ¬Fb , ¬Nb , Fa
¬Tb ∗ , Fb , ¬Nb , ¬Fa
¬Tb , Fb , ¬Nb , ¬Fa ∗
¬Tb ∗ , ¬Fb , Nb
¬Tb , ¬Fb , Nb
Tb ∗
¬Fa ∗
¬Ta , ¬Fa , Na
Tb , ¬Fb , ¬Nb , Fa ∗
¬Tb , Fb , ¬Nb , ¬Fa
Tb ∗
Fa ∗
¬Tb , ¬Fb , Nb
Tb ∗
Fa ∗
Figure 1.
The puzzle 44 of [2] states that on Island of Bahava Mr. A said that his wife is
not a normal. Mrs. A said that her husband is not a normal. We are proving that
both of them are normals on Fig. 2.
¬(Sa ¬Na0 ∧ Sa0 ¬Na ⊃ Na ∧ Na0 )
Sa ¬Na0 , Sa0 ¬Na , ¬Na ∨ ¬Na0
Ta , ¬Fa , ¬Na , ¬Ta0 , Fa0 , ¬Na0 , ¬Na0
Ta0 ∗ , ¬Fa0 , ¬Na0 , ¬Ta , Fa , ¬Na , ¬Na
¬Ta0 , Fa0 , ¬Na0 , Ta , ¬Fa , ¬Na , Na ∗
¬Ta0 , ¬Fa0 ∗ , Na0 , ¬Ta , ¬Fa , ¬Na
¬Ta , Fa , ¬Na , Ta0 , ¬Fa0 , ¬Na0 , Na0 ∗
¬Ta , ¬Fa , Na , ¬Ta0 , ¬Fa0 , Na0
¬Na ∗
¬Na0 ∗
Figure 2.
8
6
L. Aszalós
Methods of Automated Puzzle Solving
There are several methods for solving the puzzles of the Island of Knight and
Knaves.
In 1998 Piotr Rudnicki used the formalism of his proof-checker Mizar-MSE
when he formalized the puzzles in his syllabus for CMPUT 272 at the University
of Alberta. In the newsgroup comp.lang.prolog Ralph Becket showed a method
in which (in our notation) the person x says that A is written as Tx ≡ A and
our puzzles are translated to formulae of predicate logic that we can solve easily.
David Gries used a similar method, but he tests the validity of predicate logic
formulae with rewriting. In the late eighties Larry Wos formulated puzzles as a
clause-cluster form of first-order sentences [4]. He applied the first-order theorem
prover Otter to solve the puzzles. Lee Naish wrote a NU-Prolog program to test
the puzzles. He published his program in newsgroup comp.lang.prolog at 11 Feb
2000.
Independently of Lee Naish on the basis of [3] we wrote several programs to
solve the puzzles. We shall use the programming language Prolog to construct
the program to solve puzzles of the Island of Knights, Knaves and Normals. (The
programs for the other type of puzzles are similar.) For the sake of simplicity we
shall define some operators
::::::-
op(930,
op(920,
op(910,
op(909,
op(900,
op(901,
xfx,
xfx,
xfy,
xfy,
fx,
xfy,
=== ).
-> ).
\/ ).
/\ ).
~
).
s ).
/*
/*
/*
/*
/*
/*
equivalence
implication
disjunction
conjunction
not
can say
*/
*/
*/
*/
*/
*/
The canBeTrue and the canBeFalse are two predicates to test whether there is
a valuation in which the formula in the argument is true/false. For the predicate
logic the following code is well-known [3, Program 3.31]:
canBeTrue(true)
canBeTrue(~ X)
canBeTrue(X -> Y)
canBeTrue(X \/ Y)
canBeTrue(X /\ Y)
canBeTrue(X === Y)
:- !.
:- canBeFalse(X).
:- canBeFalse(X); canBeTrue(Y).
:- canBeTrue(X); canBeTrue(Y).
:- canBeTrue(X), canBeTrue(Y).
:-(canBeTrue(X), canBeTrue(Y));
(canBeFalse(X), canBeFalse(Y)).
canBeFalse(false)
:- !.
canBeFalse(~ X)
:- canBeTrue(X).
canBeFalse(X -> Y) :- canBeTrue(X), canBeFalse(Y).
canBeFalse(X \/ Y) :- canBeFalse(X), canBeFalse(Y).
canBeFalse(X /\ Y) :- canBeFalse(X); canBeFalse(Y).
canBeFalse(X === Y) :-(canBeTrue(X), canBeFalse(Y));
(canBeFalse(X), canBeTrue(Y)).
We shall write knight(X), knave(X) and normal(X) instead of Tx , Fx and Nx ,
respectively. The definition ϑT is coded as below:
A Method to Solve the Puzzles of Knights and Knaves
9
canBeTrue(knight(knight)).
canBeTrue(knave( knave) ).
canBeTrue(normal(normal)).
canBeFalse(
canBeFalse(
canBeFalse(
canBeFalse(
canBeFalse(
canBeFalse(
knight(knave) ).
knave(knight) ).
knight(normal) ).
knave(normal) ).
normal(knight)).
normal(knave)).
The definition of ϑ|= Sx B is coded as below:
canBeTrue(knight s X) :- canBeTrue(X).
canBeTrue(knave s X) :- canBeFalse(X).
canBeTrue(normal s _).
canBeFalse(knight s X) :- canBeFalse(X).
canBeFalse(knave s X) :- canBeTrue(X).
We can write down the preconditions of puzzle 40 of [2]
puzzle(w40, A s knight(B) /\ B s ~knight(A)).
and we can test models of the puzzle by the question:
?- puzzle(w40, Formula), canBeTrue(Formula)
If we want to check the validity of the formula of the puzzle then we can ask the
following:
canBeFalse(A s knight(B) /\ B s knave(A) -> ( knight(B) /\ ~ knight(A) ) \/
( knave(A) /\ ~ knight(B) ) \/ ( ~ knight(B) /\ ~ knave(A)) \/
(~ knave(A) /\ ~ knave(B) ) ).
In the program above we cheated. For the sake of simplicity we have used the
first-order predicates of Prolog. We could substitute the Horn-fact KnightA for
the predicate knight(A), and consistently in the others cases to obtain the same
program as our tableaux method, but this has no effect on the results.
7
Conclusion
We have shown a method to formulate Smullyan’s puzzles. We have proved soundness and completeness of several logics of puzzles, in meantime we introduced some
new types of puzzles that do not occur in [2]. We have shown a Prolog program
that automatizes the solution of puzzles. This program can be easily improved to
be suitable for other types of puzzles, too.
10
8
L. Aszalós
Acknowledgements
I would like to express my thanks to K. Pásztor-Varga for her many helpful criticisms during the preparation of this paper. I would like to thank late A. Kron for
his help, and for trying valiantly to make this a more readable paper.
References
[1] Raymond M. Smullyan. First-order logic. Springer-Verlag New York, Inc., New
York, 1968. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 43.
[2] Raymond M. Smullyan. What is the Name of This Book? (The Riddle of
Dracula and Other Logical Puzzles). Prentice-Hall Inc., 1978.
[3] L. Sterling and E. Shapiro. The Art of Prolog. MIT Press., 1986.
[4] L. Wos, R. Overbeek, E. Lusk and J. Boyle. Automated Reasoning: Introduction
and Applications. McGraw-Hill., 1992.