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Potts, C. 2005. The Logic of Conventional Implicatures. Oxford: Oxford University
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Searle, J. R. and D. Vanderveken. 1985. Foundations of Illocutionary Logic. Cambridge: Cambridge University Press.
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Chicago: Chicago Linguistic Society.

A simple solution to the hardest logic puzzle everBrian Rabern & Landon Rabern
We present the simplest solution ever to ‘the hardest logic puzzle ever’. We
then modify the puzzle to make it even harder and give a simple solution
to the modified puzzle. The final sections investigate exploding god-heads
and a two-question solution to the original puzzle.
1. The simplest solution to the ‘hard’ puzzle
The puzzle. Three gods A, B, and C are called, in some order, ‘True’,‘False’, and ‘Random’. True always speaks truly, False always speaks
falsely, but whether Random speaks truly or falsely is a completely random
matter. Your task is to determine the identities of A, B, and C by asking
three yes-no questions; each question must be put to exactly one god. The
gods understand English, but will answer all questions in their own language, in which the words for ‘yes’ and ‘no’ are ‘da’ and ‘ja’, in some order.
You do not know which word means which.1
1

Boolos 1996: 62. The so-called ‘hardest logic puzzle ever’ is coined as such by George
Boolos. Boolos credits the logician Raymond Smullyan as the originator of the puzzle
and the computer scientist John McCarthy with adding the difficulty of not knowing
what ‘da’ and ‘ja’ mean. Related puzzles can, however, be found throughout Smullyan’s writings, e.g. in Smullyan 1978: 149–56, he describes a Haitian island where
half the inhabitants are zombies (who always lie) and half are humans (who always
tell the truth) and explains that ‘the situation is enormously complicated by the factthat although all the natives understand English perfectly, an ancient taboo of theisland forbids them ever to use non-native words in their speech. Hence whenever
you ask them a yes-no question, they reply “Bal” or “Da” – one of which means yes
and the other no. The trouble is that we do not know which of “Bal” or “Da” means
yes and which means no’. In fact, Smullyan solves his own puzzle 162 by using an
instance of the embedded question lemma*, so he had already introduced the essential ingredient needed for a simple solution to the hardest logic puzzle ever. (Foranother related puzzle see Smullyan 1997: 114.)

Analysis 68.2, April 2008, pp. 105–112. © Brian Rabern & Landon Rabern106 brian rabern & landon rabern
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Boolos 1996 provides the following guidelines:(B1) It could be that some god gets asked more than one question(and hence that some god is not asked any question at all).
(B2) What the second question is, and to which god it is put, may
depend on the answer to the first question. (And of course
similarly for the third question.)(B3) Whether Random speaks truly or not should be thought of as
depending on the flip of a coin hidden in his brain: if the coincomes down heads, he speaks truly; if tails, falsely.
(B4) Random will answer ‘da’ or ‘ja’ when asked any yes-noquestion.
Before continuing with this article the reader may wish to pause andattempt a solution.
To solve this puzzle we introduce a function from questions to questions
and prove a lemma, which trivializes the puzzle.2 Let E be the functionthat takes a question q to the question If I asked you ‘q’ in your current
mental state would you say ‘ja’?.
Embedded question lemma. When any god g is asked E(q), a response of‘ja’ indicates that the correct answer to q is affirmative and a response of‘da’ indicates that the correct answer to q is negative.
Proof. If g is either True or False, the result follows, since both a double
positive and a double negative make a positive. Hence we may assume thatg is Random. According to (B3), when we pose E(q) to Random the hidden
coin in his brain is flipped. If the coin comes down heads, Random’s
mental state is that of a truth-teller; if tails, Random’s mental state is thatof a liar. In either case, the result again follows, since both a double
positive and a double negative make a positive.
With the embedded question lemma in our arsenal the ‘hard’ puzzle is
no more difficult than the following trivial puzzle.
The trivial puzzle. Three gods A, B, and C are called, in some order,‘Zephyr’, ‘Eurus’, and ‘Aeolus’. The gods always speak truly. Your task is
to determine the identities of A, B, and C by asking three yes-no questions;each question must be put to exactly one god. The gods understandEnglish and will answer in English.

2

Note that throughout this article we are limiting our focus to polar questions(i.e. yes-no questions), e.g. the functions that we introduce, E and E*, only takeyes-no questions as argument.

a simple solution to the hardest logic puzzle ever 107
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2. Random troubles and the Random modification
One virtue of logical argumentation is that there is not a gap between what
one means and what one says or what one says and what one means. The
puzzle was presented precisely as above and we have provided the simplest
solution to the puzzle as presented. Nevertheless, the spirit of the original
Smullyan-puzzle has certainly been lost. Most commentators on the puzzle
have assumed that Random answers randomly and that therefore nothingcan be gleaned from his answers; but that is not how Random works.
Notice what happens if we ask Random: ‘Are you going to answer thisquestion with a lie?’ If his brain-coin lands heads, he must answer negatively (since it is not true that he will lie) and if his brain-coin lands tails
he also must answer negatively (since while it is true that his answer is alie, he is lying so he will not answer affirmatively). In what sense is thisrandom? He always has to answer this question negatively!3This predictability that has been built into Random (apparently unintentionally) is precisely what we have exploited to trivialize the puzzle.
To make Random truly random, we replace (B3) with the following(and make the necessary modification to the original puzzle):(B3*) Whether Random answers ‘ja’ or ‘da’ should be thought of as
depending on the flip of a coin hidden in his brain: if the coincomes down heads, he answers ‘ja’; if tails, he answers ‘da’.
3. The simplest solution to the modified puzzle
The modified puzzle. Three gods A, B, and C are called, in some order,‘True’, ‘False’, and ‘Random’. True always speaks truly, False always
speaks falsely, but whether Random answers ‘ja’ or ‘da’ is a completely
random matter. Your task is to determine the identities of A, B, and C byasking three yes-no questions; each question must be put to exactly one
god. The gods understand English, but will answer all questions in theirown language, in which the words for ‘yes’ and ‘no’ are ‘da’ and ‘ja’, insome order. You do not know which word means which.

3 Young notes in the appendix ‘Some random observations’ to his unpublished manu-

script that if we ask Random ‘Is it true that (you are lying iff Dushanbe is in
Kirghizia)?’ Random will answer negatively when and only when it is true that
Dushanbe is in Kirghizia and will answer affirmatively when and only when Dushanbe is not in Kirghizia. But he does not note how this trivializes the puzzle. Since
Dushanbe is in Tajikistan, not in Kirghizia, Random will always answer the above
question affirmatively and it is in virtue of this unintentional predictability built into
Random that we can get useful information out of him (see the embedded question
lemma).

108 brian rabern & landon rabern
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To solve the modified puzzle we introduce another function from questions to questions and prove two lemmas. Let E* be the function that takes
a question q to the question If I asked you ‘q’ would you say ‘ja’?.4
Embedded question lemma*. When either True or False are asked E*(q),a response of ‘ja’ indicates that the correct answer to q is affirmative anda response of ‘da’ indicates that the correct answer to q is negative.
Proof. Both a double positive and a double negative make a positive.
Identification lemma. If it has been determined that a particular god is not
Random and two questions remain, then every god’s identity can be
determined.
Proof. Without loss of generality, assume it has been determined that A is
not Random. Ask A the following two questions:(1) E*(‘Is A True?’)(2) E*(‘Is B Random?’)
By the embedded question lemma*, A’s response to (1) will determine A’s
identity and then A’s response to (2) will determine the identity of both B
and C.
Using these lemmas, we can make quick work of the puzzle. Ask B thequestion E*(‘Is A Random?’). If B says ‘ja’, then either B is Random or A
is Random (by the embedded question lemma*). Hence C is not Random.
If B says ‘da’, then either B is Random or A is not Random (by theembedded question lemma*). Hence A is not Random. Whence B’s
response to our first question determines that a particular god is not
Random. Now the identification lemma finishes the job.5
4. Exploding god-heads
Since the puzzle places no restrictions on the type of yes-no questions towhich the gods will grant an answer we feel compelled (as a child in
Sunday school feels compelled) to smash it. The gods sit before us and we
ask each of them:
Are you going to answer ‘ja’ to this question?

4

Questions of similar flavour were used in Roberts 2001.

5

We note that the proof is unaffected if instead of there being a single god language,
each god has his own private language (where ‘ja’ and ‘da’ mean yes and no in someorder).

a simple solution to the hardest logic puzzle ever 109
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If ‘ja’ means no, then True will be unable to respond with the truth. If ‘ja’means yes, then False will be unable to respond with a lie. But they are
infallible gods! They have but one recourse – their heads explode.
This particular question does not seem to help us find a solution to the
puzzle, since the identity of the god with the exploded head depends on themeaning of ‘ja’ and ‘da’. By crafting our question carefully we can get
more information. Consider the question: ‘Are you going to answer thisquestion with the word that means no in your language?’ If in posing thisquestion we explode a god-head, then we know that the god was trying totell the truth. (In Boolos’s original formulation of the puzzle, it could bethat the god was Random and his coin came up heads. In this section wewill concentrate on the modified puzzle.)6
Exploding identity lemma. The identity of any god can be determined in at
most two questions.
Proof. Without loss of generality, assume that we address god A. We ask,
‘Are you going to answer this question with the word that means no inyour language?’. If his head explodes, then we know he is True and we aredone. Thus we may assume that his head does not explode. We ask ‘Are
you going to answer this question with the word that means yes in yourlanguage?’. If his head explodes, then we know he is False. If his head does
not explode, then we know he is Random.
We can now attain another simple solution of the modified puzzle using
this lemma; to wit: first determine the identity of A via the exploding
identity lemma and then depending on A’s identity, ask B either: ‘Are you
going to answer this question with the word that means no in yourlanguage?’ or ‘Are you going to answer this question with the word thatmeans yes in your language?’
5. Boolos’s original puzzle in two questions?
Prima facie, it would seem that we could prove that it requires at least
three questions to determine the identities of all the gods – there are six
possible ways for the gods to be arranged and each yes-no question
distinguishes at most two possibilities, so we need at least log2(6), i.e. 3,
questions to determine their identities. However, the assumption that each
yes-no question distinguishes at most two possibilities is in error. It is
possible to distinguish three possibilities with one question if we ask a
6

One could craft the question even more carefully to prevent Random from explodingin the Boolos version, e.g. ‘Is it possible for you to answer this question with the wordthat means no in your language?’.

110 brian rabern & landon rabern
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question that has the possibility of exploding a god-head. To illustrate wesolve the trivial puzzle in two questions.
The trivial puzzle. Three gods A, B, and C are called, in some order,‘Zephyr’, ‘Eurus’, and ‘Aeolus’. The gods always speak truly. Your task is
to determine the identities of A, B, and C by asking three yes-no questions;each question must be put to exactly one god. The gods understandEnglish and will answer in English.
Tempered liar lemma. If we ask A ‘Is it the case that: [(you are going toanswer “no” to this question) AND (B is Zephyr)] OR (B is Eurus)?’, a
response of ‘yes’ indicates that B is Eurus, a response of ‘no’ indicates that
B is Aeolus, and an exploding head indicates that B is Zephyr. Hence we
can determine the identity of B in one question.
Proof. Assume A responds ‘yes’ and B is not Eurus. Then A has answered‘yes’ to the question ‘Is it the case that you are going to answer “no” to thisquestion?’ This is impossible since A tells the truth.
Assume A responds ‘no’ and B is not Aeolus. Then A has answered ‘no’to both the question ‘Is it the case that: you are going to answer “no” tothis question AND B is Zephyr?’ and the question ‘Is it the case that B is
Eurus?’ The denial of the latter indicates that B is not Eurus and is thus
Zephyr. The denial of the former indicates either that A did not answer‘no’ or that B is not Zephyr. Contradiction.
Assume A’s head explodes and B is not Zephyr. Then B is not Eurus
either; for otherwise A would answer ‘yes’. Hence, since B is neither
Zephyr nor Eurus, A would deny both sides of the disjunction and hence
he would answer ‘no’ to the entire question. This final contradiction
completes the proof.
Now to solve the trivial puzzle in two questions, just use the tempered
liar lemma to determine B’s identity in one question and then for some god
that B is not, ask B if C is this god.7
As noted in the first section of this article, the embedded question lemma
reduces finding a three-question solution to Boolos’s original puzzle tofinding a three-question solution to the trivial puzzle. It seems reasonable
that a similar relationship would hold for two-question solutions as well.
This is indeed the case; however, care must be taken when embedding
questions that contain indexicals or demonstratives, i.e. the complex
demonstrative ‘this question’ refers to the innermost quotational block in

7 In the case that B is Zephyr, we are not able to ask A any more questions, since asking

A the first question caused his head to explode.

a simple solution to the hardest logic puzzle ever 111
02
which it is contained. We require a term that refers to the outermost
quotational block (i.e. the outermost question type) in which it is contained. This can be achieved by introducing a name.8
Let the following question be named ‘Query-1’:
E(‘Is it the case that: [(in your current mental state you would always
answer “da” to Query-1) AND (B is True)] OR (B is False)?’)Tempered liar lemma*. If we ask A Query-1, a response of ‘ja’ indicatesthat B is False, a response of ‘da’ indicates that B is Random, and an
exploding head indicates that B is True. Hence we can determine theidentity of B in one question.
Proof. Assume A responds ‘ja’ and B is not False. Then, by the embedded
question lemma, the correct answer to the question ‘Is it the case that: [(inyour current mental state you would always answer “da” to Query-1)
AND (B is True)] OR (B is False)?’ is affirmative. Since B is not False, thecorrect answer to ‘Is it the case that in your current mental state you would
always answer “da” to Query-1?’ is affirmative, but A answered ‘ja’ toQuery-1. Contradiction.
Assume A responds ‘ja’ and B is not Random. Then, by the embedded
question lemma, the correct answer to both the question ‘Is it the case that:in your current mental state you would always answer “da” to Query-1
AND B is True?’ and the question ‘Is it the case that B is False?’ is negative.
The denial of the latter indicates that B is not False and is thus True. The
denial of the former indicates that either A did not answer ‘da’ or that B
is not True. Contradiction.
Assume A’s head explodes and B is not True. Then B is not False either;for otherwise A would answer ‘ja’. Hence, since B is neither True nor False,
A would deny both sides of the disjunction and hence would answer ‘da’to the entire question. This final contradiction completes the proof.
Now to solve Boolos’s original puzzle in two questions, just use thetempered liar lemma* to determine B’s identity in one question and then

8 There are other ways to achieve this as well. One could use a definite description, e.g.

‘the question in which this question is embedded’ or one could introduce a newindexical that functions to always refer to the outermost question (or sentence) typein which it is embedded, e.g. ‘this-questionG’. Using this new indexical we could
prove a related lemma that would also provide a two-question solution to Boolos’s
original puzzle: If we ask A the question E(‘Is it the case that: [(in your current mental
state you would always answer “da” to this-questionG) AND (B is True)] OR (B isFalse)?’), a response of ‘ja’ indicates that B is False, a response of ‘da’ indicates that
B is Random, and an exploding head indicates that B is True. The proof follows thesame reasoning as the proof of the tempered liar lemma*.

112 m. j. cresswell12
for some god that B is not, ask B if C is this god (in an embedded questionof course).9
University of California
Santa Barbara, CA 93106-309, USA
brian.rabern@gmail.com
Department of Mathematics
University of California
Santa Barbara, CA 93106-3090, USA
landon.rabern@gmail.com
References
Boolos, G. 1996. The hardest logic puzzle ever. The Harvard Review of Philosophy 6:
62–65. Repr. in his Logic, Logic, and Logic, 406–10. 1998. Cambridge, Mass.:
Harvard University Press.
Smullyan, R. 1978. What is the Name of This Book? Englewood Cliffs, NJ: PrenticeHall.
Smullyan, R. 1997. The Riddle of Scheherazade. New York: A. A. Knopf, Inc.
Roberts T. 2001. Some thoughts about the hardest logic puzzle ever. Journal of
Philosophical Logic 30: 609–12.
Young C. unpublished. How I could have solved the hardest logic puzzle ever.
9 This paper would not have been possible without both the loving support and immense

tolerance of our partners Rhiannon Rabern and Jen Sorkin and the inquisitive looks of
Adisyn and Olivia Rabern. Many thanks to Tim Roberts for his encouragement andcomments on earlier drafts of this paper. We would also like to thank Ian Nance for
introducing us to the puzzle and for his helpful conversations about the complexitiesof the gods. Thanks also to Jason Sundram and Richard Chappell for looking overearlier drafts of the paper and Peterson Tretheway, who helped initiate rigorous
investigations over wine, coffee and coherent one relator groups.

Does every proposition have a uniquecontradictory?
M. J. Cresswell
If you think that a proposition can have more than one contradictory, or
can have none, then you need read no further. What I will show is that if(1) Every proposition has exactly one contradictory