American philosopher and logician George Boolos described the above riddle which was devised by Raymond Smullyan and published it in the Harvard Review of Philosophy in 1996. Boolos called it “The Hardest Logic Puzzle Ever”. You can read about making this puzzle even harder on the Physics arXiv Blog.
The original article Here :
S OME YEARS AGO, THE LOGICIAN AND PUZZLE-MASTER
Raymond Smullyan devised a logical puzzle that has no challengers I know
of for the title of Hardest Logical Puzzle Ever. 1’11 set out the puzzle here,
give the solution, and then brietly discuss one of its more interesting aspects.
The puzzle: Three gods A, R, 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, R, 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 “dam and “ja,” in
some order. You do not know which word means which2
Before I present the somewhat lengthy solution, let me give answers to certain
questions about the puzzle that occasionally arise:
- It could be that some god gets asked more than one question (and hence that
some god is not asked any question at all).
- 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
- Wheather ranom speaks truly or not should
be thought of as depending on the flip of a
coin hidden in his brain: if the coin comes
down heads, he speaks truly; if tails, falsely.
- Random will answer da or ja when asked any
The Solution: Before solving The
Hardest Logic Puzzle Ever, we will set out and
solve three related, but much easier, puzzles.
We shall then combine the ideas of their solutions
to solve the Hardest Puzzle. The last two
puzzles are of a type that may be quite familiar
to the reader, but the first one is not well
known (in fact the author made it up while
thinking about the Hardest Puzzle).
Puzzle 1: Noting their locations, I place two aces and a jack face down on a
table, in a row; you do not see which card is placed where. Your problem is to
point to one of the three cards and then ask me a single yes-no question, from the
answer to which you can, with certainty, identify one of the three cards as an ace. If
you have pointed to one of the aces, I will answer your question truthfully.
However, if you have pointed to the jack, I will answer your question yes or no,
completely at random.
Puzzle 2: Suppose that, somehow, you have learned that you are speaking not
to Random but to True or False – you don’t know which – and that whichever
god you’re talking to has condescended to answer you in English. For some reason,
you need to know whether Dushanbe is in Kirghizia or not. What one yes-no
question can you ask the god from the answer to which you can determine whether
or not Dushanbe is in Kirghizia?
Puzzle 3: You are now quite definitely talking to True, but he refuses to
answer you in English and will only say da or ja. What one yes-no question can you
ask True to determine whether or not Dushanbe is in Kirghizia?
HERE’S ONE SOLUTION TO PUZZLE 1: POINT TO THE
middle card and ask, “Is the left card an ace?” If I answer yes, choose
the left card; if I answer no, choose the right card. Whether the middle
card is an ace or not, you are certain to find an ace by choosing the left
card if you hear me say yes and choosing the right card if you hear no. The reason
is that if the middle card is an ace, my answer is truthful, and so the left card is an
ace if I say yes, and the right card is an ace if I say no. Rut if the middle card is the
Jack, then both of the other cards are aces, and so again the left card is an ace if I say
yes (so is the right card but that is now irrelevant), and the right card is an ace if I
say no (as is the left card, again irrelevantly).
To solve puzzles 2 and 3, we shall use iff
Logicians have introduced the usehl abbreviation “iff,” short for “if, and only
if.” The way “iff works in logic is this: when you insert “iff’ between two statements
that are either both true or both false, you get a statement that is true; but if
you insert it between one true and one false statment, you get a false statement.
Thus, for example, “The moon is made of Gorgonzola iff Rome is in Russia” is
true, because “The moon is made of Gorgonzola” and “Rome is in Russia” are
both false. But, “The moon is made of Gorgonzola iff Rome is in Italy” and “The
moon lacks air iff Rome is in Russia” are false. However, “The moon lacks air iff
Rome is in Italy” is true. (“Iff has nothing to do with causes, explanations, or laws
To solve puzzle 2, ask the god not the simple question, “Is Dushanbe in
Kirghizia?” but the more complex question, “Are you True iff Dushanbe is in
Kirghizia?” Then (in the absence of any geographical information) there are four
1) The god is True and D. is in K: then you get the answer yes.
2) The god is True and D. is not in K.: this time you get no.
3) The god is False and D. is in K.: you get the answer yes, because onlyone statement is true, so the correct answer is no, and the god, who is
False, falsely says yes.
4) The god is False and D. is not in K.: in this final case you get the
answer no, because both statements are false, the correct answer is yes, and
the god False falsely says no.
So you get a yes answer to that complex question if D. is in K. and a no answer if it
is not, no matter to which of True and False you are speaking. By noting the answer
to the complex question, you can find out whether D. is in K. or not.
The point to notice is that if you ask either True or False, “Are you True iff X?”
and receive your answer in English, then you get the answer yes if X is true and no
if X is false, regardless of which of the two you are speaking to.
The solution to puzzle 3 is quite similar: Ask True not, “Is Dushanbe in
Kirghizia?” but, “Does da mean yes iff D. is in K.?” There are again four possibilities:
1) Da means yes and D. is in K.: then True says da.
2) Da means yes and D. is not in K.: then True says ja (meaning no).
3) Da means no and D. is in K.: then True says da (meaning no).
4) Da means no and D. is not in K.: then both statements are false, the
statement “Da means yes iff D. is in K.” is true, the correct answer (in
English) to our question is yes, and therefore True says ja.
Thus you get the answer da if D. is in K. and the answer ja if not, regardless of
which of da and ja means yes and which means no.
The point this time is that if you ask True, “Does da mean yes iff Y?” then you
get the answer da if Y is true and you get the answer ja if Y is false, regardless of
which means which.
Combining the two points, we see that if you ask one of True and False (who
we again suppose only answer da and ja), the very complex question, “Does da
mean yes iff, you are True iff X?” then you willget the answer da ifX is true andget
the answer ja ifX is false, regardless of whether you are addressing the god True or
the god False, and regardless of the meanings of da and ja.
We can now solve The Hardest Logic Puzzle Ever.
Your first move is to find a god who you can be certain is not Random, and
hence is either True or False.
To do so, turn to A and ask Question 1: Does da mean yes i’ you are True zffB
is Random? If A is True or False and you get the answer da, then as we have seen,
B is Random, and therefore C is either True or False; but if A is True or False and
you get the answer ja, then B is not Random, therefore B is either True or False.
But what if A is Random?
If A is Random, then neither B nor C is Random!
So if A is Random and you get the answer da, C is not Random (neither is B,
but that’s irrelevant), and therefore C is either True or False, and if A is Random and you get the answer ja, B is not random (neither is C, irrelevantly), and therefore
B is either True or False.
Thus, no matter whether A is True, False, or Random, if you get the answer da
to Question 1, C is either True or False, and if you get the answer ja, B is either
True or False!
Now turn to whichever of B and C you have just discovered is either True or
False – let us suppose that it is B (if it is C, just interchange the names B and C in
what follows) – and ask Question 2: Does da mean yes iff Rome is in Italy? True
will answer da, and False will answer ja. Thus, with two questions, you have either
identified B as True or identified B as False.
For our third and last question, turn again to B, whom you have now either
identified as True or identified as False, and ask Question 3: Does da mean yes iffA
Suppose B is True. Then if you get the answer da, then A is Random, and
therefore A is Random, B is True, C is False, and you are done; but if you get the
answer ja, then A is not Random, so A is False, B is true, C is Random, and you are
Suppose B is False. Then if you get the answer da, then since B speaks falsely,
A is not Random, and therefore A is True, B is False, C is Random, and you are
done; but if we get ja, then A is Random, and thus B is False, and C is True, and
you are again done. FINIS.
Well, I wasn’t speaking falsely or at random when I said that the puzzle was
hard, was I?
A brief remark about the significance of the Hardest Logic Puzzle Ever:
There is a law of logic called “the law of excluded middle,” according to which
either X is true or not-X is true, for any statement X at all. (“The law of non-contradiction”
asserts that statements X and not-X aren’t both true.)
and philosophers have occasionally attacked the idea that excluded middle is a logically
valid law. We can’t hope to settle the debate here, but can observe that our
solution to puzzle 1 made essential use of excluded middle, exactly when we said
“Whether the middle card is an ace or not. …” It is clear from The Hardest Logic
Puzzle Ever, and even more plainly from puzzle 1, that our ability to reason about
alternative possibilities, even in everyday life, would be almost completely paralyzed
were we to be denied the use of the law of excluded middle.