**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

of nature.)

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

possibilities:

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

is Random?

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

again done.

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.)

Mathematicians

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.