NetNews Usenet Archive 1992 #18/NN_1992_18.iso/spool/sci/logic/1290

Internet Message Format | 1992-08-19 | 3.6 KB

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From: [e-mail]
Newsgroups: sci.logic
Subject: Re: The poor parachutist...
Message-ID: <[e-mail]>
Date: 19 Aug 92 09:02:12 -0600
References: <[e-mail]>
Organization: Gustavus Adolphus College, St. Peter, Minnesota
Lines: 52

In article <[e-mail]>, [e-mail] (Livy) writes:
> Ok..Now let's see you,logic gguys resolving this question:
> Let's assume that somewhere(on other planet??) there are 2 cities:a city with
> honest people ,who NEVER lie,and a city with crooks,who NEVER TELL the TRUTH.
> One day per year they are going to shoping to the other city,so they are all
> mixed together in both cities.On the very same crazy day,a poor parachutist
> jumped exactly in one of the two cities.The guy know he is in one of those
> cities,but he doesn't know in which of them.Now THE BIG QUESTION is:What
> question should he ask one of the people he sees,so he could deduce in which
> city he is.Remember:The man he is asking could be a liar or a honest man.HE (the
> parchutist) doesn't know it.
>
> NOW,let's see your brains working!!
> Livy. (I know the answer!!)

This riddle appears in Copi's "Introduction to Logic" in a variant form and is
solvable essentially by the use of truth tables..  Let A be the statement "You
(the person he is talking to) are a crook (liar)." and let B be the statement
"This city is the city of crooks."..  Now, the question that the parachutist
should ask is "Is (A <--> B) true?"..  We have 4 cases to consider..
Case 1:  A and B are both true..  In this case (A <--> B) is true but, since
         the person is a crook, he will answer "No".
Case 2:  A is true, but B is not..  In this case (A <--> B) is false but, since
         the person is a crook, he will answer "Yes".
Case 3:  A is false, but B is true..  In this case (A <--> B) is false and,
         since the person is a truth teller, he will answer "No".
Case 4:  A and B are both false..  In this case (A <--> B) is true and, since
         the person is a truth teller, he will answer "Yes".
Now notice: if the answer is "Yes", then B is false and if the answer is "No",
B is true..  Q.E.D.

Now for a riddle of a different flavor..  I first saw this in an interview
article with Paul Halmos in the Mathematical Intelligencer..  I offer it here
because it is (1) not as widely known as the riddle above, (2) more difficult
than the riddle above, and (3) its solution offers some surprises..  Here we
go.  My wife and I go to a party..  There are 4 other couples there for a total
of 5 couples.  (No funny business here..  a couple is husband and wife..)  As
is usual at a social gathering a certain amount of handshaking takes place..
After the handshaking has completed, I interview all of the people at the party
including my wife.  I ask them how many hands they shook.  (note: spouses don't
shake hands and no one shakes their own hand)..  Now each person, therefore,
could have shook as few as 0 hands or as many as 8..  Now, in fact, that range
of answers is exactly what i get for answers!  I.e., the first person I
interview responds 0, the second 1, the third 2, ..., the ninth 8..  The riddle
is "How many hands did my wife shake?"..  I'll even give you the answer..  It
is 4..  The real riddle, of course, is explaining why this is the correct
answer..  If "pigeon hole" suddenly occurs to you, you're on the right track..
Subriddle:  Who is the "meanest" person at this party married to?

enjoy,
al hart