🚀 go-pugleaf

In article <1992Aug18.131259.14667@wisipc.weizmann.ac.il> jhsegal@wiscon.weizmann.ac.il (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!!)

That's easy: he should ask "Have you ever heard the puzzle about the guy
who has to figure out whether he's in the city of all completely
truthful people, or the city of all complete liars?"

If the random person says "no" he knows he's in the city of liars.

Don't teach your grandma to suck eggs, bubela.
--
Charles R. Martin/(Charlie)/martinc@cs.unc.edu/(ne crm@cs.duke.edu) 
O/Dept. of Computer Science/CB #3175 UNC-CH/Chapel Hill, NC 27599-3175
H/3611 University Dr #13M/Durham, NC 27707/(919) 419 1754
----------------------------------------------------------------------
"I am he who walks the States with a barb'd tongue, questioning every
one I meet,/Who are you that wanted only to be told what you knew
before?/ Who are you that wanted only a book to join you in your
nonsense?"  _Leaves of Grass_ xxiii.4.

In article <1992Aug18.131259.14667@wisipc.weizmann.ac.il>, jhsegal@wiscon.weizmann.ac.il (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!!)
 
Easy. Just ask someone: "Do you live here?". If the answer is "yes", you're in
the honest city, if it's "no", you're in the liars' city. The parachutist won't
know whether he's talking to a liar or not, but he'll know which city he's in.

Jussi.

In article <1992Aug18.131259.14667@wisipc.weizmann.ac.il>, jhsegal@wiscon.weizmann.ac.il (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

You are damn right! It took me ages to think about that....
Livy

In article <1992Aug18.131259.14667@wisipc.weizmann.ac.il> jhsegal@wiscon.weizmann.ac.il (Livy) writes:
>Remember:The man he is asking could be a liar or a honest man.HE (the
>parchutist) doesn't know it.

Usually, you encourage the parachutist to ask "What name would a liar give to
the other city?". However, I am not convinced that this "works". It makes some
assumptions about lying which are not clear. For example, it assumes (which is
reasonable) that the honest people have noticed the "liars" and have come to
some understanding of what is involved. However, what lying means to the liars
is difficult to say, especially since they can't agree on it. Since the truthful
citizens know this as well, they give you an answer which a liar would give you,
but this is no use unless the liar's answer is well defined.

Right?

Later,
Andrew Mullhaupt

Maybe a group ?.logic.puzzles would be in order?

In article <1992Aug19.090212.1658@gacvx2.gac.edu>, logic@gacvx2.gac.edu writes:
> 
> 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
> 

First, lets call the nine people you (represented here by U) interviewed by the
letters A-I (A shook 0 hands, B shook 1, ... ,I shook 8 hands).

Now drawing a simple graph helps (actually, the graph could be omitted; it's
just nice to see how things turn out and to be assured that the puzzle is
soluble, that is, that U has not shaken hands with E): we see that the only way
to have the required numbers of handshakes for A-I is:
      Person	    The people with whom s/he shook hands
      	A
	B	    	I
	C		H,I
	D		G,H,I
	E		F,G,H,I
	F		E,G,H,I,U
	G		D,E,F,H,I,U
	H		C,D,E,F,G,I,U
	I		B,C,D,E,F,G,H,U
which shows us:
	U		F,G,H,I

Now we see that I has to be married to A: I has shaken hands with everyone else
(subriddle)
By the same reasoning, H is married to B: H has shaken hands with everyone but A
and B, and A is already married to I
And so on: G&C are married to each other, so U is left with E, who shook 4
hands. Q.E.D.
 
Jussi.		

Maybe a group ?.logic.puzzles would be in order?

In article <1992Aug19.090212.1658@gacvx2.gac.edu>, logic@gacvx2.gac.edu writes:
> 
> 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
> 

First, lets call the nine people you (represented here by U) interviewed by the
letters A-I (A shook 0 hands, B shook 1, ... ,I shook 8 hands).

Now drawing a simple graph helps (actually, the graph could be omitted; it's
just nice to see how things turn out and to be assured that the puzzle is
soluble, that is, that U has not shaken hands with E): we see that the only way
to have the required numbers of handshakes for A-I is:
      Person	    The people with whom s/he shook hands
      	A
	B	    	I
	C		H,I
	D		G,H,I
	E		F,G,H,I
	F		E,G,H,I,U
	G		D,E,F,H,I,U
	H		C,D,E,F,G,I,U
	I		B,C,D,E,F,G,H,U
which shows us:
	U		F,G,H,I

Now we see that I has to be married to A: I has shaken hands with everyone else
(subriddle)
By the same reasoning, H is married to B: H has shaken hands with everyone but A
and B, and A is already married to I
And so on: G&C are married to each other, so U is left with E, who shook 4
hands. Q.E.D.
 
Jussi.

P.S. Sorry for double posting if this happens. My first posting hasn't turned
up, so I sent this again.		

In article <1185@kepler1.rentec.com> 
andrew@rentec.com (Andrew Mullhaupt) writes:

>In article <1992Aug18.131259.14667@wisipc.weizmann.ac.il>
> jhsegal@wiscon.weizmann.ac.il (Livy) writes:

L
>>Remember:The man he is asking could be a liar or a honest man.HE (the
>>parchutist) doesn't know it.

AM:
>Usually, you encourage the parachutist to ask "What name would a liar give to
>the other city?". However, I am not convinced that this "works". It makes some
>assumptions about lying which are not clear. For example, it assumes (which is
>reasonable) that the honest people have noticed the "liars" and have come to
>some understanding of what is involved. However, what lying means to the liars
>is difficult to say, especially since they can't agree on it. Since the truthful
>citizens know this as well, they give you an answer which a liar would give you,
>but this is no use unless the liar's answer is well defined.
>
>Right?

I think that your worries could be summarized in an observation that the
above approach will not work in any first-order extensional (i.e. purely
truth-functional) language, regardless of any additional assumptions about
the knowledge of the liars and the truth-tellers concerning each other's
behavior.  At least so it would seem at first blush, unless we can
eliminate the strong conditional by replacing it with some sort of
dispositional idiom.  Would anyone care to try?

>Later,
>Andrew Mullhaupt

cordially,
mikhail zeleny@husc.harvard.edu
"Un de mes plus grands plaisirs est de jurer Dieu quand je bande."

P.S. The Phallogocentric Cabal has concluded its activities for this
summer.  The next meeting will occur during the week of September 18th,
with Tal Kubo speaking on infinite combinatorics.  T-shirts, syllabi, and
other paraphernalia may be requested by responding to this message.

The simplest question to ask someone is "Do you live here?".
If you are in the dsihonest city anyone you ask will say "no", but
if you are in the honest city anyone will say "yes".

--
Mark Biggar
mab@wdl1.wdl.loral.com

In article <1992Aug19.090212.1658@gacvx2.gac.edu> logic@gacvx2.gac.edu writes:
>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..

While this isn't particularly difficult, it's much easier if we are allowed
to assume that there is a unique answer.  If you just have each person shake
hands with exactly the people (s)he previously did not shake hands with, you
get the same distribution, but each person shook hands with 8-n people.  The
assumption that you know how many people one person shook hands with means
that n=8-n, so n=4.