HDTenEightyP/NetNews huggingface dataset:
From: brownea@cc.uvcc.edu
Subject: Re: Need puzzle help: eye colour enigma
Organization: Utah Valley Community College
Date: Thu, 21 Jan 1993 17:03:24 GMT

> There is an island,on which there are 100 blue eyed people and 100 brown
> eyed people.  The people, however, do not know the number of people with
> each eye colouring.  No-one knows their own eye colour, as there is no means
> of reflection, and no-one will tell anyone else their eye colour, since as
> soon as someone knows their own eye colour, they kill themselves 24 hours
> later. 
> Once every 100 years, a green-eyed guru speaks to the populus, and today's
> the day (quite by chance).  S/he stands on a rise, looking out over all these
> people (who don't know their own eyecolour), and announces:  "I can see
> someone with blue eyes"
> 
>     Who kills themselves, and when???

	I'm not sure who does it, but I think it has to do with filtered
eye-color.  Like one persons eye-color affects the way they see colors?
	Just an idea...
	Earl
Newsgroups: rec.puzzles
From: djweisbe@unix.amherst.edu (David Weisberger)
Subject: Re: Need puzzle help: eye colour enigma
Organization: large
X-Newsreader: TIN [version 1.1 PL7]
Date: Tue, 26 Jan 1993 07:21:48 GMT

THOMPSON  BENJAMIN RHINELANDER (thompson@spot.Colorado.EDU) wrote:
: In article <1993Jan19.035603.19176@CSD-NewsHost.Stanford.EDU> bhayes@Xenon.Stanford.EDU (Barry Hayes) writes:
: >In article <1993Jan19.143459.13275@waikato.ac.nz> deryl@waikato.ac.nz writes:
: >> [FAQ deleted]
: >
: >Once the guru has spoken, is there anything he or anyone else can say
: >or do that will stop the chain of events?
: >
: 
: 	If one blue-eyed person immidiately shouts "I see a blue-eyed
: person too!" doesn't that save everyone?
: 
: 
But the people don't know their own eye colors.  I suppose everyone could
yell it, then one blue-eyed person would be, but does that help?

--
Davebo                    "Don't believe I'm taken in by stories I have heard.
                           I just read the Daily News and swear by every word."
djweisbe@unix.amherst.edu
From: haddadi@sipi.usc.edu (Navid Haddadi)
Newsgroups: rec.puzzles
Subject: 27+2=30
Date: 26 Jan 1993 00:24:20 -0800
Organization: USC, LA, CA
NNTP-Posting-Host: sipi.usc.edu

Here is a very old puzzle:
Three men rent a room for $30. The inn keeper decides that he has
over charged them and gives the bellboy $5 to give back to the men.
The bellboy decides to keep $2 and give each man $1. 
So, each man has paid $9 (3x9=27) plus the $2 that the boy kept.
Hence, 27+2=30.

It is actually more funny than silly!
From: dboese@spartan.ac.BrockU.CA (Darcy Boese)
Newsgroups: rec.puzzles
Subject: Re: "map" of USA
Date: 24 Jan 93 16:34:43 GMT
Article-I.D.: spartan.1993Jan24.163443.17553
Organization: Brock University, St. Catharines Ontario
X-Newsreader: Tin 1.1 PL4

Nope: it is not necessary for cities to match.  Consider a scale map the
size of (say) a football field.  Now consider one the size of a pinhead.
(There are computer circuits sufficiently complicated that a level of
detail this fine can be assumed)

Now you must realize that there is physical space between the cities:
the United States is not one gigantic city.  Place the pinhead-sized
map in a space between cities on the larger map... then no cities can
possibly match.

This is not the same problem as a POINT matching another point on each
map, where we have mathematically that some point coincides.
From: dan@danberg.llnl.gov (Dan Bergmann)
Newsgroups: rec.puzzles
Subject: Two Circles Puzzle
Keywords: geometry
Date: 27 Jan 93 18:07:59 GMT
Organization: Lawrence Livermore National Laboratory
Originator: dan@danberg.llnl.gov


I'm new to this newsgroup, so I hope this hasn't been asked before (it
wasn't in the FAQ).

This question was asked on an SAT test about a dozen years ago.

Two circles are next to each other and touching at one point (one is NOT
inside the other). The larger circle has twice the radius of the smaller
circle. The smaller circle rolls around the large one until it comes back
to its original starting position. How many revolutions does it make about
its center?

The testing service got the problem wrong, some kid in Florida complained
that he got it right but was not given credit, and the story made it into
the newspapers.

-- 


*************************************************************************
**             							       **
**  Dan Bergmann                             (510) 423-6765            **
**  mail stop L-262						       **
**  Lawrence Livermore National Lab          dbergmann@llnl.gov        **
**  Livermore, CA    94550                                             **
**               						       **
*************************************************************************
-- 


*************************************************************************
**             							       **
Date: Wednesday, 27 Jan 1993 15:58:27 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Are you sure? NO - flame-resistant clothing ON <SPOILER>
 <1993Jan25.145759.2592@cs.cornell.edu> <1993Jan26.115545.11556@eng.cam.ac.uk>
 <1993Jan26.121406.11680@eng.cam.ac.uk>

the probability of A, given B, is the probability of A and B, divided
by the probability of B.  does anyone dare dispute me?

no? good.  to continue, let's dissect the question: "i have two children.
at least one is a daughter. what's the chance that the other is a daughter?"
is the same as "what is the chance that both of my children are female, given
that at least one is female?" so we have that A is "both of my children are
daughters", and B is "at least one of my two children is a daughter."

so the chance that both are daughters given that one is a daughter is
the chance that both are daughters AND at least one is a daughter, divided
by the chance that at least one is a daughter.

the chance that both are daughters and at least one is a daughter is
the same as the chance that both are daughters (i.e., "at least one is
a daughter" is a foregone conclusion, if both are daughters), which is
one out of four.

the chance that at least one is a daughter is one minus the chance that
both are sons, which is one minus a quarter, which is three quarters.

one quarter divided by three quarters is one third.  last i checked,
a third and a half were not the same.

to all you proponents of the so-called "common sense" argument,
remember the archaic definition of "common".

bob vesterman.


Organization: Doctoral student, Industrial Administration, Carnegie Mellon, Pittsburgh, PA
Newsgroups: rec.puzzles
Date: Tue, 26 Jan 1993 10:38:50 -0500 
From: "Bruno W. Repetto" <br0w+@andrew.cmu.edu>
Subject: Re: Are you sure?

Monty Hall revisited?  The last 2 posts sure have that flavor...

Bruno.  br0w+@andrew.cmu.edu
Date: Wednesday, 27 Jan 1993 16:16:40 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: No subject

you, eduardo, and annette are prisoners, and the warden says,
"two of you, chosen at random, are going to be executed tomorrow."
clearly, you have a 1/3 chance of living.  at least one of annette
and eduardo will die, so you ask the warden to tell you the name of
a person, besides yourself, who will be executed.  "annette's gonna
bite it," he says.

now, what's the chance of your survival? it is the chance that eduardo
will be executed, given that annette will be executed.  that is equal
to the chance that both eduardo and annette will be executed, divided
by the chance that annette will be executed.  that's 1/3 divided by
2/3.  that's 1/2.

now you wake up.  your conversation with the warden was a dream.  but,
you think, if i did ask the warden the question, and he did tell me
a name, no matter what name it is, i'd still have a 50/50 chance of
living - so i must have a 50/50 chance anyway!

well, sorry, the fact remains that you did NOT have a conversation, and
the warden did NOT tell you a name.  the only information that you know
is that at least one of eduardo and annette will die.  the chance that
you will live, given that at least one of e. and a. will die, is the
chance that you will live AND at least one of e. and a. will die, divided
by the chance that at least one of e. and a. will die.  that's one third
divided by one.  that's one third.

this is much like the children puzzle - the chance that the other kid is
a girl is different if you know that at least one is a girl than it is
if you know that the younger is a girl.

bob vesterman.


Newsgroups: rec.puzzles
From: matthew@base.cs.ucla.edu (Matthew Merzbacher)
Subject: Re: Cities with an internal border
Organization: UCLA, Computer Science Department
Date: 25 Jan 93 19:13:59 GMT

dant@techbook.com (Dan Tilque) writes:

>So far, I've found about half a dozen in North America (some are fairly
>small though), but none on other continents.

If you watch "The Third Man", you will know that Vienna *used* to be
divided into five "quarters" (Russian, American, French, British and
International).  Of course, for years Berlin was divided.

Also, Jerusalem was politically divided between 1948 and 1967.

--

Matthew Merzbacher	ARPA:	matthew@CS.UCLA.EDU
Moo - Moo Moo  		UUCP:	...!{uunet|rutgers|ucbvax}!cs.ucla.edu!matthew
               		This time, *with* the oys
From: jomp@mentor.dk (Jan Otto Munch Pedersen)
Newsgroups: rec.puzzles
Subject: Re: Cities with an internal border
Date: 26 Jan 1993 13:43:39 +0100
Organization: Mentor Informatik, Denmark
NNTP-Posting-Host: plys.mentor.dk

benha@castle.ed.ac.uk (Ben Hambidge) writes:
>Possibly in Denmark/Sweden. A large built up area exists (I believe)
>stretching from Copenhagen to Malm"o.

???. Have you looked at a map recently? There is more than 20 km's of
water between K/obenhavn and Malm"o. The bridge under construction
will not be finished before something like year 2000.

>                                      And right by the water, one on
>either side, is Helsingborg and Helsingor (same meaning, just different
>spellings in different languages.

No! borg == castle. /or == cape.

Jan
Newsgroups: rec.puzzles
From: dcarroll@oucsace.cs.ohiou.edu (Dana Carroll)
Newsgroups: rec.puzzles
Subject: Re: Are you sure?
Date: 25 Jan 93 23:48:00 GMT
Organization: Ohio University CS Dept,. Athens

In article <1993Jan25.115908.41@janus.arc.ab.ca> morgan@arc.ab.ca (Sean Morgan) writes:
>In article <1993Jan22.131719.36@janus.arc.ab.ca> morgan@arc.ab.ca (Sean Morgan)
>writes:
>>You meet the daughter of a friend of yours on the street.  You know that your
>>friend has two children.  You think to yourself, "Ignoring the occurrence of
                                                    ++++++++++++++++++++++++++

>>twins, and sex-linked differences in birth rates and infant mortality, what 
  +++++

>>are the odds of this girl's sibling also being a girl?"
>>-----------------+-------------------------------
>

And the answer...


>
>Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
>there is only a one third chance that the sibling is a girl.

Really?  Ignoring the possibility of twins rules out FF, so the chance that
the other sibling is a female is zero!


(Please insert smilies as you see fit).

--Dana

disclaimer: Ohio University ignores my opinions.

(extra lines for my poster....)
From: grue@cs.uq.oz.au (Pauli)
Newsgroups: rec.puzzles
Subject: Re: Computer Dictionary Searches
Date: 22 Jan 93 02:58:40 GMT
Reply-To: grue@cs.uq.oz.au

warwick@cs.uq.oz.au (Warwick Allison) writes:

>jpk2581@hertz.njit.edu (joseph p kisenwether apmt stnt) writes:

>>I am trying to create some original word puzzles and need to find 
>>words with certain characteristsics:
>>For example
>> two words of more than ten letters which differ by only one 
>>stictly internal letter such that there exist two other letter
>>(one to the left and one to the right) such that in word one
>>there exists no sub-word containing both the changable letter 
>>and the one to the left while in word two there exists no sub-word 
>>conatining both the changable letter and the one to the right.

Very complex problem.  The expensive bit will be the finding of the pairs
of ten letter words.  The testing if something is a word is very very fast
if you follow the scheme Warwick proposes.  Even better you can precompute
most of this stuff.  Keep a table of 10 letter words and the positions where
the changable letter satisfies the condition for left and right.  My word
list contains about 40000 10 letter words, so this isn't too excessive a task.


>>c) what the format is so that I can right the appropriate search 
>>    program

>A Trie.  You could use a DAWG, but we (Paul Dale & I) have found that
>a Trie is better, because you can store suplementary data for arcs
>in the tree.

The dictionary comes in plain text.  A DAWG or Trie happens to be a very
efficient way of searching for words given some constraints (which can
be quite general) or for determining if this word is really a word (complexity
is O(length of word) with a small constant multiplier, which should be as good
as a hash table).

Interestingly, the construction of a Trie from a plain text word list is
O(length of word list) and conversion from that to a DAWG is O(n log n) where
n is the number of nodes in the Trie.  This does assume that you have plenty
of space to hold intermediate structures before you pack everything down.

I've also implemented a best of both worlds approach:  A DAWG containing the
bulk of the word list and a secondary Trie that details additions/deletions
from the DAWG.  This gives the space savings of a DAWG with the editing
facilities of a Trie.  It doesn't allow data to be tagged with every word
like a pure Trie would.



        						Pauli

Paul Dale                       | grue@cs.uq.oz.au
Department of Computer Science  | +61 7 365 2445
University of Queensland        |
Australia, 4072                 | Did you know that there are 41 two letter
                                |     words containing the letter 'a'?

--
From: casgrain@ERE.UMontreal.CA (Casgrain Philippe)
Newsgroups: rec.puzzles
Subject: Re: A puzzle?
Date: 22 Jan 93 00:47:38 GMT
Organization: Universite de Montreal

In article <1993Jan21.115427.29195@ida.liu.se> t91matfa@odalix.ida.liu.se (Mattias Fagerlund) writes:
>A man walks into a saloon and asks the bartender for a glass of water.
>The bartender, saying nothing, pulls out a shot-gun and points it at the man.
>-Thank you, the man says and leaves the bar.
>
>Question: 
>  What really happend? Why did the man say thank you (his goal WAS actually 
> achived) and leave?
>
The man had a hiccup. He needed either to drink some water or be scared.

Philippe


-- 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Philippe Casgrain              Etudiant-Chercheur      Casgrain@ERE.UMontreal.CA
Departement des Sciences Biologiques                      Universite de Montreal
#define disclaimer(caught) (caught ? "I wasn't even there!" : "I didn't do it!")
From: kubo@birkhoff.harvard.edu (Tal Kubo)
Newsgroups: rec.puzzles
Subject: Re: Turing Machines
Date: 27 Jan 93 21:41:25 GMT
Article-I.D.: husc3.1993Jan27.164126.19870
Organization: Dept. of Math, Harvard Univ.

In article <1993Jan27.123943.2699@odin.diku.dk> 
torbenm@diku.dk (Torben AEgidius Mogensen) writes: 
>
>von Neumann constructed a TM with only 6 states that can
>simulate any other TM by coding it's "program" on the tape together
>with it's initial input. Such a TM can can indeed be called
>"universal". In this sense: yes, someone has indeed constructed a
>universal Turing machine.

Minsky has constructed of universal TM's with various small values 
of the pair (number of states, size of alphabet).  See his book 
"Computation: Finite and Infinite Machines".  Some of the values
were surprisingly small, with both numbers being less than 10.
Minsky's book is pretty old, and the state of the art may have advanced
since it was published.

>von Neumanns universal TM has incidentially been used to prove that 1
>dimensional cellular automata are universal: someone (I don't recall
>who) has constructed a 1D CA that simulates von Neumanns universal TM.

I think it was Alvy Ray Smith's PhD thesis sometime in the
1970's. This was mentioned in an old Martin Gardner column 
on the game of Life (which is itself a universal 2DCA) .  
If memory serves, Smith's universal 1DCA  had an  alphabet 
of 239 symbols.   Does anyone have more references on this
subject?

Tal   kubo@math.harvard.edu
Newsgroups: rec.puzzles
From: cooper@cbnewsg.cb.att.com (Ralph 'Hairy' Moonen)
Subject: RAAARRGGHH!!!!!  not again! (was:Re: Are you sure?)
Organization: AT&T
Date: Tue, 26 Jan 1993 09:52:18 GMT

In article <1993Jan25.115908.41@janus.arc.ab.ca>, morgan@arc.ab.ca (Sean Morgan) writes:
> 
> (Some of my mail says "not again", and some says "please post a solution".
>  Since you can't please all of the people all of the time...)
> 
> Two child families are uniformly distributed between MM,MF,FF,FM, in birth
> order.  Again, all four possibilities are equally likely.  Stop here.  Make
> sure you believe that.
> 
> Now, since at least one is F, the MM drops out, leaving MF, FF, FM, so that
> there is only a one third chance that the sibling is a girl.

Birth order? No one mentioned age here. The question was, that given one
girl, what are the chances of the other also being a girl. Given a uniform
distribution, that means for 1-child families, P(girl) = P(boy) = 0.5
Therefore, given that one child is a girl, the chance that the other child
is a boy, is 0.5, same as if it were a girl. Try this analogy, and you'll see
what's wrong with your answer.

You are throwing 2 consecutive dice. The first one turns up a six. What is
the chance of the second also being a six. It's easy to see that because dice
have no memory, the chance is also 1/6. Now, following your line of reasoning,
there are 36 different ways to throw the dice. Because 1 six already has
been thrown this leaves 11 combinations, one of which is a double six.
So, according to you, you only have 1/11 chance of throwing a second six.
Clearly this is dead wrong.

--Ralph
Date: Monday, 25 Jan 1993 16:56:18 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Need help finding a puzzle

uh, if you know what the shapes look like, cut up the side of a cardboard
box, and save yourself some cash.

bob vesterman.

Newsgroups: rec.puzzles,sci.math,rec.games.misc
From: raph@panache.demon.co.uk (Raphael Mankin)
Subject: Re: "Cut & Choose" for several players 
Distribution: world
Organization: Solvfield Ltd.
Reply-To: raph@panache.demon.co.uk
X-Mailer: Simple NEWS 1.90 (ka9q DIS 1.19)
Date: Fri, 22 Jan 1993 23:31:14 +0000

The general rule is that the player who divides does not choose. This
negative feedback prevents bias in the divisions -- assuming that there 
are no collusions.A Byznatine generals' solution would be much more
complex.

Player 1 divides the pile into a 1-portion and a (n-1)-portion. Player 2
decides wehther he or player 1 gets the 1-portion. Whoever gets it drops
out. The division the continues with n-1 players and   the (n-1)-portion.

--------------
Raphael Mankin			Nil taurus excretum
Newsgroups: rec.puzzles
From: jonwa@ida.liu.se (Jonas Wallgren)
Subject: Re: Cities with an internal border
Organization: CIS Dept, Univ of Linkoping, Sweden
Date: Wed, 27 Jan 1993 16:48:19 GMT

In article <C1DzLH.8u7@techbook.com> dant@techbook.com (Dan Tilque) writes:
>How many cities do you know with an internal border?  What I really mean
>is adjacent cities with identical names, separated by a political
>boundary.  The border can be either internal (state/provincial) or
>international.  It also can include a river.

What is adjacent? Is there an upper limit on the distance between them?
What is a city? Is there a lower limit on its size?

I've collected some names from the Swedish-Finnish and Swedish-Norwegian
borders that more or less fit your rules. At least Pello fits perfect.
My source is the "KAK Bilatlas" i.e. the car atlas published by the Swedish
correspondence to the RAC.

In the list below the following letters are used:
A: Village, farm, church, etc.
B: Mountain, hill.
C: Densely built-up and populated area (in the Nordic countries defined as at
   least 200 inhabitants and at most 200 m between the houses).

The distances given are very approximate.

Some national characters found in the names:
{ = a with dots
\ = O with dots
} = a with ring

SWEDEN-FINLAND
==============

The border between Sweden and Finland is a river.

A. Identical names.
-------------------

Pello: C-C, just across the bridge
Kuivakangas: C-A, 1-2 km, no bridge
Korpikyl{: A-A, 2-3 km, no bridge
V{h{niva: A-A, 2 km, no bridge
Luppio: B-A, 7 km

B. Different spellings depending on phonetic properties of the languages.
-------------------------------------------------------------------------

Karungi-Karunki: C-A, 2 km, no bridge
Juoksengi-Juoksenki: C-C, just across the river, no bridge
Karesuando-Karesuvanto: C-A, 1 km, bridge
K{tkesuando-K{tkesuvanto: A-A, 1 km, no bridge

C. The names have the same meaning, but some parts of them are translated.
--------------------------------------------------------------------------

Nedre(Swedish)=Ala(Finish)=Lower(English)
\vre/\ver(Swedish)=Yli(Finnish)=Upper(English)

Nedre Vojakkala-Alavojakkala: A-C, 2 km, no bridge
\vre Vajokkala-Ylivojakkala: A-C, 2km, no bridge
\vertorne}-Ylitornio: C-C, 6 km, bridge

SWEDEN-NORWAY
=============

The border between Sweden and Norway consists of mountains and forests.

A. Identical names.
-------------------

Storberget: A-B, 1 km

B. Wery small differences.
--------------------------

Skalmodalen-Skarmodalen: A-A, 3 km, road
Juberget-Juvberget: A-A, 2km, path

-------------------------------------------------------------------------------
Jonas Wallgren                                 |                 jwc@ida.liu.se
Department of Computer and Information Science |
Linkoping University                           |-------------------------------
S-581 83  Linkoping                            |Everything that is systematized
Sweden                                         |becomes aesthetic --- B. Becher
-------------------------------------------------------------------------------
From: imcc@cs.strath.ac.uk (Iain McCord)
Newsgroups: rec.puzzles
Subject: Re: A sequence. (Partial spoiler)
Date: 21 Jan 93 14:18:31 GMT
Organization: Comp. Sci. Dept., Strathclyde Univ., Scotland.
X-Newsreader: Tin 1.1 PL5

X1 = 1/(1+X0)
X2 = (1+X0)/(2+X0)

If X0 = -1,
X2= 0 , and the limit is (5**.5 -1 )/2

		~~~~~/\~~~~~
Iain McCord	~~~~/()\~~~~	Thu Jan 21 14:18:29 WET 1993
		~~~~~~~~~~~~
Newsgroups: rec.puzzles
From: cooper@cbnewsg.cb.att.com (Ralph 'Hairy' Moonen)
Subject: Re: Cities with an internal border
Organization: AT&T
Date: Tue, 26 Jan 1993 12:19:35 GMT

In article <8731@charon.cwi.nl>, dik@cwi.nl (Dik T. Winter) writes:
> 	Kerkrade(NL) - Herzogenrath(D)

There is also I belive a small village on St. Maarten which has the 
French/Dutch border running through it.

--Ralph
Newsgroups: rec.puzzles
From: dementia@cheshire.oxy.edu (Daniel Steven Reinker)
Subject: flying puzzle
Organization: Occidental College, Los Angeles, CA 90041 USA.
Date: Wed, 27 Jan 1993 23:55:38 GMT

Here is something I've been trying to figure out lately. I'm not
sure if there is actually an answer, but there must be.

Say I live in Hawaii and go to Los Angeles for college. While
I am in Los Angeles, I discover that I have the ability to fly
at a fast speed. Now, I decide that since I can fly, I should
be able to fly back home to Hawaii. Consider that I am a college
student with about $50 in cash and a library at my disposal. How
can I navigate my way to make sure that I arrive in Hawaii?

Hiz Lord Dementia
-- 
"You can't help that. We're all mad here."-The cheshire cat, Alice in WL

"A mass hysteria/a megalomania/reveal Dementia/reveal"-Metallica                
Newsgroups: rec.puzzles