HDTenEightyP/NetNews huggingface dataset:
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Newsgroups: rec.puzzles
Subject: Re: Turing Machines
Date: 26 Jan 93 14:27:26 GMT
Article-I.D.: mentor.C1GtHr.n7
Organization: Purdue University

In article <1k2ltpINNs62@gap.caltech.edu> carl@SOL1.GPS.CALTECH.EDU (Carl  
J Lydick) writes:
> In article <728019101.AA05890@csource.oz.au>,  
Ben.White@f364.n633.z3.fidonet.org (Ben White) writes:
> >Has anyone here ever constructed a universal turing machine? 
> I tried once, but I ran out of memory :-).  Seriously, by imposing the
> adjective "universal," you've required that the machine have infinite  
memory.

Wrong adjective. All turing machines, universal or not, have infinite  
memory. 

--
Dave Seaman
ags@seaman.cc.purdue.edu
Newsgroups: rec.puzzles
From: rolf@geoserv (Rolf Wilson)
Subject: Re: flying puzzle
Reply-To: rolf@geoserv.isgs.uiuc.edu
Organization: University of Illinois at Urbana
Date: Thu, 28 Jan 1993 15:05:01 GMT

dementia@cheshire.oxy.edu (Daniel Steven Reinker) writes:

>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?

   As someone else suggested, follow a jet if you can fly at 600mph.
However, even if you could, you might need more protective clothing
than you could buy with $50 to protect yourself at that speed. Also,
you would run the chance of losing the jetliner, because you would
have to fly low most of the time (not enough money for breathing
equipment) and a heavy cloud cover might separate you visually. Of
course, a quick climb could be done up through the clouds (Would you
want to fly through a storm cloud?) Then again, you might be able
to fly above most clouds (no problem breathing at 10,000 feet, maybe
up to 15,000 - you might be used to that if you lived on a mountain
in Hawaii) and watch the jet at 30,000 feet.
   I think that you could do it even without the jet. Just buy a good
compass, test your flying speed and look up the direction of Hawaii in
the library. Fly the right distance, watching the compass, then take a
quick climb up to 30,000 feet and look around. You could not breathe up
there for long, but there are 9-10,000 foot mountains that you should
be able to see from hundreds of miles away. A little calculation would
tell you how far you could see Manua Kea, and how accurately you would
have to follow the compass.
-- 
Rolf Wilson     Illinois State Geological Survey    rolf@geoserv.isgs.uiuc.edu
From: mag@dogwood.cs.uiuc.edu (Tom Magliery)
Newsgroups: rec.puzzles
Subject: Re: 27+2=30
Date: 26 Jan 93 14:25:27 GMT
Reply-To: mag@cs.uiuc.edu
Organization: University of Illinois, Dept. of Comp. Sci., Urbana, IL

haddadi@sipi.usc.edu (Navid Haddadi) writes:

>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!

Actually, to most readers of rec.puzzles, it's probably more tiresome
than funny.  I believe I've decided to add this one to the FAQ list,
which is coming Really Really Imminently Now.  Should I put in the one
about guessing the sex of the unknown sibling?  That one bores me to
tears, but there seems to be a lot of traffic on it this time.  Do
people actually enjoy that puzzle, too?

mag

P.S.  Innkeeper has 25, bellboy has 2, 3 x (man has 1).  30.
-- 
Tom Magliery ** Dept of CS ** 1304 W Springfield ** Champaign IL 61820 ** USA
Newsgroups: rec.puzzles
From: brian@maui.cs.ucla.edu (byron elbows / Brian Tung)
Subject: Re: Two Circles Puzzle
Summary: This has been done, I believe
Keywords: geometry
Organization: UCLA, Computer Science Department
Date: Wed, 27 Jan 93 20:24:54 GMT

Dan Bergmann writes:

> 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?

My only comment: how many days are there in a year?  365.25?  366.25?  (And
yes, I know that the tail is actually something like .242198+.  Stick to the
point.)

The smaller circle makes two "solar" rotations, three "siderial" rotations.

Re: the 13 spheres

Yes, I was able to track down an old copy of "Mathematical Circus," by
Martin Gardner of SA fame.  In it, Coxeter says that David Gregory and
Isaac Newton had been talking about the distribution of stars of various
magnitudes, and this had led to the discussion of whether thirteen unit
spheres could touch a fourteenth.  Gregory believed they could, and Newton
disagreed.  "One hundred eighty years were to pass before Newton was proved
correct."  It's close: I was able to place 11 down and two more which
intersected two of the 11 by just a sliver.

So, what's the answer for the fourth dimension?  I'd guess that it's
something around 19 or 20.

A last question:

Q.  Three circles of any radii r1, r2, and r3 can be drawn so that they
    are mutually tangent.  Also, a fourth circle can be drawn that is
    tangent to the first three circles, but its radius is determined
    by those of the other circles.  If the circles are "close enough"
    in radius, then the fourth circle surrounds the first three;
    otherwise, it is external to the first three.  Prove or disprove:
    in those cases where the fourth circle surrounds the other three,
    those three circles consume the maximum fraction of the area of
    the fourth circle when r1=r2=r3.

byron elbows
(mail to brian@cs.ucla.edu)
From: radcliff@csd4.csd.uwm.edu (David G Radcliffe)
Newsgroups: rec.puzzles
Subject: Re: Are you sure?
Date: 26 Jan 1993 23:19:39 GMT
Organization: UW-Milwaukee Math Department
Distribution: world
NNTP-Posting-Host: 129.89.7.4
X-Newsreader: TIN [version 1.1 PL8]

Sean Morgan (morgan@arc.ab.ca) wrote:
: 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?"

Suppose there are just four families.  Each family has two children.
One family has two boys, a second has two girls, and the other two
have one of each.  Denote these families as BB, GG, BG, and GB.

There are four girls, and two of them have sisters.  If you pick
one of the girls at random, then the probability is 1/2 that she
has a sister.

Now suppose you pick one of the mothers at random.  You ask her if
she has at least one daughter, and she says that she does.  The probability
that both of her children are daughters is 1/3.

Do you see why the answers are different?  The woman must belong to
GG, GB, or BG; each of these cases are equally likely.  The girl must
also be a member of GG, GB, or BG; but GG is twice as likely as the
other two cases. 

Let us consider another situation.  Suppose you wish to find the
average family size in a community.  If you pick children at random
and record the number of members of their families, then large families
will be overrepresented.  You would get a more accurate result if
you pick mothers at random.  (I have ignored certain sociological
realities here, of course.)

--
David Radcliffe                                
radcliff@csd4.csd.uwm.edu            
Newsgroups: rec.puzzles
From: nick@eskimo.com (Nick Adami)
Subject: Need help finding a puzzle
Organization: -> ESKIMO NORTH (206) For-Ever <-
X-Newsreader: TIN [version 1.1 PL6]
Date: Sat, 23 Jan 1993 00:35:46 GMT

Does anyone know where I can get the puzzle called pentominoes (It may be
called by other names).  It has twelve pieces and they look like this:

     X   XXX    XX    XXX  X X  XXXXX  XXXX   XXX               each made
    XXX    X     XX    X   XXX          X     XX    ..... etc   from 5 X's
     X     X      X    X   

Anyway your supposed to put them all together in a rectangle or something.
Does anyone know the name of a company that makes this?  Or where I can get
one?

Email responses preferred.  Thanks a bunch...

Nick

Newsgroups: rec.puzzles
From: camus@leland.Stanford.EDU (Theodore C. Quinn)
Subject: Re: Are you sure?
Organization: DSG, Stanford University, CA 94305, USA
Date: Tue, 26 Jan 93 22:51:24 GMT

In article <1993Jan26.204422.15093@zip.eecs.umich.edu> kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:

[Original problem deleted]

>Why don't you consider this metaphor : you have these teeming
>billions of sibling-pairs in the Earth (yeah, I mean all continents,
>islands, places hitherto undiscovered or unknown, etc. etc. taken
>together ad nauseam :-) ) ?  Now, it's quite easily seen that FF 
>pairs comprise only 25 % of _this_ population (assuming that the
>birth events of a pair of siblings are independent, thereby precluding 
>identical twins, etc.). 50 % of this population will comprise MF
>sibling pairs, and the remaining 25 %, MM pairs.

[Metaphor deleted]

>Now without any other info, 
>it's a fair assumption that you'll encounter sibling pairs on the Earth
>in proportion to their relative frequencies of occurrence.  Hence, 
>the female sibling you encounter is twice as likely to be a part of a
>mixed sex pair than of an FF pair.  Hence, the other sibling being a
>female has probability 1/3.

	Nobody seems to have noticed my original argument for 1/2, so I'll
respond to this one. Contrary to the author's intention, this is actually
a good way to see that the answer is 1/2. The key, once again, is that
there are two females in an FF pair and only one in a MF or FM pair.
Taking a cue from his argument, assume that there are 1000 sibling pairs
in the world. Of these pairs, 250 are MM, 250 are MF, 250 are FM, and 250
are FF. Now you see a female, so you know that she can't be part of an
MM. This limits you to the other 750 pairs. While it is true that there
are 500 mixed pairs remaining and 250 FF pairs remaining, there are 2 females
in each FF pair. Therefore, of the 1000 females in the mixed pairs and FF
pairs, 500 are in mixed pairs and 500 are in FF pairs. There is a 50%
chance the female you're looking at is from an FF pair, and hence her
sibling is female.
				-Ted Quinn
				 quinn@relgyro.stanford.edu
From: strgh@csv.warwick.ac.uk (J E H Shaw)
Newsgroups: rec.puzzles
Subject: Re: What's this a diagram of?
Date: 28 Jan 1993 13:00:00 -0000
Organization: Computing Services, University of Warwick, UK
NNTP-Posting-Host: violet.csv.warwick.ac.uk

In article <1993Jan28.001840.18529@sq.sq.com> dave@lsuc.on.ca (David Sherman) writes:
>                 ----__i_i
>                 |  %    /o -v\
>                  \    % \_/   \
>                  \`-----=,_--'/ o
>
>The above was constructed by a friend of mine.  He claims it's
>a good ASCII representation of .... what?
>
>David Sherman

A dead one of these:-
                       _
                 o /,-- '=-----,\
                   \   /-\ %    \
                    \^- o/ _ __%  |
                          ! !  ----

(shouldn't this be in the FAQ?)
-- 
J.E.H.Shaw,  Department of Statistics,  |  JANET:  strgh@uk.ac.warwick
             University of Warwick,     |  BITNET: strgh%uk.ac.warwick.cu@UKACRL
             Coventry  CV4 7AL,  U.K.   |  PHONE:  +44 203 523069
yacc - the piece of code that understandeth all parsing
From: cmh@eng.cam.ac.uk (C.M. Hicks)
Newsgroups: rec.puzzles
Subject: Re: Dividing apples
Date: 26 Jan 93 11:47:40 GMT
Organization: cam.eng

bcash@crchh410.BNR.CA (Brian Cash) writes:

><SPOILER>  (maybe...)

>Is it 351?  I figured it this way:

I think so, and I worked it the same way. I'm now trying to find a general
solution for dividing Q apples among P people - I'll post if I find one.

>Call the number of apples given to person 1 N, person 2 X, and person 3 Y.
>Now if N = 25, then X and Y must both equal 0. (1 solution)
>If N = 24, then {X=1,Y=0} and {X=0, Y=1} are valid. (2 solutions)
>If N = 23, then {X=1, Y=1}, {X=0, Y=2}, {X=2, Y=0} are valid. (3 solutions)
>If N = 22, then there are 4 solutions.
>If N = 21, then there are 5 solutions.
>...
>If N = 0, then there are 26 solutions (run X from 0 to 25 and Y is
>forced to a value).  
>So the total number is 1+2+3+4+...+26 or 351.

Christopher Hicks
--
 ==============================================================================
 Christopher Hicks    |    If it doesn't fit...
 cmh@uk.ac.cam.eng    |                        ...you need a bigger hammer.
 ==============================================================================
From: cmh@eng.cam.ac.uk (C.M. Hicks)
Newsgroups: rec.puzzles
Subject: Re: Are you sure? YES <SPOILER>
Date: 26 Jan 93 11:55:45 GMT
Organization: cam.eng

karr@cs.cornell.edu (David Karr) 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?"

One half - The probability of the other child being a girl is totally
independent of the sex of the first. The first child being
a girl does affect the 50-50 boy-girl distribution of the second (or any 
other) child.

Even if your friend had 10 children, and you met nine of them in the street,
all girls, then the probability of the tenth being female is still one half.

Of course, the probability of having ten girls is 1/1024 (assuming no
genetic oddities), but the probability of any one child being a girl is
one half

Christopher Hicks
--
 ==============================================================================
 Christopher Hicks    |    If it doesn't fit...
 cmh@uk.ac.cam.eng    |                        ...you need a bigger hammer.
 ==============================================================================
Newsgroups: rec.puzzles
From: rkbg2554@uxa.cso.uiuc.edu (Robert K. Brunner)
Subject: Re: flying puzzle
Organization: University of Illinois at Urbana
Date: Thu, 28 Jan 1993 01:52:48 GMT

dementia@cheshire.oxy.edu (Daniel Steven Reinker) writes:

>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

You could follow a commercial airliner, if you can keep up.  Just go
to the airport and wait for a departing flight.

Assuming you aren't up to such speeds...

Get an AM radio.  Find the frequency for some radio station in Hawaii.
It would as probably be helpful to find the approximate direction to
Hawaii, so you know which direction to start.  Go in your initial
direction until you can recieve the radio station.  Determine in which
direction your radio is most sensitive by tuning an LA station as you
leave and note which direction you must point your radio for best
reception.  When you are in the vicinity of Hawaii (within several
hundred miles at night), you should be able to follow the signal in.
As you get closer, the signal may be strong in all directions, so you
may have to switch to a low-power station so you can determine
direction more accurately.  Hopefully, this should get you close
enough to follow city lights in at night, or perhaps sea traffic
during the day.

This is just one idea. I could probably come up with several others if
I thought about it.  One problem is that the AM radio can not tell you
if you are moving towards or away from the transmitter, so you may
have to backtrack if you discover you are getting farther away and the
reception becomes worse.  Send me a postcard when you get there.

Robert Brunner
brunner@uirvld.csl.uiuc.edu
Newsgroups: rec.puzzles
From: bcash@crchh410.BNR.CA (Brian Cash)
Subject: Re: Dividing apples
Date: Mon, 25 Jan 1993 03:51:30 GMT
Reply-To: bcash@crchh410.BNR.CA (Brian Cash)
Organization: Bell-Northern Research Ltd.

<SPOILER>  (maybe...)
























































Is it 351?  I figured it this way:

Call the number of apples given to person 1 N, person 2 X, and person 3 Y.
Now if N = 25, then X and Y must both equal 0. (1 solution)
If N = 24, then {X=1,Y=0} and {X=0, Y=1} are valid. (2 solutions)
If N = 23, then {X=1, Y=1}, {X=0, Y=2}, {X=2, Y=0} are valid. (3 solutions)
If N = 22, then there are 4 solutions.
If N = 21, then there are 5 solutions.
...
If N = 0, then there are 26 solutions (run X from 0 to 25 and Y is
forced to a value).  
So the total number is 1+2+3+4+...+26 or 351.
I'm rather math-impaired, so I would like to know 1) is this right? 2)
what is a
more "scientific" way of finding the answer.

Brian /-|-\
Newsgroups: rec.puzzles
From: t91matfa@odalix.ida.liu.se (Mattias Fagerlund)
Subject: Re: A puzzle - The True Answer
Organization: CIS Dept, Univ of Linkoping, Sweden
Date: Sat, 23 Jan 1993 15:35:24 GMT


Well, well, well.
It seems a lot of people knew that one?

Nope, the true answer, given to me by Sharon Levine is

>The object is to piss of the bartender by asking for somthing
>free. It worked.

There you go.
	-Mattias Fagerlund.
Date: Monday, 25 Jan 1993 09:59:49 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Need puzzle help: eye colour enigma
 <93021.120138RVESTERM@vma.cc.nd.edu> <C19tJo.EzC@news.cso.uiuc.edu>

somebody asks why it wouldn't be more logical for the islanders to
"cancel this stupid tradition immediately".

ah, but as mr. spock also says, "logic, logic, logic... logic is the
beginning of wisdom, not the end."

bob vesterman.
From: padmanab@ecs.umass.edu
Newsgroups: rec.puzzles
Subject: Solution to dividing apples
Date: 27 Jan 93 18:21:28 GMT

>I had posted the following problem last week: 

>Suppose there are 25 identical apples. How many 
>ways can you divide them among three people? 

The reason I posted it was because I saw a vey 
interesting solution to this problem.  

Add two identical `oranges' to this collection 
of apples. The total number of ways of arranging 
this new collection (i.e., permutations) is 

FACTORIAL(25+2) / ( FACTORAL(25)*FACTORIAL(2) )

Now observe that each of the permutation is a way of 
dividing the apples among three people!! 

Of course this can be generalized into the problem of 
dividing `n' similar objects among `m' people. 

-Prasad Padmanabhan
Dept. of ECE
University of Massachusetts
Amherst MA 01003
From: chris@questrel.com (Chris Cole)
Newsgroups: rec.puzzles
Subject: Re: surgeon puzzle
Keywords: situation logic
Date: 27 Jan 93 18:21:29 GMT
Organization: Questrel, Inc.

In article <C1Gzsu.7pq@fastrac.llnl.gov> dan@danberg.llnl.gov (Dan Bergmann) writes:
>
>A man and his son are involved in an automobile accident. The man dies and
>his son sustains injuries. The son is rushed to a hospital and wheeled 
>into the operating room. The surgeon walks in and says, "I can not 
>operate on this boy. This boy is my son."
>
This question is in the the list of situation puzzle outtakes in the
rec.puzzles Archive.

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containing the line:
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From: benha@castle.ed.ac.uk (Ben Hambidge)
Newsgroups: rec.puzzles
Subject: Re: Cities with an internal border
Date: 26 Jan 93 11:39:43 GMT
Organization: Edinburgh University

Possibly in Denmark/Sweden. A large built up area exists (I believe)
stretching from Copenhagen to Malm"o. And right by the water, one on
either side, is Helsingborg and Helsingor (same meaning, just different
spellings in different languages.

Ben.

-----------------------------------------------------------------------------
Ben Hambidge    |                       |Bread is the staff of life; Toast is
4 Merchiston Ave| benha@castle.ed.ac.uk |a decadent capitalist luxury.
EDINBURGH       |-----------------------|------------------------------------ 
EH10 4NX        |Tel: +44 31 447 4047   |Smile they said, life could be worse.
U.K.            |                       |So I did, and it was.
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Newsgroups: rec.puzzles
From: ilan@leland.Stanford.EDU (ilan vardi)
Subject: Re: Reverse city names (Was Re: Cities with an internal border)
Organization: DSG, Stanford University, CA 94305, USA
Date: Tue, 26 Jan 93 21:38:31 GMT

Rome, Italy.
Date: Sunday, 24 Jan 1993 16:59:53 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: hoosier-cash-lotto-inspired puzzle.

you see the results of a lottery drawing: n numbers, i1, i2, ..., in.
what is the chance that the highest number possible in the lottery is
k? you can assume that the lowest possible number is one, that the
possible numbers are all integers, and that if j is possible and j is
not equal to one, then j-1 is possible. also, what is the chance that
there is no k such that k is the highest number possible?

bob vesterman.

Newsgroups: rec.puzzles