HDTenEightyP/NetNews huggingface dataset:
From: hsieh@math.toronto.edu (Paul Hsieh)
Subject: Re: surgeon puzzle (*SPOILER*)
Keywords: Simple puzzle.
Organization: Department of Mathematics, University of Toronto
Date: Wed, 27 Jan 93 20:14:13 GMT

In article 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 is an old one but a good one if you've never heard it before.  A good
test of your social attitudes as well ...

First hint: Suppose the boy's father was a nurse ...

Answer: ... and his wife was the surgeon in question. :)  The preceeding
hint of course is not necessary but its a useful hint.

Paul
Newsgroups: rec.puzzles
From: stein@hal.nta.no (Stein Kulseth)
Subject: Re: 21-link chain puzzle (SPOILER)
Organization: Norwegian Telecom Research
Date: Wed, 27 Jan 93 07:36:31 GMT

In article <727631103.12214@minster.york.ac.uk>, sjbuckle@minster.york.ac.uk writes:
|> Darcy Boese (dboese@spartan.ac.BrockU.CA) wrote:
|> >The other solution is this:
|> >
|> >1 1 3 5 11
|> >
|> >(yours was 1 1 3 6 10)
|> 
|> Surely there's also: 1 2 4 6 8.
|> Use 1,2,4,8 to make all numbers up to 15.
|> Then repeat how you made up 10-15 and throw in the 6 as well, to get 16-21.

How would you break links to make a 1 2 4 6 8 partition of the chain?
Look at this, here's a graphical representation of the 1 1 3 5 11 solution:

	OOO C OOOOO C OOOOOOOOOOO
(where Os are chained unbroken links, and the Cs are the unchained broken links)

And equivalently:

	OOO C OOOOOO C OOOOOOOOOO

For each break you get a single link. Your solution has only one single
link, yet the rest of the chain is broken into 4 pieces.

-- 
stein.kulseth@nta.no (Norwegian Telecom Research)
   'When murders are committed by mathematics, they can be solved by
   mathematics. Most of them aren't, and this one wasn't'
   - Nick Charles (Dashiell Hammett's "The Thin Man")
From: padmanab@ecs.umass.edu
Newsgroups: rec.puzzles
Subject: Dividing apples
Date: 24 Jan 93 16:02:37 GMT

Hi Netters, 

Here is a SIMPLE problem I came acroos long ago. 
I am posting it because I think it has a wonderful 
solution. 

Suppose there are 25 identical apples. How many 
ways can you divide them among three people? Also, 
all the apples must be divided among them. 

EXAMPLE:        person 1     person 2    person 3
                --------     --------    ---------
                   1             20          4 
                   5              0          20 
HAVE FUN! 

-Prasad Padmanabhan
Dept. of ECE
University of Massachusetts
Amherst MA 01003
Newsgroups: rec.puzzles
From: dboese@spartan.ac.BrockU.CA (Darcy Boese)
Subject: Re: "map" of USA
Organization: Brock University, St. Catharines Ontario
X-Newsreader: Tin 1.1 PL4
Date: Tue, 26 Jan 1993 13:48:49 GMT

I've already received two pieces of mail telling me I didn't read the
puzzle definition correctly... I guess what happened was the business of
calling cities points and every point a city confused me.  It doesn't 
make much sense saying "is it necessary for some city to be the same on
both maps" since cities are finite... so I just thought it was a typo...

In any case, here is a "common sense" solution to the GIVEN puzzle that
some point must be common between maps of both scales:

The smaller map need only be completely inside the bigger map to guarantee
that some point is common... it does not matter if the pieces are disjoint.

Call the big map A, and the small map B.  Consider a third map C which is
aligned with the smaller map the same way as B is aligned in A.  
ie.  A:B = B:C

If you were to continue this process (A:B = B:C = C:D = D:E = ...) then the
position of the new map which you insert into the smallest map will converge
to a point:  this is the common point between A and B.
From: barryg@hplsla.hp.com (Barry Gunn)
Date: Tue, 26 Jan 1993 15:47:30 GMT
Subject: Re: Are you sure?
Organization: HP Lake Stevens, WA
Newsgroups: rec.puzzles

>>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?"
>>-----------------+-------------------------------
>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.

But you meet the F on the street. So MF and FM are the same in this case
(but not the way the problem is usually stated). I say 50%.
Newsgroups: rec.puzzles
From: dant@techbook.com (Dan Tilque)
Subject: Cities with an internal border
Organization: Pseudopolis Yard
Date: Mon, 25 Jan 1993 01:45:49 GMT

[I'd post this to the trivia group, but they seem to have problems with
open-ended questions there.]

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.

An example is Kansas City MO/KS.  The names must be identical so
Laredo/Neuvo Laredo does not count.

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

---
Dan Tilque    --     dant@techbook.com
Newsgroups: rec.puzzles
From: mikem@ll.mit.edu
Subject: Re: Are you sure? <SPOILER>
Organization: MIT Lincoln Laboratory
Date: Tue, 26 Jan 93 19:44:18 GMT

In article <1k3f3hINN29k@gap.caltech.edu> carl@SOL1.GPS.CALTECH.EDU writes:
>
>OK, maybe if we do this by enumeration, it'll convince people.
>
... explanation deleted for brevity ...
>
>Thus there's a 50% chance that the sibling is female.  This is *NOT* the same
>as the "I've got two children and at least one of them is a girl" question. 

Excellent explanation.  Ennumerating the possibilities is the clearest way to
analyze this problem.

But, I bet a lot of people are still asking, "*WHY* isn't it the same as the 'I
have two children and at least one of them is a girl' question?"  I think this
confusion is the fundamental problem, and the reason that people still cling to
the 1/3 : 2/3 answer.

So in the original problem ("You see one of two children - the one you see is
a girl.  What's the other one?"), let's create an analog by flipping coins.
The problem becomes: you see one tail; what's the other coin?

We flip two coins, and pick one at random.  If it's a head, we throw out that
flip, and try again.  If it's a tail, we show it to you.  Over all possible
flips, here are the coins you will see (in caps):  hT, Th, Tt, tT.

The ones that get thrown out are Hh, hH, Ht, and tH.

So, of the four times you *see* a Tail, the unseen coin is a head twice, and
twice it's a tail.  Thus, the unseen coin is 50% likely to be a Head or Tail.

--

The other problem ("at least one of my children is a girl") is analogous to the
following coin-flip scenario:  We flip two coins, and if one of them is a tail,
we tell you "there's at least one tail."  We have now treated the Tt and tT
outcomes as a single event before you even hear about the throw.

If we flip HH, HT, TH, TT, then we will throw out the HH flip, and report success
on the HT, TH and TT.  In this new 'universe' of three possible outcomes, two of
them have a 'H' as the unreported coin, and only one has a 'T'.  Hence, the unseen
coin is 2/3 likely to be a 'H', and only 1/3 likely to be a 'T'.

--

To sum up, when you see one child, you may assume that child is selected at
random, and therefore the gender of the other child may be considered an
independent variable.  But when your friend tells you 'at least one child is
a girl', then the selection was *not* done at random; the gender of both
children was considered before telling you anything, so you may not treat the
gender of the other child as independent.

Hope that's clear.

-- Mike Maciolek		It is NOT an optical illusion.
mikem@ll.mit.edu		It just looks like one.
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Two Circles Puzzle <SPOILER>
Keywords: geometry
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Wed, 27 Jan 1993 20:12:58 GMT

In article <C1IyDB.2K5@fastrac.llnl.gov> dan@danberg.llnl.gov (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?

I've certainly heard this before, and it seems to me it ought to be in
the rec.puzzles archive (I couldn't find it either).

SPOILER ahead...



The answer is 3.

The way I look at it is as you roll the circle around the larger one,
turn your head at the same time so that the small circle always appears
on top.  Suppose the small circle was supposed to roll counterclockwise
around the larger one.  Then what you will see instead is one clockwise
revolution of the larger circle and two counterclockwise revolutions
of the small circle.

But now you have seen the larger circle move when it shouldn't have.
To restore the correct number of revolutions of the large circle (i.e.
0) you must turn your head back so both circles turn counterclockwise
one time.  This is the third such revolution for the small circle.

This situation is analogous to the difference between the number of
solar days in a year (~365.25) and the number of sidereal days
(~366.25).

-- David Karr (karr@cs.cornell.edu)
From: sjbuckle@minster.york.ac.uk
Newsgroups: rec.puzzles
Subject: Re: 21-link chain puzzle (SPOILER)
Date: 26 Jan 93 14:05:18 GMT
Organization: Department of Computer Science, University of York, England

Darcy Boese (dboese@spartan.ac.BrockU.CA) wrote:
>sjbuckle@minster.york.ac.uk writes:
>: Darcy Boese (dboese@spartan.ac.BrockU.CA) wrote:
>: >The other solution is this:
>: >
>: >1 1 3 5 11
>: >
>: >(yours was 1 1 3 6 10)
>: 
>: Surely there's also: 1 2 4 6 8.
>: Use 1,2,4,8 to make all numbers up to 15.
>: Then repeat how you made up 10-15 and throw in the 6 as well, to get 16-21.
>
>Actually yours is only the solution for part of the puzzle.  For the rest of
>the solution, you would need to make fewer breaks in the chain to create
>the pieces.

Sorry I've forgotten what the second part of the puzzle was.
Newsgroups: rec.puzzles
From: richard@skaro.demon.co.uk (Richard Develyn)
Subject: Re: Are you sure?
Organization: skaro
Date: Tue, 26 Jan 1993 13:56:35 +0000

I don't believe this:

> 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.

The MF also drops out (or the FM). The chance is 50:50.

Richard
--

Richard Develyn from Hastings, Sussex, England ( rad@cix.compulink.co.uk ).
Software engineer (datacomms) for Cray Network Systems. Married to Caroline.
Baby daughter Sophie (b. 3/11/92). Hobbies: Dr Who, FRPG, SF, Curry,
quality computer games, philanthropic discussions.
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: A puzzle?
Date: 22 Jan 1993 13:38:18 GMT
Organization: HST Wide Field/Planetary Camera
Distribution: world
Reply-To: carl@SOL1.GPS.CALTECH.EDU
NNTP-Posting-Host: sol1.gps.caltech.edu

In article <1993Jan21.115427.29195@ida.liu.se>, t91matfa@odalix.ida.liu.se (Mattias Fagerlund) writes:
=Well, lets see if this puzzle fits you.
=
=The solution seems me totally logical, but not every body might think so.
=You may ask me as many questions as you wish, but mail them to my person, 
=so you don't spoil it for the others.
=Questions will be answered with yes, no and irrelevant (most are).
=
=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?

Well, after we consult the frequently-asked-questions list for this newsgroup,
we conclude that the customer must've been suffering from hiccups.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: rec.puzzles
Subject: Re: Turing Machines
Date: 27 Jan 93 12:39:43 GMT
Organization: Department of Computer Science, U of Copenhagen

sobelman@cco.caltech.edu (Steven M. Sobelman ) writes:

>ags@seaman.cc.purdue.edu (Dave Seaman) writes:

>>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. 

>Actually, all turing machines have infinitely extensible memory which, in some
>implementations, is represented by a tape.  If you have the necessary splicing
>skills and can provide tape (or memory) faster than a UTM can use it, you,
>effectively, have infinite memory.

>Building a UTM is a reasonably challenging task, I doubt it's been seriously
>attempted.  I believe, however, that working non-universal TM's have been 
>constructed.  They probably aren't very fast.

There are several ways to interpret the original question, depending
one the exact meaning of "Turing machine". One can take it as a
machine with a finite number of states and an infinite tape (Note that
the "finite states" here refers to the number of "labels" or "program
points" in the TM.  Counting the tape, any TM has an infinite number
of states), or one can take it as a machine that is able to "load" any
"program" that represents a finite state TM. The latter is often
referred to as a universal TM, as it can calculate any computable
(that is, partial recursive) function.

Another question is: is there a single finite state (in the sense
above) TM that can compute all partial recursive functions? The answer
is "Yes". 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.

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.

	Torben Mogensen (torbenm@diku.dk)
Newsgroups: rec.puzzles
From: bmoore@sunfish.austin.ibm.com (Bryan Moore)
Subject: Re: RAAARRGGHH!!!! not again!
Date: Tue, 26 Jan 1993 14:49:08 GMT
Organization: IBM, Austin

From: cooper@cbnewsg.cb.att.com (Ralph 'Hairy' Moonen)
>>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
Ralph, you are dead wrong. Let's look closer at your dice analogy to see why.
To compare the dice problem to the boy/girl problem, you need to toss the
dice a bunch of times and write the results down and put them in a hat.
This is because the two children are already born, and we're not saying
that if a woman has given birth to a girl, and is pregnant, what is the
probability that the next child is a girl. We're saying that both children
have been born and you see one of them on the street, you don't know which
child). You see a girl, what's the probability the other is a girl. 

Back to the dice... Roll two dice 100 times. Get 100 pieces of paper and
write the results of the first die on one side and write the results
of the second die on the other side of the piece of paper. Put all the
pieces of paper in a hat. Now pick a piece of paper out of the hat.
You see a six on the piece of paper, what is the probability that the
other side has a six? IT IS 1/11!!!!!!! 
-- 
_/_/
_/_/_/       bmoore%sunfish.ibm.com@ibmpa.austin.ibm.com
_/_/_/_/
_/_/_/_/_/
From: jpk2581@hertz.njit.edu (joseph p kisenwether apmt stnt)
Newsgroups: rec.puzzles
Subject: wo unrelated questions
Date: 29 Jan 93 03:17:42 GMT
Organization: New Jersey Institute of Technology, Newark, N.J.

1.) Does there exist a simple proof of the existence of an NxN
    magic square for all N>2? What is it?

2.) About a year ago I came across an article which "asked"; 
    "what would have happened if  Escher knew about Penrose 
     tilings?" and printed a picture of Penrose Tiling Chickens.
    Can anyone help me find this again?

- JaKe
Newsgroups: rec.puzzles
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Subject: Re: 1/2, 1/3, boy girl the answer to life & everything
Organization: Purdue University
Date: Wed, 27 Jan 1993 15:04:50 GMT

In article <1993Jan27.112827.26276@cbfsb.cb.att.com>  
cooper@cbnewsg.cb.att.com (Ralph 'Hairy' Moonen) writes:
> 2) (The one that is subject to misinterpretation) A coin is tossed  
twice.
>    You are told that *at least* one of the tosses produced a heads. What
>    is the probablity of the other also being heads. The problem here is 
>    that people seem to interpret this question as if the sequences TH  
and
>    HT are different. But that is not how the question is worded.  
Therefore
>    the answer to this question is also 1/2. 

The reason TH and HT are treated as different is simply that they *are*  
different. There are four equally likely possibilities, one of which is  
ruled out by the statement: HH, HT and TH are all possible, and TT is not.  
Of the three equally likely cases, one results in both tosses being heads.  
Therefore, the probability is 1/3, not 1/2.

>The wording needed to come to
>    an answer of 1/3 is this: (and is not the original question)
> 
>    A coin is tossed twice. You are told that *at least* one of the  
tosses
>    produced heads. What is the probabilty that THE SECOND (and *not* the
>    first!) toss produced heads to make both tosses heads. In this case,
>    the permutations of HT and TH are distinhuished and different. the
>    answer now is 1/3. People read the original question and see the  
above
>    one, because it's easy to make this mistake. 

I don't see how anyone could possibly mistake one of these for the other.  
If THE SECOND (and *not* the first!) toss produced heads, then you are  
reduced to the single case TH. The probability of having two heads in this  
case is exactly zero.

--
Dave Seaman
ags@seaman.cc.purdue.edu
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: Turing Machines
Date: 26 Jan 1993 06:32:25 GMT
Organization: HST Wide Field/Planetary Camera
Distribution: world
Reply-To: carl@SOL1.GPS.CALTECH.EDU
NNTP-Posting-Host: sol1.gps.caltech.edu

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've got
>its order if you could be bothered (I couldn't - it goes for a few
>pages).

I tried once, but I ran out of memory :-).  Seriously, by imposing the
adjective "universal," you've required that the machine have infinite memory.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: sphere packing
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Thu, 28 Jan 1993 19:25:14 GMT

In article <1993Jan26.235433.18005@cs.ucla.edu> byron elbows writes:
>I need help on a sphere packing question.  Suppose you have a white ping
>pong ball, and 13 red ones, all the same size.  Can you glue all of the red
>ping pong balls to the white one?  [...]
>
>It is a simple matter to do it with 12.  Simply place the spheres at the
>corners of an icosahedron centered at the center of the white ping pong ball.
>It turns out this leaves some extra room for spheres to be moved around [...]

This remark caused me to wonder what is the diameter of a sphere
circumscribed about an icosahedron of side 1.  (Presumably it's less
than 2, so that the above works.)  It would be nice if there were a
simple, painless answer to this, but I couldn't think of anything that
didn't involve lots of trigonometry.  Does anyone know a very simple
method to compute this?

By the way, you also get 12 spheres around a central one if you use
either hexagonal or cubic close packing, in which 6 of the spheres lie
in one plane (along with the center sphere) with centers at the
vertices of a regular hexagon.  In this configuration all the
surrounding spheres just barely fit in around the central one with no
"wiggling room," but it's easy enough to verify by diagram that the
configuration works.

-- David Karr (karr@cs.cornell.edu)

Newsgroups: rec.puzzles