HDTenEightyP/NetNews huggingface dataset:
From: costley@solo.eng.hou.compaq.com (Brett Costley)
Subject: Re: the irrational stamp puzzle
Organization: Compaq Computer Corp.
Date: Thu, 21 Jan 1993 17:16:10 GMT

>In article <2338@sjfc.UUCP>, dmc@sjfc.UUCP (Cass Dan) writes:
>|> 
>|> You have an ink stamp which is so amazingly precise that, when inked
>|> and pressed down on the plane, it makes every circle of irrational
>|> radius (centered at the center of the stamp) black.
>|> 
>|> Question:  Can one use the stamp three times and make every point
>|> in the plane black?  [assume plane was white to begin with, and
>|> ignore the fact that no such stamp is physically possible]

You can do it in one stamp.  Put the stamp down and then without lifting
it move it around sufficiently to cover the entire area. 

Or, stamp once and use your finger to smear the ink before it dries. :)
From: Ben.White@f364.n633.z3.fidonet.org (Ben White)
Newsgroups: rec.puzzles
Subject: Turing Machines
Date: Fri, 15 Jan 1993 03:31:46

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).


 * Origin: The Little Shop of Horrors!! - [03] 702-8889 (3:633/364)
Newsgroups: rec.puzzles
From: hiraga@Csli.Stanford.EDU (Yuzuru Hiraga)
Subject: Re: "map" of USA
Organization: Stanford University CSLI
Date: Tue, 26 Jan 1993 06:16:43 GMT

In article <19692.2b617217@ecs.umass.edu> padmanab@ecs.umass.edu writes:
>Suppose you have two maps of USA of different 
>scales. Now you place the smaller map inside 
>the bigger map such that it falls completely 
>within it. 
>
>Condider each point on the map to be a distinct 
>city!! 
>
>QUESTION: Are there any city/cities on the 
>smaller map which will coincide with the 
>city/cities  on the bigger map? If so why? 

Let's not be taken by the words "USA" and "city",
as some people got worried about.

The fixed point theorem says that every contracting projection
has a fixed point (i.e. f(x)=x), so that's it (need a proof? :-)

A simple way to see this in our present case (not a rigorous proof) is:
First suppose both maps are on the XY plane with same orientation,
i.e. no rotation in the XY plane is applied.
Then move the smaller map along the Z axis in parallel.
If we connect the corresponding points in both maps, the connecting
lines will converge at a single point.  Among those lines, there will
be one which is the shortest, which happens to be perpendicular
to the XY plane (i.e. the corresponding points have the same (x,y) coordinate).

For the general case, note that any positioning of the small map
can be obtained by transposition, and rotation around a single point.
Take this point to be the fixed point above (this can be alwas done
in combination with the proper transposition).
Applying rotation will leave the fixed point intact, though none of the
others will become another fixed point.
Incidentally, this shows that there is one and only one "city" that
coincides.

-Yuzuru Hiraga
From: vivekr@osc.COM (Vivek Rau)
Newsgroups: rec.puzzles
Subject: Re: Geometry Puzzler
Date: 26 Jan 93 22:15:27 GMT
Reply-To: vivekr@osc.UUCP (Vivek Rau)
Organization: Versant Object Technology, Menlo Park, CA

Got my A's and B's mixed up there. Here's the correction in the construction
of the inverse of a point C with respect to a circle of radius AB:

	.________.________.
	A        B        C

>    Draw a circle centred at A, radius AB. Use C as centre and mark points
>    P and P' on the circumference of this circle, by drawing an arc with 
>    radius CA. Now draw arcs centred at P and P' respectively, with radius
>    PB. These arcs will intersect at 2 points, one is B itself, the
>    other is the point D which is the bisection point you are looking for.
>    For proof, consider the similar triangles APC and CPD, and remember the
>    definition of the inverse point.

Draw a circle centred at A, radius AB. Use C as centre and mark points
P and P' on the circumference of this circle, by drawing an arc with 
radius CA. Now draw arcs centred at P and P' respectively, with radius
PA. These arcs will intersect at 2 points, one is A itself, the
^^                                                ^
other is the point D which is the bisection point you are looking for.
For proof, consider the similar triangles APC and APD, and remember the
                                                  ^
definition of the inverse point.

Vivek Rau
vivekr@osc.com
Newsgroups: rec.puzzles
From: byron elbows
Subject: sphere packing
Summary: not an open question, but who knows the answer?
Originator: brian@maui.cs.ucla.edu
Organization: UCLA, Computer Science Department
Date: Tue, 26 Jan 93 23:54:33 GMT

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?  In other words, is it possible to place
13 unit spheres around one unit sphere, so that each of the 13 is touching
the central sphere, and none of the 13 intersect each other?

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, but
is it possible to move them around so that another sphere can be squashed
in there?

In the meantime, I leave you with two simpler questions:

1.  Suppose you pick a number at random from 0 to 1.  Assume uniform
    distribution (ie, pdf = 1 from 0 to 1).  Clearly, that number is
    not going to be greater than 1.  Now you pick another number at
    random from 0 to 1.  The probability that the sum of these two
    numbers is greater than 1 is simple to calculate; it is 1/2.  If
    it is greater than 1, stop; otherwise, continue picking numbers
    at random until the accumulated sum is greater than 1.  Derive the
    average number of picks required.  (What is the answer if the
    distribution is arbitrary?)

2.  Suppose I have a deck of cards numbered from 1 to 100.  You and I
    are dealt two hands of 50 cards each, randomly, from the deck.  We
    both sort the cards in increasing order.  In the first round, we
    both play our first card; the player with the higher card wins
    one dollar, and the two cards are discarded.  This routine is
    repeated for our second card, third card, etc.  What is the
    probability that I win this warped version of War by a score of
    50 to nothing?

byron elbows
(mail to brian@cs.ucla.edu)
Newsgroups: rec.puzzles
From: brian@maui.cs.ucla.edu (byron elbows / Brian Tung)
Subject: Re: Two Circles Puzzle
Keywords: geometry
Organization: UCLA, Computer Science Department
Date: Wed, 27 Jan 93 21:06:35 GMT

I wrote:

> 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.

I should add that in all cases, TWO fourth circles can be drawn.  One is
always inside the first three.  The other one is either external to the
first three, or else it encloses the first three.

byron elbows
(mail to brian@cs.ucla.edu)
Newsgroups: rec.puzzles
From: durall@nx26.mik.uky.edu (benjamin b durall)
Subject: Re: sphere packing
Reply-To: durall@mik.uky.edu
Organization: University Of Kentucky, Dept. of Math Sciences
Date: Wed, 27 Jan 1993 19:46:35 GMT

In article <C1HrDH.77t@dartvax.dartmouth.edu>  
J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
> 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?  In other words, is it possible to place
> > 13 unit spheres around one unit sphere, so that each of the 13 is touching
> > the central sphere, and none of the 13 intersect each other?
> > 
> > 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,  
but
> > is it possible to move them around so that another sphere can be squashed
> > in there?
> 
> You're asking a question about the 'kissing number'; that is, the
> greatest number of objects that can be touching another object of the
> same size.  For two dimensions (ie. circles), the kissing number is 6,
> and for three dimensions (spheres -- the question you ask), it's been
> proven that the kissing number is 12.  Therefore, there is no way to
> move those twelve spheres to make a 13th one touch the center sphere. 
> Sometime within the last year (sorry I don't have the exact date!)
> Scientific American ran a piece about the kissing number in its math
> column.
> 
> 
> --Ted Schuerzinger
> email: .zed@Dartmouth.EDU
> "I should have realized it would be bad vodka when all the label said
> was 'Russian Vodka'."

Yes there is.  Step on all 13 red ones, and glue the edges of the smushed ones  
to the white one.  You could probably get as many as 20 glued to the white one  
with this method.

(insert smiles)
--
Bryan Durall  |  durall@mik.uky.edu (preferred)  |  bbdura00@ukpr.uky.edu
"Lucky I'm sane after all I've been through,
 I can't complain but sometimes I still do.
 Life's been good to me so far."             - Joe Walsh, 1978
Newsgroups: rec.puzzles
From: s0432806@let.rug.nl (M.W. Geerligs)
Subject: Re: surgeon puzzle
Keywords: situation logic
Organization: Faculteit der Letteren, Rijksuniversiteit Groningen, NL
Date: Thu, 28 Jan 1993 11:43:26 GMT

In article <C1Gzsu.7pq@fastrac.llnl.gov> dan@danberg.llnl.gov (Dan Bergmann) writes:
>From: dan@danberg.llnl.gov (Dan Bergmann)
>Subject: surgeon puzzle
>Keywords: situation logic
>Date: 26 Jan 93 16:43:42 GMT
>
>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."
>
>
>-- 
>
>
>*************************************************************************
>**             							       **
>**  Dan Bergmann                             (510) 423-6765            **
>**  mail stop L-262						       **
>**  Lawrence Livermore National Lab          dbergmann@llnl.gov        **
>**  Livermore, CA    94550                                             **
>**               						       **
>*************************************************************************
>-- 
>
>
>*************************************************************************
>**             							       **

The surgeon is his mother... 
This one is as old as the way to Rome (literally from dutch to english, don'
t know if it's the same thing). Please send some newer puzzles!!!
Newsgroups: rec.puzzles
From: acheng@ncsa.uiuc.edu (Albert Cheng)
Subject: Re: Need puzzle help: eye colour enigma
Originator: acheng@shalom.ncsa.uiuc.edu
Organization: Nat'l Ctr for Supercomp App (NCSA) @ University of Illinois
Date: Fri, 22 Jan 1993 19:45:23 GMT


In article <93021.120138RVESTERM@vma.cc.nd.edu>, <RVESTERM@vma.cc.nd.edu> writes:
>[Detail not repeated]
>... As i posted a long time ago, being logical, they
>take a clue from mister spock: "the needs of the many outweigh the needs
>of the few, or the one".

Hm... why is it logical so?  Wouldn't it be more logical for everyone
to live?  In that case, the islanders decide logically "to nullify this
stupid tradition immediately."  Everyone lives on in peace.

Or everyone comes to the conclusion, "Well, 199 people know my eye
color.  Now this guru knows my eye-color and none of them has to die
because of that knowledge.  So, there is no logical ground for me to
die because I become one of the crowd who know my eye-color."

Another similar line of thought, "I know the eye color of 199 people
before.  I just know the eye color of the guru too.  Nice green.  If I
don't die of knowing 200 people's eye color, there is no logical ground
for me to die just because I know the eye color of one more person."

"How could this idiot be a guru when he does not know the no-no here.
Why should I listen to his baloney?"
From: dan@danberg.llnl.gov (Dan Bergmann)
Newsgroups: rec.puzzles
Subject: surgeon puzzle
Keywords: situation logic
Date: 26 Jan 93 16:43:42 GMT
Organization: Lawrence Livermore National Laboratory
Originator: dan@danberg.llnl.gov


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."


-- 


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


*************************************************************************
**             							       **
From: camus@leland.Stanford.EDU (Theodore C. Quinn)
Subject: Re: Are you sure?
Organization: DSG, Stanford University, CA 94305, USA
Date: Mon, 25 Jan 93 23:16:18 GMT

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?"

	Well, I'm new to this group and I'm sure this is "puzzle" is in the
FAQ, but I'll throw in my two cents worth, anyway.
	Of course, the common sense answer is correct: there are equal odds
of the other sibling being a female or a male. The apparent paradox lies
in an inconsistency in the counting procedure described. If one states that
there are four possible gender configurations (FF,MM,FM,MF) then clearly
the order in which the children are born is being taken into account. Otherwise
FM and MF wouldn't be listed separately.
	Therefore, since this order is being taken into account, when you
see the girl on the street there are FOUR possibilities. She is either the
youngest of a FM family, the oldest of a MF family, the youngest of a
FF family, or the oldest of a FF family. Therefore, FF gets counted twice
and the paradox is resolved. There is a 50% chance of the other sibling
being a girl.

				-Ted Quinn
				 quinn@relgyro.stanford.edu
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Newsgroups: rec.puzzles
Subject: Re: Dividing apples
Date: 25 Jan 93 23:33:20 GMT
 <C1E5Du.D3B@news.rich.bnr.ca>
Organization: Dartmouth College, Hanover, NH
X-Posted-From: InterNews1.0a5@newshost.dartmouth.edu

There is a formula (Ramanujan's Formula, I believe), that lets you
figure out the number of ways you can divide a number -- ie. the number
of different ways you can make a series of numbers that adds up to the
number in question.  I don't know the formula offhand :-) and I don't
know how useful it would be for this particular question, however.

--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have realized it would be bad vodka when all the label said
was 'Russian Vodka'."
From: cmh@eng.cam.ac.uk (C.M. Hicks)
Newsgroups: rec.puzzles
Subject: Re: Are you sure? NO - flame-resistant clothing ON  <SPOILER>
Date: 26 Jan 93 12:14:06 GMT
Organization: cam.eng

cmh@eng.cam.ac.uk (C.M. Hicks) writes:

>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.

Of course, you could say that the possible permutations for a two-child
family are:

        Male    Female
      Female    Male
      Female    Female
        Male    Male

These all have probability 1/4, and since one child is female, one permutation
is discounted (male-male), giving a final probability of a third...

Heck - flame war ahead, and all in my own mind - do I believe Bayes Theorem?

Christopher Hicks

--
 ==============================================================================
 Christopher Hicks    |    If it doesn't fit...
 cmh@uk.ac.cam.eng    |                        ...you need a bigger hammer.
 ==============================================================================
From: berninge@cs.purdue.edu (Jim Berninger)
Newsgroups: rec.puzzles
Subject: Re: Two Circles Puzzle
Date: 27 Jan 1993 13:51:17 -0500
Organization: Department of Computer Science, Purdue University
NNTP-Posting-Host: cornwall.cs.purdue.edu
Keywords: geometry

dan@danberg.llnl.gov (Dan Bergmann) writes:
>[The following question is supposedly from an SAT, typically with
> questionable answers because of smart students and not-so-smart 
> question-writers]
>
>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?

Given that this question was on the SAT, I'm not surprised.  It seems fairly
obvious to me that the larger circle has a circumference twice that of the
smaller circle.  However, the question asks for the number of "revolutions
about its center".  This raises several options:

1) Revolutions of the larger circle about the larger circle's center --
(since the larger circle is the nearest antecedent of the pronouns "it"
and "its") 0.

2) Revolutions of the smaller circle... -- 0.  "Revolution" implies motion
about a point outside a body.  The center of the smaller circle moves with
the circle, and the circle doesn't even move around the initial location of
the center point.

3) *Rotations* of the larger circle... -- 0. Again, using the nearest
antecedent, and the fact that "rotation" implies motion about a point
interior of the body in question.  Oh, and this also assumes that the frame
of reference for defining a rotation is based on the orientation of the
larger circle (i.e. the large circle doesn't move).

4) Rotations of the smaller circle... -- 3.

Consider the circles joined as: Oo, touching at the larger circle's "east"
point.  Move the smaller circle counter-clockwise.  When the smaller
circle is at "north", it is touching the larger circle with the
point originally at the smaller circle's "east".  Thus, in 1/4
revolution about the larger circle, the smaller circle has made 3/4
rotation (the "east" point has moved counter-clockwise to "south").
Completing the trip, the smaller circle does 4*3/4=3 rotations.

If the question as phrased above is accurately quoted from the SAT, I have
even less faith in the SATs than I did before.  If not, well, pick an
answer.

 ______ Jim Berninger: berninge@cs.purdue.edu|Hate is not a Family Value ______
 \ \  /   Green Lantern Fan, Extraordinaire  |     Boycott Colorado      \ \  /
  \/\/ "I don't *wanna* play Leona Helmsley!"| Think twice about Oregon   \/\/
   \/      -- Elmyra Duff, Tiny Toons Xmas   |Watch out for OH,ID,MI,ME,FL.\/..


From: barryg@hplsla.hp.com (Barry Gunn)
Date: Wed, 20 Jan 1993 17:54:15 GMT
Subject: Re: Need puzzle help: eye colour enigma
Organization: HP Lake Stevens, WA
Newsgroups: rec.puzzles

>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"

----------

Wouldn't everyone just look around and say 'wow, so do I'?


     _
    [o] Barry Gunn x2281
    /|\ barryg@lsid.hp.com
Xref: sparky rec.puzzles:8598 sci.math:18812
From: vivekr@osc.COM (Vivek Rau)
Newsgroups: rec.puzzles,sci.math
Subject: Re: Geometry Puzzler
Date: 26 Jan 93 20:27:36 GMT
Reply-To: vivekr@osc.UUCP (Vivek Rau)
Followup-To: rec.puzzles
Organization: Versant Object Technology, Menlo Park, CA


The solution to the compass construction problem can be found in
"What is Mathematics" by Richard Courant, a real classic, equally
enjoyable to the recreational mathematician or to the professional.
I don't have it now so I can't give you the publisher or year.

Being unable to draw a diagram I can't describe the construction
adequately but I'll try to cover the main points. Remember that
you are using only a compass, so whenever you "mark a point" it
is done by intersecting one arc with another arc. The straight
line joining the two points in the "diagram" is for clarity only and
is not actually drawn for you in the problem.

	.________.________.
	A        B        C

You start with two points A and B, now obtain a point C on AB extended,
so that AB=BC. Draw a circle about B with radius BA and C will be the
point diametrically opposite to A on this circle. Find this point by 
marking off points along the circumference, starting at A as centre, 
with radius BA. Call the first point A', then mark the next point 
using A' as the centre, call this A" and the next point will be C 
(like the construction of a regular hexagon inscribed in a circle).

Once you have found C, you have to find the inverse point w.r.to C on the
line AB. The inverse point is defined as a point D on AB, such that
AD/AB = AC/BC. Since in this case AC=2.AB, you will have AD=AB/2.

The construction is as follows, proof by similar triangles not included:

Draw a circle centred at A, radius AB. Use C as centre and mark points
P and P' on the circumference of this circle, by drawing an arc with 
radius CA. Now draw arcs centred at P and P' respectively, with radius
PB. These arcs will intersect at 2 points, one is B itself, the
^^
other is the point D which is the bisection point you are looking for.
For proof, consider the similar triangles APC and CPD, and remember the
definition of the inverse point.

		* End of construction *

Now that you know how to get the bisection point, it should be simple
for you to obtain the 1/n point for any integer n, using compasses
alone.

(Where I come from they spell "center" as "centre". "Spelling flames
are a hobgoblin of small minds" - misquoted from Emerson).


Vivek Rau.
vivekr@osc.com
Newsgroups: rec.puzzles