HDTenEightyP/NetNews huggingface dataset:
From: obrien@netcom.com (Robert O'Brien)
Subject: Re: Gummy Drop-bears
Date: Thu, 30 Jul 92 05:14:03 GMT
Organization: Netcom Online Communications Services (408-241-9760 login: guest)

In article <Jul.29.03.30.41.1992.18050@remus.rutgers.edu> clong@remus.rutgers.edu (Chris Long) writes:
>
>Real gummy drop-bears mass 10 grams, while imitation gummy
>drop-bears mass 9 grams.  Spike has 7 cartons of gummy drop-bears,
>4 of which contain real gummy drop-bears, the others imitation.
>Using a scale only once and the minimum number of drop-bears, how
>can Spike determine which cartons contain real gummy drop-bears?

It is necessary to believe these things are manufactured with far greater
precision than any *real* confections, but...

There'd better be 32 of the little buggers in one of the boxes...
1. number the cartons n=0,1,2,3,4,5,6
2. ignore carton 6, it can be determined once the others are known.
3. put 2^n on the scale

the difference in weight from 630g is a bit-map to which boxes are imitation.

	obrien@netcom.com	Oh!  It's ... YOU, Bob!

Newsgroups: rec.puzzles
From: jbf@wts.com (Jens B. Fiederer)
Subject: Re: 3 puzzles
Organization: Wegmans Transaction Services
Distribution: rec.puzzles
Date: Mon, 27 Jul 1992 14:08:02 GMT

In article <1992Jul26.134105.13917@ens.fr> courtes@trimaran.ens.fr (Francois Courtes) writes:
>
>In French the first one is 2 (D-E-U-X).
>Is there a language in which 1 is so ?

In German, one is E-I-N-S.

Jens Fiederer
Newsgroups: rec.puzzles
From: s290@brems.ii.uib.no ( Per Ivar Roervik)
Subject: Babble Sort (restatment)
Organization: DevNull A/S
Date: Tue, 28 Jul 92 17:52:41 GMT

>  Here is a problem for the programmers out there :)
>  Line up the numbers from 1 to 20 in the usual way.
>  Now  imagine sliding a box, which can contain two numbers,
>  along the sequence. If the numbers in the box have two or more
>  digits in common, switch the numbers.
>
>  After a number of passes with the box, whitch number is last and 
>  whitch is first.
 
   The nombers should be switched if they have ONE or more digits in
   common, ie  12&20  or 1&11

   Thx to Gary Ansok fr spottng ths. Typo I asre y.
Newsgroups: rec.puzzles
From: sethb@fid.morgan.com (Seth Breidbart)
Subject: Re: Crossing a river
Organization: Morgan Stanley & Co., New York, NY
Date: Thu, 30 Jul 1992 23:46:24 GMT

In article <1992Jul30.173556.5883@iscsvax.uni.edu>
althof5805@iscsvax.uni.edu writes:
>Three kings are travelling with three thieves.  The kings know that if, at any
>time, thieves outnumber kings (e.g. if one king leaves to take a leak behind an
>accacia tree), the thieves will kill this outnumbered kings.  The kings may
>outnumber the thieves, and the number may be equal for both sides (one king and
>one thief may go to take a leak together), but the thieves must *never*
>outnumber the kings.
>
>They come to a larger river which curves to the south.  The kings realize that
>they must cross this river (going around the bend) to reach the path they seek. 
>The only problem is this:  there is a boat available which only carries two
>people.  Thus, they will have to play "ferry" to get all of them across.
>
>How can the kings arrange things so that there are never more thieves than
>kings on either side of the river?  Swimming is out (unless they decide to pick
>up a few pet alligators), and each time the boat returns to the starting point
>there will be an uneven number of people on each side of the river.
>
>Whatcha gonna do?

Call the faq?

I don't know what you mean by "there will be an uneven number of
people on each side of the river."  There will often be even numbers
of people on both sides of the river.  After step 8, there are equal
numbers of people on both sides.

0. KKKTTT
1. Two thieves row across.  KKKT   TT
2. One thief rows back.  KKKTT     T
3. Two thieves row across.  KKK    TTT 
4. One thief rows back.     KKKT   TT
5. Two kings row across     KT     KKTT
6. One king & one thief row back.   KKTT    KT
7. Two kings row across.      TT    KKKT
8. Thief rows back.        TTT    KKK
9. Two thieves row across.   T    KKKTT
10. One thief rows back.     TT   KKKT
11. Two thieves row across.        KKKTTT

Alternatively, steps 1 and 2 can be king and thief row across, king
rows back.



Seth		sethb@fid.morgan.com

Newsgroups: rec.puzzles
From: baarniy@IASTATE.EDU (Scott A Coon)
Subject: Re: vacuuously false?
Followup-To: rec.puzzles
Reply-To: baarniy@IASTATE.EDU (Scott A Coon)
Organization: Iowa State University
Date: Wed, 29 Jul 1992 17:04:56 GMT

In article 13896 dhesi@rahul.net (Rahul Dhesi) writes

>Normally an assertion
>
>     if A then B
>
>is considered to be vacuously true if A is known to be false.  But
>consider this assertion:
>
>     if today is Monday then tomorrow is Tuesday
>
>Is this vacuously true on Friday?


Today is Friday.  "Today is Monday" is assumed true for this
vacuous proof.

If today is Friday and today is Monday, then four days after Friday
is four days after Monday.  Thus, Tuesday is Friday.  Thus Tuesday 
is Monday.  Thus Wednesday is Tuesday, and therefore Monday.  We 
can easily show that all days are the same day, and so show that 
Saturday is Tuesday, so the day after Friday is Tuesday.

Vacuous truth wins again.

Scott Coon
baarniy@iastate.edu

P.S.  I have just considered this again, and I realize that what I
have done is shown "today is monday" ALWAYS implies "tomorrow is tuesday."
I'm not altogether sure that's what you were asking, though.
If not, maybe the above could be changed suitably.  Clarify?
Newsgroups: rec.puzzles
From: kanad@top.cis.syr.edu (Kanad Chakraborty)
Subject: Re: vacuuously false?
Organization: Syracuse University, CIS Dept.
Date: Thu, 30 Jul 92 10:08:42 EDT

In article <Bs5pw4.MEp@rahul.net> dhesi@rahul.net (Rahul Dhesi) writes:
>Normally an assertion
>
>     if A then B
>
>is considered to be vacuously true if A is known to be false.  But
>consider this assertion:
>
>     if today is Monday then tomorrow is Tuesday
>
>Is this vacuously true on Friday?
>--
>Rahul Dhesi <dhesi@rahul.net>
>also:  dhesi@cirrus.com

Yes, of course.

Kanad Chakraborty (kanad@top.cis.syr.edu)
Newsgroups: rec.puzzles
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Subject: Re: number construction
News-Software: VAX/VMS VNEWS 1.41    
Organization: Texas A&M University, Academic Computing Services
Date: Thu, 30 Jul 1992 02:43:00 GMT

ajtd@st-andrews.ac.uk (Antony J. T. Davie) writes...
>Can you make the following numbers using 1,2,3,4 at most once and any
>of the following operations as many times as you like:
>+,-,*(times),/,^(raise to the power),!(factorial),square root.
>Now make 89 and 137.

*** SPOILER ***



















          .            .
89 = 3!!*.1 + sqrt(4)/.2

           .
137 = 3!!*.2 - 4! + 1

Dave Ring
dwr2560@zeus.tamu.edu
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Newsgroups: rec.puzzles
Subject: Re: Different objects
Date: 27 Jul 92 00:04:00 GMT
Organization: Texas A&M University, Academic Computing Services
News-Software: VAX/VMS VNEWS 1.41

I write...
>>Okay, how about optical trapping of atoms? Hydrogen and anti hydrogen for
>>example. Or probably any different atoms would be hard to tell apart. Any
>>experts out there?

russotto@eng.umd.edu (Matthew T. Russotto) writes...
>I don't think neutral atoms will do it-- remember that we are assuming
>infinitely sensitive instruments, and it ought to be possible to
>determine if the massive particle is positive or negative.  There
>would probably be the same problem with neutrons-- you'd have to stick
>with neutrinos, but I don't know how to enclose them...

I think they can use only neutral particles. But I have been wrong before.
Hydrogen may be out, because the species of atom is important since
the effect depends on atomic transitions. But that would not prevent
us from using atom anti-atom of an appropriate species.

Dave Ring
dwr2560@zeus.tamu.edu
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: need an algorithm
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Fri, 31 Jul 1992 13:18:48 GMT

In article <15621@star.cs.vu.nl> harmsen@cs.vu.nl (Harmsen H) writes:
>I am looking for an algoritm for determining unused ID's.
>
>Taking the next ID (1, 2, 3, .... 20, 21, .....10000, 10001 etc.) each 
>time is of course very fast but will result in an overflow sometime 
>in the future.
>
>Hans-Peter Harmsen		(harmsen@cs.vu.nl)
>Free University, Amsterdam

Here's a puzzle for you:  If you use 64-bit unsigned integers for IDs
and create one thousand million new objects every second ("one billion"
in the US), how many centuries will it take you to overflow?

>P.S. I don't want to keep a list of used ID's.

There are generally programming techniques by which you can keep a
"free list" of IDs that are below the high-water mark but not in use,
and keep this list in the memory cells of objects that have been
deleted from the list.  The maximum memory required under this scheme
is the same as if you had no free list, because it occurs when you have
exhausted the free list and need to add more objects above the high-water
mark.

Just a couple of suggestions.

-- David Karr (karr@cs.cornell.edu)
From: clong@remus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: How Many Composites? (SPOILER)
Date: 30 Jul 92 06:50:00 GMT
Organization: Rutgers Univ., New Brunswick, N.J.

In article <12435@ntdd-1>, grayt@Software.Mitel.COM (Tom Gray) writes:

> How many composite numbers with p as a factor that are:
> less than p^2

Number that are less than or equal to N is floor(N/p)-1 (N>=p); so
number that is less than p^2 is floor((p^2-1)/p)-1 = floor(p-1/p)-1
= p-2.

> less than p!

floor((p!-1)/p)-1 = floor((p-1)!-1/p)-1 = (p-1)! - 2 (for p>2).
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"In a study of schoolboys, an educator discovered a correlation between size
 of feet and quality of handwriting.  The boys with the larger feet were,
 on the average, older."        Wallis & Roberts, _The Nature of Statistics_
Newsgroups: rec.puzzles
From: dfs@doe.carleton.ca (David F. Skoll)
Newsgroups: rec.puzzles
Subject: Re: Tiling a plane
Date: 27 Jul 92 14:20:49 GMT
Organization: Dept. of Electronics, Carleton University

In article <711892126.F00001@csource.oz.au> joe@csource.oz.au (Joe Slater) writes:
>A plane can be tiled with three regular geometric shapes: squares, triangles and
>hexagons. It can also be tiled with pairs of shapes: hexagons and triangles, or
>octagons and squares.

>Are there any other pairs that will tile a plane? What about triplets? Answers
>requiring fractal-like series to fill up gaps aren't counted.

What about Penrose's kites and darts?  I don't have the exact specs for
building them, but I know they tile the plane aperiodically.

--
David F. SKoll
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: 3 puzzles
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Distribution: rec.puzzles
Date: Thu, 23 Jul 1992 18:02:28 GMT

In article <5083@transfer.stratus.com> markc@eomer.sw.stratus.com (Mark Christensen) writes:
>In article <1992Jul23.142526.9834@cs.cornell.edu> karr@cs.cornell.edu (David Karr) writes:
>><SPOILERS ahead>

>>
>>In article <5071@transfer.stratus.com> markc@eomer.sw.stratus.com (Mark Christensen) writes:
>>
>>>	What is the first number, when spelled out (e.g. O-N-E) is
>>>in alphabetical order?
>>
>>eight
>
>Ummm!  Not quite.  The letter "h" comes before "i".  Try
>again.

Oh.  The problem is your question was not an English sentence.  I thought
you meant, "What is the first number in alphabetical order, when all numbers
are spelled out?"  Apparently you actually meant, "What is the first number
whose letters all appear in alphabetical order when it is spelled out?"
In that case the answer is "forty."

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



Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: 3 puzzles
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Thu, 30 Jul 1992 15:59:41 GMT

In article <78930@bcsaic.boeing.com> lbaum@atc.boeing.com (Larry Baum) writes:
>In article <1992Jul26.233848.20722@Armstrong.EDU> water772@Armstrong.EDU (Kelly
>Waters) writes:
>
>Well EIGHT is alphabetically before FORTY, but how about:
>
>ALEPH-NOUGHT

Heh-heh.  I like this.


>(actually, there is probably a definition of ALEPH-EIGHT)

Didn't Isaac Asimov end The Realm of Numbers with a notation that
would be spelled out "aleph aleph nought"?  If this is valid, then it
is alphabetically before "aleph <blah>" for any finite <blah>, but
after "aleph aleph aleph nought" which is after "aleph aleph aleph
aleph nought" which is after ... and so forth ad nauseum.

-- David Karr (karr@cs.cornell.edu)
Newsgroups: rec.puzzles
From: tjoa@cory.Berkeley.EDU (Richard Tjoa)
Subject: Re: 3 puzzles
Summary: Alphabetically first
Keywords: numbers alphabet two different amazing answers sex
Organization: Mine
Date: Fri, 24 Jul 1992 19:06:18 GMT

In article <1992Jul24.144941.18271@fys.ruu.nl> (Timo Bootsma) writes:
>In article <5071@transfer.stratus.com> (Mark Christensen) writes:
>|> Here are three puzzles:
>|> 
>|> [1]
>|> 	What is the first number, when spelled out (e.g. O-N-E) is
>|> in alphabetical order?
>|> 
>
>In Dutch: een (1).
>
>timo

In Chinese: 0  (hey, it might as well be alphabetical...)

Also: "2" (a-b-c)


-Richard
Spelled out my telephone, that is.
Newsgroups: rec.puzzles
From: Ulf.Lunde@kvatro.no (Ulf Lunde)
Subject: Re: mini-mysteries
Organization: Kvatro A/S, Trondheim, Norway
Date: 31 Jul 92 17:31:13

>>>>> On 31 Jul 92 07:38:00 GMT, steppe@ccwf.cc.utexas.edu (Top Changwatchai) said:
> where I can find such a list?  Specifically, I'm looking for a list of those
> mini-mysteries, where the storyteller gives a short mysterious story, and
> the audience tries to figure it out, via yes-no questions.

Send a mail message to netlib@peregrine.com with "send riddle" as the whole
message body.  You will receive about 80 situation puzzles.


> there's one where a woman reads in the newspaper about a man whose wife has
> just died.  Immediately, the woman exclaims, "He killed her!"  The woman
> had never met the man or his wife.

Did she know about the death of the wife before reading the newspaper?

Ulf Lunde
From: rt@ukc.ac.uk (R.W.Thearle)
Newsgroups: rec.puzzles
Subject: Re: "Satire words".
Keywords: aster, arets, tears, stare, stear, strae, rates, reast, resat, ...
Date: 21 Jul 92 09:55:26 GMT
Reply-To: rt@ukc.ac.uk (R.W.Thearle)
Organization: Computing Lab, University of Kent at Canterbury, UK.

In article <1992Jul21.131738.5983@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:

>What word (of any length) accepts the most different single letters;
>What word (of any length) makes the most new words, after one addition;
>What nonword (of any length) accepts the most different single letters;
>What nonword (of any length) makes the most new words, after one addition.

I'd like to enter "aster" for the first two (every letter except j & q; 
over 140 new words) and "tsrea" for the second two (ditto)

					Roy

P.S. What's wrong with ASTERN instead of SANTER?
Newsgroups: rec.puzzles
From: sethb@fid.morgan.com (Seth Breidbart)
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Organization: my opinions only
Date: Wed, 29 Jul 1992 00:00:30 GMT

If two numbers are different, then there's another number between
them.  (Actually, a lot of such numbers.)

If you can't give an example of a number between 9.99999... and 10,
please don't post anymore to this thread.

Seth		sethb@fid.morgan.com
From: vreddy@das.harvard.edu (Venkatesh Reddy)
Newsgroups: rec.puzzles
Subject: Re: Proof of -1 = 1
Date: 29 Jul 92 18:11:15 GMT
Article-I.D.: das.1992Jul29.181115.26903
Organization: Aiken Computation Lab, Harvard University

In article <1992Jul24.062132.8627@uwm.edu>, radcliff@csd4.csd.uwm.edu (David G Radcliffe) writes:
|> Another proof of -1 = 1.  I hope this hasn't been posted before.
|> 
|>   1  =  3 - 2
|>      =  (2 + 1) - (3 - 1)
|>      =  2 + 1 - 3 - 1
		    ^^^
		Distribution... gets 'em every time... :)

|>      =  2 - 3
|>      =  -1     QED
|> 
|> --
|> David Radcliffe                                
|> radcliff@csd4.csd.uwm.edu            

Alternatively, there's the good ole fashioned proof :

			Assume a = b.

				a = b
				a * a = a * b
				a^2 = ab
				a^2 - b^2 = ab - b^2
				(a - b)(a + b) = (a - b)(b)
				(a + b) = b
				but, if a = b, a + b = b + b
				b + b = b
				2b = b
				2 = 1
				from here, you can get anywhere...

				2 * 2 = 1 * 2
				4 = 2
				4 - 3 = 2 - 3
				1 = -1

I think all of you can figure out the flaw in this one...

-- Venk Reddy --

PS: And then there's this one:
				sin x
				_____  = 6
				  n

Prove it.
;)
Newsgroups: rec.puzzles
From: tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Spacewarp wonders 
Date: 23 Jul 92 13:33:33 MDT
Organization: University of Utah, CompSci Dept

  Your spaceship has been sucked through a spacewarp, and then crashed
onto a place.  The place is foggy, so one can not see more than a few
kilometers clearly.  There is always a fairly bright light here.
Also, you have access to whatever science gear you might like, as it
was aboard your craft.  There are lots of life forms, all of them huge.

1) How would you be able to tell if the planet was a sphere, or a nearly flat
   plane extending in all directions?

2) How would you be able to tell if you had travelled forwards or backwards
   through time 5 billion years or more?

3) How would you be able to tell if you and your space ship were somehow
   shrunken in passage?
From: russotto@eng.umd.edu (Matthew T. Russotto)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 28 Jul 92 00:48:04 GMT
Distribution: rec.puzzles
Organization: College of Engineering, University of Maryland, College Park

In article <1992Jul27.154738.4480@cs.cornell.edu> karr@cs.cornell.edu (David Karr) writes:
>In article <BrxA8s.Ew1@news.cso.uiuc.edu> wyg9633@uxa.cso.uiuc.edu (Wooed Sun) writes:
>>karr@cs.cornell.edu (David Karr) writes:
>>>Oh.  The problem is your question was not an English sentence.  I thought
>>>you meant, "What is the first number in alphabetical order, when all numbers
>>>are spelled out [in English --sorry for omitting this condition before]?"
>>
>>billion?
>
>No, there is no number I would spell out as "billion."  I would write
>"one billion."

OK, but is that number 1E9 or 1E12?
-- 
Matthew T. Russotto	russotto@eng.umd.edu	russotto@wam.umd.edu
Some news readers expect "Disclaimer:" here.
Just say NO to police searches and seizures.  Make them use force.
(not responsible for bodily harm resulting from following above advice)
From: skataba@pbhyf.PacBell.COM (Sam Atabaki)
Newsgroups: rec.puzzles
Subject: Re: Easy prob. question
Date: 25 Jul 92 01:16:53 GMT
Reply-To: skataba@PacBell.COM (Sam Atabaki)
Organization: Pacific * Bell, San Ramon, CA

In article <1992Jul24.225745.19641@EE.Stanford.EDU> mack@isl.Stanford.EDU (mack) writes:
>
>This looks to be an easy one:
>
>Q: Two cards are picked at random from a complete deck of cards. What
>   is the probability that they are of the same suit?
>
>    --mack

Yes it is very easy. The probability is 12/51 = 4/17, since after the first
card is drawn you have 12 chances in 51 cards.

-Sam.
From: t9114145@phillip.edu.au
Newsgroups: rec.puzzles
Subject: Re: Ten equals 9.9999....
Date: 30 Jul 92 14:55:30 GMT
 <1992Jul29.000030.4839@fid.morgan.com> <1992Jul28.232532.3577@hellgate.utah.edu>
Organization: RMIT (Bundoora Campus)

In article <1992Jul28.232532.3577@hellgate.utah.edu>, tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>>If two numbers are different, then there's another number between
>>them.  (Actually, a lot of such numbers.)
> 
>>If you can't give an example of a number between 9.99999... and 10,
>>please don't post anymore to this thread.
> 
> 
> SAY WHAT?  This is the silliest bit of nonsense ever.  "If two numbers are
> different, then there's another number between them".  Let's consider the
> integers, and I will give you 2 and 3.  Now, if they are different, there
> must be an integer between them.  Well, what is it?
> 
> Now, if you assert that the above is true for REAL's then you better have
> some reason, for I can just as easily assert the opposite.  As a matter of
> fact, I can show you two numbers which are different, which do not have any
> number between them...
>   41.999999.... and 42

Hang on.... wasn't this whole thread about why x.9999... is actually equal
to x + 1? (And how wrong I was when I said it wasn't? :-)
 
Surely then 41.99999... is not different from 42?

-- 

Adam Eberbach    Computer Science student, RMIT.   t9114145@phillip.edu.au
               "We don' need no stinkin' signatures!"
From: dmccart@gomez.intel.com (D. J. McCarthy ~)
Newsgroups: rec.puzzles
Subject: Re: permutations
Date: 24 Jul 92 23:12:41 GMT
Organization: Architecture & Software Technology Group, Intel, Santa Clara CA
X-Newsreader: Tin 1.1 PL4

cdl0915@ritvax.isc.rit.edu (Christian Dylan Losinger) writes:
> 
> 	I need a good algortithm for producing all permutations of a sequence
> of items .

	There's a good (albeit recursive) one in one of Martin
Gardner's collections of Scientific Americans... if I remember
I'll find it over the weekend and post it Monday.

--
D. J. McCarthy <dmccart@gomez.intel.com>
How can we dance when our earth is turning?
Ho do we sleep while are beds are burning?
From: smneyln@ariel.lerc.nasa.gov (Michael Neylon)
Subject: Re: Ten equals 9.9999....
Summary: Hmmm...
News-Software: VAX/VMS VNEWS 1.41    
Keywords: nine ten numbers repeating digits sex
Organization: NASA Lewis Research Center
Date: 24 Jul 1992 10:02 EST  

In article <1992Jul24.102553.17411@pasteur.Berkeley.EDU>, tjoa@cory.Berkeley.EDU (Richard Tjoa) writes...
>Ok, here's something interesting...
>9.9999...  (nine point nine repeating)  =   10  (ten)
>              _
>            9.9  =  x  (yes, I mean "repeating" with the bar over it.)
>              _
>           99.9  =  10x  (multiply by 10)
>              _
>         -  9.9  = -  x  (subtract that first thing)
>          --------------
>           90    =   9x  (oooh, no more decimal)
>           10    =    x  (divide by 9)
>              _
>            9.9  =   10  (voila!  [QED])
> 
> 

If I remember correctly, there was a discussion last year that
9999999999(repeating) === 1 

so this is no fallacy (9.99999... = 9 + .999999... = 9 + 1 = 10)

Mike Neylon          aka Masem the Great and Almighty Thermo GOD
  // Only the | Univ. of Toledo - Chemical Engineering - Senior
\X/   AMIGA!  |   NASA Lewis Research Center - Summer Intern
\_____________|_____smneyln@ariel.lerc.nasa.gov________________/
       -+ How do YOU spell 'potato'?  How about 'lousy'? +- 
      "Me and Spike are big Malcolm 10 supporters."-J.S.,PLCL
From: jw7348@norway.medtronic.com (Jeff Wilkinson)
Newsgroups: rec.puzzles
Subject: Re:  Plane Division
Date: 23 Jul 92 13:57:56 GMT
Organization: Medtronic, Inc., Minneapolis, MN

In article <1992Jul22.154053.3498@alf.uib.no> s290@brems.ii.uib.no ( Per Ivar Roervik) writes:
>  This is adopted from an oldie given by Gardener.
>  
>  In a closed and bounded area of any 2-D shape, 1 million random
>  points are chosen.
>  Is it always possible to bisect the area AND the points into two
>  eqal parts with a straight line?
>
>  Proof or disproof needed.  
>    

Well, it seems that the circle argument demonstrates that the answer
is no.  However, what's the answer when a constraint is added that no
3 points are colinear?  For the more general case, what is the upper
bound on the number of points which can be colinear, for which the
bisection is still possible (assuming that the first condition is
sufficient)?

wilk