HDTenEightyP/NetNews huggingface dataset:
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: rec.puzzles
Subject: Re: Plane Division
Date: 23 Jul 92 01:24:47 GMT
Organization: U of Cambridge, England

Apologies if you get two of these. My newsreader crashed when I tried to
post a reply to this a few minutes ago...

In article <1992Jul22.114343.12817@hemlock.cray.com>, de@cray.com (Duane Eitzen) writes:
> 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.  
> >    
> 
> Since a counter-example is a disproof:
> 
> _________________________
> |                       |
> |           X           |
> ----------------------| |
>                       | |
> ----------------------- |
> |                       |
> |                       |
> -------------------------
> 
> Where "x" is the roughly the area where the 1 million points are. In this
> case the straight line will either trisect the area (not bisect), or
> will not cut the shape into two equal parts. 
> 
> This is another example of my belief that the real puzzle in most puzzles
> is in stating the problem rather than in solving it once its stated.
> 
> Duane.

Draw a line going through the X and sloping a little like \. If you get the
slope right it will bisect both the points and the area.

I don't think you will get a counter-example by bunching all the points up
together, because of the following theorem:

Given any bounded plain region, and any point in the plane, there is a straight
line through that point which bisects the region.
PROOF: imagine a line through the point sweeping round through 180 degrees.
As it goes, (area on one side) - (area on other side) goes from (say) A-B to B-A;
so it changes sign; since it varies continuously, it must be zero somewhere; so
at that point the line bisects the region.

My guess is that the answer to the original question is "yes", but I have no
proof as yet.

-- 
Newsgroups: rec.puzzles
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Subject: Re: Tiling a plane
News-Software: VAX/VMS VNEWS 1.41    
Organization: Texas A&M University, Academic Computing Services
Date: Sat, 25 Jul 1992 01:40:00 GMT

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.

Penrose tiles are a pair which can tile a plane, but the pattern cannot
be regular. I understand it is still unresolved whether there is a Penrose
'monotile'.

Dave Ring
dwr2560@zeus.tamu.edu
Newsgroups: rec.puzzles
From: rcutler@banach.WPI.EDU (Robert Cutler)
Subject: Re: Proof of -1 = 1
Originator: rcutler@banach.WPI.EDU
Organization: Worcester Polytechnic Institute
Date: Fri, 24 Jul 1992 08:03:21 GMT


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

		 - ( 3 - 1)  = - 3 + 1 x= -3 - 1
			

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



	If we follow the laws of distribution in row 3 the error 
directly follows.




				Rob
From: wilcox@vax.oxford.ac.uk
Newsgroups: rec.puzzles
Subject: Re: preposition proposition
Date: 24 Jul 92 11:57:58 GMT
Organization: Oxford University VAXcluster

In article <1992Jul23.180750.17907@col.hp.com>, cab@col.hp.com (Chris Best) writes:
>> Can someone post a sentence that ends with FIVE prepostions?
> 
>>   Time flies like an arrow.
> 
> ----------
> 
> Has it been two weeks already?  Time flies, allright.
> 
> Five is no problem.  Last posting here had seven or eight - 
> concerning a book about Australia.

I'm confused. Which of these is supposed to be a solution (if any)?
-- 

Stephen Wilcox             |   Remember what happened to the dinosaurs!
wilcox@maths.oxford.ac.uk  |    I did---and look what happened to me.
Newsgroups: rec.puzzles
From: chan@elec.uq.oz.au (Chuen Wen Chan)
Subject: Re: Proof of -1 = 1
Organization: Prentice Centre, University of Queensland
Date: Fri, 24 Jul 1992 06:43:36 GMT


>	sqrt(-1) = sqrt(-1)

>	sqrt( -1 / 1 ) = sqrt( 1 / -1 )

>	sqrt(-1)    sqrt(1)
>	-------- = ---------
>	sqrt(1)     sqrt(-1)

>	now cross multiply and we get:

>	-1 = 1  QED

Regarding the above problem, I DO know that the third line is 
wrong (ie: i != 1/i).  The point is the derivation up to that
point looks legal.

Thanks to Rob Cutler for pointing it out.
--
 ,-_|\  Chuen Chan (chan@elec.uq.oz.au)
/     * Department of Electrical Engineering
\_,--_/ The University of Queensland, St Lucia
     v  Brisbane, Australia 4072
Newsgroups: rec.puzzles
From: ph@vortex.ama.caltech.edu (Paul Hardy)
Subject: Re: carrying the torch
Organization: California Institute of Technology
	<1992Jul26.220605.4280@cco.caltech.edu>
	<26JUL199223340613@zeus.tamu.edu>
Distribution: rec
Date: Mon, 27 Jul 1992 09:11:49 GMT

In article <26JUL199223340613@zeus.tamu.edu> dwr2560@zeus.tamu.edu (RING, DAVID WAYNE) writes:

   *** SPOILER ***
















>   I write...
>   >>Set them to 'on' 'off' and 'hot'.
>
>   carl@SOL1.GPS.CALTECH.EDU writes...
>   >Close.
>   >Turn on switches 1 and 2 and turn switch 3 off.  Leave them in that state
>   >for several minutes.  Turn off switch 2.  Go into the other room.  The
>   >light that is on is controlled by switch 1.  The light that is off and
>   >warm is controlled by switch 2.  The light that is off and cold is
>   >controlled by switch 3.
>
>   Oh, sorry... Make that 'on' 'warm' and 'off'  ;-)
>

Yes, this is the answer.  I was going to name this puzzle "a flash of
inspiration" but decided that would be too much of a giveaway.

				--Paul


--
This is my address:         ph@ama.caltech.edu
This is UUCP:               ...!{decwrl,uunet}!
This is my address on UUCP: ...!{decwrl,uunet}!caltech.edu!ama!ph
Any questions?

"Does Emacs have the Buddha nature?"  --Paul Hardy   "Yow!" --Zippy
From: fxsdb@acad3.alaska.edu (Scott D. Berry)
Newsgroups: rec.puzzles
Subject: Re: Easy prob. question
Date: 25 Jul 92 00:04:43 GMT
Organization: University of Alaska Fairbanks

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?
> 

In fact, this is an easy one.  The answer depends on a couple things, tho.
What kind of deck (a pinochle deck)?  Assume a poker deck, complete.  The
chances are (52/54) * (13/54) + (2/54)^2, if two jokers are considered to 
be the same suit.
   -Scott (fxsdb@acad3.alaska.edu)
From: joshg@athena.mit.edu (Josh Gagliardi)
Newsgroups: rec.puzzles
Subject: Re: Proof of -1 = 1
Keywords: rotation proof
Date: 24 Jul 92 16:44:53 GMT
Organization: Massachusetts Institute of Technology

How about this oldie:
             1                   1
Given  that --- = 0, prove that --- = oo
             oo                  0


     1
    --- = 0             given
     oo

  -18 = 0               counterclockwise rotation

  -10 = 8               additive property of equality

   1
  --- = oo              clockwise rotation
   0

  Q.E.D.



Acknowledgement to Tau Epsilon Phi's rush mailing my frosh year...
From: water772@Armstrong.EDU (Kelly Waters)
Newsgroups: rec.puzzles
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Date: 30 Jul 92 03:49:40 GMT
Organization: Armstrong State College, Savannah, GA

>
>In any event, let us suppose that for the moment, 10 does equal 9.999...

>Then by induction, 10 also equals 9.999.....(infinity)...8
>Then by induction, 10 also equals 9.999.....(infinity)...7
>Then by induction, 10 also equals 9.999.....(infinity)...6
>
	Perhaps you could explain your 'induction.'  It does not follow from
10=9.9999... that 10 also equals 9.999.....(oo)...8 .  Additionally, what 
digits follow the 8 ie. 9.999.....(oo)...8999 or ...8888 or ....87999 ?
The basis of 10=9.999.... is that no 'last' digit exists.  With that in mind,
your arguement is 

	If we change the last digit, it does not alter the value.  

Since your premise is false, your statement is true.  But the fact is that we
cannot change the 'last' digit, and your 'inductive' step is unsupported.

>If we follow this through, than 10 = 9.  Then 10 = whatever we want.
>That is clearly silly, so I am inclined to agree that 10 != 9.99999....
>

 10 = 9 is indeed silly, but you have not established the connection between
this clearly wrong statement 10 = 9, and the statement 10 = 9.9999.... 




Kelly Waters

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
favorite nagging question : If I exist, do you? And How about Fred? Or Herman?
________________________________________________________________________________
From: jimhsu@athena.mit.edu (Jimmy M.D. Hsu)
Newsgroups: rec.puzzles
Subject: Calling on Higher Brain Functions...
Date: 30 Jul 92 17:25:06 GMT
Organization: Massachusetts Institute of Technology

Alright, you've convinced me.  No more variations!
The idea to include the simplest, base case
results in shouts of "trivial!", so this time, I'll
post the entire puzzle, and let you come up with
your own variations (lots of possibilities, here).

This is a number series that follows an
iteritive type formula that can be repeated
to infinity.

1  11  26  54  91  145...

a variation:

4  11  29  54  94  145...

These two are intimately related.

How do you generate the next numbers
in the series?  Mebbe I give too many terms?

I have the elegant solution, if
anyone asks.

I came up with this one this morning, so
don't flame me if it's too simple or if
it's been done before.

Thanks to all who replied to previous
puzzles.  Guess I shall post some more
if I think them up.

One last number series (joke, w/spoiler)

1  4  9  31  125  630...























*SPOILER*

This one is sort of unfair, but
it does generate an infinite series.

-1- *1+3 = -4- *2+1 = -9-

9 *3+4 = -31- *4+1 = -125-...

If you can't figure it out, give it
some time to sink in.  Easy as pie.

This one is not related to the previous
two, btw!


---
James Amdee   MIT

Speak to me--the pain you feel

			-Q
Newsgroups: rec.puzzles
From: ph@vortex.ama.caltech.edu (Paul Hardy)
Subject: Re: carrying the torch (HINT)
Organization: California Institute of Technology
	<1992Jul25.065442.12748@beaver.cs.washington.edu>
Distribution: rec
Date: Sat, 25 Jul 1992 21:04:36 GMT

In article <1992Jul25.065442.12748@beaver.cs.washington.edu> phaedrus@cs.washington.edu (Mark Phaedrus) writes:

> Well, if the light bulbs are fairly conventional incandescent bulbs, you
>   could do the following:

>
> Turn one of the switches (call it #1) on.  Wait about ten minutes.
> Turn another switch (#2) on.  Race into the other room.  Light #1 will be on;
> light #2 will be on, but the bulb will be noticeably cooler; light #3 will be
> off.

This is not bad, but not the exact answer.  Incandescent light bulbs heat up
quickly, so the above might not be possible.


				--Paul

--
This is my address:         ph@ama.caltech.edu
This is UUCP:               ...!{decwrl,uunet}!
This is my address on UUCP: ...!{decwrl,uunet}!caltech.edu!ama!ph
Any questions?

"Does Emacs have the Buddha nature?"  --Paul Hardy   "Yow!" --Zippy
Newsgroups: rec.puzzles
From: sethb@fid.morgan.com (Seth Breidbart)
Subject: Re: Alien colonization
Organization: Morgan Stanley & Co., New York, NY
Date: Thu, 23 Jul 1992 01:38:39 GMT

In article <1992Jul22.221102.5700@CSD-NewsHost.Stanford.EDU>
amorgan@Xenon.Stanford.EDU (Crunchy Frog) writes:
>In article <1992Jul22.153416.3564@hellgate.utah.edu> 
>  tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>
>>  The alien Ddiv lives in a universe which is three dimensional, and has
>>the planets evenly spaced throughout as in a 3D grid.
>>Ddiv is planning to have children- each child will require one full planet
>>to live on.  Ddiv is also immortal, and wishes to keep living forever.
>>He wishes to be fair to his children- so he will only have as many 
>>children as each of his children can have.  He would not have any children
>>if any single one of his offspring could not have any children, so
>>his grandkids, great grandkids, and so on must be able to have as many
>>children as he has. 
>>
>>
>>  How many children can Ddiv have? (Ddiv has children by dividing)
>
>Is the trick with this one that if Ddiv has >1 child after an infinite
>amount of time there will be an uncountably infinite number of Ddiv'lets
>and thus not enough planets for them to live on?
>
>I say one child (but I would be surprised if I am right).

You are, but for a different reason.

I'm assuming that the child must live on a planet adjacent to the
parent; otherwise, Ddiv could have infinitely many children.

Assume Ddiv has k children.  After n generations, there will be over
k**n total creatures, who will be occupying a volume of O(n**3).  If
k>1, for large enough n, this is impossible.

Seth		sethb@fid.morgan.com
From: danm@hpnmdla.sr.hp.com (Dan Merget)
Newsgroups: rec.puzzles
Subject: Re: Re: Gummy Drop-bears <SPOILER>
Date: 30 Jul 92 23:39:15 GMT
Organization: Hewlett-Packard, Santa Rosa, CA

In rec.puzzles, hhassler@iaik.tu-graz.ac.at (Hannes Hassler) writes:

> In article <Jul.29.03.30.41.1992.18050@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?
> >
> I can do it with 53 bears: Take 0, 1, 2, 4, 7, 14, and 25 bears, respectively.
> No two distinct triples over these numbers add up to the same sum.  Hence, you
> can identify all imitations.  I think that 53 bears is best possible; it is
> easy to see that 0 must occour in the optimum solution.

Well, if we cannot use any additional equipment (I posted an earlier
solution in which Spike used a bar and 6 bears) then I think you can do
it with 51 bears:  0, 1, 2, 4, 7, 13, and 24.  Unless I made a huge
mistake, this will also give you a unique sum for each triple.

I got this set of numbers by starting with {0, 1, 2} and then adding on
the minimal numbers to ensure unique triples.  (I had to backtrack a
little when I got to 3.)

       _
      | |   /\           ________________________________________________
    __| |__/ /_/\____/\__\   This note is not intended to represent the  \
   / _  |_  __/  \__/  \___  viewpoints of my god, my country, my company,\
  / / | |/ / / /\ \/ /\ \  \ or my family.  I'm not even certain that      \
  \ \_| |\ \/\ \ \  / / /  / it accurately represents my own.              /
   \____| \__/\/  \/  \/  /_______________________________________________/

	- Dan Merget
	  danm@sr.HP.COM
Newsgroups: rec.puzzles
From: Ulf.Lunde@kvatro.no (Ulf Lunde)
Subject: Re: 3 puzzles
Organization: Kvatro A/S, Trondheim, Norway
	<5084@transfer.stratus.com> <1992Jul26.134105.13917@ens.fr>
	<1992Jul27.145517.7957@cbfsb.cb.att.com>
Date: 31 Jul 92 21:03:54

 On Mon, 27 Jul 1992 14:55:17 GMT, mmokhtar@attmail.com said:
> Originator: mmdh@cbnewsf.cb.att.com
> 
> In article <1992Jul26.134105.13917@ens.fr> courtes@trimaran.ens.fr (Francois Courtes) writes:
>>>	What is the first number, when spelled out (e.g. O-N-E) is
>>>in alphabetical order?
>>>40  (F-O-R-T-Y)

Forty??   How about F-I-V-E ?   Or F-I-F-T-E-E-N?   Or E-I-G-H-T ?


>>In French the first one is 2 (D-E-U-X).
>>Is there a language in which 1 is so ?
> 
>   Eins in German.
> 
> Mazen Mokhtar

Oh yeah?  Just how do you spell e.g. 397 in German, then?  Or 8?

Have I missed something here?  I seem to disagree with everyone else...
I Norwegian I think 18 must be the first one: A-T-T-E-N.
In Esperanto it is 10 (D-E-K) if you don't count B-I-L-I-O-N-O (1E12).

How many numbers (in how many languages) tell you how many letters
they are made up of?  E.g. F-O-U-R in English.

Ulf Lunde
From: danm@hpnmdla.sr.hp.com (Dan Merget)
Date: Thu, 30 Jul 1992 04:59:37 GMT
Subject: Re: Gummy Drop-bears
Organization: Hewlett-Packard, Santa Rosa, CA
Newsgroups: rec.puzzles

In rec.puzzles, 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?

I can name that tune in 6 bears.

Take a bar of known mass.  Balance one end of the bar on some convenient
object and balance the other end on the scale.  Take one bear from 6 of
the packages (numbered 0-5) and put each bear on the bar a distance
1/(2**n) from the object to the scale.  Thus, bear #0 is on the scale,
bear #1 is halfway to the scale, bear #2 is 1/4 of the way, etc.

Now, the scale supports half the mass of the bar, and 1/(2**n) of the
mass of each bear.  Take your measurement, and subtract the bar's
influence from the measurement.  The result should be (630 - k)/32
grams.  If you express k as a 6-bit binary number and number the bits
starting from the left, then each 0 corresponds to a real bear and each
1 corresponds to a fake.

This procedure identifies 6 of the packages.  The remaining package
contains fakes if you find 2 fake bears, and it contains real bears if
you find 3 fakes.

----------------------------------------------------------------------
     Dan Merget       | This post does not represent the opinion of
     danm@sr.HP.COM   | Hewlett-Packard.  Does that surprise you?
Newsgroups: rec.puzzles
From: jchamber@ecst.csuchico.edu (John William Chamberlain)
Subject: Re: Rift in the fabric of the space-time continuum?
Organization: California State University, Chico
Date: Wed, 29 Jul 1992 05:33:32 GMT


It sounds as if you had set the watch twelve hours ahead.
Newsgroups: rec.puzzles
From: chalcraft@uk.tele.nokia.fi (Adam Chalcraft)
Subject: Re: Number of primes with n digits?
Organization: cpd
Date: Fri, 31 Jul 1992 15:22:35 GMT

In article <chepan.712595863@gsusgi1.gsu.edu> chepan@gsusgi1.gsu.edu (Parthasarathy Nambi) writes:

>    Is there an easy way to find (a formula perhaps) out
>    how many primes are there with say n digits.

Not a chance. The answer's roughly 9/10 * 10^n / (n * ln(10)), though,
or a constant times that, for large enough n.


--
| Adam      | #include <stddisclaimer.h>                                   |
| Chalcraft | You can tell how tall a story is by how many flaws it's got. |
From: wyg9633@uxa.cso.uiuc.edu (Wooed Sun)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 25 Jul 92 02:14:50 GMT
Distribution: rec.puzzles
Organization: University of Illinois at Urbana

karr@cs.cornell.edu (David Karr) writes:

>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?"  ...

billion?


-- 
email to : w-yang@uiuc.edu (wyg9633@uxa.cso.uiuc.edu)
Newsgroups: rec.puzzles
From: cbc1@Ra.MsState.Edu (christopher b carey)
Subject: Re: Proof of -1 = 1
Organization: Mississippi State University
Date: Fri, 24 Jul 1992 08:03:40 GMT

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
>     =  2 - 3
>     =  -1     QED

Almost slipped that one by me, but only because it's late.
(2 + 1) - (3 - 1)
= 2 + 1 - 3 + 1
= 4 - 3
= 1

Sorry.

Chris Carey
--
Got those wishing I was out of school so I could see my girl on
a regular basis blues.

Newsgroups: rec.puzzles
From: deghare@daisy.waterloo.edu (Dave Hare)
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Organization: University of Waterloo
Date: Wed, 29 Jul 1992 12:36:53 GMT

In article <1992Jul29.041320.8481@samba.oit.unc.edu> ajpierce@med.unc.edu (Andrew Pierce) writes:
>In article <1992Jul29.000030.4839@fid.morgan.com> sethb@fid.morgan.com (Seth Breidbart) writes:
>>If two numbers are different, then there's another number between
>>them.  (Actually, a lot of such numbers.)
>
>   I agree with you but it's not that simple.  Your statement needs
>justification.  Mathematicians:  is this an axiom of number theory for
>reals?

No, it is a simple consequence of the order relation for reals.  If a and
b are real numbers and a <> b, then either a<b or a>b.  Without loss of
generality, assume a<b.  Then c = b-a > 0, so c/2 > 0, and c/2 < c, so
a < a+c/2 < a+c = b (note a+c/2 = (a+b)/2 ).

Dave Hare
Newsgroups: rec.puzzles
From: tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Re: Ten equals 9.9999....
Date: 28 Jul 92 23:25:32 MDT
Keywords: nine ten repeating digits
Organization: University of Utah, CompSci Dept

>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
From: mk0e@ns1.cc.lehigh.edu (MAHEENDRA BABURAO KASMALKAR)
Newsgroups: rec.puzzles
Subject: Re: regrouping letter order in a sentence
Date: 23 Jul 92 22:09:40 GMT
Organization: Lehigh University


>> I'm looking for a string of characters that can be converted into two
>> different sentences by adding spaces and punctuation. Having both
>> sentences make sense would be nice, but then how do you define that?
>>
>> Any ideas as to the maximum length of string that this can be done with?
>
>
>The condemned man ate and talked an hour after he was hanged.
>
>The condemned man ate and talked; an hour after, he was hanged.
>

A telegram from a wife to her husband out of station :

Instead of writing :
     Not getting any better , come back soon .

She wrote :
     Not getting any , better come back soon .


Kassu
-- 
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Newsgroups: rec.puzzles