HDTenEightyP/NetNews huggingface dataset:
From: macster@IASTATE.EDU (Michael C Mccarty)
Newsgroups: rec.puzzles
Subject: Re: vacuuously false?
Date: 29 Jul 92 17:52:34 GMT
Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
Organization: Iowa State University

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

Ahhh....  Good question.   Well, at least, I think it is, anyway.  I am not
familiar with the term 'vacuously true', but I think I have derived a fairly
good idea of its meaning (but please enlighten me to its meaning if I am
completely off-track).  
	Logically, the first statement (often denoted, A -> B) above has a very rigid
definition.  This statement is considered 'valid' (valid is what I am assuming
is roughly what is meant by 'vacuously true') iff (iff = if and only if) when A
is true, B is true, but NOT vice versa.  This allows B to be true even if A is
not.  What causes an implication to become 'invalid' is when A is true but B is
false, thereby denying the meaning of the statement. Any other state of
condition has no effect upon the validity of the implication.
	What must be realized is that the 'natural' (ie English) interpretation of the
above statement ("if today is Monday...") includes an implicit meaning that its
logical counter-part does not.  Symbolically, it would be tempting to write the
sentence as:

	(today is Monday) -> (tomorrow is Tuesday).

	This allows (tomorrow is Tuesday) to be true even when (today is Monday) is
false.  This, of course, is wrong.  The correct and definitive way to state such
an assertion is: 'Tomorrow is Tuesday if and only if today is Monday.'
(represented by A <-> B)

					Mike

P.S.  Ugh.  Can I babble or what?!?!?   :-)
Newsgroups: rec.puzzles
From: ph@vortex.ama.caltech.edu (Paul Hardy)
Subject: Re: carrying the torch
Organization: California Institute of Technology
	<1992Jul25.080934.9218@pasteur.Berkeley.EDU>
Distribution: usa
Date: Sat, 25 Jul 1992 20:54:59 GMT

In article <1992Jul25.080934.9218@pasteur.Berkeley.EDU> tjoa@cory.Berkeley.EDU (Richard Tjoa) writes:

>   Remove the panel covering the switches and wire different ohmage resistors
>   in series with the switches.  Turn them all on, run into the other room and
>   check the brightness.  If they are not noticibly different, remove the
>   bulbs, whip out your ammeter and check the amperage.

No, you cannot remove the switch panel; you can only toggle the switches.
I guess you can remove the light bulbs, for what good that will do you.

				--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: jimhsu@athena.mit.edu (Jimmy M.D. Hsu)
Subject: Bravo!
Organization: Massachusetts Institute of Technology
Date: Wed, 29 Jul 1992 21:03:35 GMT

Congratulations to David G Radcliffe for a
correct response to the nine spot posting!

-->  James Amdee  MIT

M.C. Escher is really
Mauritus Cornelius Escher.

Now you know why he uses initials!
Newsgroups: rec.puzzles
From: herbert@let.ruu.nl (Herbert Ruessink)
Subject: Re: preposition proposition
Organization: Faculty of Arts, State university of Utrecht
Date: Tue, 28 Jul 1992 09:19:22 GMT


>>In article <2113@tymix.Tymnet.COM> filfeit@rael.Tymnet.COM writes:
>>>Everybody knows that a preposition is a word you should not
>>>end a sentence with B-).
>>>
>>>Can someone post a sentence that ends with FIVE prepostions?

marshall@force.ssd.lmsc.lockheed.com (Bob Marshall) writes:

>I remember this one :

>What did you bring that book that I didn't want to be read to out of up for?

>OK, so maybe these aren't prepositions as they're used here, but at times
>they *ARE* prepositions. I can't believe there is really a valid sentence
>ending in five prepositions that are really being used as prepositions.

The _Guinness_book_of_records_ (1978, paperback edition, Bantam Books,
p.206) gives us "a protest of a child against an Australian bedtime
story book:"

   "Mommy, what did you bring that book which I didn't want to be read
    to out of from about `Down Under' up for?"

A bit contrived, though.

Herbert.
Newsgroups: rec.puzzles
From: tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Re: SUGGESTION
Date: 21 Jul 92 01:29:02 MDT
Organization: University of Utah, CompSci Dept

>Dear Alert Puzzle Solvers,
>
>How about whenever someone makes a mistake, instead of cluttering up the newsgroup with multiple items like;
>
>	"Didn't you read the question..."
>	"Think again dope..."
>	"In case you missed the point..."
>	"Ha Ha Ha what a dork...."
>		etc etc etc
>
>	...why not just e-mail the author directly to point out the error and let them reply with their own retraction.  It would save me and many others a lot of time having to wade through countless useless items.  Also, perhaps the best way to exercise your ego is rather than pointing out other peoples mistakes publicly, why not solve the thing yourself?


Hey, what a lame idea.


But now.....

A man falls from an airplane into an unknown place .  He knows that it might be
a country where all the words are the same as english, except that "Yes" and
"No" are switched.  The people don't lie, they merely use "Yes" for "No" and
"No" for "Yes".  The two places are called "normalville" and "oppositeville"
He meets a native, and instead of asking a direct question, he outwits himself
and says:  

Our hero: Are you from oppositeville?

Stranger: No. 

Our hero: Do you use "yes" to mean "no"?

Stranger: No. Yes. I am not sure what you mean, where are you from?

Our hero: I am from normalville, are you?

Stranger: Perhaps, why do you want to know?

Our hero: Does your grandmother wear socks to bed?

Stranger: Yes.

Our hero: If I asked you a factual question to which you answered "no", would I
          answer "yes" ?

Stranger: It depends, a moment ago I would have said something different
           than I would say now.  I am not sure you would be able to answer
           any factual questions, judging by your interrogation.


Where was the stranger from?
From: mikem@ll.mit.edu
Newsgroups: rec.puzzles
Subject: Re: Plane Division
Date: 22 Jul 92 19:20:43 GMT
Organization: MIT Lincoln Laboratory


 In article <1992Jul22.154053.3498@alf.uib.no>,
 s290@brems.ii.uib.no ( Per Ivar Roervik) writes:
>  
>  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
>  equal parts with a straight line?
>    

No.  Assuming you mean "use a single line to divide the area into two
parts, each part having the same area AND the same number of points."

>
>  Proof or disproof needed.  

Not a formal proof, but an intuitively obvious counterexample:

Consider a unit circle centered at the origin.  All 1,000,000 points
are on the x-axis, between (0,0) and (1,0).

Since all the points are colinear, they can only be divided in half
by a line which passes between the 500,000th and 500,001st point.
This point will be somewhere in the right half-plane.  It will NOT
be the center of the circle (0,0).

Note:  It is assumed from the problem specification that if the
line must *divide* the points and the area, it may not intersect
any of the points.

If the line must not pass through any of the other points, neither
can it pass through the center of the circle, since all 1,000,000
points are colinear on the X axis, with the center of the circle.

But if the line does not pass through the center, it cannot divide
the circle into two equal areas.  Any straight line which divides a
circle into two equal areas must pass through its center.


                                  .
                      |           .
               xxxxxxxxxxxxxxx    .
            xxx       |       xxx .
          x           |           x
        x             |           . x
       x              |           .  x
     x                |           .    x
    x                 |           .     x
    x                 |           .     x
   x                  |           .      x
  x                   |           .       x
--x-------------------+---------o-x-o-----x--
  x                   |        .5 . .7    x
   x                  |           .      x
    x                 |           .     x
    x                 |           .     x
     x                |           .    x
       x              |           .  x
        x             |           . x
          x           |           x
            xxx       |       xxx .
               xxxxxxxxxxxxxxx    .
                      |           .
                                  .
    
-------------------------------------------------------------------------------
Michael Maciolek - mikem@ll.mit.edu --- VOICE (617)981-3174 - FAX (617)981-0189
Network Engineer - MIT Lincoln Laboratory - Computer Telecommunications Systems
-------------------------------------------------------------------------------
     "If it wasn't for disappointments, I wouldn't have any appointments"
		They Might Be Giants:  _Snowball in Hell_
-------------------------------------------------------------------------------
From: water772@Armstrong.EDU (Kelly Waters)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 26 Jul 92 23:38:48 GMT
Distribution: rec.puzzles
Organization: Armstrong State College, Savannah, GA

>>>>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?
>>>>


SPOILER ALERT !!!!!!!!!!!






	F-O-R-T-Y



Kelly Waters

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Famous Last Words Overheard in Calculus 9 :-)                     po bif asc
     ___________________________________     _________________________________
   O   |                               |   O   |                             |
  \|/  | 'You think it's on the test?' |  \|/  | 'Naw, He'd never ask that!' |
   |   |_______________________________|   |   |_____________________________|
  / \                                     / \  
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
favorite nagging question : If I exist, do you? And How about Fred? Or Herman?
________________________________________________________________________________
Newsgroups: rec.puzzles
From: de@cray.com (Duane Eitzen)
Subject: Re: Plane Division
Organization: Cray Research, Inc.
Date: 22 Jul 92 11:43:43 CDT

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.
From: karr@cs.cornell.edu (David Karr)
Newsgroups: rec.puzzles
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Date: 29 Jul 92 02:09:18 GMT
Organization: Cornell Univ. CS Dept, Ithaca NY 14853

In article <5688@osc.COM> Joe Keane <jgk@osc.com> writes:
>I'm afraid you made a mistake in your subtraction.  The problem is that
>99.999... minus 9.9999... is 89.999..., not 90 like you have.

Er--excuse me?  89.999... = 90 (assuming it makes any sense to write
89.999..., of course), so your result is technically correct :-), but
what's your justification for that result of the subtraction?

>Joe Keane, amateur mathematician
            ^^^^^^^
>jgk@osc.com (uunet!amdcad!osc!jgk)

-- David Karr (karr@cs.cornell.edu)
Newsgroups: rec.puzzles
From: amlogan@bird.Princeton.EDU (Adam M. Logan)
Subject: Re: Math puzzle
Originator: news@ernie.Princeton.EDU
Organization: Princeton University
Distribution: rec.puzzles
Date: Tue, 21 Jul 1992 20:03:18 GMT

In article <KEITH.92Jul21110018@chirality.rsa.com> keith@chirality.rsa.com (Keith Vetter) writes:
>
>As is well known the sum  1/1 + 1/2 + 1/3 + ... goes to infinity.  The 
>question is, does the partial sum  S = 1/1 + 1/2 + 1/3 + ... + 1/n
>ever equal an integer value (other than n=1)?
>
>Keith@rsa.com

No.  Proof 1:  Let 2^k be the largest power of 2 less than or equal to n.
Then when we put the fraction over its common denominator of LCM (1, 2, ...
n), all numerators will be even except for the 2^n term.  Even doesn't divide
odd.

Proof 2: By Bertrand's postulate there is a prime > n/2.  All numerators but
one will be multiples of this prime, the denominator is, so no integer.

More generally, (1/a + 1/(a + b) + 1/(a + nb)) is never an integer.

Adam
Newsgroups: rec.puzzles
From: chan@elec.uq.oz.au (Chuen Wen Chan)
Subject: Proof of -1 = 1
Organization: Prentice Centre, University of Queensland
Date: Fri, 24 Jul 1992 03:36:13 GMT


I hope this problem hasn't been posted before, but here it is anyway:

	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

ie:  Everything we learnt in classical maths is wrong
   => no more maths subjects in future
   => all mathematicians lose their jobs. Yippe!

PS:  If no one gets the solution by Monday, it will be posted.

--
 ,-_|\  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: tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Re: Easy prob. question
Date: 24 Jul 92 17:04:41 MDT
Organization: University of Utah, CompSci Dept

>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?

You pull out one card.  There are 12 left of that suit, so it is 12/51 or
12/53 depending on what type of deck you have.
From: althof5805@iscsvax.uni.edu
Newsgroups: rec.puzzles
Subject: Crossing a river
Date: 30 Jul 92 17:35:56 -0500
Organization: University of Northern Iowa

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?


Stephanie Althof                              Your mother was a hamster,
University of Northern Iowa                     and your father smelt of
Department of Teaching                          elderberrries!
Cedar Falls, IA  50613                                   --French knights
Newsgroups: rec.puzzles
From: hhassler@iaik.tu-graz.ac.at (Hannes Hassler)
Subject: Re: Babble Sort
Organization: Institut fuer Angewandte Informationsverarbeitung, TU Graz
Date: Tue, 28 Jul 92 18:51:43 GMT

In article <1992Jul28.164309.10270@alf.uib.no>  Per Ivar Roervik writes:
>
>  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 
>  which is first.

The first number is 1, the last number is 10, 12 or 20.

Argument for first number: I claim that the first eight numbers always
are 1,2,3,4,5,6,7,8 and the ninth number is 9 or 19.  Consider the first
moment where 9 is exchanged with a neighbor. The only possible neighbor
is 19, and it arrives from above. Both numbers exchange places, and in the
following round, they again are exchanged.

Argument for last number: Only 10 and 12 may cross number 20.

*****************************************************************************
 Hannes Hassler
*****************************************************************************
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Wed, 29 Jul 1992 02:23:42 GMT

In article <1992Jul28.175614.18586@hellgate.utah.edu> tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>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

I can deal with the notation 9.999... but what in tarnation is
9.999...(infinity)...8 ?  How can a sequence of infinitely many
digits have a LAST digit?

>Then by induction, 10 also equals 9.999.....(infinity)...7

This is not induction.  It's not even induction's second cousin.

>Then by induction, 10 also equals 9.999.....(infinity)...6
>
>If we follow this through, 

then (swallowing the absurd notion of an infinite terminating decimal)
every number that we generate contains an infinite number of nines
followed by one of the nine non-zero digits:

9.999...992  (just showing some more of the 9's explicitly)
9.999...991
9.999...990

But the 0 in the last number doesn't matter; we can rewrite that number as

9.999...99

i.e. an "infinite" string of 9's ending in a 9, but that's just where you
started!  The next number is 9.999...998 and so forth, repeating the first
nine numbers forever.

>than [sic] 10 = 9.

Not so, see above.

Here's an exercise:  prove that 9.999... is a real number.  Once you have
done that, we can argue about which one it is.

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



Newsgroups: rec.puzzles
From: jil@uts.uucp (Jamie Lubin)
Subject: Re: 3 puzzles
Organization: Bellcore, Piscataway, NJ
Distribution: rec.puzzles
Date: Thu, 23 Jul 92 14:12:17 GMT

In article <5071@transfer.stratus.com> markc@eomer.sw.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?
 
40  (F-O-R-T-Y)
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: rec.puzzles
Subject: Re: Plane Division
Date: 23 Jul 92 12:52:40 GMT
Organization: U of Cambridge, England

In article <1992Jul22.151943.2805@hellgate.utah.edu>, tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
> Consider all possible lines which divide a shape in half ( a regular one for
> now).  All of these lines must cross the center if it, for a circle, for a
> square, and so on.
> 
>   We now will choose all of the points so that the lie along one of these
> dividing lines, and are in one half of the figure.  For instance, the
> radius of a circle.  
> 
>  In order to break the points in half, we must strike them at some angle
> which is different than the line they all lie in.  Yet, if we did that
> then our line would no longer divide the area in half.
> 
>   It is possible to construct a set of points which can not be broken in
> half by a line which also divides the figure into two equal areas.
> 
> QED
> 

Note, however, that if we replace the points by little discs (or if we allow
ourselves to count a line through a point as cutting off any fraction of the
point), it can be done (it's a case of the Ham Sandwich Theorem, and is proved
by the sort of argument someone gave earlier).
-- 
Newsgroups: rec.puzzles
From: sethb@fid.morgan.com (Seth Breidbart)
Subject: Re: Another brain teaser
Organization: my opinions only
Date: Wed, 29 Jul 1992 00:06:36 GMT

Read the original post for the description of the game.

I assume that a "round" involves shuffling, etc.

This game is trivial to analyze:

After n turns, you must have picked up at least n cards.  Since you
want to minimize the probability that you have picked up a joker, you
should have picked up at most n cards.  Therefore, picking up one card
per round is optimal.

Seth		sethb@fid.morgan.com
Newsgroups: rec.puzzles
From: macster@IASTATE.EDU (Michael C Mccarty)
Subject: Had x 11 (revisited)
Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
Organization: Iowa State University
Date: Fri, 24 Jul 1992 02:42:49 GMT

Heh heh...  Reminds me of a quote I saw a while back of a sentence that contains
27 'and's, 21 of which are consecutive (no quotes around any of them).

Ready?

(take a deep breath...)












[spoiler]




























Wouldn't the sentence 'I want to put a hyphen between the words Fish
and And and And and Chips in my Fish-And-Chips sign' have been clearer
if quotation marks had been placed before Fish, and between Fish and
and, and and and And, and And and and, and and and And, and And and
and, and and and Chips, as well as after Chips?


(Allright, so there are quotes in the sentence, but not around the 'and's!)



				Michael McCarty
From: warwick@cs.uq.oz.au (Warwick Allison)
Newsgroups: rec.puzzles
Subject: Re: Alien colonization
Date: 23 Jul 92 01:38:45 GMT
Reply-To: warwick@cs.uq.oz.au

tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:

>Ddiv is also immortal, and wishes to keep living forever.
>He wishes to be fair to his children -

> How many children can Ddiv have? (Ddiv has children by dividing)

The obvious point is that if immortals have immortal children, then
they will fill any finite space, no matter how few children they have.  

So the Universe must be infinite for the problem to be interesting.
In an infinite space, Ddiv could have at least one child, giving
it the planet in a direction, and allowing its decendants to
propagate in that direction.

Intuitively, I think the only solution to the problem is to be able to
teleport to any empty planet in the universe instantaneously (which on
average will take you infinitely far from you starting point), then
place a son there.  Since the universe is infinite, you can do this
forever.  All generations may have infinite children.  This works in a
1 dimensional universe too (as does the single-child case).  Only an
infinitely dimensional universe could support more than 1 child, and it
would support infinite children (well, actually, probably just any
finite number of children, which is not quite the same thing).  

--
Warwick
--
  _-_|\      warwick@cs.uq.oz.au            /Disclaimer:
 /     * <-- Computer Science Department,  /  
 \_.-._/     University of Queensland,    /      void (if removed)
      v      Brisbane, Australia.        /