HDTenEightyP/NetNews huggingface dataset:
From: macster@IASTATE.EDU (Michael C Mccarty)
Subject: The Homing Pigeon Problem and more...
Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
Organization: Iowa State University
Date: Mon, 6 Apr 1992 07:38:17 GMT

	I'm new to this newsgroup, so I don't know if this has been posted before, but
I thought this was an intriguing problem (for which I am not sure there is an
answer), so I'm going to give you a try at it.  I'm not positive how the problem
was originally stated (I got this out of Scientific American, I think), but I'll
try to give you all that is necessary to know.

	-Suppose there are two armies of soldiers engaged in a war against a common
enemy.  These armies are separated by mountain range, thereby making homing
pigeon the only form of communication between them.  The commander of Army A
knows the position of the enemy, of which, both armies are able to get to.
However, the enemy is so strong that if both armies do not attack
simultaneously, it would mean certain disaster for both armies.  The commander
of Army A wishes to send Army B a message stating a time at which both armies
may attack.  Unfortunately, the homing pigeons are a rather unreliable form of
communication (they tend to get lost, killed by predators, etc.).  Therefore
Commander A requests that Commander B send a reply message stating the approval
of the plan.  Commander B also realizes the limitations of the homing pigeon,
and requests a return message acknowledging his message.  Well I think you can
see where this is leading...  What message, if any, can the commanders send to
each other that will ensure that an answer to whether the message was received
or not can be implied?

I, honestly, do not know a definite answer to this problem.  I thought maybe if
you people worked on it, that we all may gain a little insight into the
solution.


	INTERESTING-YET-SIMPLE BAR TRICK: Take two straws, melt them together at the
ends at approximately 90 degree angle.  Take a third straw and prop the fused
straws up to form a pyramid (Note: the third straw is not fused to the other
straws, it merely supports them).  Now, taking a pencil, pick up all three
straws SIMULTANEOUSLY, USING ONLY THE PENCIL, NOTHING ELSE!

I find about 1 in 4 people will solve this puzzle on the first try.  The rest
either give up or try something completely off the mark. 

And finally, name a six-letter word that when the first letter is capitalized,
changes the pronunciation of the word.

					Mike

P. S.  Sorry, I haven't come up with a neat .signature yet.
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Number puzzle (dollars per hour)
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Mon, 13 Apr 1992 15:48:56 GMT

In article <1992Apr11.173533.10303@hellgate.utah.edu> tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>
>Put it this way, if someone offered you XXX dollars to die in five years, how
>much money would this have to be for you to consider it?  They pay you now,
>and you must die in five years.

No amount XXX would be adequate, but not because I'm "priceless."  Simply
put, no amount of money in the bank would enable me to do all the things
I want to do in life within the next five years.  So any figure you could
come up with this way would be meaningless.

How about an alternative method:  You are offered a sum in payment for a
year of your life.  (There are two ways you could pose this:  either your
life is shortened by a year at the end, which would require prescience on
the part of the other party; or they just put you to sleep for a year
solid, so you miss the kids' birthdays, a year's career development, etc.,
while your body ages a full year so you die just as soon as you would 
have otherwise.)  Now what is an adequate sum?  You might very well be
able to put an exact price tag on it if you thought hard enough.

Let's hear from some of the "dying young" people on this one.
From: emulcahy@unix2.tcd.ie (QUETZALCOATL              MONTEZUMA OAXACA TENOCHTITLAN INTI RAYNI RA CHICHEN ITZA TLAXCALA MACCHU PICCHU CUZCO AKRANES REKJAVIK AKUYEYRI KEFLAVIK HAFNAFJIORDHUR GODTHAAB THULE MINNESOTA NORTHSTARS SEATTLE SEAHAWKS KON-TIKI)
Newsgroups: rec.puzzles
Subject: farms
Summary: ha
Keywords: gch
Date: 8 Apr 92 14:12:48 GMT
Organization: Trinity College, Dublin


     


    If a Horse costs #10,

    a goat costs &1,

    & 8 chickens cost #1,

    
    how many of each animal must you buy to spend exactly #100,

    & have 100 animals ,with at least one of each kind?
From: tom@ulysses.att.com (Tom Smith)
Newsgroups: rec.puzzles
Subject: Re: Sequence puzzle, retry after mailer problems
Date: 9 Apr 92 02:19:19 GMT
Reply-To: tom@ulysses.UUCP (Tom Smith)
Organization: AT&T Bell Laboratories, Murray Hill

In article <1992Apr8.210658.971@spectra.com> morgan@spectra.com (Mike Morgan) writes:
>
>This is supposedly pretty simple, but.....
>
>What are the next two numbers and why?
>1,11,21,1211,111221, ..........

* SEMI-SPOILER ALERT *


	I've seen this one before, but I'll post my response since
	I solved it unaided when I first saw it:

	1,11,21,1211,111221, 312211, 13112221 ..........

	The inventor of this puzzle (John Conway of Princeton) gives
	enormously entertaining lectures about number theory and
	some of his pet sequences (of which this is one).  He also
	invented "Life" (not the board game).

	A little hint:  Conway wrote a paper about this in which
	he coined the term "audioactive decay."

>Later daze!
>
>-- 
>Michael L. Morgan 
>morgan@spectra.com

	Tom Smith
	tom@ulysses.att.com
From: mblum@chaos.cs.brandeis.edu (Cerebus)
Newsgroups: rec.puzzles
Subject: Re: A simple sequence of numbers ...
Date: 9 Apr 92 16:51:29 GMT
Article-I.D.: chaos.mblum.702838289
Organization: Brandeis University

meng@idt.unit.no (Tore Mengshoel) writes:

>In article <1992Apr7.220710.27512@Cadence.COM>, hemmady@Cadence.COM (Shankar Hemmady) writes:
>|> 
>|> 
>|> Continue this sequence:
>|> 
>|> 	3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...
>|>
>e.g. 3, 5, 4, 4, 3, 3, 5, 5, 4, 3, .... 

What do you mean, "e.g. . . ." ??
The answer is:

6, 6, 8, 8, 7, 7, 9, 9, 6, . . .

Isn't it??

-- Matt
Matt "Matt" Blum  		mblum@chaos.cs.brandeis.edu

The city, 1:00am. People in their rooms, asleep, lonely, depressed.
One of them gets the urge to pet a small, furry animal.  That's where
I come in.  My name's Friday.  I carry a badger.
From: hxie@grad1.cis.upenn.edu (Hong Liang Xie)
Newsgroups: rec.puzzles
Subject: Re: A simple sequence of numbers ...
Date: 8 Apr 92 21:32:03 GMT
Organization: University of Pennsylvania

In article <1992Apr7.220710.27512@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:
>
>Continue this sequence:
>
>	3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...
>
>-Shankar Hemmady

(possible spoiler)








 	3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 3, 4, 4, 4, 5, 5, 5, 3. 


-Hong, CIS Dept, U of Penn

Organization: Freshman, Math/Computer Science, Carnegie Mellon, Pittsburgh, PA
Newsgroups: rec.puzzles
Date: Tue, 14 Apr 1992 18:11:31 -0400 
From: "Daniel S. Efran" <de0t+@andrew.cmu.edu>
Subject: Re: trvia (alternative spoiler)
	<1992Apr12.062432.20383@rice.edu>

Joseph De Vincentis writes:
>But, your solution can still work if you allow only one wall.
>Make your house round!

The problem specifies a "square house."  If you can make a round square house,
please explain how.


	---Daniel Efran
----------------------------------
An it harm none, do as you Will.
Practice random kindness and senseless acts of beauty.

Newsgroups: rec.puzzles,soc.culture.indian
From: bs@wucs1.wustl.edu (Bala SWAMINATHAN)
Subject: Re: A simple number puzzle
Organization: Washington University, St. Louis MO
Date: Wed, 8 Apr 1992 04:20:47 GMT

In article <1992Apr08.021647.17896@eng.umd.edu> russotto@eng.umd.edu (Matthew T. Russotto) writes:
>In article <1992Apr7.202400.26466@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:
>>
>>The puzzle goes as follows:
>>
>>"My child's age raised to itself gives my age.
>>If my wife is younger than me by my child's age
>>and if we are LIVE humans, then what are our ages?"
>
>child	wife	you
>1	0	1 	naa
>2	2	4	probably not
>3	24	27	sure
>4	252	256	naa
>
>Sometimes, there is no substitute for exhaustive search.
>
>
>-- 
>Matthew T. Russotto	russotto@eng.umd.edu	russotto@wam.umd.edu

Did the original article imply any where that LIVE humans can have only
integral ages? OR did the very use of the word *age* imply an integer?

If the age can be counted in months (say 26 years and 5 months etc.)
then there are more solutions.

AAAARRRRRRRRRRGGGGGGHHHHH !!!!
chill out
S_Bala
From: rose@fsu1.cc.fsu.edu (Kermit Rose)
Newsgroups: rec.puzzles
Subject: f(a)=b and f(b)=c
Date: 3 Apr 92 15:32:24 GMT
Article-I.D.: mailer.1992Apr3.103224.21334
Reply-To: rose@fsu1.cc.fsu.edu
Organization: Florida State University
News-Software: VAX/VMS VNEWS 1.3-4


Yes, this was a good puzzle.
f(a)=b. f(b)=c.
let f(x)=dx+e
da+e=b
db+e=c

d=(b-c)/(a-b)
f(x)=[(b-c)/(a-b)](x-a) + b


Kermit@FSU or rose@fsu1.cc.fsu.edu.bitnet
If you want to be sure I see your response, please e-mail me.
-----------------------------------------------------------------------
This is the signature file of Kermit Rose.

Believing in dichotomies leads to hardening of your logic.

What determines the existence of Free Will?
I strive to believe only what is true. I have faith that whatever is 
true will also be logically consistent and knowing truth will increase my 
capacity to be kind.
Peace is the way to peace.  Hostility is the way to hostility.
From: jerryd@hpgrla.gr.hp.com (Jerry Donovan)
Date: Fri, 10 Apr 1992 14:22:29 GMT
Subject: Re: Brick Problem (crude ASCII)
Organization: Hewlett-Packard, Greeley, CO
Newsgroups: rec.puzzles

>As has been said, and it is obvious once it has been pointed out, any
>brick with an odd number of walls must contain one end of the line.
>Hence, counting the outside space as another "brick", such puzzles can
>only be solved without a trick if the number of bricks with odd numbers
>of walls is 2 or 0. (In the latter case both ends of the line must be in
>one brick.) 

1 will also work.  It just makes a reqirement for one end of the line.
From: earutter@athena.cas.vanderbilt.edu (Ethan Rutter)
Newsgroups: rec.puzzles
Subject: Re: Monday morning brain energizer April 6th
Keywords: A weekly series in the making?
Date: 10 Apr 92 01:37:26 GMT
Reply-To: earutter@athena.cas.vanderbilt.edu (Ethan Rutter)
Distribution: na
Organization: Mathematics, Vanderbilt University, Nashville

In article <1992Apr6.160440.28526@clipper.ingr.com> kontos@clipper.ingr.com (Thorne Kontos) writes:
>
>        The Deception Test
>
>A,B,C,D, and E have just finished taking a test for Deception.
>In their usual chatty way, they are making remarks about their
>places. The remarks of the first and second place finishers
>are false, while the others are true. Lets listen in now.
>
>A: "D was third"
>B: "E was not first"
>C: "I was not last"
>D: "C was lower than B"
>E: "B was second"
>
>What were their places?

I came up with two solutions which seem to fit the statements.
1) A E C D B
   A is false (D = 4)
   E is false (B = 5)
   C is true (3 < 5)
   D is true (3 < 5)
   B is true (E = 2)

2) A E C B D
   A is false (D = 5)
   E is false (B = 4)
   C is true (3 < 5)
   B is true (E = 2)
   D is true (3 < 4)

If "C is lower than B" means C *placed* lower than B, the order is
A E B C D.


-- 
-------------------------------------------------------------------------
Ethan Rutter                       | "If I claim to be a wise man,
earutter@athena.cas.vanderbilt.edu |  it surely means that I don't know."
Math Dept.                         |      - Carry On Wayward Son, by Kansas
From: ncramer@bbn.com (Nichael Cramer)
Newsgroups: rec.puzzles
Subject: Re: A question of time
Date: 17 Apr 92 04:22:11 GMT
Reply-To: ncramer@labs-n.bbn.com (Nichael Cramer)
Organization: Bolt Beranek and Newman Inc., Cambridge MA
NNTP-Posting-Host: bbn.com

fca00@chennai.ccc.amdahl.com (A. Francis Chandran) writes:
>Well here's a wierd possibility ;-)

>The clock at the top of the stair is really an anti-clock that
>goes anti-clockwise and whose digits are arranged anti-clockwise[...]

Ah, interesting.

But you don't have to stretch _quite_ that far...

Maybe the prof sees the clock in a mirror, after all, he _does_ have bad
eyesight.  (One could make this slightly fancier by supposing glowing/
radium hands in an otherwise dark hallway.)

Is there anything we can do with a mirror-image digital clock?

N
Newsgroups: rec.puzzles
From: holt@lambda.msfc.nasa.gov (Tom Holt)
Subject: Ladder puzzle
Summary: A geometry problem
Reply-To: holt@lambda.msfc.nasa.gov
Organization: New Technology, Inc.
Date: Fri, 17 Apr 1992 17:54:21 GMT


 I have an interesting little puzzle.

     -   |       |
     |   |\     /|  -
     |   | \   / |  |
         |  \ /  |
    15   |   x   |  10
         |  / \  |
     |   | / 6 \ |  |
     |   |/  |  \|  |
     -   ---------  -

         |-  ?  -|


 Two ladders cross one another in an alley. Their bottom edge is 
 butted into the corner. The first one hits the far wall 15 ft. up.
 The second one hits the other wall 10 ft. up. The point at which 
 they intersect is 6 ft. high. What is the width of the alley?

 I'd like to see how a solution is reached, also.
From: hansz@cs.ruu.nl (Hans Zantema)
Newsgroups: rec.puzzles
Subject: Re: Yet another number problem
Date: 8 Apr 92 15:45:06 GMT
Organization: Utrecht University, Dept. of Computer Science

In <1992Apr7.215858.22701@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:


>
>This puzzle is not as simple as the last one I posted.
>
>There is something unique about the unordered pair of numbers (2, 4):
>
>There are no other two DISTINCT positive integers x and y such that
>	
>	x ** y = y ** x (x to the power y equals y to the power of x)
>
>
>Can you prove this uniqueness property?
>
>
>I have not been able to prove it without using differential calculus.

Assume x<y.

Let p be any prime factor of x, write x = a*p^n  and  y = b*p^m with a and b
not divisible by p.

Then the number of factors p in x^y is ny and the number of factors p in y^x 
is mx. So ny = mx, since x<y and n>0 we conclude m>n.

This holds for all prime divisors of x, so x is a divisor of y. Write y=kx.
Then (kx)^x = x^(kx) = (x^k)^x. So kx = x^k, so k = x^(k-1).

By induction one shows that k < 2^(k-1) for k > 2.
We have k = x^(k-1) >= 2^(k-1) >= k.
The first ">=" is equality if and only if x=2,
the second ">=" is equality if and only if k=2.

So x=k=2, so y=4.

I am not aware of any differential calculus in this proof.

				Dr Hans Zantema
				Department of Computer Science
				University of Utrecht
				P.O. Box 80.089 
				3508 TB  Utrecht
Newsgroups: rec.puzzles
From: jwd2@elf.owlnet.rice.edu (Joseph William DeVincentis)
Subject: Re: Fill in the blanks.
Organization: Rice University
Date: Sun, 12 Apr 1992 05:49:46 GMT

In article <11APR199215071585@zeus.tamu.edu>, dwr2560@zeus.tamu.edu (RING, DAVID WAYNE) writes:
> In article <1992Apr10.203558.34277@watson.ibm.com>, surya@rothko.watson.ibm.com (Surya Mohanty) writes...
> >What are the next three numbers in the sequence and why ?
> > 
> >3 5 11 17 31 41 67 - - - ?
> 
> Did you forget 59?

No, Dave, his sequence is correct as written.

Spoilers follow:

The next three numbers are 83, 107, and 113.

"Why?" follows:

Let p(n) be the nth prime. Then the nth element of this sequence
is p(p(n)).

-- 
\  ,------------------------------------------------------------.  /
 \/     Joseph  De Vincentis            jwd2@owlnet.rice.edu     \/
 /\           Help! My .sig just flipped umop-ap!sdn !           /\
/  `------------------------------------------------------------'  \
Newsgroups: rec.puzzles
Subject: Geographical puzzle.
From: wft@math.canterbury.ac.nz (Bill Taylor)
Date: 15 Apr 92 17:06:12 +1200
Distribution: world
Organization: Department of Mathematics, University of Canterbury

What property do the following southern-hemisphere cities and towns have 
in common? As far as I can check, it is an exhaustive list (for this property).

Buenos Aires
Melbourne
Mar del Plata      (Argentina)
Bahia Blanca       (Argentina)
Christchurch       (New Zealand)   ........[ yay ! home team !]
Comodoro Rividavia (Argentina)
Punta Arenas       (Chile)
Ushuaia            (Argentina)


There is a similar list for the Northern hemisphere. I'll leave it to
some native to work that one out.

---------------------------------------------------------------------
  Bill Taylor                             wft@math.canterbury.ac.nz 
---------------------------------------------------------------------
               " Pretentious ?  MOI !? "
---------------------------------------------------------------------
From: weemba@libra.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: rec.puzzles
Subject: Re: Dying young
Date: 7 Apr 92 15:17:36 GMT
Reply-To: weemba@libra.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman)

In article <1992Apr6.175229.23741@hellgate.utah.edu>, tolman%asylum (Kenneth Tolman) writes:
>
>You are put in a room by evil Nazi's.
                          ^^^^ ^^^^
Pleonasm alert!
-- 
-Matthew P Wiener (weemba@libra.wistar.upenn.edu)
Newsgroups: rec.puzzles
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Subject: Re: Isaac Asimovs mysteries puzzle (hint and comment)
Reply-To: carl@SOL1.GPS.CALTECH.EDU
Organization: HST Wide Field/Planetary Camera
Date: Sat, 11 Apr 1992 04:08:07 GMT

In article <1992Apr9.172711.2748@anasaz>, morris@anasaz (Bob Morris) writes:
>In article <3847@shum.huji.ac.il> yehuda@shuldig.huji.ac.il (Jonathan Berlinger) writes:
>>Donn't know if anyone heard this one yet, so:
>>
>>Why does>	Christmas = Halloween
>>
>Comment:  This works in the US and some other places, but not everywhere.
>And no, I don't mean places where they don't celebrate Christmas.  It
>doesn't work in some places where Christmas is celebrated.  (Does anybody
>outside of the US celebrate Halloween?)
>
>Puzzle #1:  Why doesn't it work everywhere?  Where does it work?
>
>Puzzle #2: Why does Christmas = Thanksgiving (sometimes but not always)
>
>Puzzle #3: Why does Halloween = Thanksgiving (sometimes but not always)
>
>(The answer to #2 and #3 is the same as the original.)
SPOILER

1)  It works only in countries where the name of the 10th month is abbreviated
    OCT and the name of the 12th is abbreviated DEC, since we all know that
    31 OCT(al) = 25 DEC(imal).
2) and 3)  I don't know of ANY numeric base abbreviated NOV, so haven't a clue
    as to what you're talking about.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXes and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
From: ts@cup.portal.com (Tim W Smith)
Newsgroups: rec.puzzles
Subject: Re: Puzzles from an ordinary life
Date: 6 Apr 92 09:07:31 GMT
Organization: The Portal System (TM)

> 3) You are interested in driving home without ever stopping at a light.  
> Also, you hope to live on.  How can you achieve this?  (if lights abound
> in your city)

The control boxes that control the lights often have provisions for
manual control.  For example, when an event ends at Flint Center
in Cupertino, CA, someone goes out and plugs a contol box into the
light controller, and manually regulates traffic flow, because
the default timings aren't set up to deal with the abnormal
traffic flow.

So the obvious solution is to make a radio control unit that you
can plug into the manual control inputs on the traffic control
boxes along your route home.  Control it from a unit in your
car.

						Tim Smith
Newsgroups: rec.puzzles
From: jmarlan@titan.nmt.edu (Jon Marlan)
Subject: Re: Sequence puzzle, retry after mailer problems
Organization: New Mexico Institute of Mining and Technology
Date: Wed, 8 Apr 92 22:42:59 GMT

In article <1992Apr8.210658.971@spectra.com> morgan@spectra.com (Mike Morgan) writes:
>
>This is supposedly pretty simple, but.....
>
>What are the next two numbers and why?
>1,11,21,1211,111221, ..........
312211,13112221

I think this might be on the FAQ.  I remember seeing a progression 
like this (if not the same one) a few weeks ago.  Each number gives
information about the previous number.  Take 11 for example. 11 
indicates that the number before it is 1 (one 1 = 11).

>Later daze!
>
>-- 
>Michael L. Morgan 
>morgan@spectra.com

Jon

--------------------------------------------------
"You are what goes into your mind" -- Zig Ziglar
--------------------------------------------------



Newsgroups: rec.puzzles
From: mitchell@cs.uchicago.edu (Mitchell Marks)
Subject: Re: Another Isaac Asimov puzzle -- geography
Reply-To: mitchell@cs.uchicago.edu
Organization: University of Chicago Computer Science
	<9471@wrgate.wr.tek.com>
Date: Thu, 16 Apr 1992 00:26:19 GMT

>>>>> "DT" == Dan Tilque <dant@logos.wr.tek.com> writes:

  DT> ed@wente.llnl.gov (Ed Suranyi) writes:
>
>Here's Asimov's version:  What do New York City, and Xenia, Ohio have
>in common?
>
>Here is my version:  There are a set of cities in the world, numbering
>less than fifty, that share a certain characteristic.  Exactly two of
>these cities are in the United States:  New York City and Xenia, Ohio.
>What is this characteristic?

Spoiler and comments on it follow


  DT> At one time I would have guessed that they're the largest cities with
  DT> names starting with their respective first letters.  But I wouldn't be
  DT> too sure of that since placenames starting with X are not uncommon in
  DT> Central America.

  DT> Now though, China has adopted the Pinyin romanization where X has a
  DT> sound roughly like 'sh' (or possibly like 'zh') and so there are
  DT> probably several there that move Xenia down the list.

This is the correct answer.
This is one of my favorite puzzles of this sort.

Like Ed Suranyi, I also remember reading it in Asimov's F&SF science
column.   And I also join Ed in rephrasing it just about the way he
did.

I also usually make a point of saying something to the effect that the
puzzle dates from the seventies, and the answer might not hold today.
Besides the factors Dan points out, there have been population changes
(and two U.S. censuses) since the time this puzzle was first
formulated.  An "out of date" disclaimer seems to give people an extra
clue, too: the characteristic is something that is subject to change.

--
Mitch Marks    mitchell@cs.UChicago.EDU
"Oh God comma I abhor self-consciousness" --John Barth