HDTenEightyP/NetNews huggingface dataset:
From: potamian@hpl.hp.com (Spyros Potamianos)
Subject: Killers & Pacifist (NEW VARIATION)
Date: Mon, 21 Dec 1992 22:23:25 GMT
Organization: Hewlett-Packard


First, let me state the original puzzle:

    You arrive in a village (or island, or whatever) inhabited by
    K killers and P pacifists. Every day two persons are selected at
    random. If both of them are pacifists nothing happens. If one is
    a killer and the other a pacifist then the killers kills the pacifist.
    If both are killers, then they kill each other. You can join the
    "game" as a pacifist or a killer. Which choice maximizes your chance
    of survival?

The answer is that it doesn't matter. If K is odd (NOTE: 'K' is the number
of killers *before* you join) you die anyway. If K is even you have a
1/(K+1) chance of survival (no matter what you decide to become)

Now the new variation: Everything remains the same, except for a slight
change in the algorithm for selecting the two persons:

    Randomly choose a killer. Then randomly select another person
    (killer or pacifist).

If you don't believe that that's different from the original problem try
K=2, P=0. If you become a killer your chance of survival is 1/3, but if
you become a pacifist it's 1/2

Spyros Potamianos
potamian@hpl.hp.com

:-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-)

P.S. BTW, I encoded a message in this post, and it has a pretty good chance
of surviving a billion years (yes, USENET will still exist...)

:-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-)
From: leon@unixg.ubc.ca (Leon ter Beek)
Newsgroups: rec.puzzles
Subject: Re: Help me with MH puzzle, also, a new puzzle.
Date: 4 Jan 1993 03:28:35 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
NNTP-Posting-Host: unixg.ubc.ca

In article <1i7cb6INNe0@mirror.digex.com> adric@access.digex.com (William Johnson) writes:
>When playing Russian Roulette, each member has a revolver with 6 chambers
>and 1 bullet (may have been 2 bullets, I don't think it changes the answer,
>just the likelihood of death).  Each member then fires the gun twice into
>his/her own head.  Members may spin the revolver before each shot.  Which
>option gives the best chance of survival:
>	A) Spin shoot spin shoot
>	B) Spin shoot shoot
>	C) Shoot spin shoot
>	D) Shoot shoot.
>
>Ok, it seems obvious to me that the two options where you shoot twice
>without a spin in the middle are worst.  Because assuming you survive
>the first shot (because if you didn't, it's over), you have now reduced
>your chances of surviving the next chamber.
>
>But I don't see the difference between A and C, unless in case C you know
>in advance that where you put the bullet either was or was not in the
>next chamber to be fired.  This may have been the answer, but I don't think
>so, because reading their answer sheet, the probabilities weren't as clear
>cut as that.  Any suggestions?

A=C if you didn't spin before you start the next round. 
Spinning makes the events independent, ie. next shot same probability as
first shot if you survive the first one.

A: P(survive) = 5/6*5/6= 25/36
B: P(survive) = 5/6*4/6= 20/36 Second shot only 4 empty chamber left.
C: P(survive) = 3/6*3/6= 9/36  Don't spin after preceding round.
D: P(survive) = 3/6*2/6= 6/36

P(A) > P(B) >P(C) > P(D)

Reasoning with 2 bullets is indeed the same.

Regards,
Leon
----------------------------------------------------------------------
"De wereld is mijn vaderland          Email: leon@nematic.chem.ubc.ca
 en wetenschap mijn religie"          Fax:   +1-604-82-22847
   -Christiaan Huygens-               Phone: +1-604-82-22603
From: phs503e@vaxc.cc.monash.edu.au (Scott Shannon)
Newsgroups: rec.puzzles
Subject: Re: The Rifleman Problem
Date: 26 Dec 92 18:18:33 +1100
Organization: Dept. of Physics, Monash Uni.


I remember reading this puzzle in an old Scientific American, probably long
before the question was published in Omni. Have no idea in what edition 
unfortunately, one somewhere in the 70's I think. I recall that they said
the exact solution for the 2d plane (all riflemen in a square boundary I think)
was found by two japanesse math professors about 1 year after the question 
was first published in a magazine as a mathematical puzzle. Obviously a
rather hard problem, but apparently solvable. I can't remember the exact
solution, but I did some computer simulations a long time ago with 100 
riflemen and on average got about about 27 survivors (or somewhere there abouts)
which I recall was about the correct answer. 

Newsgroups: rec.puzzles
From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Subject: A nice New Year puzzle.
Organization: University of Michigan EECS Dept., Ann Arbor, MI
Date: Fri, 1 Jan 1993 00:20:31 GMT

This puzzle was given to me by a friend Mr. L.Chandar who
is with the Physics dept., Syracuse University. I've solved it
and have really enjoyed spending several hours of the New Year's
eve working on it. Here it goes :

Given a plane and exactly 3 colors, prove whether or not it is possible
to assign a color to each point of the plane in such a way that no two points
exactly 1 inch apart have the same color.  If the assignment is possible,
suggest a way to make it; if not, explain why. (Note : The points of the
plane form one continuum; they are not discretely placed as in a grid.) 
-----------

Have a nice New Year.

Kanad
Newsgroups: rec.puzzles
From: arromdee@jyusenkyou.cs.jhu.edu (Ken Arromdee)
Subject: Re: Direction of the sun
Organization: Johns Hopkins University CS Dept.
Date: Sun, 27 Dec 1992 23:35:25 GMT

In article <1hjv7nINN34t@gap.caltech.edu> carl@SOL1.GPS.CALTECH.EDU writes:
>>Two boys, A and B, are watching the sunset.
>>A points to the sun and says:
>>"This is the direction where the sun is right now."
>>B replies:
>>"But don't you know that the rays take over 8 minutes to reach the Earth?
>>The sun is actually already below the horizon."
>>Which is correct?  Why? (Make it understandable to a ten year-old.)
>The one who says that the sun is below the horizon.

Wrong.  The rays from where the sun is now will reach us in 8 minutes, and
will at that time be below the horizon, but that's because the horizon has
moved in the intervening time, not because the sun has moved.
--
"the bogosity in a field equals the bogosity imported from related areas, plus
the bogosity generated internally, minus the bogosity expelled or otherwise
disposed of."  -- K. Eric Drexler

Ken Arromdee (arromdee@jyusenkyou.cs.jhu.edu, arromdee@jhunix.hcf.jhu.edu)
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Anagram Program
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Mon, 28 Dec 1992 05:04:58 GMT

In article <mannj.725306509@acf9> mannj@acf9.nyu.edu (Jordan Mann) writes:
>"John R Kennedy" <p00200@psilink.com> writes:
>
>>Does anyone know where I can get the source code to an anagram program?
>>I mean the kind that anagrams whole phrases, not just words.
>
>>P.S.  Please don't just say "yes".
>
>No.

Why not write one?  This could well be easier than trying to understand
someone else's source.  A former boss of mine used to present this
problem as an interview question.  A brute-force approach is easy;
a little thought should show you how to be more efficient.

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


Newsgroups: rec.puzzles
From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Subject: Re: A nice New Year puzzle.
Organization: University of Michigan EECS Dept., Ann Arbor
Date: Sun, 3 Jan 1993 19:20:18 GMT

In article <1993Jan1.053639.27111@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
>In article <1i081uINN6k5@cascade.cs.ubc.ca> kvdoel@cs.ubc.ca (Kees van den Doel) writes:
>>In article <1993Jan1.002031.5763@zip.eecs.umich.edu>
>>kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:
>>
>>>Given a plane and exactly 3 colors, prove whether or not it is possible
>>>to assign a color to each point of the plane in such a way that no two points
>>>exactly 1 inch apart have the same color.
>>
>>2 points sqrt(3) inch apart have the same color (take 2 points of
>>colors, say, 2 and 3, 1 inch apart. Form 2 equilateral triangles with
>>those 2 points, you get 2 new points (sqrt(3) inches apart) which must
>>have color 1, i.e. the same).  Pick a point, say it has color 1. The
>>circle with radius 1 around it has color 2 or 3. The circle with radius
>>sqrt(3) around it has color 1. On this big circle, pick an arbitrary
>>point and draw a circle of radius sqrt(3) around *it*, which must also
>>have color 1. The last and the first circles intersect at a point which
>>therefore must have color 1 *and* color 2 or 3, which is impossible.
>
>Why not pick any point on the sqrt(3) circle and draw a radius 1 circle
>around it?
>The center and intersecting points must be of the same color.
>
>-YH

Excellent !  Or easier still, a solution based on the same idea as
above : draw an isosceles triangle ABC with AB =
AC = sqrt(3) inches and  BC = 1 in.  (This triangle is possible because
the sum of any two sides is > the third). Then A and B must have the
same color, and so must A and C. Ergo, B and C must have the same color.
Ergo, the distance BC != 1 in.  A contradiction.       

Hope you guys enjoyed this puzzle.

Kanad
From: clong@romulus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: Killers & Pacifist (SPOILER)
Date: 2 Jan 93 06:47:19 GMT
Article-I.D.: romulus.Jan.2.01.47.19.1993.24988
Organization: Rutgers Univ., New Brunswick, N.J.

In article <Bzq92E.9As@hplabs.hpl.hp.com>, Spyros Potamianos writes:

>> we are the only pacifist; it's clear now that P=1/(k+1) since
>> we survive iff we don't take part in the first k interactions.

As pointed out, this isn't right.  Doing the multiplication
correctly, I get that P=(2k)!/(4^k*(k!)^2), where 2k is the number
of killers.  For large k this is approximately 1/sqrt(pi*k), which
is substantially better than the 1/(2k+1) odds you get by being
a killer.
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"S.B., with an I.Q. of 161, failed to complete his course of study,
running away instead with his professor's wife."
H. J. Eysenck, _Know Your Own I.Q._
Newsgroups: rec.puzzles
From: timh@igc.ethz.ch (Tim Harvey)
Subject: Lost solution
Organization: Computational Chemistry, ETH, Zuerich
Date: Tue, 22 Dec 1992 19:30:53 GMT

Dear Netters,
	A log while ago I collected the following brainteaser;

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;


Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
any truth from any set of axioms.  Two integers (not necessarily unique) are
somehow chosen such that each is greater than 1 and less than 100.  Mr. S.
is given the sum of these two integers; Mr. P. is given the product of these
two integers.  After receiving these numbers, the two logicians do not
have any communication at all except the following dialogue:
   Mr. P.:  I do not know the two numbers.
   Mr. S.:  I knew that you didn't know the two numbers; I do not know the
            two numbers.
   Mr. P.:  Now I know the two numbers.
   Mr. S.:  Now I know the two numbers.
"The two numbers" in this dialogue refers to those two integers that were
mysteriously chosen.

Given that the above statements are absolutely truthful, what are the two
numbers?
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
Anyway (and I hope it's not a FAQ) I didn't save the solution!

Can anyone supply a definitive answer? (Then we can all get back to work)
Many thanks,
Tim
From: joe@csource.oz.au (Joe Slater)
Newsgroups: rec.puzzles
Subject: Guessing at random numbers - spoiler
Date: Sat, 26 Dec 1992 23:04:05

ePE> I pick two numbers, randomly, and tell you one of them. You are
ePE> supposed to guess whether this is the lower or higher one of the two
ePE> numbers I picked. Can you come up with a method of guessing that does
ePE> better than picking the response "low" or "high" randomly (i.e.
ePE> probability to guess right > .5) ?

If you don't pick negative numbers, then I'd do best to *always* assume the
other number is higher. There are an infinite number of higher numbers, after
all.

If you allow negative numbers then I'd pick that your number is the lower of
the two if it's negative, and the higher of the two if it's positive.

jds

 * Origin: What horrors wait for me in this, the Phantom's Opera? (3:632/351)
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: Guessing at random numbers.
Date: 27 Dec 1992 10:18:56 GMT
Organization: HST Wide Field/Planetary Camera
Distribution: world
Reply-To: carl@SOL1.GPS.CALTECH.EDU
NNTP-Posting-Host: sol1.gps.caltech.edu

In article <1992Dec24.015625.3450@csservices.Princeton.EDU>, egs@cs.Princeton.EDU (Emin Gun Sirer) writes:
=Here's a paradoxical (at least to me) puzzle that I could not locate
=in the FAQ. Enjoy.
=
=I pick two numbers, randomly, and tell you one of them. You are supposed
=to guess whether this is the lower or higher one of the two numbers I
=picked. Can you come up with a method of guessing that does better than
=picking the response "low" or "high" randomly (i.e. probability to guess
=right > .5) ?
=
=Nothing is known about my random number generation algorithm. I give you
=the higher or lower number with equal probability, so trying to find patterns
=in my behaviour is useless (unless you believe that past coin tosses affect
=the outcome of future tosses :-).
=
=Thanks to Steve Gortler for this mind-twister.

Well, ask a silly question, get a silly answer.  Here's the silly answer:
You've picked a finite number as the first number.  Now, the measure of all the
real numbers greater than the one you picked is infinite, while the measure of
all the positive reals less than the one you picked is finite.  Hence, the
second number has probability 1 of being greater than the first number.

You see, unless you tell us what distribution you're using, you've got that
out.  And, of course, there's no way to "randomly" pick a number from all the
positive reals, which is the crux of the paradox.  So when you specify "nothing
is known about my random number generation program," if you're presenting this
puzzle as it has been presented in the past, you're making a fundamental
mistake.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen 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.
Newsgroups: rec.puzzles
From: lordSnooty@cup.portal.com (Andrew - Palfreyman)
Newsgroups: rec.puzzles
Subject: Re: Help Needed for Bob Puzzle
Date: Wed, 30 Dec 92 23:11:56 PST
Organization: The Portal System (TM)

: Has anyone seen this puzzle before? What's the answer to it?
: 
: Bob opens up the newspaper and reads a story about a woman (Bobbi)
: who skiied over a cliff and died while on a vacation with her
: husband (Rob). Bob calls up the police and tells them to arrest
: Rob for murder. Why?

Bob knows that Rob gets all she owns from the will he helped witness
and that she came into a large sum of money before the holiday and
that she was blind. So Bob just let her go.

Or something like that.
 --------------------------------------------------------------------------
| lord snooty @the giant |  inceptus clamor frustratur hiantes             |
| poisoned electric head |              andrew_-_palfreyman@cup.portal.com |
  --------------------------------------------------------------------------
From: clong@romulus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: Guessing at random numbers. (SPOILER)
Date: 2 Jan 93 07:02:28 GMT
Organization: Rutgers Univ., New Brunswick, N.J.

In article <1992Dec24.015625.3450@csservices.Princeton.EDU>, Emin Gun
  Sirer writes:

> I pick two numbers, randomly, and tell you one of them. You are supposed
> to guess whether this is the lower or higher one of the two numbers I
> picked. Can you come up with a method of guessing that does better than
> picking the response "low" or "high" randomly (i.e. probability to guess
> right > .5) ?

This is an old chestnut.  Pick any cumulative probability function P(x)
such that a>b ==> P(a) > P(b).  Now if the number shown is y, guess
"lower" with probability P(y) and "higher" with probability 1-P(y).  This
strategy yields a probability of > 1/2 of winning since the probability
of being correct is 1/2*( (1-P(a)) + P(b) ) = 1/2 + (P(b)-P(a)), which
is > 1/2 by assumption.

Notice that the fact that the lower and higher numbers are given with
equal probability is important, as otherwise the giver has a strategy
that keeps the probability of success to 1/2 or less (what is it?).
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"S.B., with an I.Q. of 161, failed to complete his course of study,
running away instead with his professor's wife."
H. J. Eysenck, _Know Your Own I.Q._
From: clong@romulus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: Killers & Pacifist (SPOILER)
Date: 22 Dec 92 13:29:08 GMT
Organization: Rutgers Univ., New Brunswick, N.J.

In article <BzMrJ3.JHH@hplabs.hpl.hp.com>, Spyros Potamianos writes:

> Now the new variation: Everything remains the same, except for a slight
> change in the algorithm for selecting the two persons:

>     Randomly choose a killer. Then randomly select another person
>     (killer or pacifist).

Let there be 2k killers, p pacifists.  If you choose to be a killer,
P(survival)=1/(2k+1) by symmetry.  If you choose to be a pacifist,
P(survival)=1/(k+1) since as before we may ignore any interaction
between a killer and a different pacifist, hence we may pretend
we are the only pacifist; it's clear now that P=1/(k+1) since
we survive iff we don't take part in the first k interactions.
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"S.B., with an I.Q. of 161, failed to complete his course of study,
running away instead with his professor's wife."
H. J. Eysenck, _Know Your Own I.Q._
Newsgroups: rec.puzzles
From: rodney@garfield.cs.mun.ca (Rod Hutchings)
Subject: Re: Mini-crossword-acrostic-whatever.
Organization: CS Dept., Memorial University of Newfoundland
Date: Wed, 30 Dec 1992 19:28:31 GMT

In article <1hqb2eINNjn8@controversy.math.lsa.umich.edu> elnitsky@math.lsa.umich.edu (Serge Elnitsky) writes:
>  The 7 words given by the clues form a well-known mathematical fact.
>
>  1. Cogito, ergo ___ .      1: SUM
>  2. The New York ___ .      2: TIMES
>  3. Vive la ___ !           3: DIFFERENCE
>  4. First among ___ .       4: EQUALS
>  5. See 3.                  5: DIFFERENCE
>  6. Baseball pos.           6: OF
>  7. Dull, boring people.    7: SQUARES
>
>


-- 
--
regards,
the Rodster
rodney@garfield.cs.mun.ca
Newsgroups: rec.puzzles
From: boll@CS.ColoState.EDU (dave boll)
Subject: Re: The Rifleman Problem
Date: Thu, 24 Dec 1992 22:19:01 GMT
Organization: Colorado State University, Computer Science Department

  (suggestion about placing points on the surface of a sphere deleted)
>
>It also flies in the face of the problem statement!  "...on a plane"
>does not, to me, seem to permit using a sphere.
>
>Also, what about distance ties?  Or are they ignored as a set of
>probability zero?
   yes

  Perhaps I should have said 'placing points randomly in 2D'. The
  whole point is that for large N, is doesn't matter if you place them
  on a square or on a circle, since only ~SQRT(N) will be affected by
  the boundary anyway. And, an EZ way to simulate this is to use small
  N on a surface with no boundary problems; a sphere seems the easiest
  choice.

  At least, if the goal is to write a program to do a bunch of random
  runs, (as I did a couple years ago) this is a pretty good solution.
  If the goal is to mathematically show that, on average, each point
  has a certain chance of survival, all this doesn't really matter.

  BTW, Omni briefly discussed the 3D problem as well, and asked the
  readers to imagine it being carried out among the stars in our
  galaxy. They said (in spirit, anyway): "Our sun would be a survivor in
  such a galactic holocaust, since its nearest neighbor, Proxima Centauri,
  is a member of a multiple star system" ....

  Puzzle: What is wrong with this reasoning?
-- 
     ------------------------------------------------------------
     Dave Boll  boll@handel.cs.colostate.edu
                "The speed of time is 1 second per second"
     ------------------------------------------------------------
Newsgroups: rec.puzzles
From: joelw@convex.com (Joel Williamson)
Subject: Re: New Year's problem
Date: Thu, 31 Dec 1992 21:09:08 GMT
Organization: Engineering, CONVEX Computer Corp., Richardson, Tx., USA
X-Disclaimer: This message was written by a user at CONVEX Computer
              Corp. The opinions expressed are those of the user and
              not necessarily those of CONVEX.

In article <1992Dec31.192849.4576@ll.mit.edu> mikem@ll.mit.edu writes:
>In article <1htrk2INNrql@mirror.digex.com> kfl@access.digex.com (Keith F. Lynch) writes:
>>What property did 1987 have, that no year after it will have until 2013?
>>
>>What property did 1961 have, that no year after it will have until 6009?
>>
>>Keith Lynch, kfl@access.digex.com
>
>Well, 1961 can be turned upside-down and it's still 1961.  The same can be
>said for 6009.  But, if you assume a font equivalent to a 7-segment digital
>display, the same can also be said for 2002.  I guess it just depends on
>your choice of fonts.  Many computer fonts display the digit '1' so that
>it is quite obviously not the same upside-down as right-side-up.
>
>
>for 1987, it seems you're looking for a year in which it's possible to express
>the date and time using the digits 0-9 exactly once - this first happened in 1987
>on February 3 at 4:56 AM.  2/3/1987 04:56
>
>or, if you don't like leading zeroes, 3/20/1987 4:56.
>
>This feat is not possible for subsequent years, since there is always a repeated
>digit in the year, until you get past 2012.
>
>However, even in 2013, with no repeated digits, we have a problem.  You have to
>somehow use the digits 4-9 in a time and date.  Try it.  You'll quickly see that
>you can't compose a date and time which use all six digits between 4 and 9.
>
>So it seems this isn't *really* a valid answer to the 1987/2013 puzzle.  Anyone
>else care to offer a "better" solution?

Perhaps it's just the simple:  1987 and 2013 have no repeated digits,
and all years in between do.

Joel Williamson
-- 

Newsgroups: rec.puzzles