HDTenEightyP/NetNews huggingface dataset:
Date: Monday, 14 Dec 1992 13:33:33 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Ask Marilyn

i didn't see marilyn's article, nor did you post the question, but i
can guess what the question is, and tell you that you are wrong.

here's my guess as to the question: you've got two pancakes.  one has
both sides brown, the other has one side brown.  you pick a pancake at
random.  the side that you see is brown.  what's the chance that the
other side is brown?

the answer is 2/3.  there are three brown sides that you could be seeing.
two of them have brown on the other side.

where you went wrong is here: you assume that, given the fact that you
see a brown side, the chance that you picked the all-brown pancake
equals the chance that you picked the not-quite-all-brown pancake.
this is a fallacy.  think back to your high school (or college, whatever)
probability class: the chance of A, given B, is the chance of A and B
divided by the chance of B.

here we have that A is "the pancake i picked is brown on both sides," and
B is "the side that i see is brown."

"A and B" translates to "the pancake i picked is brown on both sides, and
the side that i see is brown," which is 1/2.

the chance that the side you see is brown, after picking a pancake and a
side at random, is 3/4: there are three brown sides, and four sides.

so the chance that the pancake you picked is brown on both sides, given
the fact that you see a brown side, is (1/2)/(3/4), which is 2/3.

this is a very common problem given in puzzle collections or elementary
probability texts, to try to show the reader that intuition is not always
right (intuition, here, being, "there are only two pancakes.  either you
picked the one with both sides brown, or you didn't.  50/50.").

bob vesterman.
From: mouse@thunder.mcrcim.mcgill.edu (der Mouse)
Newsgroups: rec.puzzles
Subject: Re: The Worm and the Blanket Problem
Date: 21 Dec 92 01:05:32 GMT
Organization: McGill Research Centre for Intelligent Machines

> Distribution: rec

Please don't misuse newsgroup hierarchy names as distributions.

In article <92352.064038RVESTERM@vma.cc.nd.edu>, <RVESTERM@vma.cc.nd.edu> writes:

> i really don't understand this problem.  why can't the worm curl
> away?  if you want the answer to be finite, you need restrictions on
> the movements of the worm.  what are they?

None.  But the blanket gets to move to follow it.  (Blanket motion is
rigid, purely translation and rotation.)

Blanket motion may or may not be allowed to be discontinuous; all the
answers I've seen so far are for the discontinuity-allowed case.

To be precise, for each position of the worm, there must be a
rotation/translation of the blanket that covers it.  Nobody has talked
about the possibility of asymmetrical blankets with reflection allowed
yet, at least not that I've seen.

					der Mouse

			    mouse@larry.mcrcim.mcgill.edu
Newsgroups: rec.puzzles
From: jayc@hubcap.clemson.edu (Jay Crawford)
Subject: Crossword Puzzle Generation
Keywords: crossword
Organization: Clemson University
Date: Thu, 17 Dec 1992 15:14:05 GMT

Where can I find an algorithm to use in generating crossword puzzles?  Any 
software available to do this?                       
  

-- 
Jay Crawford          
jayc@hubcap.clemson.edu
From: jerryd@hpgrla.gr.hp.com (Jerry Donovan)
Date: Fri, 11 Dec 1992 22:50:55 GMT
Subject: Re: Christmas riddle
Organization: Hewlett-Packard, Greeley, CO
Newsgroups: rec.puzzles

>What's the perfect Christmas gift for the man who has everything ???
>What's the perfect Christmas gift for the woman who has everything ???

Some common sense.
From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Subject: Re: Integer solutions of 2^m + 3^n = k^2
Organization: University of Michigan EECS Dept., Ann Arbor
Distribution: usa
Date: Fri, 18 Dec 1992 01:53:06 GMT

In article <1992Dec18.014545.17332@zip.eecs.umich.edu> kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:
>In article <1992Dec17.233055.836@news.eng.convex.com> dodson@convex.COM (Dave Dodson) writes:
>>Find all solutions of the equation 2^m + 3^n = k^2 with m, n,
>>and k in the integers, and show that you've found them all.
>>
>>----------------------------------------------------------------------
>>
>>Dave Dodson		                             dodson@convex.COM
>>Convex Computer Corporation      Richardson, Texas      (214) 497-4234
>
>Case 1 :  m > 0.
>==================
>(1) 1st observation : k must be odd.  (Easy!)
>(2) 2nd observation :  n must be even.  (Express 3 as 1+2, perform
>the binomial expansion and cancel out terms on both sides.)
>(3) Then 2^m = (k + 3^(n/2))(k - 3^(n/2))
>(4) From (3) it follows quite easily that
>p and q are the largest positive integers such that :
>
>k mod (2^p) = 1     ... (i)  
>k mod (3^q) = 2     ... (ii)
>
>This is seen as : k = 3^(n/2) + 2;  i.e. q = n/2  and  k = 2^(m-2) + 1; hence, 
>p = (m-2). As an aside, this proves that m >= 2.
>
>Based on the above observations, we notice that the problem boils down
>to finding a power of 2 which is exactly 1 greater than
>a power of 3.
>
>Obviously,  p=q=1 does the job; hence set k to 3, which produces
>m = 3 and n = 0.
>
>Likewise, any (and only) pairs of positive integers that differ by 1
>and happen to be of the form (power of 2, power of 3) produce a valid
>m,n,k tuple.
>
Case 2 : m = 0.

This case is very easy to develop. (k+1) and (k-1) must be 2 different
powers of 3 -- this admits only one possibility, namely k=2.
Therefore, m=0, n=1 and k=2.

Kanad
Newsgroups: rec.puzzles
From: bcash@crchh410.BNR.CA (Brian Cash)
Subject: Re: Serial representation of arbitrary data
Reply-To: bcash@crchh410.BNR.CA (Brian Cash)
Organization: Bell-Northern Research Ltd.
Date: Tue, 15 Dec 1992 04:02:06 GMT


Oops!  I messed up...


In article <1992Dec15.033040.9985@bnr.ca>, bcash@crchh410.BNR.CA (Brian
Cash) writes:
|>From: warwick@cs.uq.oz.au (Warwick Allison)
|>The problem is this:
|>
|>	Represent a sequence of arbitrary length binary values.
|>
|>Do it in as short a number of bits as possible.
|>
|>For example, one solution is to represent
|>
|>	10010010
|>	1001
|>        1010010
|>	11
|>	110111
|>==================================================
|>How about 4 bit binary coded length with an extra "and" bit.
|>
|>Like this:
|>
|>1000 0 10010010 0100 0 1001 0111 0 1010010 0010 0 11 0110 0 110111
|>  L  +    V       L  +  V     L   +   V      L  +  V   L  +   V
|>
|>the "and" bit (the lone zeros above) would concat. the 4-bit segments
|>for lengths longer than 16 bits (code 0000 to equal 16 in a single 
|>4-bit length but keep it as zero for longer lengths).

This is to make the ave. length smaller.  This doesn't need the "and"
bit, either. (a long string of 4-bit bytes wouldn't start 0000 1,
since it is not fixed length.)  There is no "0001 1 0000 0" length.

|>
|>ex.  10110101011010010101101110010101
|>would be 
|>
|>0001 1 0000 0 10110101011010010101101110010101
|>  L  +   L  +       V
|>

0010 1 0000 0 ...

is right

|>
|>and 1010001011010101 
|>would be
|>
|>0000 0 1010001011010101 
|>  L  +      V
|>
_This_ would be

0000 1010001011010101
  L        V

|>
|>How's that?
|>
|>Brian /-|-\ (I didn't recheck bit lengths or binary values, so no flames
|>on those please!)
                                             
Is that better?
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: Christmas songs
Date: 14 Dec 1992 02:20:47 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 <101567@netnews.upenn.edu>, greg@hx2.med.upenn.edu (Gregory Connelly) writes:
>How about these?

Spoilers

>Small aromatic tree fruits being energized on accessible briquets

Chestnuts Roasting on an Open Fire.

>A Wendy's dessert is a masculine pile of frozen precipitation

Frosty the Snowman?
--------------------------------------------------------------------------------
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.
From: klas@harry.uh.edu ()
Newsgroups: rec.puzzles
Subject: Mr. Singer has a brainstorm
Date: 17 Dec 1992 00:34:14 GMT
Organization: University of Houston
Distribution: world
Reply-To: klas@harry.uh.edu ()
NNTP-Posting-Host: harry.imb.uh.edu

Here is a puzzle from the real world:

Standard sewing machines, i.e. the type you find in homes, make a seam
consisiting of two threads according to the diagram below.


                                                 to needle
                                               /
                     A              B         /
                                             /
            ...____      _________      ____/
fabric layer------ |    | ------- |    | ---------------------
                   \ __ /         \ __ /
                    \__\           \__\
                   /    \         /    \
fabric layer------ |    | ------- |    | ---------------------
            ...____|    |_________|    |____
                                            \
                                             \
                                              \ to lower spool


The top thread runs along the top fabric layer and dips into the
fabric with loops which are locked in place by corresponding loops
coming from the bottom thread. The two threads never switch sides.

Now imagine that the seam has been finished to A. The left end of both
the top and the bottom threads are thus attached to the fabric at A.
The right end of the top thread goes through the needle and then on to
the top spool. The right end of the botton thread goes down a little
hole and to the lower spool (this is the same little hole that the
needle goes through). Now you let the machine go one more cycle and
examine what has happened. The two threads have now formed a knot with
each other at B, but both threads were attached at both ends so this
shouldn't be possible. 

Topologically this is the same as if two persons, each holding a piece
of string between their hands, could make the strings cross without at
least one of them having to let go of one end of his or her piece of
string. 

I realized this miraculous appearance of the new knot a couple of
years ago as I was using my sewing machine. It stunned me for several
minutes. (I did eventually figure it out.)

I now submit the same question to the readers: How does the machine
pull off this violation of basic topology?

Cheating, i.e. observing a machine while it's working, isn't allowed
as a method of solving this puzzle of course, and I don't think it
helps either because the crucial parts aren't visible while the
machine is running. Only under exceedingly intense scrutiny of the
pertinent parts will the sewing machine reveal its secrets, so I think
careful and balanced application of imagination and logic is just as
effective.


Klas Hedenstierna
klas@harry.imb.uh.edu
From: mnewman@comlab.ox.ac.uk (Matthew Newman)
Newsgroups: rec.puzzles
Subject: DISCRETE PROBLEM / CIRCULAR GRAPH
Date: 19 Dec 92 11:32:04 GMT
Organization: Oxford University Computing Laboratory, UK
Originator: mnewman@client24.comlab


Choosing 1,2,3,7,8,9,13,14,15,...,97,98,99 gives 51 counters from
the hat.

In the circular graph problem, one of the graphs had a typo,
it should be  (1)--(3)
               |    | 
              (7)--(2)

Where 1=(1); 2=(2) ; 3=(3) ; 4=(1)--(3) ; 5=(3)-(2) ; 6=(1)--(3)-(2);
      7=(7); 8=(1)-(7); 9 = (7)--(2); 10=(1)-(7)--(2)
     11=(7)-(1)--(3); 12=(7)--(2)-(3) ; 13=(1)--(3)-(2)--(7)

And yes, the 8 case is 57 (an upper bound is n(n-1)+1 - i.e. the
number of connected subgraphs - however this bound cannot be
reached for n=7 - and I want to know why ! )

Hope all's clear now (unless I've made another mishtake).
Newsgroups: rec.puzzles
From: wft@math.canterbury.ac.nz (Bill Taylor)
Subject: Re: Answers to Hard Analogies
Organization: Department of Mathematics, University of Canterbury
Date: Wed, 16 Dec 1992 04:01:25 GMT

>2. Mountain : Plain :: Acrocephalic : ?

>2. Homalocephalic

>Acrocephalic is having a pointed head; homalocephalic is having a flat head.

What about PLATYCEPHALIC ?
Newsgroups: rec.puzzles,sci.math
From: amorgan@csd-d-3.Stanford.EDU (Crunchy Frog)
Subject: Re: Circular graph problem
Originator: amorgan@csd-d-3.Stanford.EDU
Organization: /usr/lib/news/organizatoin
Date: Thu, 17 Dec 1992 02:55:15 GMT

In article <1992Dec17.003342.13658@dartvax.dartmouth.edu> 
  J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:

>> >mnewman@comlab.ox.ac.uk (Matthew Newman) writes:
>> >
>> >> Consider circular graphs with N nodes, such that there is an integer
>> >> at each node. Consider those integers K such that for some graph G
>> >> each integer in [1..k] is the sum of integers at the nodes of some
>> >> connected sub-graph G. For example with N=4 I can get K=11 with the
>> >> following graph :     (1)--(2)
>> >>                        |    | 
>> >>                       (3)--(5) 
>> >> 
>> >> and I can do better than this and get the maximum possible k=13 like
>> >> this :     (1)--(2)    and like this:   (1)--(3)
>> >>             |    |                       |    |
>> >>            (4)--(6)                     (2)--(7)

>I don't understand why these last two count as valid -- in the first,
>there's no combination that totals 4

What about the node 4?  It meets the above requirements.  Your objection
to 7 is handled in the same way.

>--Ted Schuerzinger

C Frog

Newsgroups: rec.puzzles
From: stein@hal.nta.no (Stein Kulseth)
Subject: Re: Monte Hall II
Organization: Norwegian Telecom Research
Date: Fri, 18 Dec 92 15:38:08 GMT

|> 1.  If doors A, B, and C (door A leftmost, door C rightmost) have the
|> probabilities a, b, c with a+b+c=1 of having the good prize behind
|> them, what is your best strategy?

choose door with least probability (say a), and switch (for a winning
probability of b+c >= 2/3).

|> 2.  If when Monte is faced with two empty doors instead of choosing
|> uniformly he chooses the leftmost door with probability p, the
|> rightmost with probability q and p+q=1, what is your best strategy
|> now?

pick any door and switch. The winning probability is 1/(1+p) when the left
door is opened, or 1/(1+q) when the right door is.
This probability may range from a solid 1 down to 1/2 dependant on p,q.
Your average win probability will equal 2/3, though

|> 3.  What if both conditions hold?

choose door with least probability(again, say a), and switch.

And again the actual winning probabilities when switching may vary,
they are c/(c+ap) or b/(b+aq), but the average probability will
equal b+c.

-- 
stein.kulseth@nta.no (Norwegian Telecom Research)
   'When murders are committed by mathematics, they can be solved by
   mathematics. Most of them aren't, and this one wasn't'
   - Nick Charles (Dashiell Hammett's "The Thin Man")
Xref: sparky rec.puzzles:7993 sci.math:17079
Date: Wednesday, 16 Dec 1992 15:53:12 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles,sci.math
Subject: Re: Circular graph problem
 <1992Dec16.172145.21848@dartvax.dartmouth.edu>
 <1992Dec16.195725.11131@zip.eecs.umich.edu>

kanad, you have a rather odd definition of "first" and "second".

regardless, though, i think that the puzzle assumes that one node
counts as a combination of nodes.

bob vesterman.
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: rec.puzzles
Subject: Re: Godel, Escher, Bach cypher
Summary: B'reshit bara Elohim...
Date: 11 Dec 92 23:07:05 GMT
Article-I.D.: husc3.1992Dec11.180706.18413
Organization: Harvard Math Department

In article <1992Dec6.211953.20486@leland.Stanford.EDU>
duizie@leland.Stanford.EDU (Ken Duisenberg) writes:
>In Douglas Hofstadter's epic "Godel, Escher Bach: An eternal Golden Braid",
>at the end of the list of figures (page xviii) is what seems to be a cypher,
>perhaps a cryptogram.

Since no one else answered this: yes, it is a cryptogram of sorts --- in Hebrew.

>Does anyone know what it says?

Yes.  It is explained a few pages back, in the beginning of the list of figures.

--Noam D. Elkies (elkies@zariski.harvard.edu)
  Dept. of Mathematics, Harvard University
From: dsbb@ns1.cc.lehigh.edu (D. SPENCER BEECHER)
Newsgroups: rec.puzzles
Subject: Re: Five Box Puzzle
Date: 11 Dec 92 21:52:26 GMT
Organization: Lehigh University

Dan Morera for Sean O'Rourke) writes:

>       Draw a single connected line that goes through each wall of the
>       figure.  You may not go through a wall twice.  Each wall is defined
>       by a line, with at least two other walls connected to it.
>
>                       THE PUZZLE
>
>          -------------------
>          |     |     |     |
>          -------------------             (rough representation in keyboard
>          |        |        |                 characters)
>          -------------------


Can't be done - draw a map.

The long and short of it is, there are three rooms of the box with
an odd number of walls.  If you start inside one, after drawing
through each wall you will end up outside that room.  Now, looking at
the two remaining rooms (even though they may have some walls transversed
at this point), you may realize that, by the time you transect all walls
of either of the two, you will be on the inside, having yet left an
odd number of walls on the third room unscathed.
If you start outside any of the three, you will find yourself stuck
after having transected all the walls of only one of the three odd-wall
rooms.

It can be done if you 'cheat' by tracing along one of the walls and
considering this to be a transection.
-- 
                   _ _     _ _     _ _
  -Spence         [O.O]   )x.x(   )O.O(
                    O       O      -*-
Newsgroups: rec.puzzles
From: merlyn@digibd.digibd.com (Merlyn LeRoy)
Subject: Re: Monty Hall Puzzle RIP
Organization: DigiBoard, Incorporated, Eden Prairie,MN
Date: Sat, 19 Dec 1992 16:24:09 GMT

<SHERE@SLACVM.SLAC.STANFORD.EDU> writes:
>You will note that out of the twenty-four permutations of this game
>you can win exactly twelve ways, six ways by staying with your
>original choice (ACKMUW)...

Bzzt.  Any analysis must show that staying with your original choice
results in a win 1/3 of the time, not 1/4 of the time.

---
Merlyn LeRoy
Newsgroups: rec.puzzles
From: jkenton@world.std.com (Jeff Kenton)
Subject: Cute problem
Organization: Kenton Systems Corporation, Weston MA
Date: Sun, 13 Dec 1992 16:04:54 GMT

Here's a problem I haven't seen before that I thought folks
might enjoy:

	Given the numbers 101, 10101, 1010101, ... (decimal, not binary)
with 'n' 1's separated by 'n-1' 0's.  101 is prime.  Prove that none of
the rest is prime.


-- 
-------------------------------------------------------------------------
=			Jeff Kenton	(617) 894-4508			=
=			    jkenton@world.std.com			=
-------------------------------------------------------------------------
Date: Wednesday, 16 Dec 1992 12:31:12 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Big Numbers

the question is, "what is the smallest number which cannot be described
using less than 150 letters?"

the answer is, "there is no such number."

proof (by contradiction): assume that there is a smallest number which
cannot be described by 150 letters or less.  then "the smallest number
which cannot be described by one hundred fifty letters or less"
is a description of said number which uses less than 150 letters.

bob vesterman.
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: Bit Counting
Date: 17 Dec 92 01:56:50 GMT
Organization: CWI, Amsterdam

In article <Yf=bQj200Vp=IOqm9s@andrew.cmu.edu> court+@CMU.EDU (Court Demas) writes:
 > Er- Have you ever heard of an ALU which doesn't support ANDs?  I can't
 > see how people don't consider boolean functions to be arithmetic.
 > 
I did an earlier follow-up where I told that I checked the manuals for the
machines I have at home.  (It did not yet arrive here, probably it did
already arrive elsewhere; the vagaries of the news system are interesting.)
But of course my intention was that I had the manuals at home, not the
machines.  (I would face bankruptcy if I had to face the electricity bill
for all those machines.)
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: Bit Counting
Date: 14 Dec 92 22:26:15 GMT
Organization: CWI, Amsterdam

In article <1992Dec14.140459.17126@cs.cornell.edu> karr@cs.cornell.edu (David Karr) writes:
 > A stupid but effective method when "non-arithmetic" operations are
 > considered to take no time.
Where did I say *that*?
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
From: irw@bharh58.bnr.co.uk (Ian Woollard)
Newsgroups: rec.puzzles
Subject: Re: The Worm and the Blanket Problem: minimum = 0.217 = poss max too
Date: 17 Dec 1992 20:38:20 GMT
Organization: BNR Europe Ltd, London Road, Harlow, England.
NNTP-Posting-Host: 47.131.128.36

In article <1go973INNn49@bHARs12c.bnr.co.uk> irw@bharh58.bnr.co.uk (Ian Woollard) writes:
>In fact I cannot find a shape that will not fit in the right angle
>triangle with sides 1, 1/2 and ~0.866, at present. I've just finished
>considering a slightly kinked line lying along the side of length 1,
>but I *think* it *does* fit, very surprisingly. I think this is the
>upper limit... but proving it is a real b*tch.
>

As I worked out within an hour or two of posting, this is of course
complete and utter rubbish, and indeed there is even a two kink refutation
(kindly supplied to me by dmi@peregrine.com) as well as the three kink
refutation that I found. Such is life. (I thought that I had a four kink
refutation for the half circle, but using a calculator for a while
proved me wrong, that would have been interesting...)

-Ian.
From: rapp@pecan.cns.udel.edu (Brian Rapp)
Newsgroups: rec.puzzles
Subject: Re: The Rifleman Problem
Date: 16 Dec 92 16:17:27 GMT
Organization: University of Delaware, Newark

In article <Dec15.235132.63842@yuma.ACNS.ColoState.EDU> boll@CS.ColoState.EDU (dave boll) writes:
>
>  N riflemen are situated randomly on a plane. At the same instant, each
>  shoots and kills his nearest neighbor. How many are left?
>

	Before or after the bullet-riddled plane crashes?
	Sorry for the lame pun, but my initial interpretation
	really involved homicidal maniacs on an airliner.
	Given the cramped seating on most planes, it might
	prove awkward for the gunmen to point a rifle at their
	nearest neighbor.  But then, these are idealized riflemen.

	ObPuzzle:  If you knew that you were going to be on
	one of these lethal flights, where would you sit?

--------------------------------------------------
| Brian Rapp | rapp@freezer.cns.udel.edu | Ferret | "and many times confused"
--------------------------------------------------