HDTenEightyP/NetNews huggingface dataset:
From: DICKG@VM.TEMPLE.EDU (Dick Grant)
Newsgroups: rec.puzzles
Subject: Re: Cute problem(generalized) [spoiler]
Date: 20 Dec 92 03:33:06 GMT
Organization: Temple University - Computer Services

In article <1ghg9cINNp6k@aludra.usc.edu>, rmurphy@aludra.usc.edu (Bob Murphy)
says:
>>       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.
>QUESTION:  For which bases is 101 prime?  This can be rephrased
>           as: For which values of b is b^2 + 1 a prime?
>
>I don't know the answer to this and I believe that it
>may be a difficult problem.

   As noted this is indeed a very difficult problem. It is interesting to
rewrite the original problem in base 100 (the square of whatever base it
was in originally). Then it would be asking what numbers of the form
111111... are primes. These numbers can be called "repunits". As was shown
if the base is a square integer only 11 has a chance at being prime. For
most bases the answer is unknown. In particular for base two these are
exactly the Mersenne primes. In base 10 many are known (there is a book
called "Repunits" by Samuel Yates, 1982 that discusses some of these
issues). What bases besides perfect squares can we characterize?
Newsgroups: rec.puzzles
From: Dave Dodson <dodson@convex.COM>
Subject: Re: Aces and Queens (P.S. to: The paradox
Originator: dodson@bach.convex.com
Date: Fri, 18 Dec 1992 16:48:21 GMT
Reply-To: dodson@convex.COM (Dave Dodson)
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 <BzFoFt.E7A@unccsun.uncc.edu> sbardhan@uncc.edu writes:
>X+ X+ X+ X+ X+ X .... X times = X^2
>
>differntiating both sides w.r.t. X we obtain
>
>1+1+1+1+.......X times = 2X
>or, X = 2X
>as X may not be zero,
>1 = 2.

Doing the left side differentiation by the limit definition:

f(X) = X + X + X + X + X + X .... X times


f(X+h) = X+h + X+h + X+h + X+h + X+h + X+h .... X+h times

Hmmm.  I know how to add X terms when X is an integer, but if X+h
is not an integer, how do you add X+h terms?

----------------------------------------------------------------------

Dave Dodson		                             dodson@convex.COM
Convex Computer Corporation      Richardson, Texas      (214) 497-4234
Newsgroups: rec.puzzles
From: joneja@pluto.ecn.purdue.edu (Ajay Joneja)
Subject: Mr. Singer ...(HINT).. (SPOILER)
Keywords: sewing machine, stitches
Organization: Purdue University Engineering Computer Network
Distribution: usa
Date: Thu, 17 Dec 1992 08:36:55 GMT




 ahhh.. kinematics puzzles.. thank you, thank you..
 I would love to see more of these puzzles. I guess part of the
 trouble arises from poor media (ascii) for necessary figures.
 For those still working on this one, a hint follows



 Didja notice the arm sticking out of the side of the sewing machine, with
 a hole at its free end ? The upper-thread runs through that hole before
 passing through the hole in the needle. Each stitch sees this arm go down
 as the needle penetrates the cloth, to provide a _large_ slack to the
 thread (large enough to account for the downward displacement of the
 needle and then some.. to provide for Mr. Singer's inspiration*.)
 Of course, as the needle moves up, so does the arm, taking back the slack
 it gave.


 * I think it was Mr Elias Howe, not Singer. Both were contemporary U.S.
   inventors, though.

 To complete the solution...



 The loose end of the lower-thread shown in the puzzle goes directly into
 the spool. One arm of the U-shape formed by the upper-thread-thru-needle
 is pulled (using the extra slack mentioned above) and looped around the
 spool  (yeah, it's a "floating" spool) to create the  "B" stitch.


 That's the best verbal description I could contrive.. for a problem which
 was very well expressed, IMHO.
--
-------------------- esse quam videri --------------------

Ajay Joneja, School of IE and NSF ERC,
Purdue University, W. Lafayette, Indiana 47907
Newsgroups: rec.puzzles,sci.math
From: pgm@jolt.att.com (Patrick G. Maggiulli)
Subject: 3D motion solution of Rubic's Cube?
Followup-To: poster
Originator: news@cbnewsg.cb.att.com
Keywords: Rubic's Cube, Graphics
Organization: AT&T
Date: Fri, 18 Dec 1992 16:01:51 GMT

Greetings,

Some time ago I read that there is available
software that will present Rubic's Cube in
3D space, scrambled, and on user command
graphically solve the puzzle showing the
rotations of the planes/slices of the puzzle until
a solution is presented.

My questions are:

1) does software exist to do this?
2) is there an ftp site where I can get is from?
3) does it run on a SUN SPARC operating under SUNOS 4.1?

Can someone share with me, and others, your comments,
advice and/or experiences about this topic?


	ADVthanksANCE,
	Patrick

                                       +---------+---------+
                                      /         /         /|
                                     /         /         / |
    Patrick G. Maggiulli            /         /         /  |
    AT&T Bell Laboratories         +---------+---------+   |------+
    200 Laurel Avenue              |         |         |   |     /|
    Middletown, NJ 07748-4801     /|         |         |   +    / |
    USA                          / |    +---------+    |  /    /  |
                                +--|   /         /|    | /----+   |
                               /   |  /         / |    |/    /|   |
                              /    +-/         /  +----+    / |   +
    att!probe!pgm            /      +---------+  /         /  |  /
    pgm@jolt.att.com.us     +-------|         | +---------+   | /
    ph:  + 1 908-957-6418   |       |         | |         |   |/
    fax: + 1 908-957-7227   |       |         | |         |   +
                            |       |         | |         |  /
                            |       |         |/|         | /
                            |       +---------+ |         |/
                            +-------|         | +---------+
                                   ||         |   +    |  /
                                   ||         |  /     | /
                                   ||         | /      |/
                                   +|         |/-------+
                                    +---------+
Newsgroups: rec.puzzles
From: amorgan@csd-d-4.Stanford.EDU (Crunchy Frog)
Subject:  Big hairy numbers
Originator: amorgan@csd-d-4.Stanford.EDU
Organization: Computer Science Department, Stanford University
Date: Thu, 17 Dec 1992 21:18:24 GMT

All this discussion has interested me.  Is there any one source that
talks about all the really gonzo huge numbers and compares them?  Sounds
like it could be a really fun recreational math book if it exists.

C Frog
Organization: The American University - University Computing Center
Date: Mon, 14 Dec 1992 12:02:50 EST
From: <HERSCH@auvm.american.edu>
Newsgroups: rec.puzzles
Subject: Ask Marilyn

Does everyone agree that Marilyn vos Savant's answer to the pancake
probability problem in yesterday's Parade magazine was wrong? The
answer is really 50-50. Either you've picked the pancake with 2 brown
sides, in which case the other side is brown, or you've picked the
pancake with one brown side, in which case the other side is not
brown. 50-50.

H.

Herschel Browne
"The" American University
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Big Numbers
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Thu, 17 Dec 1992 01:38:49 GMT

In article <1gnli9INN12c@cat.cis.Brown.EDU> ST102315@brownvm.brown.edu (Jake) writes:
>
>What is the smallest number that cannot be specified by a description
>containing less than one hundred fifty letters?
>
>Answers are much appreciated.

You're trying to pull a fast one on us, aren't you?  "The smallest
number that cannot be specified by a description containing less than
one hundred fifty letters" contains less than one hundred fifty
letters.  In English, it contains 92 letters.

I'm reminded of Martin Gardner's discussion of "uninteresting" numbers.

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


Newsgroups: rec.puzzles
From: agarwal@jans.Stanford.EDU (Amit N. Agarwal)
Subject: Re: Interesting Calculus Puzzle
Organization: Stanford University
Distribution: su
Date: Fri, 18 Dec 92 20:32:06 GMT

> X+ X+ X+ X+ X+ X .... X times = X^2
> 
> differntiating both sides w.r.t. X we obtain
>
> 1+1+1+1+.......X times = 2X
> or, X = 2X
> as X may not be zero,
> 1 = 2.


We have X + X + ... + X ( X times ) = X^2

On the left hand side, we have X coming in "two" places, so we should
differentiate in the following manner,


1 + 1 + ... + 1 ( X times )  +   X + X + ... X ( 1 time ) = 2*X

or 

	X 		     + 		      X 	  = 2*X

or 
	2*X = 2*X
	 2 = 2 

No Contradiction. 
Newsgroups: rec.puzzles
From: rodney@riemann.math.mun.ca (Rod Hutchings)
Subject: Re: Christmas songs
Organization: CS Dept., Memorial University of Newfoundland
Date: Fri, 11 Dec 1992 21:26:58 GMT

In article <101567@netnews.upenn.edu> greg@hx2.med.upenn.edu (Gregory Connelly) writes:
>How about these?
>
>Small aromatic tree fruits being energized on accessible briquets
>
	Chesnuts Roasting on an Open Fire

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

	Frosty the Snowman.

Regards, and Yultide Greetings
rodney@riemann.math.mun.ca
From: mouse@thunder.mcrcim.mcgill.edu (der Mouse)
Newsgroups: rec.puzzles
Subject: Re: Mr. Singer has a brainstorm [SPOILER]
Date: 21 Dec 92 00:40:35 GMT
Organization: McGill Research Centre for Intelligent Machines

In article <1gohu6INNksh@menudo.uh.edu>, klas@harry.uh.edu () writes:

> Standard sewing machines [...] make a seam consisiting of two threads
> according to the diagram below.
> [diagram deleted - threads are effectively twisted, with the cloth
> layer(s) between]

> 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.

More precisely, it requires that one thread loop around an end of the
other, which is in fact exactly what happens: a loop of the needle
thread is pulled down and passed around the bobbin.  (The mechanism for
doing this while still maintaining tension on the bobbin thread,
without losing control of the bobbin, is (part of) the reason sewing
machines are as complex as they are.)

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

This *apparent* violation... :-)

> [O]bserving a machine while it's working, isn't allowed [...], and I
> don't think it helps either because the crucial parts aren't visible
> while the machine is running.

Well, it might help if you happen to notice the huge length of needle
thread that disappears beneath the fabric during each cycle and is
pulled back up again later.

					der Mouse

			    mouse@larry.mcrcim.mcgill.edu
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Bit Counting
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Mon, 14 Dec 1992 14:04:59 GMT

In article <8325@charon.cwi.nl> dik@cwi.nl (Dik T. Winter) writes:
>[...]  So if
>that is your definition of arithmetic operations the problem can be
>solved in O(1) arithmetic operations.  Otherwise you had better inform
>us what your definition of arithmetic operations is.  (I would say:
>addition, subtraction, multiplication, division, exponentiation,
>root finding, etc., etc., but not clearing specific bits, that is a
>logical operation.)

In that case I can count the bits in 0 arithmetic operations.  I just
have to simulate an increment operation using only logical operations
(e.g. shift bits, bitwise AND, bitwise OR) and then I can shift all
the input bits through one at a time, incrementing a counter.
A stupid but effective method when "non-arithmetic" operations are
considered to take no time.

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


Xref: sparky rec.puzzles:7967 sci.math:17018
From: mnewman@comlab.ox.ac.uk (Matthew Newman)
Newsgroups: rec.puzzles,sci.math
Subject: Circular graph problem
Date: 16 Dec 92 11:35:00 GMT
Organization: Oxford University Computing Laboratory, UK
Originator: mnewman@client24.comlab

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)

and for N=8 I can get K=53 with 1-4-22-7-3-6-2-12- (where notationally
the above graphs are 1-2-5-3- ; 1-2-6-4- ; 1-3-2-7- i.e. read round
clockwise) which again is maximal.

The question is, what is the largest K I can get for a given N ?
From: naras-r@acsu.buffalo.edu (Ramprasad Narasimhan)
Newsgroups: rec.puzzles
Subject: ELEMENTAL STATES
Date: 20 Dec 92 16:41:25 GMT
Organization: UB


This should be a simple one.

First - GUESS the number of STATES (as in USA) whose 2-letter abbreviations 
are also the 2-letter abbrevs of ELEMENTS.

Next - check it out.

Ramprasad T N.
From: shamrake@ccwf.cc.utexas.edu (Danil Suits)
Newsgroups: rec.puzzles
Subject: Re: Christmas songs
Date: 15 Dec 92 08:15:21 GMT
Organization: Zansellquasure Flamarion Bridge Club

You wanna know who said what? look it up.

::How about these?

::Small aromatic tree fruits being energized on accessible briquets

--Chestnuts Roasting on an Open Fire
  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>This is actually the first line of "The Christmas Song".

A great accomplishment in marketing.  If I remember the 
tale correctly, this little ditty was released in July,
giving it plenty of air time before the shopping season
even hit (Christmas shopping started much later back
:-)

Tom Lehrer tried something similar with his tune. His
royalties, alas, were far less.

Danil
``It doesn't matter how sincere it is, nor how heartfelt
the spirit. Sentiment will not endear it, what's important
is the price....''



-- 
-- The Shamrake --    You just can't top the
@ccwf.cc.utexas.edu      Copper Top (tm)
From: klausf@rama.informatik.rwth-aachen.de (Klaus Frank)
Newsgroups: rec.puzzles
Subject: Re: Typing teaser
Date: 17 Dec 92 22:24:55 GMT
Organization: Rechnerbetrieb Informatik - RWTH Aachen
Distribution: world
NNTP-Posting-Host: rama.informatik.rwth-aachen.de

J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:

>What's the longest possible word where you can have such typos and end
>up with a different word?  What if you only make one typo per word?

>Remember, I'm not talking about switching just any letter.  It has to
>be one of the key adjacent to the letter you're trying to type. (eg.
>for 'g' the four 'typos' would be f, t, h, and b.)

It would be interesting if the ergonomic keyboard layouts (e.g. the
dvorak keyboard, which substitutes asdfghjkl; by aoeuidhtns) increase
this maximum length. The original qwerty keyboard layout was designed
to prevent the typewriter types from getting stucked by separating
frequently used keys with seldomly used ones. Perhaps this belongs
to comp.human-factors.

--
Dvorak keyboard:
see soda.berkeley.edu (128.32.149.19), pub/typing-injury/xdvorak.c
or look in the GNU Emacs LCD (cmp. emacs/etc/FAQ, 18) for dvorak
From: Tony Lezard <tony@mantis.co.uk>
Newsgroups: rec.puzzles
Subject: Re: 8-queen problems (revisited?)
Date: Thu, 17 Dec 92 11:59:20 GMT
Distribution: world
Organization: Mantis Consultants, Cambridge. UK.

chris@questrel.com (Chris Cole) writes:

> >Has anyone heard of the 8-queens problem?
> >What is the algorithm?
> 
> This question is in the rec.puzzles Archive.
> [...]
> The Archive has been posted to news.answers.

When? I never saw it.

> [...]
> From the Archive, the answer is:
> ********
> games/chess/queens.s
> ********
> 92.  The following program uses a backtracking algorithm to count positions:
> 
> #include <stdio.h>
> 
> static int count = 0;
> 
> void try(int row, int left, int right) {
>    int poss, place;
>    if (row == 0xFF) ++count;
>    else {
>       poss = ~(row|left|right) & 0xFF;
>       while (poss != 0) {
>          place = poss & -poss;
>          try(row|place, (left|place)<<1, (right|place)>>1);
>          poss &= ~place;
>          }
>       }
>    }
> 
> void main() {   
>    try(0,0,0);
>    printf("There are %d solutions.\n", count);
>    }
> --
> Tony Lezard IS tony@mantis.co.uk OR tony%mantis.co.uk@uknet.ac.uk
> OR EVEN arl10@phx.cam.ac.uk if all else fails.

Cor! Fame at last! Enshrined in the rec.puzzles FAQ list! But can anyone
elegantly solve the puzzle I posed when I posted this:

Generalize this program to n queens on an n x n chessboard.

--
Tony Lezard IS tony@mantis.co.uk OR tony%mantis.co.uk@uknet.ac.uk OR things
like tony%uk.co.mantis@uk.ac.nsfnet-relay OR (last resort) arl10@phx.cam.ac.uk
      "The two most common things on Earth are hydrogen and stupidity."
              -- Geoff Miller (geoffm@purplehaze.Corp.Sun.COM)
Date: Thursday, 17 Dec 1992 06:50:33 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Serial representation of arbitrary data
 <92350.111945RVESTERM@vma.cc.nd.edu>

i doubt anyone cares, and probably no one knows what i'm talking about,
and it's quibbling anyway, but just in case...

in my method of encoding arbitrary binary strings, change "for no
terminal string E of B is any initial string D of C such that E
concatenated with D equal to C" to "if E is a terminal string of B
and D is an initial string of C such that E concatenated with D
equals C, then E is the null string and D is C."

bob vesterman.
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Newsgroups: rec.puzzles
Subject: Typing teaser
Date: 15 Dec 92 16:42:47 GMT
Organization: Dartmouth College, Hanover, NH
X-Posted-From: InterNews1.0b1@newshost.dartmouth.edu

A lot of times when I type I hit the wrong key.  I'll either hit one of
the neighboring keys on the same row, or I'll hit one of the
neighboring keys the same finger would hit.  Most of the time, I end up
with nonsense words that the spell check discovers, but I began
wondering:

What's the longest possible word where you can have such typos and end
up with a different word?  What if you only make one typo per word?

Remember, I'm not talking about switching just any letter.  It has to
be one of the key adjacent to the letter you're trying to type. (eg.
for 'g' the four 'typos' would be f, t, h, and b.)

So far the longest word I've gotten is 'fathers' for 'gathers'.  But
I'm sure the net can do much better.

--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have known it would be bad vodka when all the label said was
'Russian Vodka'."
Date: Wednesday, 16 Dec 1992 13:26:01 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Big Numbers
 <92351.123112RVESTERM@vma.cc.nd.edu>

hmm... as i proved in my last post, there is no smallest number which
cannot be described by 150 characters or less.

however, i just realized that only finitely many things can be described
by 150 characters or less.

paradox? i'd like to name this "vesterman's paradox", thank you.

bob vesterman.
From: katz@guvax.acc.georgetown.edu
Newsgroups: rec.puzzles
Subject: Re: Ask Marilyn
Date: 14 Dec 92 15:54:34 -0500
Distribution: world
Organization: Georgetown University

In article <1992Dec14.190942.15297@CSD-NewsHost.Stanford.EDU>, snewman@Xenon.Stanford.EDU (Steven Newman) writes:
>> Does everyone agree that Marilyn vos Savant's answer to the pancake
>> probability problem in yesterday's Parade magazine was wrong? The
>> answer is really 50-50. Either you've picked the pancake with 2 brown
>> sides, in which case the other side is brown, or you've picked the
>> pancake with one brown side, in which case the other side is not
>> brown. 50-50.
> 
> I don't read Parade, but this sounds like a standard puzzle: you have
> two cards, one red on both sides, one red on one side and white on the
> other.  You put both cards into a bag and then pull one out at random,
> looking only at one side of the card.  The side you see is red.  What
> is the probability that the other side is red?
> 
> The answer is 2/3, not 1/2.  Many people find this *very* hard to
> believe, but it is true.  The argument above fails because the odds
> of having picked the red-red card are GREATER THAN 50%.  This is
> because we have tossed out some of the cases where you pick the red-
> white card, i.e. those cases where you happened to pull it out with
> the white side on top.
> 
This problem, by the way, is isomorphic to the infamous Monty Hall 
problem, which is why it may seem unintuitive.  The odds that the 
bottom side is brown would be 50-50, if you couldn't see the top --- 
just as the odds that the prize was behind the second door would be 
50-50, if you didnt know that Monty opened the third door in 
preference to the second.  But - once you know the top side is brown 
[once you know Monty deliberately opened door 3 instead of door 2]
the odds rise to 2/3.
Newsgroups: rec.puzzles
From: rolls@cis.umassd.edu (Frank Pinto)
Subject: 8-queen problems (revisited?)
Organization: University of Massachusetts Dartmouth
Date: Mon, 14 Dec 1992 23:54:15 GMT

Hello all

If this is old, forgive-me.

Has anyone heard of the 8-queens problem?
What is the algorithm?

Thanks in advance

Frank Pinto

Newsgroups: rec.puzzles
From: jjchew@math.toronto.edu (John J. Chew III)
Subject: Re: Bit Counting Again
Organization: The Poslfit Committee
Date: Sat, 12 Dec 92 02:50:17 GMT

In article <1992Dec11.112641.894@thunder.mcrcim.mcgill.edu> 
  mouse@thunder.mcrcim.mcgill.edu (der Mouse) writes:
> Only with O(n) hardware, and even then, it takes O(n) time.  You need
> O(n^2) hardware, I think, to add two n-bit numbers in O(1) time.

If you consider the practical nature of the problem for sufficiently
large n, signal propagation delay will keep you from bettering O(n)
time no matter what your addition algorithm is.

John
-- 
john j. chew, iii / department of mathematics / university of toronto
poslfit@gpu.utcs.utoronto.ca / poslfit@utorgpu.bitnet / jjchew@math.toronto.edu
"My Canada includes... a Lebesgue non-measurable subset."