HDTenEightyP/NetNews huggingface dataset:
From: comptec91065@camins.camosun.bc.ca
Subject: dividing 100 digit nos.?
Organization: Camosun College
Date: Wed, 26 Aug 1992 01:40:16 GMT

Hello,

	Can you help me locate the easiest method of dividing
	integers of at least 100 digits (possibly up to 4300 digits) ?

	I have a 386-DX at home linked to a VAX at school 
	(in case that matters).

	Thank you ...

Sincerely,

	Allan Curry (in beautiful B.C. / CANADA)
From: jpc@pascal.msri.org (Joe Christy)
Newsgroups: sci.math
Subject: electronic NSF proposals
Followup-To: sci.math
Date: 28 Aug 1992 17:43:18 GMT
Organization: Mathematical Sciences Research Institute, Berkeley
Distribution: usa
NNTP-Posting-Host: 128.3.188.66

Last year there was an experimental program for electronic submittal
of NSF proposals. Unfortunately, when I moved this summer, I lost the
info on it. Does anyone know if it still exists, and if so how one
accesses it?
-- 
Joe Christy   |      Head of Mathematical Computing      | Two wrongs
jpc@msri.org  | Mathematical Sciences Research Institute | don't make a right,
(510)643-6069 |          1000 Centennial Drive           | but three lefts do
**************|           Berkeley, CA 94720             |********************
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Newsgroups: sci.math
Subject: Re: Abel's proof of the insolubility of the quint
Date: 3 Sep 92 20:13:36 GMT
 <MARTIN.92Sep2212731@lyra.cis.umassd.edu> <87834@netnews.upenn.edu> 
 <1992Sep3.122450.15337@husc3.harvard.edu> <87849@netnews.upenn.edu>

In <87849@netnews.upenn.edu> weemba@sagi.wistar.upenn.edu (Matthew P Wiener) writes:

>In article <1992Sep3.122450.15337@husc3.harvard.edu>, elkies@ramanujan (Noam Elkies) writes:
>>>I came across the proof in Rotman: identify the conjugacy classes of
>>>A_5, and notice the impossibility of any of them forming a partition
>>>for a size that non-trivially divides 60.

>>This nice proof is reasonably well-known, but can one get a proof
>>of the simplicity of all A_n (n>4) from such ideas?

>I don't think so, although I don't know of a particular A_n where the
>partition is possible, of course without forming a subgroup.  Rotman
>redoes the partition proof for A_6, and then uses the result for A_6
>to prove the result for A_n, n>6.

I think it would be simpler to show that:  the group generated by any
conjugacy class includes a specific element ( (123)  (12345) ?), and to
show that all conjugates of that generate A_n.
--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
My interaction with our news system is unstable; please mail anything important.
From: roth@3d.enet.dec.com (Jim Roth)
Newsgroups: sci.math
Subject: Re: Polygon point enclosure
Date: 1 Sep 92 00:31:25 GMT
Organization: Digital Equipment Corporation


In article <thompson.714934114@daphne.socsci.umn.edu>, thompson@atlas.socsci.umn.edu (T. Scott Thompson) writes...
>gjm11@cus.cam.ac.uk (G.J. McCaughan) writes:
> 
>>In article <7122@bigbird.hri.com.hri.com>, mps@sparc68.hri.com (Mark Stockley) writes:
> 
>>> I'm looking for a method to determine if a point is enclosed
>>> by a polygon. Any help would be greatly appreciated.
> 
>>Well, here's one approach. Imagine a ray (=half-line) coming off from the point
>>in some direction. Count how many of the sides of the polygon it goes through.
>>(If it goes *along* any of those sides you're in trouble and should try another
>>direction.) If it's an odd number the point is inside the polygon; if it's an
>>even number the point is outside the polygon.

It is much better to choose your ray parallel to the X axis in this approach.

It is then easy to handle all "degenerate" cases such passing exactly
thru a vertex (which you can detect precisely without incurring any roundoff
error), passing coincident to a side (which again is easily and precisely
checked), these cases handled with a simple convention about half open
intervals in the x and y directions.

You'll be pleasantly surprised at how nicely the algorithm comes
out when you code it up this way.  It's foolproof.

- Jim
Newsgroups: sci.math
From: steve@hubcap.clemson.edu ("Steve" Stevenson)
Subject: Re: Proof of God's Existence
Organization: Clemson University
Date: Tue, 1 Sep 1992 12:37:17 GMT

rsilver@world.std.com (Richard Silver) writes:

>In article <17ui6kINNsft@matt.ksu.ksu.edu> bubai@matt.ksu.ksu.edu (P.Chatterjee) writes:
>>I was told by somebody that there is a mathematical proof of God's existence;
>>was wondering if somebody could shed some light on the same. 
>>Thanks. Looking forward to the response.

The concept of proving God's existence goes back to Aquinas (13th Century).
(Eventually, the Reformation would finger Aquinas and the Roman Catholic
Church for "secular humanism" on this account) Aquinas' preoccupation with
this question lead to the advancement in scholastic logic---the logic of
mathematics until the late 19th century. Descartes put forth several proofs,
all of which were refuted. The arguments involved were instrumental in
strengthening the logic of the day. The difficulties in "proving" the
existence of God made it a seminal problem in logic---both philosophical and
mathematical.

>You do not need proof. Just ask those who die in his name. 
>And please find other newsgroups for such lame queries. 

Far from being a lame query, the problem helps to point out the limitations
to deduction. You should do a bit of reading.

-- 
===============================================================================
Steve (really "D. E.") Stevenson           steve@hubcap.clemson.edu
Department of Computer Science,            (803)656-5880.mabell
Clemson University, Clemson, SC 29634-1906
Newsgroups: sci.math
From: takriti@engin.umich.edu (samer Takriti)
Subject: Re: Combinatorial Problem
Date: Thu, 27 Aug 92 16:16:53 EDT
Organization: University of Michigan Engineering, Ann Arbor
Followup-To: I have a Problem, which I think is very easy
Keywords: combinatorics

I will restate your problem first:
Find the integers x(i), i=1, ..., m
such that
x(1) + x(2) + ... + x(m) = K
l(i) <= x(i) <= u(i), i=1, ..., m

I don't think that you can compute the exact number of feasible
solutions. You may obtain some bounds on the number of possible
solutions though.
If you need to exhaust all the solutions you may want to do the 
following:
Create m for loops as follows:
for (x(1)=l(1); x(1)<=K && x(1)<=u(1); x(1)++)
 for (x(2)=l(2); x(2)<=K-x(1) && x(2)<=u(2); x(2)++)
  for (x(3)=l(3); x(3)<=K-x(1)-x(2) && x(3)<=u(3); x(3)++)
   ..........
    for (x(m-1)=l(m-1); x(m-1)<=K-x(1)-x(2)-...-x(m-2) && x(m-1)<=u(m-1); x(m-1)++)
     {
     x(m) = K - x(1) - x(2) - x(3) - ... - x(m-1)
     if x(m) <= u(m)
        printf ("%....", x(1), x(2), ..., x(m))
     }
As you see, the previous code or algorithm gives an upper bound 
on the number of the possible solutions.
If you want to implement this algorithm on a computer, there is 
no need for m-1 loops. You can use recursion.

If you are interested in one solution only, you can use integer
programming to do so.
-Samer
From: kevin@vaxc.cc.monash.edu.au
Newsgroups: sci.math
Subject: Re:Proof of God's Existence
Date: 1 Sep 92 21:53:30 +1000
Organization: Computer Centre, Monash University, Australia

If one reads through philosophy, one will find many attempts at mathematical
proofs of God's existence. All these 'ontological proofs' have very obvious
and silly errors. It is quite clear that mathematics is not capable of making
a statement on the ontological status of God. Sure, maths can make lots of 
comments about the hyperbolic plane, the Zariski Topology, the lattice of r.e
sets and so forth. But God? No way!
Asides from limitations in the expressive power of mathematics, the following
will show any proof of the ontic status of God futile.
1. It is possible to imagine a world, completely agreeing with the current one
on all sensory data, in which God exists. 
2. It is possible to imagine a world, completely agreeing with the current one
on all sensory data, in which God doesn't exists.
Call a world satisfying condition 1 M, and a world satisfying 2 M'.
Then the existence of these epistemically indistinguishable worlds proves that
any ontological statement about God is INDEPENDENT of any epistemically 
knowable axioms on the real world. So no proof either way exists. Bravo.
Notes 1. Some people say world M' is impossible, because the beauty of the
world we live in necessarily entails God's existence. There are many similar
arguments, basically content isomorphic to this. In fact, this argument is
content isomorphic to the trivial or empty argument, so not much need be said.
I assume my response to such a person is implicitly made clear in the previous
comments.
2. It will strike many as odd that even Godel believed he had a proof, using
modal logic. If you can bother spending the time reading it (it is quite
complex), I'm sure you'll spot the error. It's more sublte than most other
ontological errors, but it's there, unavoidable, and unrectifiable. I suppose
even great mathematicians can go loony every now and again.
Well, there you have it. No further comment necessary.
Kevin Davey
Monash University 
Australia.
Newsgroups: sci.math
From: ara@zurich.ai.mit.edu (Allan Adler)
Subject: Re: Godel and the US constitution
Organization: M.I.T. Artificial Intelligence Lab.
	<1992Sep02.134107.33391@watson.ibm.com>
	<ARA.92Sep3053902@camelot.ai.mit.edu>
	<1992Sep3.114520.21990@ms.uky.edu>
Date: Fri, 4 Sep 1992 07:55:46 GMT

Charles Yeomans says that the president has the right to suspend habeas corpus
during martial law and that this right is in the constitution. I don't remember
that being there. Can someone supply a specific reference?

I also mentioned that the Bush Administration tried to convince the Supreme
Court recently that the constitutional ban on the establishment of
religion did not mean there could not be a state religion. Charles seemed
to think that I was referring to individual states, since if one of the states
has its own religion, like its own bird, that would be a state religion, like
the state bird. Actually, what I was saying was that the Bush administration
was trying to convince the Supremem Court that it would be constitutional
for the US government to institute a national religion. When a country has
a state religion, it does not mean the individual states in that country;
it means that the state (i.e. the government) institutes a religion.

I believe Italy has a state religion, for example, and everyone is taught
religion in the public schools. A particular religion.

I believe wholeheartedly in the separation of church and state and I
am dismayed that we have a president who doesn't. Dan Quayle denounces
Ross Perot for suggesting that there could be room for improvement in the
constitution, but actually the Bush Administration does not believe in the
rights guaranteed by the constitution and is hard at work trying to 
weaken them. I can't imagine a greater disaster for this country than
for Bush to be reelected and allowed to appoint a replacement for the
next Supreme Court justice to retire, which will be fairly soon.

Allan Adler
ara@altdorf.ai.mit.edu

Disclaimer: These are my own opinions, not those of MIT, as far as I know.
Newsgroups: sci.math
From: drobot@sjsumcs.sjsu.edu (Vladimir Drobot)
Subject: Re: Proof of God's Existence
Organization: San Jose State University - Math/CS Dept.
Date: Thu, 3 Sep 1992 02:41:31 GMT

If, God willing, God does not exist, then thank God.
If however, God forbid, God does exist, then God save us all.
Newsgroups: sci.math
Subject: mixed-field arithmetic
From: moon@moon.ee.usu.edu (Todd Moon)
Date: 1 Sep 92 11:16:04
Distribution: sci
Organization: Utah State University Dept EE Logan, Utah



Here is a problem in mixed field arithmetic.  In what follows, + will
denote addition over the field of reals, and | will denote addition
over GF(2).  Multiplication is over the reals.

consider a system of equations such as

a_1 c_0               + a_2 c_3               = s_0
a_1 c_1               + a_2 c_4               = s_1
a_1 c_2               + a_2 c_5               = s_2
a_1 (c_0 | c_1)       + a_2 (c_3 | c_5)       = s_3
a_1 (c_1 | c_2)       + a_2 (c_3 | c|4 | c_5) = s_4
a_1 (c_0 | c_1 | c_2) + a_2 (c_3 | c_5)       = s_5
a_1 (c_0 | c_2)       + a_2 (c_4 | c_5)       = s_6

where a_1 and a_2 are real and known, the s_i are real and known, and
the unknowns c_i are in GF(2) (that is, either 0 or 1).  Although the
problem has been written similar to a set of linear equations, it is
easy to verify that this is not a linear set of equations.

1) Find the set of c_i that satisfy this set of equations.  For this
problem, it can be accomplished simply by combinatorial search
techniques.  However, there are larger problems of interest similar to
for which searching is impractical.

2) Now consider the case of finding the "best" solution to the set of
equations where the equality does not hold -- there is unknown "noise"
in the equations:

a_1 c_0               + a_2 c_3               = s_0 + n_0
a_1 c_1               + a_2 c_4               = s_1 + n_1
a_1 c_2               + a_2 c_5               = s_2 + n_2
a_1 (c_0 | c_1)       + a_2 (c_3 | c_5)       = s_3 + n_3
a_1 (c_1 | c_2)       + a_2 (c_3 | c|4 | c_5) = s_4 + n_4
a_1 (c_0 | c_1 | c_2) + a_2 (c_3 | c_5)       = s_5 + n_5
a_1 (c_0 | c_2)       + a_2 (c_4 | c_5)       = s_6 + n_6

As before, a_i and s_i are known, but the noise n_i is not known.
Find the best (perhaps in least squares sense) set of c_i in GF(2).

What would be nice would be to have an algebra for these mixed field
problems, from which exact and pseudo-inverse solutions coulde be
obtained.

Any suggestions, comments, references, etc. would be greatly
appreciated.


Todd Moon
Electrical Engineering Dept., UMC 4120
Utah State University
Logan, UT  84322
801 750 2970
moon@moon.ee.usu.edu
--

Todd Moon
Electrical Engineering Dept., UMC 4120
Utah State University
Logan, UT  84322
801 750 2970
moon@moon.ee.usu.edu
From: shallit@graceland.uwaterloo.ca (Jeffrey Shallit)
Newsgroups: sci.math
Subject: proof of quadratic reciprocity theorem
Date: 31 Aug 92 20:01:21 GMT
Organization: University of Waterloo

Jerry Tunnell once showed me the following neat proof of the
quadratic reciprocity theorem.   It goes as follows:

	Let p and q be distinct odd primes.

	(a) p is a square (mod q)

	(b) left multiplication by p gives an even permutation of Z/(q)

	(c) the discriminant of X^q - 1 is a square in Z/(p).

	(d) ((-1)^(q-1)/2) * q is a square (mod p).

	Then (a) <==> (b) <==> (c) <==> (d).

Anybody know who this is originally due to?  Anybody know where
Tunnell is currently?  I couldn't find his address in the
CML.

Jeff Shallit
University of Waterloo
Newsgroups: sci.math
Subject: Re: Problematic smurfs!
From: kleber@husc10.harvard.edu (Gwydden)
Date: 30 Aug 92 22:44:17 GMT
Keywords: smurfs

Quoth Godfrey Degamo:

>              The wearer of the hat has no means whatsoever for obtaining the
> color of his hat.  It is never the case that all 1000 smurfs will wear
> the same color hat.

Amusingly, this is a paradox.  From this we can determine that there can
never be one smurf with color A and 999 with color B, since then the odd
smurf out would know his hat color.  But then you can never have 2 A's
and 998 B's, since each A would know there can't be only one A, so he
must be the other.  But then... etc.

Rewriting to get "The mayor says they're not all wearing blue hats,
and so would the ones with the red hats please stand up," though...

> find:
>      The exact amount of smurfs that rise.  
>          (Hint: This number will be greater than one.)
>      The color of the hats of the smurfs that rise.

Well, the obvious answer would be the 3 red-hatted smurfs... except that
that assumes all smurfs are perfect logicians.  The last question,
though, implies that it's possible for a non-red-hatted smurf to stand,
so they're *not* all perfect logicians.  So the correct answer is:

Nine hundred ninety-eight smurfs all stand-- all the ones that had
blue hats on.

Why?  Well, the two red-hatted smurfs were lazy and grouchy.  Each one could
see exactly one other red hat at the beginning, so when the other didn't
rise after Mayor Smurf's first request, they each knew their hat was red.
But then lazy fell asleep, and grouncy was so annoyed that he was one of only 
two red hats that he sat and fumed, and so when Mayor Smurf asked a 2nd time, 
neither one stood.  Each of the 998 blue-hat smurfs remaining saw this, 
concluded (incorrectly) that it meant the two red-hats could see a third
one-- obviously, themselves-- and so the third time, all the rest of them
stood up.  Then they realized what had happened, and all laughed and started
singing insipidly cute songs, all 998 at once, and the combined noise
let Gargamel find them, and he stepped on all 1001 of them at once.

--Michael Kleber                 I don't have an overactive imagination...
  kleber@husc.harvard.edu        I have an underactive reality...   --EG
From: amnell@klaava.Helsinki.FI (Marko Amnell)
Newsgroups: sci.math
Subject: Re: Looking for faulty proof that 0=1.
Date: 2 Sep 92 17:22:20 GMT
Organization: University of Helsinki

I looked in Spivak's _Calculus_ and found a similar bogus proof:

Let x=y.  Then
 
       x^2=xy,
   x^2-y^2=xy-y^2,
(x+y)(x-y)=y(x-y),
       x+y=y,
        2y=y,
         2=1.

So, given your proof, I conclude that 2=0 !  ;-)

Marko.
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Problematic smurfs!
Keywords: smurfs
Date: 30 Aug 92 23:50:52 GMT
Organization: U of Cambridge, England

I shall assume that all the smurfs are perfect logicians (else anything could
happen!), and that "the mayor is not apart of the 1000 smurfs" means that the
mayor is not one of the 1000 smurfs, rather than that he is not separate from
them.

1. The first time, a smurf would only stand up if he saw that all the other
   hats were blue (for then his must be red).
   So after the first call, all the smurfs know that no smurf sees only
   blue hats.

2. The second time, a smurf will stand up if (and only if) he sees exactly
   one red-hatted smurf (if he did, he would reason "X would have stood up
   if my hat were blue too").
   So after the second call, all the smurfs know that every smurf can see
   at least 2 smurfs with red hats.

3. The third time, a smurf will stand up if (and only if) he sees exactly 
   two red-hatted smurfs. (Same sort of logic as before.) And indeed, at
   least one smurf does see exactly two red-hatted smurfs.
   So, there must have been exactly 3 smurfs with red hats (if only two,
   they would have stood up at the second call).
   How many smurfs stand? Exactly those three.
   And of course all the smurfs who stand have red hats.
Newsgroups: sci.math
From: hporopudas@tnclus.tele.nokia.fi
Subject: Re: Proof of God's Existence
Organization: Nokia Telecommunications.
Date: Fri, 4 Sep 1992 16:20:46 GMT



In article <ARA.92Sep4030328@camelot.ai.mit.edu>,
ara@zurich.ai.mit.edu (Allan Adler) writes:
> Matt Wiener has apparently missed my point. Without some agreement as
> to what one is supposed to be proving, there is no possibility of
> mathematical proof.
>
> Allan Adler
> ara@altdorf.ai.mit.edu

 It seems to me that what is supposed to be proving is the following
 theorem below the definition:

 Definition:  God is Existence.

 Theorem:     Existence exists.


 I think that Existence can never defined completely and theorem is
 something self-evident.

              Hannu.
Newsgroups: sci.math.symbolic
From: aicklen@utdallas.edu (Greg Aicklen)
Subject: Re: Mma Solve[] oddity (a much shorter version)
Organization: Univ. of Texas at Dallas
Date: Tue, 1 Sep 1992 13:14:40 GMT

In article <PARKES.92Aug26173241@ptah.itp.ethz.ch> parkes@itp.ethz.ch (Andrew Parkes) writes:
>
>David Petry writes:
>> In article <28914@option.GBA.NYU.EDU> jcao@option.GBA.NYU.EDU (Jingbin Cao) writes:
>> >The following log documents some weired Solve[] oddities. They are:
>> >
>> >	1. Solve[] cannot find roots in some cases, at least not within
>> >	   reasonable period of time. But if you divide the equations
>> >	   by something, Solve[] finds some roots!

	... deletions ...

>> 
>> eqs  =  {  a*x*y + b*x + c*y ==0,  d*x*y + e*x + f*y + g == 0}
>> 
>> Solve[ eqs, {x,y}]

	... deletions ...

>> David Petry
>> 
>Here's an even simpler version of the above 
>
>eq  = (a x^2 + b x + c)/(d1 + d2 x) 
>answer = Solve[ eq == 0 , x]
>
>Mma 1.2 gets the answer immediately of course.
>
>But I find that our  Mma 2.0  gets  gets nowhere. 
>I killed the job after 40mins when it got to 30MB.

	... deletions ...

>In case this is version dependent, our system is:-
>$Version =  SPARC 2.0 (July 8, 1991)
>$VersionNumber = 2
>
>Andrew Parkes,              | E-mail: parkes@itp.ethz.ch 
>Theoretische Physik,        | 
>ETH-Honggerberg,            |                           
>CH-8093 Zurich, Switzerland.|                           

With Mma 2.1 on a SPARC, I get quick solutions to both problems given above.
For Andrew Parkes "even simpler version" I get

/tmp_mnt/home/tulia/aicklen % math
Mathematica 2.1 for SPARC
Copyright 1988-92 Wolfram Research, Inc.
 -- X11 windows graphics initialized --

In[1]:= eq  = (a x^2 + b x + c)/(d1 + d2 x)

                     2
        c + b x + a x
Out[1]= --------------
          d1 + d2 x

In[2]:= answer = Solve[ eq == 0 , x]

                            2                               2
                 b    Sqrt[b  - 4 a c]           b    Sqrt[b  - 4 a c]
               -(-) + ----------------         -(-) - ----------------
                 a           a                   a           a
Out[2]= {{x -> -----------------------}, {x -> -----------------------}}
                          2                               2


The original problem posed by David Petry is also solved readily by Mma 2.1,
but the answer is way too long to post.

This leads to the question ... how can I find out what the changes were from
version 2.0 to 2.1?

Greg Aicklen
University of Texas at Dallas
Erik Jonsson School of Engineering and Computer Science
Organization: Mathematics, Carnegie Mellon, Pittsburgh, PA
Newsgroups: sci.math.symbolic
Date: Wed,  2 Sep 1992 15:28:10 -0400 
From: Michael Albert <ma1q+@andrew.cmu.edu>
Subject: Todd-Coxeter algorithm

I was wondering if there is a public domain implementation of the Todd-
Coxeter coset enumeration algorithm anywhere? 

I know the "correct" answer to this is "get Cayley", but the $$$ are not
currently available.

The "group" package in Maple dies when there are more than 1000 or so
cosets on my machine (a NeXTStation). 

I can imagine doing some C coding myself but I thought I'd try the
lazy way out first! (By the way I tried an "archie" request, but it
doesn't seem to have been very successful).

Thanks in advance

Michael Albert
Newsgroups: sci.math.symbolic
From: ppc1@cec2.wustl.edu (Peter Pui Tak Chiu)
Subject: Probability
Organization: Washington University, St. Louis Mo.
Date: Sat, 29 Aug 1992 13:54:32 GMT



Hi everyone!


Some of my friends and I have been discussing a question and
we came up w/ an answer and would like to verify if its correctness.

Please give any comments freely!!!

Thanks...

Question:

Assume there are m kinds of balls in an infinitely large pool
(i.e. there are infinitely many balls)
and the distributions of different kinds of balls are equal.
Find the probability of getting all kinds of balls if n balls
are picked.

Solution:

First group the n balls you picked into m groups according to
their kinds using m-1 sticks.  So the number of possible
arrangements for the sticks and balls is (n+m-1)C(m-1).

Then we are to find the number of possible arrangements of:
(m balls of different kinds) + (n-m balls of whatever kinds) + (m-1 sticks)
since the m balls are fixed (to satisfy the constraint that there
exists at least one ball of each kind) and what we are trying to
find is the number of arrangements of n-m balls and m-1 sticks.
So, the possible arrangements is:    ((n-m)+(m-1))C(m-1) = (n-1)C(m-1)

Therefore, the probability is:    (n-1)C(m-1)
				---------------
                                 (n+m-1)C(m-1)
Newsgroups: sci.math.symbolic