HDTenEightyP/NetNews huggingface dataset:
From: volpe@bart.NoSubdomain.NoDomain (Christopher R Volpe)
Newsgroups: sci.math
Subject: Re: Problematic smurfs!
Keywords: smurfs
Date: 31 Aug 92 18:54:04 GMT
Reply-To: volpe@ausable.crd.ge.com
Organization: GE Corporate Research & Development

In article <94952@bu.edu>, spacefox@acs.bu.edu (Godfrey Degamo) writes:

|>      A group of 1000 smurfs wear either a red or blue hat on their head.
|> They can only wear one color. The color hat on their head is known to all, but
|> the wearer.  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.
|>      One day the mayor, not apart of the 1000 smurfs, decides to call a
|> town meeting.  All 1000 smurfs are gathered into the town hall.  The mayor
|> is not wearing a hat.  During the meeting, the mayor asks that all the
|> smurfs wearing red hats to stand up.
|>      There is a bit of commotion in the crowd, but no one stands up.
|>      The mayor requests the same demand a bit more sternly.
|>      Again, commotion, but no one stands up.
|>      The mayor, irate, demands the red hatted smurfs to stand up.
|>      Then, a certain amount of smurfs rise.
|> 
|> 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.
|> 
|> more importantly,
|>      Explain how you deduced your answer.

They can't all be wearing blue, therefore at least one is wearing red.
After the first demand by the mayor, no one stands up. If ONLY ONE
had been wearing read, he would see that no one else was wearing read,
and he could deduce that he himself was wearing red, and he would stand up.
Since no one did, there must be at least two red hatters among them, and
every smurf now knows this.

After the second demand by the mayor, no one stands up. If there were
exactly two red hatters, each of them would see only one red hatter
among the rest of the group, and could deduce that he had the other 
red hat. Again, since no one stoop up, there could not have been 
exactly two. Every smurf now knows that there are at least three
red hatters.

After the third demand by the mayor, some of the smurfs stood up. The only
way a smurf could deduce that he had a red hat on at this point, is if he
saw only two red hats among the others. Thus, there are exactly three red 
hatters, and exactly those three stood up.

Did I make a mistake somewhere?
-Chris
-- 
==================
Chris Volpe
G.E. Corporate R&D
volpecr@crd.ge.com
Newsgroups: sci.math
From: torkel@sics.se (Torkel Franzen)
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Organization: Swedish Institute of Computer Science, Kista
	<1992Aug11.190119.23518@ariel.ec.usf.edu>
	<1992Aug16.172642.13398@sics.se>
	<1992Aug24.153555.1298@ariel.ec.usf.edu>
Date: Thu, 27 Aug 1992 14:11:49 GMT

In article <1992Aug24.153555.1298@ariel.ec.usf.edu> mccolm@darwin.math.usf.
edu. (Gregory McColm) writes:

   >And what, precisely do you mean by "and so on"???  You may think 
   >that it is obvious what you mean, but a number of people, from 
   >Aristotle to Brouwer, have had doubts about the murkey depths of 
   >that "and so on".

  Judging by the responses that have appeared, my simple point - as I thought -
is difficult to grasp. To repeat:

   Any obscurity or ambiguity that you
   find in this explanation carries over to any explanation of what is
   meant by "PA".

  In other words, I have not claimed that my explanation of what is
meant by "the natural numbers" is clear. On the contrary, it is perfectly
compatible with my claim that that explanation is meaningless,
unintelligible, or a piece of insane raving. What I *am* claiming is
that it is exactly as clear as any explanation you may come up with of
what the phrase "the formal system PA" refers to.
Newsgroups: sci.math,misc.misc
From: ghica@fig.citib.com (Renato [Titanium Cranium] Ghica)
Subject: value of pie
Originator: ghica@duck
Organization: Citibank IBISM
Date: Wed, 2 Sep 1992 16:09:54 GMT




I can probably find this in a math book somewhere, but
I don't have one handy.

I seem to remember that PI was calcucalte with an infinite summation
such as ....X + 1/3 - 1/5 + 1/7 ... 

does any one know, offhand, what the summation really is?
-- 
--------------------------------------------------------------------------
"These views are my own. / Mail me or Flame me"

>@)
From: bubai@matt.ksu.ksu.edu (P.Chatterjee)
Newsgroups: sci.math
Subject: Topology:HELP
Date: 31 Aug 1992 23:47:40 -0500
Organization: Kansas State University
NNTP-Posting-Host: matt.ksu.ksu.edu

I'm taking a course in Topology and Homotopy this semester. Any recommendations on what book to use as a beginner's reference? It doesn't look too difficult at first blush but, I'm sure, there'll be times when I'll be scrambling in search of a good reference. 

Would really appreciate any help in this regard.
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: e-mail address in published papers
Date: 25 Aug 92 15:20:46 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <BtHpGM.FoG@watserv1.uwaterloo.ca> shallit@graceland.uwaterloo.ca (Jeffrey Shallit) writes:
>This has happened on more than one occasion.  What is the rationale
>behind this?  How can we convince editors to stop doing this?

Unfortunately, the best way to deal with both questions, is to become an
editor.

Later,
Andrew Mullhaupt
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: sci.math
Subject: Corrections Re: Abel's proof of the insolubility of the quintic
Date: 4 Sep 92 16:17:45 GMT
Article-I.D.: husc3.1992Sep4.121746.15366
Organization: Harvard Math Department

In article <1992Sep4.120349.15365@husc3.harvard.edu> I wrote:
>The identity and simple transpositions together give a union of
>conjugacy classes of size (n^2+n+2)/2,
                              ^^^
of course this should read (n^2-n+2)/2.  (typo)

>which divides |A_n|=n!/2 at least for n=11,18,27,37,38,46.

These values are correct.  Unfortunately there are no simple
transpositions in A_n.  So let's use 3-cycles instead; then
instead of (n^2-n+2)/2 we have (n^3-3n^2+2n+3)/3.  Here
the first sufficiently smooth value occurs at n=68, when
(n^3-3n^2+2n-3)/3 = 100233 = 3*3*7*37*43.

--Noam D. Elkies (elkies@zariski.harvard.edu)
  Dept. of Mathematics, Harvard University
Newsgroups: sci.math
From: ruberman@binah.cc.brandeis.edu
Subject: Abel's proof of the insolubility of the quintic
Reply-To: ruberman@binah.cc.brandeis.edu
Organization: Brandeis University
Date: Wed, 2 Sep 1992 20:42:29 GMT

Does anyone know a reference for a modern treatment of Abel's
proof of the insolubility of the quintic?  A high-school student 
recently came to me, having learnt the solutions to the lower degree
equations, and hoping to understand the higher degree case as well.
He tried to read Abel's memoir (translated into English in the venerable
Source book in Mathematics).  Understandably, he didn't get too far
since Abel's paper is rather dense and obscure to an untutored reader.
(For instance he doesn't take any pains to explain what the meaning
of `a solution in radicals to the general quintic' would actually mean.)

One suggestion to this student is to read enough about groups and fields
to learn the rudiments of Galois Theory.  There is a book of H. Edwards
which more or less travels this route.  On the other hand, Abel's proof
is quite elementary, and can in principle be explained to a high-school
student.  (I said in principle!)  I am looking for a source for such an
explanation; any pointers would be greatly appreciated.

Daniel Ruberman
Department of Mathematics
Brandeis University
Waltham MA 02254

ruberman@binah.cc.brandeis.edu
ruberman@brandeis.bitnet
From: frandag@math.ksu.edu (Francis Fung)
Newsgroups: sci.math
Subject: Re: p prime, p divides ab => pdivides a or b
Date: 2 Sep 1992 16:50:38 -0500
Organization: Dept. of Mathematics, Kansas State University
NNTP-Posting-Host: hilbert.math.ksu.edu

rusin@mp.cs.niu.edu (David Rusin) writes:

>In article <Btyo8q.E63@ux1.cso.uiuc.edu> ceblair@ux1.cso.uiuc.edu (Charles Blair) writes:
(stuff deleted)
>(Actually I think I remember some quadratic extensions of  Z  for
>which you can show no division algorithm exists but which are still
>PID's. Readers?)

>dave rusin@math.niu.edu

You can show that {(a+b(\sqrt(-19)))/2 | a and b have the same parity} is
a PID which is not Euclidean. Details can be found in Grove's Algebra.
Francis Fung
PS Hi Dave! Sorry I didn't get to think about spectral sequences much yet :)
Newsgroups: sci.math
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
Subject: Re: Proof of God's Existence
Organization: Rose-Hulman Institute of Technology
Date: Wed, 2 Sep 1992 12:26:14 GMT


In article <1992Sep1.213206.18901@griffin.itc.gu.edu.au>  
jchen@tomahawk.me.gu.edu.au (Jinghong CHEN) writes:
> Proof of God's inexistence:
> 
> God can do anything.
> 
> So he can make a stone which he cannot lift.
                               ^^^^^^^^^
oops.  You can't redefine your terms in the middle of a theorem 
and expect to make any sense.  This is equivalent to
"This statement is false."
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: Polygon point enclosure
Date: 31 Aug 92 00:33:20 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <1992Aug27.180024.2843@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
>In article <thompson.714934114@daphne.socsci.umn.edu> thompson@atlas.socsci.umn.edu writes:
>:gjm11@cus.cam.ac.uk (G.J. McCaughan) writes:
>:
>:>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.

>... This requires computing O(N) equations and intersections.

Not hardly. You can eliminate a lot of possible edges from consideration by
using a half ray parallel to the x-axis and sorting the endpoints of the
edges by y-coordinate. It's pretty typical to do even fancier stuff like using
segment trees, (see for example Preperata and Shamos).

Later,
Andrew Mullhaupt
From: matt@amsaa-cleo.brl.mil (Matt Rosenblatt)
Newsgroups: sci.math
Subject: Re: Calendar (NEW)
Keywords: Vernal Equinox
Date: 25 Aug 92 21:23:18 GMT
Organization: U.S. ARMY MATERIEL SYSTEMS ANALYSIS ACTIVITY (AMSAA), APG, MD.

In article <1992Aug24.191238.20599@linus.mitre.org> 
chaloux@accordion.mitre.org (Dave Chaloux) writes:

>Hannu writes:

>>  I have not the Bible at my hands now but I remember that it was in some
>>books of Mooses, where all holy days are defined according to the old Moon
>>Calendar. Year's first month called nisa month is also defined in the Bible.
>>The use of vernal equinox in Easter Rule is based also on the old time
>>counting of old Moon Calendar in the Bible.  

>>   The form of the Easter Rule in the Bible was something like:

>>   "Easter Sunday is 14.th day of nisa month"   [Hannu]

>. . . I believe you are thinking of the Passover. The celebration of the 
>resurrection is commonly linked to the celebration of the Passover.  
>[Dave Chaloux]

That's right.  Numbers 28:16 provides for the Passover sacrifice to
be made on the fourteenth day of "the first month," and for the
seven-day Feast of the Passover to commence on "the fifteenth day
of this month."  Exodus 12:2 and 13:4 make it clear that "the first 
month" is the month that was known as "Abib" (= "Spring") in 
pre-Exilic times, such as the times when the "books of Moses" --
including Exodus and Numbers -- were written.  After the Babylonian 
Exile, the Jews started using the Babylonian names for the lunar months.  
See Esther 3:6:  "In the first month, which is the month Nisan, . . ."  
Because lunar months begin with the New Moon and last 29 or 30 days, 
the "fifteenth day of this month" is the day of the Full Moon -- the 
first Full Moon after Spring begins at the Vernal Equinox.

Whatever it is called, the month in which the Passover feast occurs
has to come in the Spring ("Abib"), i.e., *after* the Vernal Equinox.  
If it became apparent to the Sanhedrin that twelve lunar months would go
by since the previous year's "first month" and it would still be Winter,
why, they'd just proclaim the addition of a thirteenth lunar month to
that year, so that the new year's "first month" would begin during the
Spring.  

>The celebration of the 
>resurrection is commonly linked to the celebration of the Passover.  
>It should also be noted that not all sects of Christianity make 
>this connection at the same time.  [Dave Chaloux] 

This, too, is right, which is why they have "Russian Easter" in the
Orthodox Church.  In the West, Easter Sunday usually does indeed
fall close to the Passover.  But not *too* close: the "Easter Rule"
ensures that if the 15th day of Nisan -- the first Full Moon after
the Vernal Equinox -- occurs on a Sunday, Easter itself will be 
relocated to another Sunday.  In this way, the rule keeps the 
celebration of the resurrection from coinciding with the beginning 
of the Jewish Passover feast.

Hope this helps.

					-- Matt Rosenblatt
					(matt@amsaa.brl.mil)
Newsgroups: sci.math
From: frank@Cookie.secapl.com (Frank Adams)
Subject: Re: Primes
Date: Tue, 25 Aug 1992 22:48:28 GMT
Organization: Security APL, Inc.

In article <1992Aug22.040442.17832@mailer.cc.fsu.edu> rose@fsu1.cc.fsu.edu writes:
>In article <1992Aug18.091619.10152@waikato.ac.nz>, bill@waikato.ac.nz writes...
>>Does anyone know how Mathematica stores its table of primes, and how efficient
>>it is in terms of space and access time? What's the minimum space a table
>>of primes can be compressed into if constant access time to find the nth prime
>>is required?
>> 
>In general, one can use a bitmap to represent a prime number table.  Some
>compression can be obtained by representing only odd numbers, or only 
>odd numbers not divisible by 3.  
>Then the computer can count primes by using the count bits function of the 
>hardware amazingly fast.  If the computer does not have a count bits 
>function, then this technique is lost.

No, it isn't.  Software bit-counting, using a table lookup method, is also
very fast.  You can't do it directly in most higher-level languages,
however, without giving up a lot of the speed.

However, this does not answer the original question, which required constant
access time to find the nth prime.  This can be provided by including a list
of the (n*k)th primes for some (fairly large) k.

This method, like just listing the primes, requires O(n) space to store the
primes (with or without the auxilliary table).  The difference is the
constant.  I doubt you can do better than this maintaining O(1) access time
for the nth prime; with O(log n) time it might be possible.
Newsgroups: sci.math,sci.math.stat,sci.math.num_analysis
From: shankx@cbnewsi.cb.att.com (n.k.shankaranarayanan)
Subject: Fitting a few coefficients to a 'peaky' function
Organization: AT&T
Date: Fri, 28 Aug 1992 18:56:22 GMT

SUMMARY:
This question is about extracting a few 'coefficients'
from a complex (peaky) funtion, so that changes in the
function can be tracked from sample to sample.
Would appreciate general advice and suggestions,
in the hope of uncovering some methods unknown to me.
Please also recommend books/references.

PROBLEM:
I have a datsets - sets of points, representing a function
 y = f(x). The points are not equally-spaced along the
x-axis. Some data sets can contain a very sharp peak 
around an x-value that varies from dataset to dataset. 
The datasets are probability distributions.
The shape of the function changes slightly from 
dataset to dataset; the nature (height, width, location) of the 
peak also changes. There is also some noise.

I need to extract a FEW coefficients
to describe the significant parts of the function, to
enable further investigation - sort of looking for
a pattern dependency of some kind.

APPROACH: 
I tried cubic splines - that gives many coefficients, and
requires some personal inspection of each data-set.

A Fourier analysis was better. The first few coefficients
are very useful to track some low-frequency information from
dataset to dataset.

The problem of the peak remains. The Fourier (FFT) approach is
not good at measuring the position and nature of the peak with
a few parameters. I am now thinking in terms of a hybrid method -
use the FFT for some low-frequency stuff and study the 
peak separately. Ideally, the algorithm should be robust and
should and detect and locate a peak, and branch off accordingly.

This is in the early stages of a project. So, it is 
acceptable to learn and apply a new technique altogether,
and maybe change the dataset creation process.

Any suggestions? 

Thank you for your time.

-Shankar

Newsgroups: sci.math
From: gross@math.utah.edu (Fletcher Gross)
Subject: Re: n doesnt divide 2^n-1
Date: Thu, 3 Sep 1992 21:48:07 GMT
Organization: Math Dept, University of Utah
Followup-To: sci.math

In article <Sep.3.16.29.43.1992.1193@yoko.rutgers.edu>,
gore@yoko.rutgers.edu (Bittu) wrote:
> 
> From kedlaya@b124.super.org (Kiran Sridhara Kedlaya) Fri Aug 28 16:31:44 1992
> 
> > Sure. Let n be the smallest integer > 1 such that n divides 2^n - 1.
> > Let d be the smallest integer such that n divides 2^d - 1 (i.e. the order
> > of 2, with respect to n). Then d divides n (otherwise n = rd + s, where
> > 0 < s < d, and since 2^n and 2^d are congruent to 1 mod n, so is
> > 2^(n - rd) = 2^s, whence n divides 2^s - 1, which is impossible
> > by the choice of d). Also 2^d - 1 divides 2^n - 1, and d > 1 (otherwise
> > n divides 2^1 - 1 = 1). So d is an integer > 1 but smaller than n, such
> > that d divides 2^d - 1. Contradiction.
> 
> But what if d = n? I don't see the contradiction in that case. Am I
> missing something? How does one show that d < n? There is probably a
> theorem that says that the order of 2 w.r.t an odd integer n is always
> less than n.
> 
> -Bittu

Since n divides 2^n - 1, n and 2 are relatively prime. If m is the Euler
phi function of n, then n must divide 2^m - 1. Hence d <= m. Since m is the
number of positive integers less than n and relatively prime to n, m <= n -
1. It follows that d < n.

Fletcher Gross                  |   (801) 581-6567
Department of Mathematics       |   gross@math.utah.edu
University of Utah              |
Salt Lake City, Utah   84112    |
From: bwhite@oucsace.cs.ohiou.edu (William E. White )
Newsgroups: sci.math
Subject: Need help on convex 4-gon to convex 4-gon xform algorithm
Date: 2 Sep 92 07:00:41 GMT
Distribution: usa
Organization: O.U., Harvard on the Hocking, Berkeley in the Boondocks!

Where I work we need to transform an arbitrary convex four-sided polygon
to another arbitrary convex four-sided polygon (well actually we know the
coordinates of the source and destination 4-gon; what we want is to find
a mapping that goes from one to the other).  In the end, we should be
able to apply this mapping to, say, a grid of lines within the source 4-gon 
and have a similar (but "squashed") grid appear on the other 4-gon.

The only thing I could think of off hand was using two affine mappings and
dividing the 4-gon into two triangles; this works for our original needs
(which was that the transform would only be used on points on the sides of
the 4-gon itself), but now we need to be able to transform any point, inside
or outside of the original 4-gon.  I know there's a way to do this but I'm
having enough trouble with linear algebra as it is.

Please email all responses.  This is going into software; exact answers take
precedence over computational time.  I have code to invert a matrix if the
solution can be expressed as a linear equation.

William White						bwhite@oucsace.cs.ohiou.edu
-- 
|  Bill White   +1-614-594-3434     |  bwhite@oucsace.cs.ohiou.edu (normal)   |
|  `finger bwhite@view..' for more  |  bwhite@view.cs.ohiou.edu (alternate)   |
|  --> I'm a replicating .signature virus!  Copy me and join in the fun! <--  |
From: baez@ucrmath.ucr.edu (john baez)
Newsgroups: sci.math
Subject: Re: Matrix Logarithms
Date: 28 Aug 92 18:42:03 GMT
Article-I.D.: galaxy.22036
Organization: University of California, Riverside

In article <1992Aug26.050544.15141@newshost.anu.edu.au> rwc124@rsphy7.anu.edu.au (Roderick Vance) writes:
>Let U be a unitary, symmetric matrix; U is expressible as
>
>U = exp(i H), where H is real and symmetric.
>
>I should like to know whether there is a simple method of finding
>H given U, aside from diagonalising U. 

Just one point - H is not at all unique.  Sorry if this is obvious to
you; if one didn't notice this it could cause a lot of sorrow.
Newsgroups: sci.math
From: torkel@sics.se (Torkel Franzen)
Subject: Re: Godel and the US constitution
Organization: Swedish Institute of Computer Science, Kista
Date: Tue, 1 Sep 1992 18:14:00 GMT

In article <a_rubin.715364793@dn66> a_rubin@dsg4.dse.beckman.com 
(Arthur Rubin) writes:

   >This reminds me of a story I heard at a memorial seminar on Godel, on which
   >I would like confirmation.



   The story is told as follows by Hao Wang, in Reflections on Kurt Godel.
Wang refers to H.Zemanek and E.Kohler.


        Even though the routine examination G. was to take was an easy
        matter, G prepared seriously for it and studied the US
        constitution carefully. On the day before the interview G told
        Morgenstern that he had discovered a logical-legal possibility
        of transforming the United States into a dictatorship. Morgenstern
        saw that the hypothetical possibility and its likely remedy involved
        a complex chain of reasoning and was clearly not suitable for
        consideration at the interview. He urged G to keep quiet about his
        discovery.
          The next morning Morgernstern drove Einstein and G from Princeton
        to Trenton. Einstein was informed; on the way he told one tale after
        another, to divert G from his Constitution-theoretical explanations,
        apparently with success. At the office in Trenton, the official in
        charge was Judge Philip Forman, who had inducted Einstein in 1940
        and struck up a friendship with him. He greeted them warmly and
        invited all three to attend the (normally private) examination of G.
          The judge began, 'You have German citizenship up to now'. G
        interrupted him, 'Excuse me sir, Austrian.' 'Anyhow, the wicked
        dictator! but fortunately that is not possible in America.' 'On
        the contrary,' G interjected, 'I know how that can happen.' All
        three joined forces to restgrain G so as to turn to the routine
        examination.

Newsgroups: sci.math
From: kedlaya@metropolis.super.org (Kiran Sridhara Kedlaya)
Subject: Re: n doesn't  divide .......
Organization: Supercomputing Research Center
Date: Thu, 3 Sep 1992 19:50:25 GMT

Incidentally, I posted a slightly different solution a few messages back. (I
forgot to email it. Mea culpa.) But one solution is quite enough.

One comment on Logan's comment: he says that the problem "prove that n^2 | 2^n
+ 1 iff n = 1 or 3" is "similar but harder". Similar, yes; harder is perhaps
an understatement, considering that the original problem can be solved in a
few lines, while the latter was Problem 3 on the 1990 International Math
Olympiad.

Kiran Kedlaya
kedlaya@super.org

P.S. Bart-say hello to Adam for me.
From: nps1@ukc.ac.uk (N.P.Smart)
Newsgroups: sci.math
Subject: Re: Godel and the US constitution
Date: 3 Sep 92 18:14:47 GMT
Reply-To: nps1@ukc.ac.uk (N.P.Smart)
Organization: Computing Lab, University of Kent at Canterbury, UK.

As a Brit could someone post a summary of what the amendments to the 
constitution are.
IE So when I read this discussion, which is very interresting, if I
see 14th Amendment I know what your talking about.

Nigel
Newsgroups: sci.math