HDTenEightyP/NetNews huggingface dataset:
From: dak@messua.informatik.rwth-aachen.de (David Kastrup)
Subject: Re: Easy (?) probability question
Organization: Rechnerbetrieb Informatik  /  RWTH Aachen
Date:  4 Oct 92 11:10:06 GMT

dak@tabaqui.informatik.rwth-aachen.de (David Kastrup) writes:

>davidb@triangle.Berkeley.EDU (David T. Blackston) writes:

>>Hi all,

>>Here's a problem that came up recently in an analysis of a sorting
>>algorithm.  It seems really obvious, but I can't come up with a nice
>>proof and I was wondering if anyone out there had one.

>>Let X be a random variable, and define the random variable Z by
>>Z=min{X, k} where k is some constant.  Show that var(Z) <= var(X).

>>I have shown this in the case where X is a bounded, integer valued r.v.,
>>but the proof is messy and I feel that there has to be a better way.
>>Is there?
>Well suppose X is realvalued. Then let's arbitrarily set k to 0
>(variance does not change under shifting, so we could always shift
>our original distribution so that this is the case, and shift back
>afterwards). Let's call the new variable X'. Now we introduce
>the variable X^+, which shall be X under the constraint X>0,
>and X^-, which shall be X-X^+.

>Some manipulation suffices to show that
>var(X') = var(X) - var(X^+) + 2 E(X^+) E(X^-)

Oops, forgot some factors:
var(X') = var(X) - Prob(X>0) var(X^+) + 2 Prob(X>0) Prob(X<=0) E(X^+) E(X^-)

>Now var(X^+) is positive, E(X^+) positive, E(X^-) negative, so
>var(X') must be smaller than var(X).

>Since I am not familiar with complex variances, I cannot make any
>guesses.
From: cjh@tinton.ccur.com (Christopher J. Henrich)
Newsgroups: sci.math
Subject: Re: Trisect an angle???
Date: 1 Oct 92 18:28:25 GMT
Organization: Concurrent Computer Corp., Tinton Falls, NJ

In article <1992Sep16.164530.8636@nosc.mil> bell@nosc.mil writes:
>My wife is a high school advanced math teacher.  One of her students asked her
>the other day if there was any way to trisect an angle by geometric
>construction.  We did not believe that this was possible.  Are we mistaken or
>is there no way to do this by construction?
>
It cannot be done using straightedge and compass, which are the tools
generally assumed for "geometrical construction."

_What Is Mathematics_, by Courant & Robbins, explains why. (This is
a very good book for any high-school teacher of math to have.)
The bare bones of the story are:
1. Use of straightedge and compass allows you to add, subtract,
   multiply, divide, and take square roots.  Conversely, every
   ruler & compass construction can be described with these
   algebraic tools.  Thus ruler and compass amount to being able
   to solve linear and quadratic equations.

2. Trisecting an angle would amount to solving a cubic equation.

3. There is just no way to get an exact solution of a cubic
   equation from asny amount of solving quadratic equations.


If you relax the restriction to straightedge and compass, then
there are several ways to trisect an angle.  Courant & Robbins
describe one, which uses a _marked_ straightedge.  Another neat
one is to use an instrument which looks like a cross between a
T square and a tomahawk:



                                                       P
                                                       |
                                                       |
                                                       |
------------------------------------------------------_|Q
                                                    /  |
                                                   /   |
                                                   |   |R
                                                   |   |
                                                   \   |
                                                    \__|S

(I hate ASCII graphics!)  Here, the intention is that
PQ = QR = RS, and a semicircle, centered at R, is part of
the instrument.  To trisect an angle, place P on one leg of it,
pass the long arm thru the vertex, and adjust it so that the
semicircle is tangent to the other leg - say, and point T.
Prove to yourself that PQ, QR, and RT subtend equal angles at the
vertex.

There is a recent book, titled _A Budget of Trisections_, by
(I think) Underwood Dudley, describing the author's encounters
with numerous people who think they have discovered trisections.
There was also an article in the _American Mathematical Monthly_
a few years ago, by Isadore Singer (maybe I've got the author's
name correct), which explores what can be constructed with
straightedge, compass, and trisector.  

Regards,
Chris Henrich
From: tao@fine.princeton.edu (Terry Tao)
Newsgroups: sci.math
Subject: Re: INTERESTING TOPOLOGY QUESTION
Keywords: Dense sets
Date: 6 Oct 92 01:00:27 GMT
Distribution: usa
Organization: Princeton University

The answer is () yes of course! 

take A to be the set of irrational numbers between 0 and 1/2, and the
rational numbers between 1/2 and 1.  Conversely for B.

Terry
From: zlsiida@fs1.mcc.ac.uk (dave budd)
Newsgroups: sci.math
Subject: Re: Warm and Fuzzy?  Koalas vs. Math
Date: 30 Sep 92 11:33:47 GMT
Organization: Manchester Computing Centre

In article <1992Sep14.012850.17387@nuscc.nus.sg> matmcinn@nuscc.nus.sg (Mcinnes B T (Dr)) writes:

>By the way, John, what IS the best flavour of ice cream?

Ice cream has only one flavor: cream.
All else is additives.


              Have another day    [Armageddon, the musical] 
  Dave Budd, MCC, Oxford Rd, Manchester, England   (44|0)61-275-6033
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Re: Platonic solids: history?
Date: 26 Sep 92 19:24:24 GMT
Organization: M.I.T. Artificial Intelligence Lab.

Heath has some discussion of this in his edition of Euclid's elements.

Allan Adler
ara@altdorf.ai.mit.edu
From: wjcastre@magnus.acs.ohio-state.edu (W.Jose Castrellon G.)
Newsgroups: sci.math
Subject: Re: Path connectedness
Date: 29 Sep 92 23:54:03 GMT
Followup-To: sci.math
Organization: The Ohio State University,Math.Dept.(studnt)

In article <1992Sep28.233653.19230@galois.mit.edu> "John C. Baez" <jbaez@BOURBAKI.MIT.EDU> writes:

>I am posting this for Allen Knutson, aknaton@math.Princeton.EDU,
>who had trouble posting.

>
>I'm curious about the differences between connectedness and path-connectedness
>If X is a connected topological space with two marked points, say that Y is
>X-connected if for all y,y' \in Y one can map in X so as to have X's two
>marked points hit y,y' respectively.
>
>Now define a pre-partial order on connected spaces: "X connects more
>than X'
>does" if being X-connected implies being X'-connected. So for instance, all
>R^n's (with whatever two marked points you like) are quickly seen to be
>equivalent, but the topologist's sine curve connects strictly more than [0,1]
>does.
>
>This partial order (once we mod out by equivalence) plainly has a least
>element: spaces where the two marked points coincide. Does it have any
>maximal elements? Any canonical choice of representative from a given
>connectedness class? Should one restrict to a smaller class of spaces
>(say, compact Hausdorff) for some reason? How tall is it? (If people are
>bugged by the category of connected spaces not being a set, I'm happy
>to just take the ones of cardinality \leq 2^{\aleph_0} or whatever.)
>
>References would be good too.                                   Allen K.
>

I'm posting this to sci.math, rather than to .research, as what follows
only deals with a particular case.
Let me first say that I didn't quite understand the wording of the question
(it could be my english), so here is my guess:

For X and J topological spaces, a,b two distinguished (I would add different)
points of J, say that  X  is  J-connected if for all pairs of points x,y in
X, there is a continuous function  f: J --> X  that sends  a,b  to  x,y  (not
necessarily in that order).  [so [0,1]-connected spaces are the path-connected
spaces].  Say that  S > J  if  the set of S-connected spaces properly contains
the set of  J-connected spaces. [so if S is the topologist sine curve, then
S > [0,1] ].

The question (I think) is : 
Given a class of connected topological spaces decide whether there is a maximal
element under that ordering. (i.e. is not smaller than any other elemtn of the
class).

I believe that (as the proposer pointed out) one has to place restrictions on 
the class in order to have a non-trivial answer; it turns out (by repeated
applications of Zorn's lemma) that for the class of connected subsets of the 
plane (or any R^n, n>1) there is no maximal element. (I won't spoil it, but 
I'll e-mail the proof upon request). 
Now a question, whose answer I think would be messy: for these classes, how 
many inequivalent connected subsets are there? (2^2^Aleph_0 ?)
Newsgroups: sci.math
From: cmenzel@tamsun.tamu.edu (Christopher P Menzel)
Subject: Re: ZFC etc. (was Re: Report on Philosophies of Physicists)
Organization: Texas A&M University, College Station
Date: Sun, 27 Sep 1992 05:24:08 GMT

In article <22846@optima.cs.arizona.edu> gudeman@cs.arizona.edu (David Gudeman) writes:
>In article  <TORKEL.92Sep26103351@bast.sics.se> Torkel Franzen writes:
>]  [Stuff deleted]
>]  The only person who stated an unrestricted comprehension axiom was
>]Frege, who did not do set theory, and whose system had had no impact
>]whatever on mathematics. Neither Cantor nor any other mathematician
>]took the view that the set-theoretic paradoxes invalidated even
>]set-theoretical reasoning. But this, of course, is a different topic.
>
>As for the unrestricted comprehension axioms, set theory was not
>axiomatized until many years after Russell's paradox was first
>published...

Only if 5 counts as many.  Russell first published his paradox in The
Principles of Mathematics in 1903; Zermelo's initial axiomatization of
set theory (which included the restricted comprehension axiom, better
known as Separation) appeared in 1908 in the paper "Untersuchungen
"uber die Grundlagen der Mengenlehre, I."  It's translated in van
Heijenoort.

--Chris Menzel
  Department of Philosophy
  Texas A&M University
Xref: sparky sci.math:12568 sci.physics:15758
From: marc@vieta.math.uni-sb.de (Marc Conrad)
Newsgroups: sci.math,sci.physics
Subject: Re: Another Cute Math Joke  (explanation of)
Date: 1 Oct 92 13:56:09 GMT
Organization: U of Saarbruecken, Dept. of Computational Linguistics

dms@krang.vis.citri.edu.au (David Mcgregor SQUIRE) writes:

>Q: What do you get if you cross an elephant with a grape?

>A: |elephant||grape|sin(theta)
I'm stupid, but can someone explain this to me ??
Marc

--
     \   /                     | Marc Conrad, Universitaet des Saarlandes 
      \ Luxemburg              | marc@math.uni-sb.de  phone: 0681 302-2297
France \|   Germany            | these opinions are not necessarily these   
        \x <---- you are here! | of the SIMATH-group (and maybe even not mine).
Newsgroups: sci.math
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: Mattel says "Math is tough."
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Tue, 6 Oct 92 01:54:34 GMT

In article <1aq70rINN1oe@roundup.crhc.uiuc.edu> hougen@vision.csl.uiuc.edu (Darrell Roy Hougen) writes:
>matmcinn@nuscc.nus.sg (Mcinnes B T (Dr)) writes:
>
>*>What would you have her say?: 
>
>How about, "Math is fun!" or "I like mathematical puzzles!" or
>anything else that portrays in a positive light a subject that
>children should have a positive attitude about.  
>
>Darrell

I don't see that Barbie dolls should be in the business of
propagandizing for mathematics.  The notion of a doll that said "Math is
fun!" seems laughable to me - a blatant attempt at brainwashing.  But
the notion of a doll that says "Math class is hard!" seems like 
equally blatant brainwashing, as well as a really boring thing to have a
doll say.  I emphasized the latter point in my letter to Mattel; I'm
sure one of the last things girls want to have their dolls talking about
is the difficulty of certain subjects.  

As I noted in my last post on the subject, it seems that there has been
enough of a reaction on this one to get Mattel to invent a more
intelligent Barbie.  A friend called the number and the operator
immediately knew what she was complaining about, so I guess they are
getting bombarded on this.
From: wft@math.canterbury.ac.nz (Bill Taylor)
Newsgroups: sci.math
Subject: Re: consequences of the Axiom of Choice
Date: 5 Oct 92 05:06:22 GMT
Organization: Department of Mathematics, University of Canterbury

In article <TORKEL.92Oct5044312@lludd.sics.se>, torkel@sics.se (Torkel Franzen) writes:
|> 
|>   Let me repeat my suggestion: let S be the first (or, if you like, the
|> fifteenth) ordinal definable countably infinite set of pairs of infinite
|> sets of real number.

I don't think this will really work, will it ?  The first ordinal definable
set (of the type given) is some specific set given by a class formula with no
free variables; at least that's the first definable set. And the 15th is
some other specific one. I'm not sure what the ordinal-definable sets are,
other than definable ones that may refer to some specific ordinal, which won't
change the 1st, or 15th very much, but may change the omega'th somewhat.

Of course we can't be certain that these sets will have specific, definable
choice functions, without seeing the specific formulas involved, which in
turn will depend on the precise language and axiomatization; but it seems
likely that the first and the 15th would have fairly simple definable choice
functions, and so could the omega'th, probably.

To find one that definititely won't have a definable choice function, we
must be more cunning, I think.

Specifically:

(1) call N and phi,   0-constructible.

(2) if X is m-constructible, call P(X), U(X), and {x in X | prop(x)},
                               and {y | exists x in X : y = fn(x)},
    (m+1)-constructible, for all prop and fn with no other free variables.

(3) for limit ordinal l, call X  l-constructible if it is m-constructible for
                                 some m<l.

Now look at the set of:    pairs of infinite sets of real numbers that are
                 z-constructible, but not m-constructible for any m<z.

If we choose z to be the first non-constructive ordinal, then the set mentioned
will be countable, and definable in ZF, but there will be no "uniform" way of
specifying a choice function for it,  i.e. no definable choice function.

But we can't be sure of this, for any smaller ordinal.
------------------------------------------------------------------------------
            Bill Taylor              wft@math.canterbury.ac.nz
------------------------------------------------------------------------------
                The universe is its own fastest simulator.
------------------------------------------------------------------------------
From: kwong@mary.cs.fredonia.edu (Harris Kwong)
Newsgroups: sci.math
Subject: Re: a graph problem
Date: 6 Oct 92 15:39:24 GMT
Organization: Math / CS, State Univ. of N. Y. College at Fredonia
X-Newsreader: Tin 1.1 PL3

nakada@ntttsd.ntt.jp (Hiroshi Nakada) writes:
: In article <MARCOS.92Oct5145828@iota.ece.ucsb.edu> marcos@nu.ece.ucsb.edu writes:
: >I came across this problem and I believe it must have been already solved
: >by someone. Please send me the solution and/or references if you know it.
: >
: >What is the maximum number of nodes possible in a graph if all nodes have
: >degree at most d (in my case d=16) and the distance between any two nodes
: >is at most dist (i.m.c. dist=4) ??
: 
: Let n(d,D) be a number of nodes of a graph with a given maximum degree
: d and diameter D (D is the maximum distance between any two nodes).
: A theoretical upper bound on n(d,D) is giben by Moore (see [1]) as
: 
:                  n(d,D) <= (d(d-1)^D -2)/(d-2)
: 
: for d>2. This bound is called the Moore bound.  The graphs 
: satisfying the equality are called the Moore graphs.
: Unfortunately, the Moore graphs can exist only if D=2 and 
: d=3, 7 or 57.  
: 
: [deleted]
: 

I believe the case of D=2, d=57 is still open.

--
Harris Kwong

Dept. of Math. & Comp. Sci.
SUNY College at Fredonia
Fredonia,  NY   14063
Email: kwong@mary.cs.fredonia.edu
Newsgroups: sci.math
From: lew@cbnewsd.cb.att.com (lewis.h.mammel..jr)
Subject: Re: various number systems
Organization: AT&T
Date: Wed, 30 Sep 1992 15:17:26 GMT
Summary: found it

In article <1992Sep29.180205.26078@galois.mit.edu>, jbaez@riesz.mit.edu (John C. Baez) writes:
> 
> By the way, I repeat my challenge for someone to find where the
> word "quaternion" appears in the bible (King James ed.)!


I found it but I cheated. "quaternion" is in Oxford's Bible Concordance
( title approximate. )  It's interesting that the reference is to
"four quaternion's" even though quaternion already means "a tetrad."

SPOILER:



























































Acts 12:4
Newsgroups: sci.math
From: frank@Cookie.secapl.com (Frank Adams)
Subject: Re: A Problem to Amuse You
Date: Wed, 30 Sep 1992 19:35:52 GMT
Organization: Security APL, Inc.
Keywords: Mathematics,Problem

In article <BvEBA5.KyH@world.std.com> rjk@world.std.com (robert j kolker) writes:
>Let D be an enumerable subset of the Euclidean plane. Show that D can be
>written as D_x U D_y where D_x intersects all the lines parelell to the x
>axis in a finte number of points, and D_y intersects all the lines
>parallel to the y axis in a finite number of points.

(This should really say "each line", not "all the lines".)

This is easy.  You don't even need the axiom of choice.  And it doesn't have
to be RxR; SxS for *any* set S will work just as well.
Newsgroups: sci.math,rec.puzzles
From: jgk@osc.COM (Joe Keane)
Newsgroups: sci.math
Subject: Re: Z <-> Q formula
Summary: Here's another one.
Keywords: bijection
Date: 30 Sep 92 05:53:59 GMT
Reply-To: Joe Keane <jgk@osc.com>
Organization: Versant Object Technology, Menlo Park, CA
Weather: sunny, high 81, low 55
Moon-Phase: waxing crescent (18% of full)

In article <1992Sep27.085952.7534@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU
(Yuzuru Hiraga) writes:
>The set of integers Z and the set of rationals Q have the same cardinality,
>i.e. there exist a bijection between them.  So we all know that.  But then,
>is there a "simple" formula that expresses any such mapping?

How about this:

Define q(0) = 0 and q(1) = 1.
If K = 2^k and K <= n < 2*K, then
q(K + n) = q(n) / (q(n) + 1), and
q(2*K + n) = q(n) + 1.
Also, q(-n) = -q(n).

In English, for positive n, express n in binary, rotate the initial `1' to the
end, count the lengths of runs of digits, put them in a vector, interpret that
as a continued fraction, and that's q(n).

I claim that this function enumerates the rationals.

The function is about as continuous as you could expect from such a thing.
The graph of q(n) versus log n is a fractaloid curve which keeps repeating but
gets more filled in as n gets larger.

--
Joe Keane, amateur mathematician
jgk@osc.com (uunet!amdcad!osc!jgk)
From: cxm7@po.CWRU.Edu (Colin Mclarty)
Newsgroups: sci.math
Subject: Re: consequences of the Axiom of Choice
Date: 3 Oct 1992 14:47:22 GMT
Organization: Case Western Reserve University, Cleveland, OH (USA)
Reply-To: cxm7@po.CWRU.Edu (Colin Mclarty)
NNTP-Posting-Host: slc5.ins.cwru.edu


In a previous article, holmes@opal.idbsu.edu (Randall Holmes) says:


>>
>>I certainly don't think you should _not_ mention AC when it's used.  I
>>just think, a la Gunning's comment, the decision on _whether_ to use AC
>>or some other assumption is pretty much a stylistic matter.
>>-- 
>>-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
>
>It's not entirely a matter of style.  It is legitimate, for instance
>(although not usually done) to assume that all sets of real numbers
>are Lebesgue measurable.  But then one _cannot_ assume AC in
>unrestricted form...
>
>
>-- 
>The opinions expressed		|     --Sincerely,
>above are not the "official"	|     M. Randall Holmes
>opinions of any person		|     Math. Dept., Boise State Univ.
>or institution.			|     holmes@opal.idbsu.edu
>

	This is something I have asked around about but never got
much of an answer.  Is there any use doing measure theory this way?
Or is the theory so cut up by the loss of unrestricted choice that
it is harder than the classical theory?

Colin McLarty

Nous savons lire les guillemets, et nous savons lire a travers eux.
Xref: sparky sci.math:12374 sci.philosophy.meta:2043 sci.logic:1603
From: dseal@armltd.co.uk (David Seal)
Newsgroups: sci.math
Subject: Re: "Solvability" in Chess
Date: 30 Sep 92 14:28:13 GMT
Distribution: sci
Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK

In article <1992Sep29.230338.22552@ccu.umanitoba.ca>
sbloch@silver.cs.umanitoba.ca (Stephen Bloch) writes:

>What's this "chance" showing up in the game?  I was under the
>impression chess, whether taking turns or simultaneous, was
>deterministic.  Or are we assuming each player's strategy is
>randomized?  If so, I believe it's a theorem that if there is an
>optimal randomized strategy then there is one with distribution
>[0,0,0,...,1,0,0,0...], i.e. not actually using the randomness.
>(This is illustrated, though not proven, in the following paragraph.)
>
>However, there is a problem when you introduce simultaneity.  (Gee, I
>started this post intending to show that there WASN'T...)  Suppose we
>simplify the rules immensely to a one-round game in which each player
>outputs a bit; I win if they're equal, the other player wins if
>they're not.  If I have no knowledge of my opponent's strategy, and
>if (for example) I assume a uniform distribution over {0,1}, then no
>strategy is any better than any other for me.  (Using randomness
>myself would neither help nor hurt.)  On the other hand, if I assume
>anything other than a uniform distribution in what my opponent is
>doing right now, my optimal strategy is obvious: do (with probability 1)
>whatever I think my opponent is most likely to do.

You've chosen a perfect example of why the optimal randomized strategy is
not necessarily [0,0,0,...,1,0,0,0,...]. But you've got to look at it the
other way around.

If (in your bit-matching game) you choose 0 and 1 randomly and with equal
probability, there is *nothing* your opponent can do to win more than 50% of
the time on average (other than install a tap on your random number
generator :-). If you have any determinism in your algorithm, your opponent
has the possibility of latching onto it and defeating you when it crops up.
If you choose purely randomly, but on a p:(1-p) basis with p not equal to
1/2, your opponent can get his winning rate up to MAX(p,1-p) > 1/2 by
choosing the correct value consistently.

In other words, your optimal strategy is [1/2,1/2] because, even if your
opponent works out what your strategy is, he cannot exploit this knowledge.

David Seal
dseal@armltd.co.uk

All opinions are mine only...
From: cxm7@po.CWRU.Edu (Colin Mclarty)
Newsgroups: sci.math
Subject: Re: Topos theory for mere mortals (?)
Date: 26 Sep 1992 17:47:57 GMT
Organization: Case Western Reserve University, Cleveland, OH (USA)
Reply-To: cxm7@po.CWRU.Edu (Colin Mclarty)
NNTP-Posting-Host: slc12.ins.cwru.edu


In a previous article, holmes@opal.idbsu.edu (Randall Holmes) says:

>In article <1992Sep25.152726.2075@maths.tcd.ie> tim@maths.tcd.ie (Timothy Murphy) writes:
>>cxm7@po.CWRU.Edu (Colin Mclarty) writes:
>>
>>>	There is just no way that what Halmos calls naive set theory
>>>can be regarded as the common heritage of all rational humans.  It
>>>takes training to learn to think that way.
>
>As the poster below says, finite set theory is part of the common
>heritage.  


	Oh, the theory of finite sets of people, or shoes, et c. is
fairly implicit in all human rationality.  But if you think the theory
of hereditarily finite pure sets is common property, or even the theory
of sets of sets of shoes, you have been studying too long.

Colin 

 
Xref: sparky sci.math:12444 sci.math.symbolic:2584