Subject: Re: nonsense sequences
From: hillman@math.washington.edu (Christopher Hillman)
Date: 20 Jan 1997 11:27:38 GMT
In article <32E26319.18DA@marvel.conecta.it>,
massimo ales writes:
|> I'm studing (binary) sequences without "nonsenses".
|>
|> It means that in these sequences, every subtring (from a length
|> n),cannot appear reversed; or with some parts reversed ; or like a
|> combination of (few) other substrings; or is "codified" with small
|> codewords, and so on.. .
|> In few words without every transformation that applied i.e. to a text or
|> a flow of data received from a probe, or from a cd-rom or to DNA, would
|> give us a nonsense.
This description wasn't clear to me. By "substring" you probably mean
what symbolic dynamicists usually call "words" or "blocks"; for
instance in an infinite binary sequence, one word might
be 011, which could occur again and again a semi-regular manner,
or might only occur once, etc. It's not at all clear to me what
you mean by "[not] codified with small codewords"; maybe you want
to consider only (infinite?) binary sequences having positive entropy
(which would mean that the sequence cannot be produced by a short
program)? Or maybe you have in mind that there are no short range
correlations apparent? You also appear to demand that if a word w appears,
then its "reversal" w* cannot, where 011* = 110. If so, you should
probably look at the classic work of Morse and Thue on a sequence x
with the property that if x contains a word w, then it does not
contain the word ww, where if w=011, ww=011011. This turns out
to lead to the "Thue-Morse sequence" formed from words of form w(w*) !
See the review paper "Axel Thue's Work on Repetitions in Words"
by Jean Berstel, available at the URL
http://www-igm.univ-mlv.fr/~berstel/
You may also want to look at his paper on Sturmian words. For
an introduction to symbolic dynamics, see the new textbook
Lind and Marcus, Introduction to Symbolic Dynamics and Coding,
Cambridge: Cambridge University Press, 1995.
Hope this helps!
Chris Hillman
Subject: Re: Performance Index for Covering Designs, Wheels
From: bm373592@muenchen.org (Uenal Mutlu)
Date: Mon, 20 Jan 1997 10:27:32 GMT
>>|> The WCD is a number only (ie. no percent value). The higher the WCD,
>>|> the better the wheel. It should be used for comparison of 2 or more
>>|> wheels of same type with possibly differing number of tickets.
>>|>
>>|> (cf. also the other postings on this which also include some examples)
>>|>
>>
>> I'm not sure it's that easy to compare wheels of differing number of
>>tickets by this method. Let's take a very simple example to make this
>>clear. Consider a 6/14 lottery with the winning criteria being a
>>3+ match.
>> The number of combinations is 3003 and the best wheels I can think of
>>are, for 1 to 4 tickets:
>>
>>
>> Number of tickets = 1: Numbers 1, 2, 3, 4, 5, 6
>>
>> 1589 100
>> Coverage is 1589, Therefore WCD = ---- * --- = 52.91 %
>> 3003 1
>>
>>
>> Number of tickets = 2: Numbers 1, 2, 3, 4, 5, 6
>> 7, 8, 9, 10, 11, 12
>>
>> 2778 100
>> Coverage is 2778, Therefore WCD = ---- * --- = 46.25 %
>> 3003 2
>>
>> Number of tickets = 3: Numbers 1, 2, 3, 4, 5, 6
>> 7, 8, 9, 10, 11, 12
>> 1, 2, 3, 4, 13, 14
>>
>> 2988 100
>> Coverage is 2988, Therefore WCD = ---- * --- = 33.17 %
>> 3003 3
>>
>> Number of tickets = 4: Numbers 1, 2, 3, 4, 5, 6
>> 7, 8, 9, 10, 11, 12
>> 1, 2, 3, 4, 13, 14
>> 5, 6, 7, 8, 13, 14
>>
>> 3003 100
>> Coverage is 3003, Therefore WCD = ---- * --- = 25.00 %
>> 3003 4
>>
>> Which would you say is the best wheel, and how does your
>>judgement compare to the WCD rating?
>>
>> The point I'm trying to make is that the WCD is always going to go
>>down with increasing number of tickets, because you can't avoid
>>coverage redundancy as you introduce more tickets. The same applies
>>to the various wheels in 6/49. Consider again a 'wheel' of 1 ticket.
>>Its coverage is 260524 out of 13983816, giving a WCD of 1.863%. This
>
>maybe a typo: should be 260624; it is the maxCover value, ie. the
>nbr of combinations 1 block ideally should cover.
>
>I'm currently working on an new method, but the relations for the
>given examples remain similar I fear. The advantage of the new formula
>is that it gives a true p-value (ie. percent value if multiplied by 100).
>
>>is to be compared with the best current 27 ticket wheel with a
>>coverage of about 44.5%, ie a WCD of 1.648%. Then the 168 wheel has
>>a WCD of just 100/168 or 0.595%. Again, does the WCD give a good
>>guide to which is the best wheel? Essentially the WCD will _always_
>>give the best result for a wheel of 1 ticket, and will favour wheels
>>of lower number of tickets. Now I'm not a great expert or user of
>>wheels, but I'm fairly sure that most people who are keen on wheels
>>do not consider a wheel of one ticket to be a good wheel!
>>
>> So what I'm really trying to say is that I don't think the WCD
>>is a good means to compare two wheels of different numbers of tickets..
>>However, I can't say I've got any better suggestion.
Below is an example output of my new method called "DCPI".
I'll also post these DCPI-values for the above examples of the
6/14 case. The program is short and I'll try to post it here.
====
ANNOUNCEMENT: program DCPI.EXE for Win95/WinNT
DCPI.EXE is a 32-bit commandline program for Win95/WinNT for the
calculation of the "Design Coverage Performance Index (DCPI)"
(and some other indices) for Covering Designs, Wheels etc.
THIS VERSION IS AN EARLY BETA VERSION. COMMERCIAL USAGE OR
REDISTRIBUTION STRICTLY PROHIBITED (ie. VERBOTEN! :)
DOC/Text not yet finished; will follow soon along with the C++ source.
Below is an example calculation given.
Please try it out, and let me know of any bugs you find etc.
Author:
Uenal Mutlu (bm373592@muenchen.org)
Munich/Germany
----------------------------------------------------------------------------
"DCPI" without any arguments gives a short help:
Commandline: DCPI
DCPI v1.00 970118Sa (c) 1997 U.Mutlu (bm373592@muenchen.org)
Design Coverage Performance Index (DCPI); see Text
Usage : DCPI vAll vSub k minWin b winCombs [b2]
Example: DCPI 49 6 6 3 27 6215835
Output should be: pWin=44.450205867% DCPI=88.3326597%
b2 is optional: will give the approximated pWin for b2
(for mathematicians only: b and b2 can be float-values)
----------------------------------------------------------------------------
And here an example calculation:
Commandline: DCPI 49 6 6 3 27 6215835
DCPI v1.00 970118Sa (c) 1997 U.Mutlu (bm373592@muenchen.org)
Design Coverage Performance Index (DCPI); see Text
Input : vAll=49 vSub=6 k=6 minWin=3+ b=27.000000 winCombs=6215835
Other Indices:
all 27.00 blocks cover 6215835 combinations of 13983816 (= 44.450205867%)
ideally, 1 block should cover 260624 combinations (MaxCover)
actually, 1 block covers on avg 230216.111111111 winning combinations
actually, 1 block covers on avg 3.703703704% of winning combinations
actually, 1 block covers on avg 1.646303921% of total combinations
WCD = winPerc / nBlocks = 1.646303921
Output: WinComb (ie. pWin) = 44.450205866553% DCPI = 88.332659736291%
This design with 27.00 blocks covers (only) 88.332659736% of what an
ideal design for the same nbr of blocks would cover (ie. DCPI=100%),
and wins (>= 3) in 44.450205867% (=pWin) of the possible cases (ie. of
all combinations), whereas the ideal design, if existed, would assure
the same in 50.321371505% of all the possible cases.
More than 50.321371505% pWin can not be achieved by this nbr of blocks.
The ideal design would cover...
100% of all combinations in 53.655135367426 blocks
90% of all combinations in 48.289621830683 blocks
80% of all combinations in 42.924108293941 blocks
70% of all combinations in 37.558594757198 blocks
60% of all combinations in 32.193081220456 blocks
50% of all combinations in 26.827567683713 blocks
40% of all combinations in 21.462054146970 blocks
30% of all combinations in 16.096540610228 blocks
20% of all combinations in 10.731027073485 blocks
10% of all combinations in 5.365513536743 blocks
27.000000 blocks can cover max (=ideal) 50.321371505% of all combinations
* The DCPI is between 0.0% and 100.0%. The higher the DCPI, the better
the performance of the wheel in respect to the game it's used for (ie.
it depends also on vAll, vSub, k etc.). The DCPI is mainly based on
the fixed theoretic LowerBound and so on the MaxCover (cf. above).
And, simple Linear Regression (transformation) is applied.
* The DCPI is recommended for performance comparisons of designs of
same type in same environment.
-- Uenal Mutlu (bm373592@muenchen.org)
* Math Research * Designs/Codes * SW-Development C/C++ * Consulting *
Loc: Istanbul/Turkey + Munich/Germany
Subject: Re: 360 degees and divisiblity
From: Milo Gardner
Date: Fri, 17 Jan 1997 15:44:27 -0800
I agree with Ben.
On 17 Jan 1997, Ben Z. Tels wrote:
> As far as I know it is because this number is easily divisible by several
> other numbers, including 1,2,3,4,5,6,15, etc., which gives you nice numbers
> in lots of basic divisions.
>
From the Egyptian, Greek and Roman side of numeration the denominator of the
first partition of well known 2/p unit fraction series (1/a listed below) -
beginning with the RMP - was abundant for p > 11. The abundant
divisbility property allowed conversion of p/q to exact unit fraction
rules like the following to be used for over 3,000 years:
n/p - 1/a = (na -p)/ap
na/ = 1/a + (na-p)/ap
where the divisors of a were additively used to find alternative na -p
values. For example, for 2/19 the first partition usually used 1/12,
with (2a -p) = 24 -19 = 5; with two choices for using the divisors of
12 (6, 4, 3, 2, 1), 3 + 2 and 4 + 1 to equal 5.
Since 3 + 2 offers the smallest last term, and the shortest series,
Egyptian, Greek and Roman alike accepted this exact version of 2/19.
Regards,
Milo Gardner
Sacramento, Calif.
Subject: Re: Eigenvalues
From: Dmitry Tikhonov
Date: Mon, 20 Jan 1997 09:54:33 -0800
John Hench wrote:
>
> Dmitry Tikhonov wrote:
>
> > I'm afraid it does'nt possible in simple way.
> > The only Tr() functional of matrix are linear.
> > Tr(A+B)=Tr(A)+Tr(B)
> > Tr(AB)=Tr(A)*Tr(B)
>
> Uh, no. Let A=[0,1;0,0] and B=A' (in MATLAB
> notation). Trace(A)=Trace(B)=0, but trace(AB)=
> 1. Note that trace(AB)=trace(AA')=frobenius norm
> of A in this case. Since A is not identically
> zero, trace(AB) must be greater than zero, right?
>
> On the other hand: det(A)*det(B)=det(AB).
>
> -------------------------------------------------
> Dr. J.J. Hench
> Dept. of Mathematics, Univ. of Reading, England
> Institute of Informatics and Automation, Prague
> -------------------------------------------------
Of course, I was wrong writting:
Tr(AB)=Tr(A)*Tr(B),
Really I mean:
Tr(AB)=Tr(BA)
Sorry, for my mistake.
Dmitry Tikhonov.
Subject: Re: Why do Black Holes Form at all? (A slight deviation)
From: hillman@math.washington.edu (Christopher Hillman)
Date: 20 Jan 1997 12:07:01 GMT
In article <01bc05fe$2ccbb130$4e0597cb@JX>,
"John DeHaven" writes:
|> What an interesting thread! Makes me want to ask some related questions.
|>
|> 1) The point of view of almost everyone in almost every non-suicidal
|> situation is for sure "outside" the S-radius of a black hole, and hovering
|> isn't necessary. Looking thru a telescope would be good enough to observe
|> it.
The formation of the black hole from a collapsing star? Or the hole
after formation? Looking through a telescope you could follow the
early stages of the collapse, but (by definition) you couldn't see what
happens to anything after it passes through the event horizon.
|> If I understand this thread right, it is hard to see how a black hole
|> would seem ever to "grow" from this viewpoint. Time would "stop" (from our
|> viewpoint) for the thing the instant the developing black hole condensed
|> enough to "close up" and form an S-radius.
This sounds confusing to me. My point is that if you just want to convince
yourself that the concept of a black hole is not internally inconsistent,
you should concentrate on understanding the geometry of Schwarzschild space
under the interpetation of unit tangent vectors to timelike curves
("world lines") outlined in an earlier post. Then the event horizon
appears as a single two dimensional sphere, where the coordinates
"behave badly" in the same sense that radial coordinates "behave badly"
at r=0 (the "badness" here is that r=0, theta = pi/2 represents the
same point as r=0, theta = any other angle; in the same way the event
horizon appears as a "cylinder" r= R_0, t = anything,
angular coords = anything, when in fact this is an ordinary two sphere.
|> This would seem to imply that
|> every black hole is some kind of fossil of the instant of its creation.
|> (Being really simplistic about it and excluding Hawking radiation, etc.)
|> Infalling matter (from our point of view) wouldn't _reach_ the radius. And
|> if time (from our point of view) has stopped at the radius, it necessarily
|> would be (from our point of view) changeless. That means it could not
|> expand (from our point of view).
We're not really talking about the point of view of a distant observer
so much as the Scharzchild coordinates, which are fine for discussing
events far from the hole, but which break at the event horizon. Near
the event horizon other coordinates (which do not break down there) are
more useful, because they represent more directly and simply what particles
passing through that region actually experience. What a distant observer
SEES them experience is pretty irrelevant because as we said, such an
observer would in fact see very little about what's going on near the
event horizon (neglecting such effects as accretion disks).
|> 3) Everyone talks about the singularity at the middle, of zero size, and
|> the very much non-zero S-radius that defines the "surface." What's in
|> between? If we managed to dive through, we surely could not cover that
|> distance instantly, FTL-wise. What _would_ our (or some godlike-being's)
|> experience be as we (or it) fell further towards that singularity? What is
|> the spacetime picture inside the hole? _Is_ there even any "distance?" Why
|> _isn't_ the volume between the singularity and the S-radius populated with
|> falling stuff? Or would it be?
You really should read the appropriate chapters in Misner, Thorne and Wheeler,
Gravitation, Freeman, 1970. The so-called Schwarzchild radial coordinate
has the following meaning: thin spherical shells around the hole have the
same geometry as thin spherical shells in ordinary space. In particular,
from measuring the area or circumference (both quantities make sense without
"going off the shell"; we say they are INTRINSIC) we can define a "radius"
by assuming that the shell is sitting in ordinary flat space. In the
Scharzchild universe, this radial coordinate does NOT measure "distance
to the center" (this is an EXTRINSIC notion), unlike the case where the
spherical shell is in fact sitting in ordinary flat space. You can picture
spacelike slices of the collapsing star as the upper half of a paraboloidal
surface of revolution with a "spherical cap" pasted onto the bottom. See
the pictures in Misner, Thorne, and Wheeler, which also has a very good
discussion (with pictures) of how complex OTHER space like slices can look.
At the risk of causing further confusion, I mention that inside the event
horizon (which appears as a cylinder in the Schwarzchild coordinates, but this
cylinder is really only a sphere) the radial coordinate becomes TIMELIKE,
while "longitude lines" are now SPACELIKE. If you take the timelike "disk"
for t=0, 0 <= r <= R_0, then the intrinsic geometry turns out to be exactly
that of the Friedmann universe (with "maximal" radius R_0 and with one
space dimension suppressed; see MTW for "suppression" of coordinates).
This emphasizes the point that there is in principle nothing "strange"
about space time inside the event horizon; this region simply has the
property that all timelike curves have finite "length" and have one
endpoint at a singularity of infinite curvature.
|> 4) I am skeptical (at the moment) about singularities, at least as they are
|> usually characterized. A singularity is a purely relativistic idea. But as
|> a blob of stuff collapses towards a limiting size of zero, it seems to me
|> that it must (eventually) collapse through epochs where _quantum_ spacetime
|> effects completely overwhelm relativistic spacetime effects. I don't think
|> you can even speak coherently about a "point in space" in quantum spacetime
|> at very small scales. Is that right? If so, then there would be some
|> super-intense, and very "singular," quantum-foam
Indeed, most physicists agree that there is every reason to expect the
"classical" gtr picture to break down at r ~ the so-called Planck length,
precisely because gtr no longer approximates the (unknown but presumably
more generally correct) theory of quantum gravity. In addition, as I
mentioned in my previous post, even in classical relativity, you would
not expect a real collapsing star to experience PERFECTLY spherically
symmetrical collapse, and under the assumption that the collapse is
only partially symmetric, you can be confident that inside the horizon,
the classical predictions would very quite a bit from the Schwarzchild
universe at scales well above the Planck length. So the Schwarzchild
universe should be viewed as the grt equivalent of the 1/R Newtonian
gravitational potential induced by a "point mass", which is quite
accurate as a first approximation for spacetime near the sun, the
earth, etc., but not very accurate for what actually happens to the
surface of a collapsing star after passing through the event horizon.
Also, the Schwarzchild solution does not actually include any model
of a collapsing star--- that has to be "pasted" into the model in a
fashion well described in MTW.
Hope this helps!
Chris Hillman
Subject: Re: Why do Black Holes Form at all?
From: schmelze@fermi.wias-berlin.de (Ilja Schmelzer)
Date: 20 Jan 1997 14:05:37 GMT
In article <32DD9A27.36B2@epix.net> Mick Spencer writes:
>> Let me ask a follow up question. In Schwarchild's coordinates the
>> particle passes through the event horizon at time t = infinity.
>> However, Hawking has shown that a black hole cannot last until
>> t = infinity, it will evaporate first. If by the time the particle
>> enters the black hole it does not exist any more, how can it do it?
>> Someone told me that the "paradox" comes from an improper mixture of
>> classical and quantum physics, but I would appreciate any more detailed
>> explanation about how matter can fall inside an evaporating (non rotating
>> and non charged) black hole.
>I'm not really a black hole expert, but I'll take a stab. The particle
>appears frozen on the horizon (to an outside observer, if they could
>actually 'see' a particle on the horizon) because time at the horizon is
>infinitely dilated in the observer's frame. From the particle's
>viewpoint however, it plunges right on through to the singularity in
>very short order. If the black hole were to eventually evaporate and an
>outside observer was nearby to see it, I don't think they would *ever*
>see the particle fall through the ever diminishing event horizon. I'm
>guessing (though I admit I haven't thought about this one much) that the
>observer would see the particle disappear at the same moment the horizon
>disappears. This way "t = infinity" might be said to be the same as
>"the moment will never come". Okay, so I'm out on a limb here. :)
There is simply a problem in GR which doesn't allow to make a unique
meaningful prediction. It is the problem of time. GR doesn't predict
what is the "correct" time. To derive Hawking radiation, we have to
add a lot of common sense about the correct time which is not part of
GR.
My theory of gravity, PG, is roughly speaking GR with a fixed time,
and in this theory Hawking radiation can be predicted uniquely. The
"correct time" in my theory is the Schwarzschild time, and that's why
the black hole evaporates without having been even formed. The
collapsing body is - like in Schwarzschild coordinates - always a very
very little bit greater than the horizon, and does no longer collaps
because of time dilation. In this state, Hawking radiation radiates
and the mass of the collapsing body is decreasing. Thus, in my theory
the black hole never really forms, even if we cannot distinguish this
from outside from a GR black hole.
But GR itself is not such unique in the prediction. I have a citation
that my scenario was considered in GR too, but the leading scenario
seems to be a different one. There is no quantum theory of GR we can
use to derive it, that's why all such guesses are only guesses.
Ilja
Subject: Re: 360 degees and divisiblity
From: olson@ici.net (Jeremy J. Olson)
Date: Mon, 20 Jan 1997 10:00:30 -0500
In article ,
Milo Gardner wrote:
… I agree with Ben.
…
… On 17 Jan 1997, Ben Z. Tels wrote:
…
… > As far as I know it is because this number is easily divisible by several
… > other numbers, including 1,2,3,4,5,6,15, etc., which gives you nice numbers
… > in lots of basic divisions.
Slightly off topic to the divisibility question of 360° (did that print
right? -- it's a degrees sign), but why do we still use this system of
360°? Considering that other scientific/mathematic units have been switched
over to metric, why aren't "grads" more commonly used now? 400 grads per
circle, which allows for 100 (i.e., metric!) grads per right angle. Eh,
just my 2¢...
Jeremy J. Olson
\ / http://www.ici.net/cust_pages/olson/olson.html
U
\n/ If you send any unsolited advertisements to this
address, they will be sent to your postmaster
and hundreds of copies will be bounced back to you.
Now, do you still want to email me your spam crap?
Subject: Re: Simple (I hope) math problem
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 20 Jan 1997 09:30:20 -0500
In article <32E30057.2A16@midusa.net>, Jim Dolenz wrote:
:Gene Koesling wrote:
:>
:> Hi all,
:> I've got something thats driving me nuts. I have to determine whether a
:> point is within a polygon (its for use in a computer program). My
:> apologies in advance if I'm using inappropriate terminology or notation...
:> its been about 18 years since my last math class.
:>
:> I have a polygon with n vertices and I know the x,y coordinates of the
:> vertices. I'll be given a point with x,y coordinates. I need to
:determine >> whether the point is inside the polygon or not.
:>
:> e.g. a four sided polygon with vertices at A(1,1), B(2,5), C(3,3), D(7,3)
:> and a series of points a(3,2), b(5,2), c(3,4). By drawing this on paper I
:> can see that only 'a' is within the polygon ABCD but how would I determine
:> that mathematically. (In the program I'd be testing one point at a time)
:>
:> Please note that for the purposes of my program, my polygons will have
:> (potentially) between 3 and 256 vertices.
:>
:> Any help at all would be appreciated.
:>
:> Gene Koesling
:> AKA cst0415@nait.ab.ca
:> --
:> -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
:> Semper in excreta -- solum profundum variat
:> -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
:> http://www.connect.ab.ca/~geak
:A sharp guy I work with, Mark Beyer, showed me this tecnique several
:months ago: Let the polygon vertices, in order, around the perimeter
:be: N(1), N(2), N(3),... and the point to be P. Further, let N(1,2) be
:the vector from N(1) to N(2), N(2,3) from N(2) to N(3) etc. Just form
:the vector cross products N(1,2)XN(1,P); N(2,3)XN(2,P); ...
:N(n-1,n)XN(n-1,P); N(n,1)XN(n,P) where n is the number of vertices. If
:these cross products all have the same sign, the point P is in the
:interior of the polygon N(1),N(2),...,N(n).
Doesn't it fail for non-convex polygons and a point P in the places where
the convexity of the polygon is violated?
(The above test confirms only points with respect to which the polygon
is "star-shaped".)
Example - L-shaped room; some cross-products are of different signs even
for an interior point indicated as P.
|---------|
| P |
| |
| |
| L---------
| |
| |
| |
| |
L-------------------
Cheers, ZVK (Slavek).
Subject: Re: Clinton pardons Ohair on condition she converts to buddhism!
From: David Kastrup
Date: 20 Jan 1997 17:10:16 +0100
"John P. Boatwright" writes:
> Your Name wrote:
> > But I guess God must approve of incest, as only a short time later -
> > Biblically speaking of course - He pares the population down to Noah,
> > Noah's wife and their children. So from this small, inter-related
> > coregroup, the world is repopulated.
>
> Well again you're missing the whole point:
>
> God makes everyone and everything. Do you demand Intel build
> processors by your method? Nah, they do it the way they want to.
>
No, you haven't actually seen *what* point he missed. Noah went into
the ark with his wife, and his sons, and their wives. That's
sufficient. The cousins from his son's kids are far enough in
relation to legally intermarry. While you might still want to call a
marriage of cousins somewhat intercestuous, it is not disallowed by
the biblical law.
And we are starting with the gene pools of Noah, his wife, and the
wives of his sons. Not very good, but also much better than pure
incest.
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Re: paradox
From: a.hunger@hulmecc.nwnet.co.uk
Date: Mon, 20 Jan 1997 16:19:07 GMT
kkumer@desdemona.phy.hr (Kresimir Kumericki) wrote:
>weasel (weasel@televar.com) wrote:
>: kkumer@desdemona.phy.hr (Kresimir Kumericki) writes:
>: > : Joseph H Allen wrote:
>: > : > Thus all of the non-black things you find which
>: > : > aren't ravens (your red coat, the white ceiling, etc.)
>: > : > also support your generalization that "all ravens are black".
>: >
>: > Hm, isn't there a name for this paradox? 'Hempel paradox' or
>: > something? I really like it: Some lazy ornithologist on a rainy
>: > day could investigate whether all ravens are really black just
>: > by going around his room and noting objects which are not ravens
>: > and which are not black.
>
>: This is not a paradox. The statement "all non-black things aren't
>: ravens" is the contrapositive of the hypothesis "all raven's are black.
>: Mr. Allen is correct; the two statements are logically equivalent.
>
>: I'd not call the ornithologist lazy, however. In order to prove the
>: hypothesis (all ravens are black) by demonstrating the contrapositive,
>: he'll have to examine each and every non-black item and show that
>: none of them are ravens. I'd call that a more daunting task even than
>: checking up on all the ravens.
>
>
> I am well aware that these statements are logically equivalent.
>Perhaps I was insufficiently precise, but what I meant is not
>that these statements constitute *logical* paradox, but the one
>which is *epistemological* in nature.
> The thing is that you want to know generally how knowledge is
>acquired and what constitutes a 'proof' of some scientific
>theory. Theories that are most difficult to handle are those
>which state that all members of some class have some property.
>E. g. why are we so sure that 'all ravens are black'. It is
>obvious that it is impossible to prove this because only proof
>that would be generally acceptable is something like collection
>of photographs off all ravens in the world (for all times in the
>history). On the other hand, we are pretty sure that all ravens
>are really black and now we want to find a scientific rationale
>for believing in such, unproven, statement.
> One direction in solving this is to say that the more black ravens
>we see the more we are sure that our theory is correct. We can
>quantify it by saying that each theory has 'correctness value' between
>0 and 1, and with each black raven we see the value is closer to 1
>(never reaching exactly 1). But then aforementioned 'paradox' comes
>to the game because by noting that my computer screen is green I also
>move this number closer to 1. I know that I move it by a very very
>small amount, but it is paradoxical that I can move it at all.
> This is how I understand this. I'd like to here from someone who
>thought about it more than I did. (Or read about it - I'm sure
>that this is a very famous paradox in modern philosophy of knowledge,
>and I just cannot remember his exact name or what was response
>of philosophers to it.)
> But to tell you the truth, now, when I thought about it a bit
>writing this article, it doesn't seem much paradoxical to me
>either, even in the epistemological sense. Perhaps I'm
>missing something. (Physicists often tend to think that most
>of philosophy is no good :)
>
>Kresimir Kumericki
>--
>------------------------------------------------------------
>Kresimir Kumericki kkumer@phy.hr http://www.phy.hr/~kkumer
>Department of Physics, University of Zagreb, CROATIA
>------------------------------------------------------------
The paradox can be resolved simply by using Bayes' Theorem.
suppose f(x) is a Boolean function on the natural number x.
We have many instances of f(x) for different values of x, and we want to
investigate the bearing of these data on the hypothesis:
forall x in Y : f(x) (1)
where generally the number of different x where we have verified f(x) is
small compared to the cardinality of Y. On the other hand, no
counterexamples have been found.
Clearly, for each x in Y that we have found f(x) to be true - e.g. x is the
index number of a raven, f(x) is "raven x is black" and cardinality of Y is
the size of the raven population - f(x) is implied by hypothesis (1). In
fact, the disjunction for all x tested of f(x) is implied by (1). Thus if
A is hypothesis (1) and B is the disjunction of all the f(x) for known x,
P(B|A) = 1
This is always true when A implies B as is easily found from the formula:
A implies B is equivalent to not A or B
We are interested in P(A|B) - the probability of hypothesis (1) given the
data.
We use Bayes' Theorem :
P(A|B) = P(B|A) P(A) / P(B)
P(B|A) = 1 so P(A|B) = P(A) / P(B)
P(B) can be interpreted as the initial probability of hypothesis B, which
may be caused by many things. Since A implies B, P(B) >= P(A) as a is a
cause of B. If any other reason for the truth of B is highly unlikely,
then P(B) will not be much greater than P(A) and thus P(A|B) will be not
much less than 1. On the other hand, if there are many other credible
hypotheses that imply B (or suggest B with high probability) then P(A|B)
will be considerably less than 1.
To sum up, if the only credible reason for all the ravens that you have
seen are black is that all ravens are black, then this hypothesis has a
high probability by Bayes' Theorem.
In fact, this hypothesis is wrong due to the existence of albino ravens!
It can be replaced with: "by far the vast majority of ravens are black".
A no longer strictly implies B, but P(B|A) is very close to 1 since albino
ravens are very rare.
Thus we have resolved this so-called paradox by using Bayes' Theorem, which
is fortunate, since inductive reasoning from special cases (experiments) to
the general case (the formulation of a natural law) is central to all
science.
Interestingly, the philosopher Hume in the eighteenth century rejected all
such inductive reasoning as invalid. He evidently had never heard of his
French near-contemporary Laplace, who actually discovered Bayes' Theorem in
its modern form.
"Buses are bosons!"
a.hunger@hulmecc.nwnet.co.uk
Subject: Re: Gaussian primes
From: HJSmith@ix.netcom.com (Harry J. Smith)
Date: Mon, 20 Jan 1997 09:24:20 GMT
On Sun, 19 Jan 1997 16:45:08 GMT, HJSmith@ix.netcom.com (Harry J.
Smith) wrote:
>If you are interested, check out my definition of Gaussian Primes at
>URL http://www.netcom.com/~hjsmith/GPrimes/GPriWhat.html
>
>Gaussian primes are the Gaussian integers that are in the following 3
>classes:
>
>(a) The 4 divisors of 2: 1 + i, 1 - i, -1 + i, and -1 - i.
>
>(b) The prime integers that cannot be factored even in Z[i]: Prime
>integer of the form |p| = 4k + 3. p = +/-3, +/-7, +/-11, +/-19, +/-23,
>etc.
>
>(c) The 8 divisors of each the odd prime integers that can be factored
>in Z[i]: m + ni, m - ni, -m + ni, -m - ni, n + mi, n - mi, -n + mi,
>and -n - mi, where m^2 + n^2 = 4k + 1 = |p| an integer prime.
>
>-Harry
Correction:
Gaussian primes are the Gaussian integers that are in the following 3
classes:
(a) The 4 non-trivial divisors of 2: 1 + i, 1 - i, -1 + i, and -1 - i.
(b) The 4 trivial divisors of each of the prime integers > 2 that
cannot be factored even in Z[i]: p, -p, pi, and -pi, where p = 4k + 3
= 3, 7, 11, 19, 23, etc.
(c) The 8 non-trivial divisors of each of the prime integers > 2 that
can be factored in Z[i]: m + ni, m - ni, -m + ni, -m - ni, n + mi,
n - mi, -n + mi, and -n - mi, where m^2 + n^2 = p = 4k + 1 = 5, 13,
17, 29, etc.
-Harry
--
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