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Rainer Rosenthal science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 114

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: A day of a CAS superhero



"Alec Mihailovs" wrote
Quote: 
That is a good demonstration of one specific
weakness of automated bug testing. Most of the
bugs that were found that way are such bugs
that nobody meets in real calculations and
nobody cares about (except automated bug
testing system creator).

.... and here we enter the recreational part!
There is an old joke of mine, which I am quite
proud of:
Two programs meet.
Says the one:
"Do you still believe in programmers?"
Best regards,
Rainer Rosenthal
r.rosenthal@web.de 

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mensanator@aol.compost science forum Guru
Joined: 24 Mar 2005
Posts: 826

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: JSH: Big collapse on surrogate factoring



fiziwig wrote:
Quote:  Your example is trivial. Solving for one unknown certainly does not
need to iterate with all posible values of a.

But I din't know whether _you_ knew it.
Quote:  Nor did I make such a preposterous claim.

But you made the preposterous claim that you can't seperate
p and q without knowing one of them in advance.
Quote: 
Finding p and q, factors of N, is not so trival. As for "that I know
of" I invite you to supply a counter example of a factoring method
that
does not involve either O(p) discrete calculations or repeatedly
computing some intermediate value and performing some test on that
value.

Absence of evidence is not evidence of absence.
Quote:  In fact, no such factoring method exists.

That you know of.
Quote:  If it did factorization would be trivial.

I believe I already said that.
Quote:  I can make this claim simply because
any noniterative factoring method, if it existed, would compute the
factors in a single step. Since no singlestep factoring method
exists, they are all, by defintion, iterative. (You may double check
this assertion by enumerating the factorization methods listed in
"Prime Numbers and Computer Methods for Factorization" by Hans
Riesel.On page 221 he discusses possible lines of research that might
someday lead to noniterative factoring methods, but no such method
exists as yet.)

I reiterate, show me a proof that a noniterative factoring
method CANNOT exist. Obviously Hans Riesel is not in posession
of such a proof or he wouldn't be proposing lines of research.
When you claim surrogate factoring is IMPOSSIBLE, you must have
such a proof or your claim is simply bogus.
Quote: 
What makes one factoring method better than another is the size of
the
search space. Trial division uses the search space of all primes
sqrt(N) while Pollard's rho, for example, has a search space no
larger
than O(sqrt(p)) where p is the smallest prime factor of N. (Reisel,
pp223) 


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fiziwig science forum beginner
Joined: 03 Apr 2005
Posts: 36

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: JSH: Big collapse on surrogate factoring



Your example is trivial. Solving for one unknown certainly does not
need to iterate with all posible values of a. Nor did I make such a
preposterous claim.
Finding p and q, factors of N, is not so trival. As for "that I know
of" I invite you to supply a counter example of a factoring method that
does not involve either O(p) discrete calculations or repeatedly
computing some intermediate value and performing some test on that
value. In fact, no such factoring method exists. If it did
factorization would be trivial. I can make this claim simply because
any noniterative factoring method, if it existed, would compute the
factors in a single step. Since no singlestep factoring method
exists, they are all, by defintion, iterative. (You may double check
this assertion by enumerating the factorization methods listed in
"Prime Numbers and Computer Methods for Factorization" by Hans
Riesel.On page 221 he discusses possible lines of research that might
someday lead to noniterative factoring methods, but no such method
exists as yet.)
What makes one factoring method better than another is the size of the
search space. Trial division uses the search space of all primes <
sqrt(N) while Pollard's rho, for example, has a search space no larger
than O(sqrt(p)) where p is the smallest prime factor of N. (Reisel,
pp223) 

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mensanator@aol.compost science forum Guru
Joined: 24 Mar 2005
Posts: 826

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: JSH: Big collapse on surrogate factoring



fiziwig wrote:
Quote:  Why just suppose you had qr?

Simply an example. Having pr would do just as well.
Quote:  Why not suppose you had q?

Because you assumed you don't have either p or q. Duh.
Quote:  Why not suppose you had a million dollars?
Why not suppose anything you want?
That's not the point.

Yes, it is the point. If there was some easy way to find qr,
then the factoring problem becomes trivial. Isn't that kind of
what JSH is proposing, that there's some magical way he can
get the answer without doing the work? Has it been proved that
no such process exists such as the case with circle squaring?
Quote:  The point is you can't separate p from q
algebraically. That does not say that you can't invent some other
_iteration_ that searches for p, or for q or for some qr.

Why do you assume it _has_ to be an iterative search?
Quote:  But reagrdless, you have to _search_ for some number that
fullfills those conditions.

So if I wanted to find the first interger value of 'a' such
that the following rational number is also an integer
604462909807314587353088*a  890771586760950746471897

79766443076872509863361
I'd have to iterate through all integers from 1 to
79766443076872509863361 until I found one that fullfills
the conditions?
Quote:  You can't simply compute that number directly.

In this case, I can. a = 56303391467108934933739 and the rational
expression reduces to 426662021339354613506535. Feel free to
verify it. How long do you think an iterative search would take?
Quote:  In fact every workable factoring scheme that I know of

"That I know of" isn't proof.
Quote:  either _searches_ for p,
or some pr, or some pair (a,b) such that gcd(N,ab) gives the factor.
The only thing that makes one factoring algorithm different from any
other is the nature of the search and the manner in which the size of
the search space is reduced.
The only possible exception

That you know of?
Quote:  would be algoprithms that generate very
large numbers in the hope of collecting some factor if N. But even to
do that requires iterating through all the possible divisors
_searching_ for the magic one that finally reveals p or q. 


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fiziwig science forum beginner
Joined: 03 Apr 2005
Posts: 36

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: JSH: Big collapse on surrogate factoring



Why just suppose you had qr? Why not suppose you had q? Why not
suppose you had a million dollars? Why not suppose anything you want?
That's not the point. The point is you can't separate p from q
algebraically. That does not say that you can't invent some other
_iteration_ that searches for p, or for q or for some qr. But
reagrdless, you have to _search_ for some number that fullfills those
conditions. You can't simply compute that number directly. In fact
every workable factoring scheme that I know of either _searches_ for p,
or some pr, or some pair (a,b) such that gcd(N,ab) gives the factor.
The only thing that makes one factoring algorithm different from any
other is the nature of the search and the manner in which the size of
the search space is reduced.
The only possible exception would be algoprithms that generate very
large numbers in the hope of collecting some factor if N. But even to
do that requires iterating through all the possible divisors
_searching_ for the magic one that finally reveals p or q. 

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mensanator@aol.compost science forum Guru
Joined: 24 Mar 2005
Posts: 826

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: JSH: Big collapse on surrogate factoring



fiziwig wrote:
Quote:  This is not really a "proof", but it seems like a reasonable claim
just
the same.
Let M be written pq. Is there any function not involving either p or
q
explicitly that will yeild a term that has either p or q separate
from
the other? In fact, no algebraic operation can pry p away from q
without explicitly using p or q. In other words, no surrogate
factoring method can work without explicitly knowing p or q in
advance.
kpq; multiplication does not separate p from q.
pq/k; division does not separate p from q.
pq+k; addition does not separate p from q.
pqk; subtraction does not separate p from q.
(pq)^k = p^k * q^k; exponentiation does not separate p from q.
X mod (pq) does not separate p from q.
pq mod X where X does not already explicity separate p and q does not
separate p from q.
No combination of the above operations will separate p from q. Thus
p
cannot be separated from q algebraicly. Thus surrogate factoring is
not possible.
Disclaimer: I'm just an amateur, so there may be holes in my
reasoning.
If so, feel free to give a counterexample.

Suppose you had another number qr (and you don't know what
r is either). Wouldn't GCD(pq,qr) return q?


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Robin Chapman science forum Guru Wannabe
Joined: 25 Mar 2005
Posts: 254

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Is anything less valued that the Turkish lira?



Opinicus wrote:
Quote:  matt271829news@yahoo.co.uk> wrote
I was playing around with a currency exchange rate
calculator and
punched in $1 US to see how many Turkish Lira I would
get.
1,347,800 !!!
Over a million lira for one buck?
It reached a high of 1,850,000 or so at one point...

I found a 500,000 lira note lying in the road the other day.
Thought I might get a few quid for it, but checking exchange rate,
it was worth less than 20p :(

Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_ 

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Opinicus science forum beginner
Joined: 24 Mar 2005
Posts: 2

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Is anything less valued that the Turkish lira?



<matt271829news@yahoo.co.uk> wrote
Quote:  I was playing around with a currency exchange rate
calculator and
punched in $1 US to see how many Turkish Lira I would
get.
1,347,800 !!!
Over a million lira for one buck?

It reached a high of 1,850,000 or so at one point... ;)
Quote:  I don't think this is true any more. Turkey redenominated
its currency
on Jan 1st 2005, and lopped off six zeros. The "New
Turkish Lira" is
worth one million of the old ones.

It's not true any more. And people are having a hard time
getting used to it.

Bob
Kanyak's Doghouse
http://www.kanyak.com 

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Mommio2 science forum beginner
Joined: 24 Mar 2005
Posts: 4

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Mathcounts Question 2



THANKS, everyone! (And sorry my picture didn't come through so well!) You
have really helped!
Mommio (who wants to WIN this year!)
"Mommio2" <mommio2@insightbb.com> wrote in message
news:HhjKd.23118$ox3.3730@attbi_s04...
Quote:  Hello again,
Here's the next question. What I really need to know is if there is a
pattern or method to solve this without just counting.
"How many rectangles are in the array below?"
The answer is 36.
Thanks!



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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Given a plane convex closed curve



Bob Pease wrote:
Quote:  astanoff@yahoo.fr> wrote in message
news:1111663342.606914.272940@f14g2000cwb.googlegroups.com...
Thank you for clearing what was confused :
"circumference/greatest diameter <= pi "
that was exactly what i meant but could not state
correctly maybe due to my poor english !
v.a.
My first approach would be to make it the class of all curves
which could be inscribed in a unit circle.
This seems to be wlog.
Then show that by calculus that L( r (theta) ) 0,2PI is
minimized iff r = a constant .

I think you mean maximized.
Quote:  It seems to be a difficult problem to me.

I also thought about using a calculus approach, but the problem becomes
how to put in the convexity requirement. Certainly the result is not
true if the curve is not convex. (In fact, the ratio can even be
infinite if the curve's not convex; i.e., a Koch snowflake.)
 Christopher Heckman 

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Larry Hammick science forum Guru Wannabe
Joined: 24 Mar 2005
Posts: 112

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Does .999999999..... equal 1?



"Oppie"
Quote:  I almost hate to ask this, but I need to settle an argument between two
of my friends. Here's the debate:
Danny says that .999999999.... EQUALS 1.
It does. This thing: 
..9999....
(with trailing dots to indicate that infinitely more terms follow)
is merely notation for the _limit_ of the sequence
..9, .99, .999, .9999, ....
and lots of different sequences can have the same limit; e.g.
1.1 , 1.01 , 1.001 , 1.0001 , ...
has limit 1. 

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Bob Harris science forum beginner
Joined: 24 Mar 2005
Posts: 5

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Does 8.999999999..... equal 9?



Rich Holmes wrote:
Quote:  Bob Harris <plasticnitlion@wrappermindspring.com> writes:
Rich Holmes wrote:
... can you prove the assertion that 9.999...  0.999... = 9? Or is it
8.999...?
Yes
Perhaps you misunderstood my point. ...

(chuckle) Nope, but you misunderstood mine. I was answering an either/or
question with a "yes". Was I trying to make an intelligent contribution to
the discussion? Or was I just trying to be a smart ass?
Bob H 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Does 8.999999999..... equal 9?



In article <u4acqmg42d.fsf@mep1.phy.syr.edu>,
Rich Holmes<rsholmes+usenet@mailbox.syr.edu> wrote:
Quote:  Virgil <ITSnetNOTcom#virgil@COMCAST.com> writes:
The usual interpretation of a nonterminating decimal is the value to
which the corresponding infinite series converges, for example
0.333... = lim_{n > oo} SUM_{k = 1..n} 3/10^k, which limit is 1/3.
Exactly. In which case proving that 9.999...  0.999... = 9 entails
showing that 0.999... = lim_{n > oo} SUM_{k = 1..n} 9/10^k = 1 (and
similarly for 9.999... = 10)  but since this is what the "proof" in
question purports to do, that proof then fails due to circularity.
Again, the assertions are correct, but the proposed proof doesn't
stand up to scrutiny.

I do not see your objection.
If there is to be a nonstandard iterpretation, it is the interpreter's
obligation to make his interpretation clear. Otherwise, the analysis
holds as given.
If you want a more high powered proof, we can use the well known Cauchy
sequaence model, one can represent each real with any representative of
the equivalence class of Cauchy sequences of rationals converging to it,
two such sequences being equivalent if their difference is a sequence
converging to zero in the rationals.
Since the difference between sequence A_n = SUM_{k = 1..n} 9/10^k and
sequence B_n = 1 is C_n = B_n  A_n = 1/10^K, which clearly converges to
zero, it follows that A_n and B_n represent the same real number in the
'Cauchy' model of a complete ordered Archimedian (real) field.
And since it has been proven that Complete ordered Archimedean fields
are ispmorphic....
But this approach is overkill. 

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Rich Holme science forum beginner
Joined: 06 Jun 2005
Posts: 45

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Does 8.999999999..... equal 9?



Virgil <ITSnetNOTcom#virgil@COMCAST.com> writes:
Quote:  The usual interpretation of a nonterminating decimal is the value to
which the corresponding infinite series converges, for example
0.333... = lim_{n > oo} SUM_{k = 1..n} 3/10^k, which limit is 1/3.

Exactly. In which case proving that 9.999...  0.999... = 9 entails
showing that 0.999... = lim_{n > oo} SUM_{k = 1..n} 9/10^k = 1 (and
similarly for 9.999... = 10)  but since this is what the "proof" in
question purports to do, that proof then fails due to circularity.
Again, the assertions are correct, but the proposed proof doesn't
stand up to scrutiny. 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Thu Mar 24, 2005 7:53 pm Post subject:
Re: Does 8.999999999..... equal 9?



In article <u4r7jzf0sc.fsf@mep1.phy.syr.edu>,
Rich Holmes<rsholmes+usenet@mailbox.syr.edu> wrote:
Quote:  Bob Harris <plasticnitlion@wrappermindspring.com> writes:
Rich Holmes wrote:
... can you prove the assertion that 9.999...  0.999... = 9? Or is it
8.999...?
Yes
Perhaps you misunderstood my point. The suggested proof makes use of
decimal arithmetic (to infinite precision) on infinitely repeating
decimal fractions. Conventional arithmetic procedures on decimal
numbers are algorithms that work on one decimal place at a time; they
will never terminate, meaning they will give no final result, when
applied to infinitely repeating fractions. In order for the proof to
work, one needs to supply proof, in some form other than an arithmetic
algorithm, that 9.999...  0.999... = 9. Of course this is a true
statement (as is the statement 9.999...  0.999... = 8.999...), but
one cannot simply assert it to be true.

If you will tell us what YOU mean by 9.999... and 0.999..., etc., we can
tell you whether 9.999...  0.999... = 9.
The usual interpretation of a nonterminating decimal is the value to
which the corresponding infinite series converges, for example
0.333... = lim_{n > oo} SUM_{k = 1..n} 3/10^k, which limit is 1/3.
Given the usual interpretations for 9.999... and 0.999..., they
represent the same numbers as 10 and 1, respectively, so their
difference is 9.
If one does not choose use the usual interpretation, one should explain
in some detail what interpretation one is using. 

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