Re: Digest for competition-corner-participant-discussion@googlegroups. com - 5 updates in 1 topic

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Lin Gold

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Oct 10, 2021, 2:47:36 PM10/10/21
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OK, I've regrettably asked for "No email" status from this group.   Although I was willing to
contribute some money to George's project, like George, I've had to change my mind when
someone thinks it takes $500,000 or so per year to do the project and nobody thinks
differently.   As for the math problems, I don't do them anymore (maybe some AI discussion
would be interesting, but that's probably another group.)
 
Let's see if my "No email" status works.
 
Lin


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From: competition-corner-p...@googlegroups.com
To: Digest recipients <competition-corner-p...@googlegroups.com>
Subject: Digest for competition-corner-p...@googlegroups.com - 5 updates in 1 topic
Date: Sun, 10 Oct 2021 06:54:14 +0000

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Walter Effross <eff...@wcl.american.edu>: Oct 09 11:21AM

I'm still getting the messages.
Several times, I've tried to unsubscribe by sending an e-mail to the address listed for that; each time, I get back a message saying that it's invalid because I'm not a member.
Best regards,
Walter
________________________________
From: 'David Ash' via Competition Corner Participant Discussion <competition-corner-p...@googlegroups.com>
Sent: Friday, October 8, 2021 10:53 PM
To: Competition Corner Participant Discussion <competition-corner-p...@googlegroups.com>
Subject: Re: finite field problem
 
It might be good if Walter verified this--or at least if he verifies if it did not work since he wouldn't see this if it did! I'm still seeing the total number of members as 116 which is the number it has been for a little while now.
 
Also if you click on 'My membership settings' you can adjust the amount of email you get from GG. If someone wants to occasionally check in on how this discussion is going but wants to minimize spam, they can adjust the settings to get less or no email when people make postings.
 
On Friday, October 8, 2021 at 6:23:26 PM UTC-7 igl...@gmail.com wrote:
I do think it worked. You are not in the member list any more.
 
On Friday, October 8, 2021 at 4:38:00 PM UTC-5 Walter Effross wrote:
Could someone please unsubscribe me?
I've tried the instructions at the bottom of these e-mails, but I'm still in the loop.
I'm still happy to talk about the project, going forward.
Thank you!
Walter
________________________________
From: 'David Ash' via Competition Corner Participant Discussion <competition-corner-p...@googlegroups.com>
Sent: Friday, October 8, 2021 5:29 PM
To: Competition Corner Participant Discussion <competition-corner-p...@googlegroups.com>
Subject: Re: finite field problem
 
There are, I believe, many possibilities for p(x) that will work, but some are easier to prove than others. Yes, x^10+x^3+1 is irreducible over GF(2) and proving that is one important step. It does then follow that 1, x, x^2, ..., x^9 form a basis but unfortunately this basis is not in the form that we are asked to find a basis. If p(x)=x^7 works, we'd have to somehow prove that x^7, x^14, x^28, x^56, ..., x^3584 form a basis. I'm not sure that you've proved that. It may very well be true--I haven't worked out the details but do know that there are a significant number of polynomials p(x) that work--there is not just one unique solution. However I believe there may be other polynomials p(x) which work and which require much less rote arithmetic (computation) to prove.
 
On Friday, October 8, 2021 at 1:33:11 PM UTC-7 t...@bellefleurbooks.com wrote:
Dear David:
 
My inclination was wrong. Step 1 was a waste of effort. When you skip
step 1, step 2 amounts to computing p(x) assuming that 1, x, x^2, . . .
, x^9 form a basis. There are multiple possibilities for p here, the
simplest being p(x) = x^7. Nevertheless, you still have to prove that
1, x, x^2, . . . , x^9 form a basis (step 3) which is equivalent to
proving that x^10+x^3+1 cannot be factored.
 
You have much the same issue (step 3) in your approach. You state that
x^9+x^7+x^6+x^3+1 is not 1, but it could be equal to 1 if
x^9+x^7+x^6+x^3 = 0, i.e., if the two polynomials, x^9+x^7+x^6+x^3 and
x^10+x^3+1 shared a common factor. In fact, they do not, but you have
to prove as much.
 
Proving that x^10+x^3+1 cannot be factored, or something like this, is
something to look for in the proofs submitted. It is a good problem.
 
Sincerely,
 
Tom
 
On 2021-10-08 12:28, 'David Ash' via Competition Corner Participant
Discussion wrote:
 
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Istvan Lauko <igl...@gmail.com>: Oct 09 10:02AM -0700

I did communicate with Walter. He is still a member, but now he does not
receive any emails.
I wander, if we should do this as a default for all, who are not active
here (would hate to generate unwanted emails for anyone). Unfortunately
there is no option for weekly or monthly "digest", only daily.
 
On Saturday, October 9, 2021 at 6:21:16 AM UTC-5 Walter Effross wrote:
 
Istvan Lauko <igl...@gmail.com>: Oct 09 10:05AM -0700

I mean I wonder.... (a case in point)
 
On Saturday, October 9, 2021 at 12:02:01 PM UTC-5 Istvan Lauko wrote:
 
t...@bellefleurbooks.com: Oct 09 05:02PM -0500

Dear David:
 
I proved that p(x) can equal x^7, but I did not provide the proof. You
know how to do it.
 
By the way, every solution, p(x), has an x^7 term. I can prove this too
but have not provided the proof.
 
The original problem is a good problem--a little more challenging than
the Sierpinski tower.
 
Sincerely,
 
Tom
 
On 2021-10-08 16:29, 'David Ash' via Competition Corner Participant
Discussion wrote:
David Ash <david_...@yahoo.com>: Oct 09 04:31PM -0700

Tom,
 
Yes--it appears to be true that every solution must have a (nonzero) x^7
term. To see this, first observe that it is fairly easy to show that for
any element t in GF(2^10), t^1024=t. So if we repeatedly square any element
of the field, we eventually get back to the original element. Note that the
basis we are asked to construct basically involves taking an element and
then squaring it nine times in succession to produce the other nine
elements of the basis. Next we look at the squares of each of 1, x, x^2,
..., x^9:
 
square of 1 is 1
square of x is x^2
square of x^2 is x^4
square of x^3 is x^6
square of x^4 is x^8
square of x^5 is x^3+1
square of x^6 is x^5+x^2
square of x^7 is x^7+x^4
square of x^8 is x^9+x^6
square of x^9 is x^8+x^4+x
 
Note that the operation of squaring never introduces an x^7 term unless we
already have an x^7 term to start with. Since a basis, by definition, must
span the entire space, the only way a basis generated by squaring can span
x^7 terms is if the initial element includes an x^7 term.
 
This might be a good alternative problem to the original problem
BTW--instead of asking people to give an example of a p(x) that works
instead ask them to prove that p(x) must include an x^7 term.
 
On Saturday, October 9, 2021 at 3:02:37 PM UTC-7 t...@bellefleurbooks.com
wrote:
 
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David Ash

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Oct 10, 2021, 5:44:47 PM10/10/21
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Hi Lin,

It's rather strange, given the stated focus of this group, that as soon as we actually start talking about real math problems, people start losing interest.

These days AI is of interest to me only to the extent that it is rooted in rigorous mathematics. I find that other branches of computer science have done a far better job than AI in rooting their disciplines on a solid mathematical foundation. An example of this was when I worked in finance. The various options pricing models, like Black-Scholes and proprietary variants thereof that we developed, were rigorous and ready for the cutthroat environment of the trading floor. The AI models that we looked at, by comparison, were smoke and mirrors and not ready for prime time.

AI does raise some fascinating ethical and philosophical questions but, yes, those probably are best discussed elsewhere.

With regard to the costs, I also was surprised that the costs suggested by István were as high as they were--and I actually think his numbers added up to closer to $1M, not $500K, per year. However I don't have enough information to propose an alternative budget. Before István provided his estimates on here, I had been working from George's estimate of $50K in startup costs from page xlvi of The Book in the Aftermath. Granted, that figure was from 1980, but adjusted for inflation that is $166K in 2021. That's still quite a bit lower than $500K or $1M per year. I'm not ready to opine on which way we should go, but if there are ways of doing this on more of a shoestring budget, we should explore all alternatives.

It is also a bit unclear as to what capacity the former CC participants are being considered for participation in the journal. Would we be being paid for our participation (I noted that István's numbers include a number of paid roles); would we be paying for our participation (i.e. we would donate to fund this effort); or would we be volunteers, contributing mathematically but neither giving nor receiving money?

I personally might be willing to donate to the effort, but the amount of funds I could donate would be in the low five figures. So I can't do $1M or $500K or $166K on my own. Even $50K would be a stretch for me. However if I were donating I'd want to be involved mathematically as well. I don't need to be paid but I do need to be involved mathematically to consider donating.

I think we are going to need a Zoom (or many) to nail this stuff down, but I also think it needs to wait until István and Gabi are ready, and I'm sure they are really busy, like probably everyone else. So I can be as patient as needed. However a lot of people--like Walter, Lin, and perhaps myself--seem to be adopting a bit of a wait and see approach. There is interest but there isn't enough detail yet to commit. I know we are waiting in part in hope that we will generate more interest, but my sense is there won't be a lot more interest or commitment until we formulate the structure a bit more. I already ran the idea by one person I know. His strong suggestion is that we need buy-in from multiple strong universities throughout the US--Wisconsin is a good start but it won't get us all the way to where we want to be. It is tough to ask people for help, though, until we know what we are asking them for.
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