| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I attended a lecture by Paul Erdos at the New York Geometry Seminar. Here are some notes: --------- For every n, there is an f(n) such that f(n) points in the plane always determine a convex n-gon. For example, f(4)=5. That is, given any 5 points in the plane, some 4 of them are the vertices of a convex quadrilateral. It is known that f(5)=9 and believed that f(6)=17. The main conjecture is that f(n)=2^(n-2)+1. It is known that n-2 2 + 1 <= f(n) <= binomial(2n-4,n-2) . --------- Same problem, but now we want to determine a convex n-gon with no other point of the set interior to it. In this case it is known that f(4)=5 and f(5)=10, but f(6) is unknown. It has been proven that f(7) does not exist. --------- Szekeres has shown that 2^n points in the plane determine some angle of measure greater than or equal to pi(1-1/n) radians. --------- If you have n points in the plane that determine the minimum number of different distances, must some 4 of the points determine 2 distances? --------- Can you have n points in the plane in general position (no 3 on a line, no 4 on a circle) such that 1 distance occurs 1 time, some other distance occurs 2 times, some other distance occurs 3 times, ..., some distance occurs n-1 times? Configurations are known for n up to and including n=7. --------- Given n points in the plane, no 3 on a line, no 4 on a circle, it is conjectured that there are more than n distances determined.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 265.1 | LATOUR::AMARTIN | Wed Apr 24 1985 19:52 | 3 | ||
Say, Stan, do you know what your Erdos number is? /AHM/THX | |||||
| 265.2 | HARE::STAN | Wed Apr 24 1985 20:33 | 3 | ||
Having never published a paper with a collaborator, my Erdos number is therefore 0. I ate dinner with Erdos, though. Does that qualify me for anything? | |||||
| 265.3 | LATOUR::AMARTIN | Thu Apr 25 1985 11:15 | 8 | ||
No, only he has a number of 0. Since you have never coauthored a paper, then your number would probably be undefined. There was a story about this in AMM(?) (an old one). It was on a discard pile. I noticed the article, and photocopied it. I will see if it is short enough to type in. /AHM | |||||
| 265.4 | AURORA::HALLYB | Thu Apr 25 1985 15:32 | 3 | ||
I have an Erdos number of 3. Anybody wanna collaborate? John | |||||
| 265.5 | LATOUR::AMARTIN | Thu May 02 1985 11:52 | 33 | ||
The American
MATHEMATICAL MONTHLY
Volume 76 Number 7
August-September 1969
MATHEMATICAL NOTES
Edited by David Drasin
AND WHAT IS YOUR ERDOS NUMBER?
Casper Goffman, Purdue University
The great mathematician Paul Erdos has written joint papers with man
mathematicians. This fact may lend some interest to the notion of
Erdos number which we are about to describe.
Let A and B be mathematicians, and let A(i), i=0, 1, ..., n, be
mathematicians with A(0)=A, A(n)=B, where A(i) has written at least
one joint paper with A(i+1), i=0, ..., n-1. Then A(0), A(1), ...,
A(n) is called a chain of length n joining A to B. The A-number of B,
nu(A;B), is the shortest length of all chains joining A to B. If
there are no chains joining A to B, then nu(A;B)=+inf. Moreover,
nu(A;A)=0. Then nu(A;B)=nu(B;A) and nu(A;B)+nu(B;C)>=nu(A;C).
For the special case A = Erdos, we obtain the function nu(Erdos; . )
whose domain is the set of all mathematicians.
I was told several years ago that my Erdos number was 7. It has
recently been lowered to 3. Last year I saw Erdos in London and was
surprised to learn that he did not know that the function nu(Erdos; . )
was being considered. When I told him the good news that my Erdos
number had just been lowered, he expressed regret that he had to leave
London the same day. Otherwise, an ultimate lowering might have been
accomplished.
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