| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 629.2 | Apparent improvement on R(4) | SQM::HALLYB | Are all the good ones taken? | Fri Dec 12 1986 18:00 | 13 |
| > R(4) = SQRT(2) = 1.414... square - is this minimal?
Correct me if I misunderstand, but wouldn't an equilateral triangle
plus center point have a ratio of 1/SQRT(1/4 + 1/3) = 1.3093...
Suppose we have an equilateral triangle based at (0,0), (1,0) and
(1/2, SQRT(3)). Draw the three medians which are also the three
interior bisectors and also perpendicular to the opposite sides.
The three lines intersect at one point (old theorem from 10th grade),
and the coordinates of that point can be seen to be (1/2, tan(pi/6)),
so the distance from there to the origin is SQRT(1/4 + 1/3), since
tan(pi/6) = sqrt(1/3). The max distance is 1 by construction, hence
the result.
|
| 629.3 | understanding is OK, numbers are off | SSDEVO::LARY | | Fri Dec 12 1986 18:21 | 13 |
| > Suppose we have an equilateral triangle based at (0,0), (1,0) and
> (1/2, SQRT(3)). Draw the three medians which are also the three
> interior bisectors and also perpendicular to the opposite sides.
> The three lines intersect at one point (old theorem from 10th grade),
> and the coordinates of that point can be seen to be (1/2, tan(pi/6)),
> so the distance from there to the origin is SQRT(1/4 + 1/3), since
> tan(pi/6) = sqrt(1/3). The max distance is 1 by construction, hence
> the result.
The top point of the triangle would be at (1/2, SQRT(3)/2); the intersection
point is at (1/2, tan(pi/6)/2) so the distance is SQRT(1/4+1/12) = SQRT(1/3)
and the ratio is SQRT(3) > SQRT(2).
|
| 629.4 | an experiment - no results | ENGINE::ROTH | | Mon Dec 15 1986 08:46 | 19 |
| I tried an empiricist approach of running a small program that
drew 6 random points and then each iteration, nudged the furthest
ones together and the closest apart by a small random amount
Each iteration I plotted the network of lines after centering the
figure and normalizing the longest line to a length of one.
Most converged to a pentagon, but a few went to a figure which only
had symmetry about a line, but none attained the lower bound in the
base note. It is hard to believe.
The program is nothing to write home about, its very inefficient, but
it was interesting that it converged to some plausable minima.
I have some so called 'stochastic gradient' optimization procedures
that I once used for circuit optimization, and may try that on the
problem, just to see if it works any better...
- Jim
|
| 629.5 | | CLT::GILBERT | eager like a child | Mon Dec 15 1986 13:38 | 18 |
| I tried the following approach.
Draw two points at distance sqrt(3), and draw circles of radii 1
and sqrt(3) centered at these points. Now the circles of radius
sqrt(3) form a lozenge in which the other four points must lie,
and the four additional points may not lie within the circles of
radius 1. Thus, there are two 'diamond-shaped' areas that must
contain the four remaining points.
A little playing should convince you that three points (separated
by a distance of at least 1) will not fit in one of diamond-shaped
regions. Hence, the four additional points must be distributed
two per diamond. However these four points are placed, two of them
(from opposite sides) will be separated by more than sqrt(3).
Thus, the answer of sqrt(3) appears to be wrong. Could someone
verify or duplicate this approach?
- Gilbert
|