Re: The Unsolved Problems web site

[Graham Cooper]

On May 16, 2:26 pm, timro21 wrote:
> The Unsolved Problems web site at
>
> http://unsolvedproblems.org/
>
> has recently been updated.
>
> There are prizes of US$500 for many of the problems listed.
>
> Tim

Hi Tim,

I think I solved one!

http://unsolvedproblems.org/index_files/LonelyRunner.htm

Lonely Runner Conjecture


Suppose there are k runners, all lined up at the start of a circular
running track of length 1. They all start running at constant, but
different, speeds.

The Lonely Runner conjecture states that for each runner, there will
come a time when he or she will be a distance of at least 1/k from
every other runner.

The conjecture has been proved for small values of k (<=7).

The problem is to prove or disprove the conjecture for the general
case, or for cases where k > 7.




SOLUTION
--------

Express the k runners speeds as rational fractions.

n1 n2 .. nk
-- -- .. --
d1 d2 .. dk

e.g. runner 1 runs n1 laps in d1 seconds

Where all denominators are odd
and all but one numerator are even.

Let n1 be the only odd numerator.

e.g. for k=4 runners

ODD EVEN EVEN EVEN
--- ---- ---- ----
ODD ODD ODD ODD

This still allows arbitrarily high precision
to represent every runner's speed accurately.

After t = d1 X d2 X ... X dk seconds
All runners are back at the starting line.

PROOF: at time t each of the runners runs their
time for ni laps (di) multiplied by all other
denominators which is a whole factor.

The number of laps finished by each runner after
time t can be calculated by the following equations.

laps1 = n1 X d2 X d3 X ... X dk
laps2 = n2 X d1 X d3 X ... X dk
laps3 = n3 X d1 X d2 X ... X dk
..
lapsr = nr X d1 X d2 X ... X dk-1 laps

WHERE

laps1 is ODD
laps2 is EVEN
laps3 is EVEN
..
lapsk is EVEN

PROOF: all factors in the equation for laps1 are odd.
1 factor in each of the other laps equations is even.

When the race is at time t/2

At time t/2, r1 has run laps1/2 laps.
At time t/2, r2 has run laps2/2 laps.
At time t/2, r3 has run laps3/2 laps.
...
At time t/2, rk has run lapsk/2 laps.

laps1/2 has remainder 1/2
laps2/2 is a whole number
laps3/2 is a whole number
..
lapsk/2 is a whole number

This puts runner 1 half way around the track at time t/2
with every other runner back at the starting position.

From this it follows that runner 1 is more than
the track length / k distance from all other runners.


G. Cooper
BINFTECH UQ

[jdawe]

Using simple Action-Reaction logic we can prove that there can be no
lonely runner in the race.

The umpire of the race who stands in the dead centre of the circular
track is the only lonely one in that race.

Alone - Crowded

Singular - Plural

Centre - Perimeter

Because there is only one spot (singular) in the dead centre of a
circle, it is the only location where someone can be all alone. There
is not enough room for 2 (plural) things.

Official - Competitors

Alone - Crowded

Absolute - Relative

Zero - Value

Lonely is always absolute fixed that equals 0. Implying you have zero
people keeping you company.

Relative to the Absolute lonely centre is a degree of crowding present
on the perimeter that goes around that centre.

So, if you are to umpire a running race. The first thing you do is
walk into the middle of a field where the race will be held and stand
in the dead centre. Then you draw a circle around yourself on the
ground so that only you can fit inside. You will be all alone during
the race, a neutral observer.

Neutral - Engaged

Then you draw another circle around your neutral centre which will be
active and crowded with mulitple competitors.

Absolute - Relative

The degree of crowding around the outside is a relative value to the
absolute isolation and loneliness of the neutral centre.

-----------------------

Answer - Question

So, next time you have a question about our universe such as:

"Does the moon of our Earth spin on its own axis?"

To solution is easy.

Solution - Problem

Always look to the centre to solve any problem.

You must be a neutral observer in the dead centre of the moon to
answer without prejudice whether or not the moon actually spins.

Impartial - Biased

Agree - Disagree

We can only agree with an impartial observer. We always disagree with
biased relative opinion.

The moon doesn't spin relative to the fixed stars. The stars (plural)
is biased and should be ignored. We can only regard the view of a
'singular' impartial point of reference that is only in the dead
centre of the moon. The umpire of the moon.

[Graham Cooper]

HAHA! T.L.R. Conjecture wasn't much of a hard problem, took me 1/2
hour!! Surprised it was around since 1967.

Lonely Runner Conjecture

Suppose there are k runners, all lined up at the start
of a circular running track of length 1.  They all start
running at constant, but different, speeds.

k different frequencies must must have a common beat frequency.

The hard part was getting your runner to run an odd number of laps!
No prime coefficients needed though, just state the runner's speed you
want to isolate as

ODD Laps
---
ODD Seconds

E.g. 5 laps in 33 seconds.

Then multiply the denominators for everyone to do a full lap.

Divide by 2 and he's the only one half way across the Track!


http://en.wikipedia.org/wiki/Lonely_runner_conjecture

http://upload.wikimedia.org/wikipedia/commons/thumb/3/33/Lonely_runner.gif/220px-Lonely_runner.gif

Great Race Graphic with different speed runners.



Herc

--

IS DOG+1 a NUMBER ?

http://blockprolog.com/nat-s-dog.png