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Topic: The Unsolved Problems web site
Replies: 4   Last Post: May 19, 2013 1:35 AM

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Graham Cooper

Posts: 4,321
Registered: 5/20/10
Re: The Unsolved Problems web site
Posted: May 18, 2013 5:35 AM
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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






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