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




Re: The Unsolved Problems web site
Posted:
May 18, 2013 5:35 AM


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 dk1 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



