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Re: The Unsolved Problems web site
Posted:
May 19, 2013 1:35 AM


On May 19, 1:12 pm, jdawe <mrjd...@gmail.com> wrote: > On May 18, 7:35 pm, Graham Cooper <grahamcoop...@gmail.com> wrote: > > > > > > > > > > > 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 > > Using simple ActionReaction 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. >
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/220pxLonely_runner.gif
Great Race Graphic with different speed runners.
Herc

IS DOG+1 a NUMBER ?
http://blockprolog.com/natsdog.png



