... spread out along a one-metre line. You don't know how many but there is at least one. At time t=0 they all start moving along the line at 1 m/min. Some move in one direction, some in the other. When 2 ants meet they instantaneously turn and move in the opposite direction at 1 m/min. When an ant gets to the end of the line it also instantaneously turns and moves off at 1 m/min. There is an ant an the centre point of the line at t=0.
(a) Prove that there must be an ant at the centre point at t=1 min.
(b) what are the necessary and sufficient conditions for this ant to be the same one that was at the centre at t=0.
Originally posted by 24en1KDexd29I'm not much for proofs, but this is my thought process:
... spread out along a one-metre line. You don't know how many but there is at least one. At time t=0 they all start moving along the line at 1 m/min. Some move in one direction, some in the other. When 2 ants meet they instantaneously turn and move in the opposite direction at 1 m/min. When an ant gets to the end of the line it also instantaneously tur ...[text shortened]... essary and sufficient conditions for this ant to be the same one that was at the centre at t=0.
For a single ant, starting at x=0.5 and moving in either direction for one minute, the ant would reach of of the ends (x=0 or x=1) at t=30s, turn around, and reach x=0.5 at 1 minute.
If this ant meets another ant, that ant will take the place of the original ant, but still be on the exact same trajectory as the original ant had they not collided. Therefore, this replacement ant would now be on pace to reach x=0.5 at t=1m.
Any additional collisions will cause the new ant to be on pace to reach x=0.5 at t=1m.
Originally posted by forkedknight"I'm not much for proofs ..." , "Thinking very briefly ..."
Thinking very briefly, I believe the necessary and sufficient conditions for the original ant to be at the origin at 1 minute is an equal number of ants on either side of him at t=0.
They should be able to be heading in either direction.
I think you are being overly modest Mr Knight, or may I call you Forked. Either that or you have seen the problem, and its solution, before.
Originally posted by 24en1KDexd29I just thought about the case where there are 3 ants, one on either side of the center ant.
Could you explain why you believe that?
In this case, it is guarateed that the original ant (A0) will collide with the ant in whichever direction he is heading (A1), and the the distance he has traveled at this collision point is < 0.5m. Therefore, he will turn around and head back across the center point.
The ant on the opposite side (A2) must cross the center point, therefore, he will also collide with A0 after it turns around.
If this second collision (A0:A2) happens after A0 crosses the center point, then A0 will turn around, and be at the center at t=1m.
If the (A0:A2) collision happens before A0 crosses the center point, then there must be a second collision between A0 and A1, in which case A0 must turn around an be at the center point at t=1m
That's as far ahead as I thought, but it seemed like symmetrically adding ants wouldn't change the outcome.
*edit* Looks like if A1 and A2 and fairly near A0 and heading toward the center (at t=0), then there can be a few more collisions, but I think it still works out. I have only really thought about the common cases and didn't take the time to draw it out, which is why I wasn't sure If I was leaving out any corner cases to begin with.
My late night thoughts...
a)
You are better off ignoring the turning round when they meet condition. Imagine ghost ants that pass through each other, in effect this is the same as the two ants turning round. The ant in the middle then walks to an end turns round and back to the middle.
b)
From a similar argument to a) after each minute the arrangement of ants will reverse. So the ant in the middle will be the same ant if it has equal numbers of ants to the left and the right at the start