Originally posted by 24en1KDexd29
Could you explain why you believe that?
I just thought about the case where there are 3 ants, one on either side of the center ant.
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.