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$N$ ants are situated at $N$ different points on the circumference of a circle of circumference length $1$ cm. At $t=0$ each of them start moving at speed $1$ cm/min in either clockwise or anticlockwise direction. Suppose ant Bob is at point $A$ at $t=0$. Further suppose that ants, when they meet, change direction but continue to travel at same speed. Find the condition for which at $t=1$ the ant at point $A$ is Bob.

I am able to prove that there will be an ant at $A$ at time $t=1$ by the following argument. Suppose Bob carries a flag and passes the flag to the next ant it meets, who in turn passes the flag again to the next ant it meets. In this way the flag travels at $1$ cm/sec in either clockwise or anticlockwise direction and will be at point $A$ at $t=1$. The last flag bearer will be the ant at $A$ at $t=1$. But i cannot find the condition for which that ant will be Bob. Please help.

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    https://www.reddit.com/r/mathriddles/comments/l3ieni/ants_in_a_circle/?rdt=33902 – Intelligenti pauca Aug 13 '24 at 20:05
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    Well, one condition is that there is only one ant. Alternatively all ants move in the same direction. Or there are two ants starting at antipodes and heading in opposite directions. Or four ants equal distance apart pair wise heading toward each other. Seem there's gazillions of ways to do it. It's hard to know what specifically they want in an answer. – fleablood Aug 13 '24 at 22:42
  • Not the same question, but related: Modified Ants on a Stick Probability Question. – David K Aug 14 '24 at 00:10
  • Just to clarify, does $t$ refer to minutes? Additionally, are we intended to find the condition for the problem as a whole or just for $N$ (e.g. can we say "all ants move clockwise" or does it have to be something like $N = 1$)? – Mathemagician314 Aug 14 '24 at 12:47
  • @fleablood (I disagree with your assessment) The answer can be classified as: When either $N/2$ or $N$ ants are moving in the same direction as Bob. – Calvin Lin Aug 18 '24 at 17:36
  • @CalvinLin if the answer is that simple you are correct. But I find the asking for "the condition" to be oddly worded and I'm without faith (for the time being) that is is clear that there a "the condition" that always works and is ambiguous and no other work. – fleablood Aug 19 '24 at 16:51
  • @fleablood Right, I'm thinking of it as "a necessary and sufficient condition". The phrasing is unclear, though usually that's the intended interpretation. There could be several equivalent conditions, like if you were able to fully capture the gazillions of ways to do it. $\quad$ As for the proof, I'm happy to provide some hints, or you can look at the reddit hint. $\quad$ (JK) Maybe think about them as fleas instead of as ants? – Calvin Lin Aug 19 '24 at 17:23
  • I have an argument which I am convinced is wrong, but cannot figure out the problem. First of all the positions of all the ants overall (rather the set of positions which have an ant) is the same. Also, at each step the cyclic order of the ants relative to each other remains the same as whenever two ants collide they both change direction. So unless there is some sort of symmetry in the arrangement, the arrangement is exactly the same and Bob is at position A. – Shreya Mundhada Sep 03 '24 at 04:36
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    @ShreyaMundhada Play with a few explicit examples, esp where my condition of "N/2 or N ants moving in the same direction" is not satisfied, and you'd see that at time $t = 1$, the ant's position is cyclically rotated around the circle. (There exists a $k$ such that for all $i$, ant $a_i$ will be where ant $a_{i+k}$ was located at time $ t = 0 $.) This preserves the cyclic order of the ants, but doesn't require any symmetry in the arrangement. – Calvin Lin Sep 07 '24 at 20:32
  • @CalvinLin That makes a lot of sense. Thank you :) – Shreya Mundhada Sep 09 '24 at 04:25

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