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You must place 7 robot ants on a long rod and set each of them to move left or right starting at time 0. You can set any positive speed for each ant. When 2 or more ants meet, they turn around. When an ant reaches the end it falls off.

Show that you can place and set the ants in such a way that none of them ever falls off.

Bonus question: Can you do this for other odd numbers, too?

Eric
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    Since someone tried to wiggle out of the intended wording with the last problem, perhaps clarify that "left or right" is along the direction of the rod/parallel to its run/equivalent to moving along a long but not infinite axis – bobble May 29 '22 at 04:40
  • What happens if I put two ants with the same starting position and direction and speed? – justhalf May 29 '22 at 04:56
  • @justhalf Starting positions must be different. – Eric May 29 '22 at 05:14
  • I have found that for 3 ants is impossible. Interesting problem. – justhalf May 29 '22 at 05:32
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    @justhalf how? can you elaborate a bit – I'm Nobody May 29 '22 at 06:56
  • The general idea is to look at ant configuration (position and direction) at the beginning, and after a cycle (to limit the range, there must be a cycle). They must be exactly the same. With 3 ants, after 1 cycle you can calculate that the position will not be exactly the same (in order to be exactly the same, one of the ants' speed need to be 0). – justhalf May 29 '22 at 08:44
  • The case for 7 ants still has no answer. – Florian F May 30 '22 at 08:33

1 Answers1

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Picture showing how to place 4,7 or 10 ants:

enter image description here

x-axis is time, y-axis is position. The 4 black ants alone are viable as are the blacks together with the 3 blues or all 10.

Ants here move either at speed 1 or 1/3. Except for the first and last ones, the slower ants can be left out.

Therefore this construction yields solutions for 4, 6, 7, 8, 9, 10, 11, ... ants.

P.S.: 5 is also possible:

enter image description here

Albert.Lang
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