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Bumper Cars (Posted on 2004-12-30) Difficulty: 3 of 5
Some bumper cars are moving around a circular track at the same constant speed. However, they are not all going in the same direction. Collisions are perfectly elastic, so that two colliding cars instantaneously change directions (and continue at the same speed).

Show that at some point in the future, all the cars will be back to their starting positions and directions. Assume that each car has no length.

See The Solution Submitted by David Shin    
Rating: 2.5000 (4 votes)

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Solution re(2): Solution | Comment 6 of 14 |
(In reply to re: Solution by Vee-Liem Veefessional)

But Larry's entire post is a proof: consider the cars going through one another, just switching their identities.  After one period has gone by, the same configuration of cars exists again (position and direction), but with the identities different.

Larry then makes the observation that as in fact the cars do not go through one another, the permutation of the cars' identities is in fact one cycle, or a group of cycles of equal length.  For example, if the cars are identified as A, B, C, D, E and F in clockwise order, the new configuration, according to exact position and direction of travel, has one of these sets of identities: BCDEFA, CDEFAB, DEFABC, EFABCD, FABCDE (I don't know if ABCDEF is a possibility immediately, but if so, it's solved already) As each unit period of time would permute the cars identically, after 6 time periods, the BCDEFA case or the FABCDE case would have the cars return to ABCDEF (again, position and direction being identified by the location within this string). CDEFAB or EFABCD would require 3 time periods to return to the original configuration; DEFABC would require only 2.

Regardless of how many cars, N, there are, the return to the original identities of the cars occupying the same position/direction states would be accomplished in some divisor of N time periods.

  Posted by Charlie on 2004-12-31 13:32:58
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