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Posted 26 November 2013 - 04:08 AM
Posted 27 November 2013 - 05:27 AM
Let A be the slowest car in the group. Since every car behind A will eventually catch up to A and be stuck behind it, there will be one traffic jam with A and all subsequent cars, and howevermany traffic jams involving the cars in front of A. If there are M cars in front of A then there can be no more than M+1 traffic jams, and that's only in the special case where every car in front of A is traveling slower than the car in front of it (if you count a single unimpeded car to be a traffic jam). In general there will be less than M+1. So if car A is randomly distributed among the pack, M will on average be about N/2 and the total number of traffic jams should be less than that.
Posted 27 November 2013 - 05:05 PM
yes, i tihnk i ran my simulation wrong. trying to think of a way to fix it.
Posted 02 December 2013 - 03:34 PM
Spoiler for apparently
I agree. Nice!
Is there a clever derivation?
Vidi vici veni.
Posted 03 December 2013 - 01:47 AM Best Answer
Posted 03 December 2013 - 01:49 PM
Spoiler for Here's the gist of my not-very-clever method of solving this:
Vidi vici veni.
Posted 03 December 2013 - 07:22 PM
Edited by BMAD, 03 December 2013 - 07:23 PM.
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