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If there were one single lane, the problem would be to determine how many times a smaller realisation and it has been solved in a much older Riddler. Namely
In the two-lane system, the slow one only gathers cars with speeds lowest than the current speed, i.e. a decreasing sequence of speeds, creating single car convoys. The fast lane thus captures cars that are above the current minimum in the sequence, which, as it converges to the global minimum at some points, means that all following cars are found in the fast lane. i thought this would bring the number of convoys close to twice the above logarithmic sum (which sealed my destiny during an entrance oral exam 40 years ago!), but there are actually more of them for N large enough , which may be due to the possibility of the slow lane to capture more moderate speed cars in the beginning… The above compares the number of convoys for one lane (in red) and two (in gold), as well as the remarkable fit when regressing these numbers against log(N).
Here is the R code I used for that simulation
convoy=function(N,T=1e5){ for(t in 1:T){ speed=runif(N) slow=fast=NULL slow=speed[1] for(i in 2:N){ if(speed[i]<min(slow))slow=c(slow,speed[i]) else fast=c(fast,speed[i])} F=F+length(slow) if(length(fast)>0)F=F+1+sum(fast[-1]<cummin(fast)[-length(fast)])} return(F/T)}
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