Mark, I think your answer is very well thought out.
I saw it this way. The cop can walk one of the corridors to the end and ensure that the bum is not there. Now he has two corridors - A and B - and he knows that the bum is farther than a0=1 in A and farther than b0=1 in B. The cop runs x0=b0+1=2 meters into A and back in x0 seconds. Now a1 = (x0+1) - x0/2 = x0/2 + 1 = 2 (the cop inspected x0+1 meters of the corridor A, but in x0/2 seconds that he ran back, the bum could reclaim x0/2 meters).
In the corridor B, the bum could make it to the room in b seconds and run into the "clean" corridor for another 1 meter, but the cop is already in the room and he will see the bum, so he can be sure that the bum is still in B, so b1=1. Next, the cop runs x1=a1+1=3 meters into B and back in x1 seconds. Now b2 = x1/2 + 1 = 5/2 and a2 = 1. Thus xn+1 = an+1 + 1 = (xn/2 + 1) + 1 = xn/2 + 2 which converges to 4, thus the maximum corridor length is 5(ish)
Who said they were easy?
DIGGING
No such thing as half a hole.
COWBOY
Well, Sunday is two days after Friday; so he didn't have to resort to riding horses named Friday and Sunday.
SONS
There was a third child. They were triplets.
Whim
Aorals:
0 = 1
01 = 1
You either get it in a nanosecond or look at it for ages.
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