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        riddles >> hard >> Circular Jail Cell
        
(Message started by: NickH on Jul 28^th, 2002, 6:50am)

Title: Circular Jail Cell
Post by NickH on Jul 28^th, 2002, 6:50am

Ten prisoners found their door open after 100 rounds of the drunken jailor. They were the ones whose cell number is a perfect square.

Each door is opened or closed once for each factor of its cell number, including 1 and the cell number itself. For example, for cell 12: 1, 2, 3, 4, 6, 12.

Most whole numbers have an even number of factors. This is because each factor f can be paired with n/f. The exceptions are the perfect squares, where f = n/f for precisely one factor.

Title: Re: Circular Jail Cell
Post by horseisahorse on Aug 7^th, 2002, 6:00pm

Actually, I think the real-world answer would be zero, since each prisoner would have booked out of the prison as soon as the jailor opened his door during the first round.

Title: Re: Circular Jail Cell
Post by Ivan on Dec 18^th, 2002, 10:20pm

I got the same answer as NickH. Since I like doing things the hard way, I wrote a python program (the formatting here ain't great, look at it at http://linuxhelp.hn.org/jail-cell.py if you're interested. (with proper block spacing)):
Code: #!/usr/bin/python doors = {} # create doors 1 - 100 for x in xrange(101): doors[x] = 0 del doors[0] # first, the jailer goes to each cell # this number will be increased in each round sep = 1 # repeat until jailer falls down while 1: doorcount = 1 for x in doors: # check if the door by the current door skipping factor if doorcount % sep: if doors[x] == 0: # if it's closed, open it doors[x] = 1 else: # if it's open, close it doors[x] = 0 doorcount += 1 sep += 1 if sep > 100: # jailer falls down break # count how many doors are open in the end doors_open = 0 for x in doors: if doors[x] == 1: doors_open += 1 print doors_open