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 The flooble question (Posted on 2003-02-18)
At flooble there are 40 problems in the queue. (this may not be true but lets just pretend it is.) A few crazy hackers somehow manage to promote themselves to scholars. On the first day the first hacker will vote thumbs up on all problems displayed.(The 10 most recent) On the second day the second hacker votes thumbs down on every second problem. On the third day the third hacker votes thumbs up on every third problem. And so on and so on. (When it gets to the eleventh day the eleventh hacker will do what the first hacker did)

How many days will it take for every problem in the queue to be live on the site?

Note: For those who don't know there are only 10 problems that can be voted thumbs up or thumbs down every day and these problems are the 10 least recent. Also a problem with three thumbs up will be posted to the site and taken out of queue. Only one problem can be posted to the site per day. Also if a problem gets 3 thumbs down it is deleted.

Btw: for those who like an extra challenge what if one problem is submitted every 3 days?

Also: A hacker will always vote before a problem becomes live.

 See The Solution Submitted by Alan Rating: 4.0000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution? | Comment 39 of 51 |
Noting fwaff's method, and reconsidering the statement "a hacker will always vote before a problem becomes live," I see that fwaff has probably correctly interpreted this to mean that on any given day, any posting, even of previously +3 or higher, waits until after the voting for that day. I have modified my program and do indeed get closer to fwaff's results: only 1 rejection and the last post on day 181, being question #38. The only difference is that the one rejected is #35 rather than #30. Perhaps it has to do with the choice of which of several eligible puzzles to post. For example if day 10n+2 has left, say, currently voted positions 4, 6 and 8 at +3, and position 9 at +2, the next day's vote will push position 6 to +4 and position 9 to +3, so positions 4, 6, 8 and 9 have +3, +4, +3 and +3 respectively. In this situation I have chosen to post the one in position 4, as the earliest in the queue to meet the criterion. Another strategy might be to post the one in position 4, as having the highest net. Another might be to post position #9 as it just achieved +3 that day. Again, I used the strategy of posting the earliest in the queue meeting the posting criterion, and get the last post on day 181, with 1 rejected along the way.
 Posted by Charlie on 2003-03-04 03:33:43

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