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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.)
 Looking back at the rules. | Comment 43 of 51 |
I was looking back at Alan’s comment 32:”If there is more than one problem that is trying to become live the deciding factor is date at which point the most recent one (Which would be problem, if it has 3) would be chosen.”

As fwaff’s and my solution always posted the oldest eligible puzzle, I redid the simulation, this time posting the latest puzzle to be voted up to +3 or higher on the given day, if such exists. If not, and a prior-voted puzzle has 3 or higher, the first such in the queue would be posted. Using this strategy, all 40 in the queue do eventually get posted with no rejections, with the last, still #38, being posted on day 201.

For the extra challenge, however, we still get rejections, the first one being on day 306, after 72 have been posted, with the queue having built to 69. A repeating cycle is not entered until day 364 with current votees having status 21XX011X00, where X, again, means –1, and 74 are in the queue, 86 having been posted and still only that one rejection. On day 524 the same set of statuses appears, and the 160-day difference is a multiple of 10, the voting pattern will repeat in this cycle also. During the 160-day cycle, 38 puzzles get posted and 1 rejected. However, 160 is not a multiple of 3, so the full cycle of queue growth is 480 days, during which time of course 38x3=114 puzzles get posted, 3 get rejected, and the queue grows by 42, so by end of day 844 (the start of the second cycle), 117 puzzles are in the queue, 200 puzzles have been posted, and 4 have been rejected.

 Posted by Charlie on 2003-03-05 05:00:25

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