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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.)
 re: First guess | Comment 2 of 51 |
(In reply to First guess by fwaff)

I get a different result using fwaff's assumptions. I find that only 4 are rejected, and that the last day is day 64 when a previously accepted puzzle is posted. Perhaps fwaff and I made different other assumptions. I assumed that "every second problem" means skip 1, vote on 2, skip 3, vote on 4, etc. If I were a C programmer I might consider the first in the queue to be the zeroth, but that's another story. Likewise "every third problem" skips over the first two, etc. Also, I assume we're counting (and skipping) only the puzzles currently being voted on, not their original position in the queue.

I will present my findings in one or more replies to this message (to fit within the size limits -- Levik, What's the limit to the size of one of these comments?).

But, as there are rejections (7 or 4 depending on who's counting) using the assumption that ups and downs don't cancel, and the original problem called for every one being posted, I think this is the wrong assumption. Especially since thumbs up are said to count +1 and thumbs down -1. What would be the point of assigning these signed values if they didn't cancel each other?

To conserve space on the future posts, the key to these is:
q=on queue but not yet voted on, V=currently being voted on, a=accepted but not yet posted, r=rejected, P=posted. On the line following these 40 codes will be the up-votes and down-votes for each of the ten current votees. They never reach 3 because on that day they'd be shown as accepted, posted or rejected. Thus the vote count columns lose puzzles and new ones travel in from the right, so it can be a little confusing. List to begin on the next post...
 Posted by Charlie on 2003-02-18 05:20:24

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