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.

(In reply to How the voting works by levik)

Hmmm... I don't know how this affects what we said previously except that the vote cancelling is the correct way. The problem states that the 11th hacker votes with the same strategy of the first. It was only a matter of convenience to call this the return of the first voter Presumably there are as many hacker/voters as needed to see this task through. Then as the twelfth hacker/voter acts like the second, etc. there is no question of any voting a second time on the same problem.

The only impediment I can see is the fact that several puzzles become eligible for posting on the same day. If they are not all actually posted, I think what Levik is saying here is that no more than one puzzle starts being considered for voting that day. In effect, if ten puzzles all have 3 thumbs up, then only one gets posted that day, and only one will start to be considered in the voting.

If that is taken into consideration, indeed the umpteenth hacker (acting like the "umpth" hacker) could very well reduce a +3 to a +2. Levik is also saying that the problems with a -3 stick around also, but then wouldn't ten problems with -3 eventually cause voting to come to a halt? I thought they were forever out of contention.

I'm afraid that I still don't know exactly how the voting works. In order for all -3 problems to not take up all the voting slots, there'd have to be an allowance for 10 viable (that is neither +3 nor -3 nor presumably higher absolute values) puzzles. I don't know if it is clear that there is such an allowance.

Posted by Charlie on 2003-02-18 09:54:09