[seqfan] Re: puzzle suitable to make a sequence ?
Dion Gijswijt
dion.gijswijt at gmail.com
Tue May 11 09:12:24 CEST 2010
This is a very nice puzzle! I've seen it before in K\"omal, the Hungarian
math and physics journal for high school students. Certainly worth to check
out :)
http://www.komal.hu/verseny/1999-02/mat.e.shtml
Best regards,
Dion Gijswijt.
On Tue, May 11, 2010 at 1:28 AM, <franktaw at netscape.net> wrote:
> I can't quite tell whether you already know the answer or not. If not,
> I don't see how you can suggest it as a puzzle.
>
> Answer below.
>
> Franklin T. Adams-Watters
>
> -----Original Message-----
> From: Alexander P-sky <apovolot at gmail.com>
>
> Hello,
>
> Here is the puzzle suitable IMHO to make a sequence ....
> Given are the "n" story building and two balls, which are made of some
> breakable material (say glass). The balls are such that they break
> being thrown down from the certain height (and above), while being
> thrown at lesser height, the ball remains to be in tact.
> What is the minimum number of throwing attempts "k" is needed to
> experimentally but systematically and successfully determine (for the
> the worst case scenario ) starting from which floor of the "n" story
> building (if from any) the ball will start to break while being
> thrown down. Again it is important to stress that only two balls are
> available for experimental testing and that obtaining the definite
> result is required.
>
> I am leaving to the audience (recipients of this list) to come up
> with the strategy of determining the minimum number of throwing
> attempts "k" for given "n".
>
> Is k(n) sequence in OEIS ?
>
> Regards,
> ARP
> _______________________________________________
>
> The answer is A002024, n repeated n times. This is both the minimum
> number of throws, and the first story we throw from.
>
> First, consider the case where only 1 ball is available. The only
> viable strategy is to throw the ball from floors in increasing order, 1
> to n; and so it will take n throws in the worst case.
>
> Now with 2 balls, throwing the ball from story i, if the ball breaks,
> we are reduced to the previous case for the first i-1 stories, hence
> the worst case is i-1+1 throws (i-1 remaining throws, +1 we just did).
> If the ball does not break, we can ignore the first i stories, and
> proceed as above with the remaining n-i stories. To get the optimum
> result, we must take a(n-i) < i, hence the "repeated n times".
>
>
>
>
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>
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>
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