Linda, why do you eschew @ (and also, presumably @: )
but allow [: ?
You've probably been asked before....
MIke
On 15/01/2012 10:40 AM, Linda Alvord wrote:
> I have been reluctant to include the interesting fd created by Henry Rich
> and Ric Sherlock because I was unable to write the explicit version that
> produces their tacit version. I finally got it , so we now have another fd.
>
>
>
> fdrs=:[: /:~ ({. , #)/.~
>
> fdrs
>
> [: /:~ ({. , #)/.~
>
>
>
>
>
> a=:3 2 3 3 1 3 4 1 1 2
>
> fd=: 13 :'/:~(~.y),.#/.~y'
>
> fd
>
> [: /:~ ~. ,. #/.~
>
> fd a
>
> 1 3
> 2 2
> 3 4
> 4 1
>
>
>
> From: Linda Alvord [mailto:[email protected]]
> Sent: Monday, January 09, 2012 10:45 AM
> To: 'Linda Alvord'
> Subject: Challenge 4 Bountiful Birthdays
>
>
>
> Chalenge 4 Bountiful Birthdays PLEASE DO NOT RESPOND UNTIL 1/16/2012
> 12 am EST
>
>
>
> For this challenge, as usual, do not use @ but you may use whatever style
> and strategy you like to accomplish the task.
>
>
>
> The problem is to simulate the classic birthday problem.
>
>
>
> A single trial works this way. People enter a room one by one and declare
> their birth date. Suppose the 29th person is the first person to match a
> birthday of someone in the room. The result of the first trial is 29.
>
>
>
> Repeat for 500 trials.
>
>
>
> If you use Kip's frequency distribution of the results you have lots of
> information.
>
>
>
> fd=:[: /:~ ~. ,. [: +/"1 ~. =/ ]
>
>
>
> But the mean gives a more concise summary of the data. So, the final part
> of the problem is to obtain a list of 10 means of 500 trials and the
> mean of the means.
>
>
>
> 'til next week.
>
>
>
> Linda
>
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>
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