I'm glad to be rid of the problem.  Thanks.

Linda

-----Original Message-----
From: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of km
Sent: Sunday, February 24, 2013 1:40 AM
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] The farmer's fence

The problem goes away if you use  {:  instead of  }.  in your definition of
r .

    r1 =: [: > [: {: p.
    r1 0 100 _2
 50 0
    a r1 0 100 _2
 25

The problem you had is caused by the fact that  }.  applied to a vector
always has a vector result and is solved by the fact that  {:  applied to a
vector always has a scalar result.  Someone else will have to explain why
opening  , <50 0  has a different result from opening  <50 0  , and also
explain the relevance of the difference.  At least I can make the problem go
away!

Kip

Sent from my iPad


On Feb 23, 2013, at 10:34 PM, "Linda Alvord" <lindaalv...@verizon.net>
wrote:

>  My first line select the two roots:
> 
>   r=: 13 :'> }. p. y'
>  r 0 100 _2
> 50 0
> 
> These are the roots:
> 
> Now find the axis of symmetry  x  value  (the average of the roots,
> 
>   a=: 13 :'(+/y)%#y'
>   a 50 0
> 25
> 
> So if I apply  a  to  r 0 100 _2  I expect  25.  I get 1 2
> 
> 
> 
> 
> -----OriginalaMessage-----
> From: programming-boun...@forums.jsoftware.com
> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of km
> Sent: Saturday, February 23, 2013 9:53 PM
> To: programm...@jsoftware.com
> Subject: Re: [Jprogramming] The farmer's fence
> 
> Linda, if you enter  50 0  and then enter  1 2 $ 50 0  the printed 
> results appear the same, but the first has two items and the second only
one item.
> What do you expect the average of a one-item array to be?  Can you 
> figure out how to use rank to get what you want from  r 0 100 _2  ?
> 
> Sent from my iPad
> 
> 
> On Feb 23, 2013, at 6:21 PM, "Linda Alvord" <lindaalv...@verizon.net>
wrote:
> 
>> A should give an average.  The roots are  50 0 .
>> 
>> Shouldn't the average be 25 rather than 1 2  ?
>> 
>> Linda
>> 
>> -----Original Message-----
>> From: programming-boun...@forums.jsoftware.com
>> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of km
>> Sent: Saturday, February 23, 2013 6:45 PM
>> To: programm...@jsoftware.com
>> Subject: Re: [Jprogramming] The farmer's fence
>> 
>>   r 0 100 _2
>> 50 0
>>   $r 0 100 _2
>> 1 2
>> 
>> Sent from my iPad
>> 
>> 
>> On Feb 23, 2013, at 4:35 PM, "Linda Alvord" <lindaalv...@verizon.net>
> wrote:
>> 
>>> The roots of a polynomial:
>>> 
>>> r=: 13 :'> }. p. y'
>>> r 0 100 _2    :
>>> 50 0
>>> r
>>> [: > [: }. p.
>>> 
>>> 
>>> The average of the roots or  x  coordinate of axis of symmetry: 
>>> a=: 13 :'(+/y)%#y'
>>> a 50 0
>>> 25
>>> a   
>>> +/ % #
>>> 
>>> Find the maximum:
>>> 
>>> 0 100 _2 p. 25
>>> 1250m
>>> 
>>> Sadly this doesn't work:
>>> 
>>> a r 0 100 _2
>>> 50 0
>>> 
>>> Any idea why not?
>>> 
>>> Linda
>>> 
>>> 
>>> -----Original Message-----
>>> From: programming-boun...@forums.jsoftware.com
>>> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Alex 
>>> Giannakopoulos
>>> Sent: Saturday, February 23, 2013 3:28 PM
>>> To: J Programming forum
>>> Subject: Re: [Jprogramming] The farmer's fence
>>> 
>>> Nice little gotcha there, assuming that the shape will be a square, 
>>> since a square maximizes the contained area for a rectangle, while 
>>> forgetting that the wall gives you extra perimeter for free, 
>>> depending on
>> the shape.
>>> 
>>> By the same analogy I'd tackle Roger's version of the problem, i.e.  
>>> find ANY shape that will maximize the area:
>>> Again, I suspect that going for a (semi)circle might be essentially 
>>> the same gotcha.
>>> 
>>> I haven't got time to code it at the moment, but I'd investigate an
>>> (half) ellipse and also a parabola.
>>> Will need some integration though, to find the expression for the 
>>> length of their curves.
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