Without getting into calculus of variations, we can think of combining the
rectangular chicken yard with its mirror image on the other side of the barn
wall. The result is a rectangle of perimeter 200. We saw that the
corresponding largest area rectangle was a 50 by 50 square. So perhaps the
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
Checked the ellipses, turns out the maximum area is a circle after all,
provided the wall is at least 200/π m.
The area would be ±= 1591.549 m², unless anyone else knows better.
--
For information about J forums see http://www.jsof
Right! u Area returns
abscissa, area
That, along with incorrectly placed parens were errors.
I'll replace the 1 dimensional amoeba with Brent's method,
relax the tolerances, and see what happens. But not now.
Thank you!
NB. don't do this! too much output.
cubic =: (p.&([smoutput)~ _4&{.)~
(-@
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---
I was thinking that raindrop formation on windows would suggest some
shape that is approximately circular.
If we have a half circle, the circle's radius should be
R=: 100%1p1
and the half circle's area should be:
(o.R^2)%2
1591.55
So if there's a better shape it must have a larger area.
I
Old and senile as I am, this looks to me like a problem in calculus of
variations.See, e.g., en. wikipedia.org/wiki/Calculus_of_variations. You
are not likely to get the solution by guessing that the shape is
elliptical, or catenary, or parabolic, etc. I am too old and lazy to try
to solve it m
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
In amoeba, minindx comes from
minindx =. tryerr i. minval =. <./ tryerr =. uverb simptry
which should be bulletproof if uverb returns a scalar. You could try
putting something like
assert. (,3) -: $tryerr
after that line to see if uverb is returning junk. (It's not clear to
me that Area r
Kip,
Alternative formulations for your adverb that require fewer calculations of
u y.
Max1 =: 1 : 0
((= >./)@:u # ] ,. u) y
)
Max2 =: 1 : 0
fnres=. u y
where=. (= >./) fnres
where # y ,. fnres
)
I'd be interested in a tacit implementation of one of the adverbs Max
above. I came up wit
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
Barn/fence problem, which shape holds maximum area?
See Mouse on moon problem
http://projecteuler.net/problem=314
which I haven't solved.
Let's build a system to test an arbitrary function.
A symmetrical figure has maximal area. Proof---a
preferred figure would be preferred in reflection
and the
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" 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
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 1
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
Borrowing ideas from Raul, I like
Max =: 1 : 0
max =. >./ u y
where =. max = u y
where # y ,. u y
)
which identifies the max and where it occurs:
*: Max i:2
_2 4
2 4
(4 - *:) Max i:2
0 4
Sent from my iPad
On Feb 23, 2013, at 1:04 PM, Jose Mario Quintana
wrote:
> I did not
I did not see your second post!
area=. ] * 50 - %&2
area(max=. (>./) @:) 0 to 100
1250
max
>./@:
On Sat, Feb 23, 2013 at 1:46 PM, km wrote:
> Can we have an adverb Max so that f Max y finds the maximum of f on
> the list y ?
>
> Sent from my iPad
>
>
> On Feb 23, 2013, at 12:03
Can we have an adverb Max so that f Max y finds the maximum of f on the
list y ?
Sent from my iPad
On Feb 23, 2013, at 12:03 PM, Aai wrote:
> With some math in J
>
> x : length barn side
> -->the other rectangle side is then
> -: 100 - x
>
> i.o.w. we have
> A = x (50 - x/2)
>
> w
Nicely done. The "Harvard Calculus" recommends tables, graphs, formulas, and
words as problem-solving tools, and J can help with tables and graphs. About
Roger's question, I am inclined to guess a half-ellipse for maximum area, but
it's just a guess. --Kip
Sent from my iPad
On Feb 23, 2013
With some math in J
x : length barn side
-->the other rectangle side is then
-: 100 - x
i.o.w. we have
A = x (50 - x/2)
with J
0 1 +//.@(*/) 50 _0.5
0 50 _0.5
first derivative
0 50 _0.5&p. d. 1
50 _1&p.
second derivative
0 50 _0.5&p. d. 2
_1"0
--> a maximum for
1{:: p. 5
When I read it I thought the problem was to find the largest area as a
function of the size of the existing barn wall. If size is zero or
less then the area is (*:100%4); if it is (100%3) or larger the the area is
(*:100%3); and the interesting part is in between. The function, I think,
is:
(
On Sat, Feb 23, 2013 at 12:37 PM, Raul Miller wrote:
> Plotting these values, we see that 50 meters corresponds to the apex
> of a parabola, and there's no need to consider further where the
> maxima is at.
More specifically, 1250 square meters corresponds to the apex of a
parabola when plotted a
As a first pass, let's consider integer valued lengths for the lengths
of one of the fences. For simplicity, let's include cases where the
area is zero.
L1=: i.101
Since I allowed this fence to be 100 meters in length, that means that
the other two sides are equal length and use the remainder
You can get a larger area 30*40 by using 30 for the east-west dimension.
Sent from my iPad
On Feb 23, 2013, at 9:40 AM, Roger Hui wrote:
> The largest area obtains when it's a square and that area is:
>
> *: 100%3
> .11
>
> i.e. the length of the barn wall is 100%3 meters and so are th
When I read this, the first thing that popped into my mind was "I thought the
rule of thumb was that for a given perimeter, a circle encloses the maximal
area". This led me down a path of thought which lasted throughout morning
shower, and has left me with three things: (1) a realization that m
The largest area obtains when it's a square and that area is:
*: 100%3
.11
i.e. the length of the barn wall is 100%3 meters and so are the lengths of
the other 3 sides.
Exercise for the reader: What is the largest area that can be enclosed
with 100 meters of fence and one side of the bar
Use J to solve the farmer's fence problem:
A farmer with 100 meters of wire fence wants to make a rectangular chicken yard
using an existing barn wall for one of the north-south sides. What is the
largest area he can enclose if he uses the 100 meters of fence for the other
three sides, and wha
27 matches
Mail list logo
