countRects=: */@(2 ! >:) NB. How many pairs each of vertical
* horizontal lines
getSizes=: ,@(>:/~) # [: ,/ ,"0/~ NB. All pairs of i. y
idxClosest=: 4 : '(i. <./)@(x |@:- ])y'"(0 2) NB. Index of mat y to
value x
({~ *2e6*&idxClosest@:(countRects"1)) getSizes >: i.200 NB. Closest
to 2e6
77 36
({~ *1e6*&idxClosest@:(countRects"1)) getSizes >: i.200 NB. Closest
to 1e6
63 31
countRects"1 ] 63 31,:77 36 NB. How close is each?
999936 1999998
On Fri, Oct 10, 2014 at 1:14 PM, Linda Alvord <[email protected]>
wrote:
> What is the correct answerfor this problem?
>
> Linda
>
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]] On Behalf Of Stefano
> Lanzavecchia
> Sent: Friday, October 10, 2014 11:47 AM
> To: [email protected]
> Subject: Re: [Jprogramming] Project Euler 85, Python and J
>
> Actuary the use of ravel and antibase is common practice to solve
> certain problems in APL and isn't considered cheating. So I wouldn't
> say it's "not nice" but I would definitely go for antibase instead of
> a combination of floored-divide and modulus. As a bonus, a solution
> based on antibase would scale to problems of any rank and not just 2d
> matrices.
>
> Have fun!
> --
> Stefano
>
> > On 10/ott/2014, at 17:35, Sebastiano Tronto <[email protected]
> >
> wrote:
> >
> > Hi,
> > A dirty trick to get the job done would be to ravel the matrix ( , ),
> solve
> > the 1d version of the problem and then get the "true" indexes with
> > something like (<.@%&200 , 200&|).
> > For example, if you needed to just find the max:
> > (<.@%&200 , 200&|) (i. >./) , m
> > where m is your matrix.
> >
> > I know this isn't a nice way to solve the problem, but it should work.
> >
> > Sebastiano
> >
> > 2014-10-07 6:37 GMT+02:00 Jon Hough <[email protected]>:
> >
> >> Project Euler 85: https://projecteuler.net/problem=85
> >> This problem is not really conceptually hard, but I am struggling with a
> J
> >> solution.I have solved it in Python:
> >> =============================================
> >> def pe85(larg, rarg): count = 0 llist = range(1, larg+1)
> >> rlist = range(1, rarg+1)
> >> for l in llist: for r in rlist: count +=
> >> l*r
> >> return count
> >>
> >> if __name__ == "__main__": # test for 2x3 grid, as in question.
> k
> >> = pe85(2,3) print "Test value: "+str(k) l1 = range(1,200)
> #
> >> 200 lucky guess l2 = range(1,200) bestfit = 10000 # just a big
> >> number area = 0 for i in l1: for j in l2:
> >> diff = abs(2000000 - pe85(i,j)) if diff
> <
> >> bestfit: area = i*j
> >> bestfit = diff
> >> print "AREA is "+str(area)
> >>
> >>
> >> ================================================The above script will
> give
> >> the final area of the closest fit to 2 million. (The python code may not
> be
> >> the best). Also I tested all possibilities up to 200x200, which was
> chosen
> >> arbitrarily(~ish).
> >> Next my J. I go the inner calculation ok (i.e. see the function pe85
> >> above). In J I have:
> >> pe85 =: +/@:+/@:((>:@:i.@:[) *"(0 _) (>:@:i.@:]))
> >> NB. I know, too brackety. Any tips for improvement appreciated.
> >>
> >>
> >> But from here things get tricky. If I do the calculation over 200x200
> >> possibilities I end up with a big matrix, of which I have to find the
> >> closest value to 2 million, of which then I have to somehow get the
> (x,y)
> >> values of and then find the area by x*y.
> >>
> >> The main issue is getting the (x,y) from the best fit value of the
> array.
> >>
> >> i.e. If I do pe85"(0)/~ 200, I get a big array, and I know I can get the
> >> closest absolute value to 2 million but then I need to get the original
> >> values to multiply together to give the best fit area. Actually I have
> >> bumped into this issue many times. It is easy enough in a 1-d array,just
> do:
> >> (I. somefunc ) { ])
> >>
> >> or similar to get the index. But for two indices the problem is beyond
> me
> >> at the moment. Any help appreciated.Regards,Jon
> >>
> >>
> >>
> >> ----------------------------------------------------------------------
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> > ----------------------------------------------------------------------
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>
--
Devon McCormick, CFA
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