Using (2 ! >:) is clearly better than doing my double for-loop. I'm embarrassed
I missed that.
The real meat of my confusion with multidimensional arrays is in not just
finding the indices but doing something with the elements at the indices.
e.g. For a single dimension array, there could be a function to fund the primes
less than 100.
primelist =: (I.@:(1&=)@:(1&p:)) { ]
and primelist >: i. 100
should spit out
2 3 5 ... ...97
But what if instead of having >: i. 100
I had ( for some reason )
arr =: 10 10 $ >: i. 100
So I have a 10 by 10 matrix of all positive ints up to 100.
primelist clearly will not work on arr. But if I want to return the list of
primes, as a single dimensional list I'm not sure how to do that.
Or, for example, if I want to change the elements of arr to 1 if and only if
the sum of the (i, j) indices are prime (just a random example).
In procedural python this could be quickly done with a double for-loop and a
prime test. In J this type of problem still escapes me.
> Date: Fri, 10 Oct 2014 19:35:26 -0400
> From: [email protected]
> To: [email protected]
> Subject: Re: [Jprogramming] Project Euler 85, Python and J
>
> 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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> >
>
>
>
> --
> Devon McCormick, CFA
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
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