Idot =: $ #: I.@:, NB. the I. for higher dimensions
NB. there are (1/2 * N * N+1) = -: (* >:) N possibilities on a side with
length N
NB. and the product of two of those (*/~) should be (closest to) 18.
Idot (= <./@:,) |18 - */~ -: (*>:) i.200
2 3
3 2
NB. now with two million, and multiplying the two sides:
*/"1 Idot (= <./@:,) |2e6 - */~ -: (*>:) i.200
2772 2772
________________________________________
From: [email protected]
[[email protected]] on behalf of Tikkanz
[[email protected]]
Sent: Tuesday, October 07, 2014 13:41
To: Programming JForum
Subject: Re: [Jprogramming] Project Euler 85, Python and J
Note that 200 x 200 is a bit of an overkill given 3x2 = 2x3
The following choses the lower triangular of a matrix of the different
sized rectangles to investigate.
getSizes=: ,@(>:/~) # [: ,/ ,"0/~
getSizes >: i. 5
Given the sides of a rectangle you can count the number of rectangles as
follows:
countRects=: 4 %~ */@(, >:)
countRects 2 3
Now get the index of the rectangle size with a count closest to 2million
idxClosest=: (i. <./)@(2e6 |@:- ])
Putting it together
*/@({~ idxClosest@:(countRects"1)) getSizes >: i.200
On Tue, Oct 7, 2014 at 5:37 PM, Jon Hough <[email protected]> wrote:
> 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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