Here is another version of countRects
countRects=: */@(2 ! >:)
On Wed, Oct 8, 2014 at 9:07 AM, Tikkanz <tikk...@gmail.com> wrote:
> Sorry, yes that is a leap.
> (x * (x + 1)) * 0.5 is the number of ways to choose two horizontal lines
> to make 2 sides of the rectangle.
> (y * (y + 1)) * 0.5 is the number of ways to choose two vertical lines to
> make the other 2 sides of the rectangle
> ((x * (x + 1)) * 0.5) * ((y * (y + 1)) * 0.5) is the number of ways to
> choose the lines to make a rectangle
> refactoring:
> 4 %~ x * (x+1) * y * (y+1)
> 4 %~ */ x,(x+1),y,(y+1)
> 4 %~ */ x,y,(x+1),(y+1)
> 4 %~ */ (, >:) x,y
>
> HTH
>
> On Wed, Oct 8, 2014 at 4:30 AM, Devon McCormick <devon...@gmail.com>
> wrote:
>
>> Hi -
>>
>> "countRects" seems like a bit of a leap. I think I understand "4 %~"
>> because you're overcounting by 4 rotations, but I don't comprehend the
>> magic behind "*/@(,>:)".
>>
>> I see that "(,>:)" concatenates the shape to its increment, e.g. 2 3 3 4
>> for the input 2 3, but what's the rationale behind this?
>>
>> Thanks,
>>
>> Devon
>>
>> On Tue, Oct 7, 2014 at 7:41 AM, Tikkanz <tikk...@gmail.com> wrote:
>>
>> > 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 <jgho...@outlook.com> 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
>> > >
>> > >
>> > >
>> > > ----------------------------------------------------------------------
>> > > For information about J forums see
>> http://www.jsoftware.com/forums.htm
>> > >
>> > ----------------------------------------------------------------------
>> > For information about J forums see http://www.jsoftware.com/forums.htm
>> >
>>
>>
>>
>> --
>> Devon McCormick, CFA
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
>>
>
>
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