Would this work for Euler 73?
load 'strings'
f =. x:1%2+10000%~i.10001
+/12001<".(": f) rplc ' ';',';'r';','
Skip Cave
Cave Consulting LLC
On Sun, Jun 15, 2014 at 2:39 PM, Mike Day <[email protected]>
wrote:
> FWIW, my solution (called pp for some reason) runs in about 3 seconds and
> 0.3MB
> 1 ts'pp 12000'
>
> 3.10239 366720
>
> on a Samsung notebook running 64bit J802 under Windows 8.
> It's a pretty simple function involving one implicit loop within one
> explicit loop (!?).
>
>
> It's some years since I solved it. My script shows I was exploring ways
> of speeding it up, including use of the Moebius function,
> but I seem to have moved on to something else. A promising function which
> looks almost correct takes about 0.5 seconds, but uses about 2MB.
>
>
> Please note that the Euler Project people don't encourage open discussion
> of solutions. I don't think I've given too much away here.
>
> Mike
>
>
> On 15/06/2014 16:59, Jon Hough wrote:
>
>> Another Project Euler... (apologies)
>> #73 http://projecteuler.net/problem=73
>> I found this one more tricky than it first seems.
>> My attempt fails.
>> My reasoning of solution. Trying to find all reduced fractions with
>> denominator and numerator in range 1... 12000 that are greater than 1/3 but
>> less than 1/2.The most naive way would be to make a 12000x12000 grid of
>> fractions, nubbing out duplicates. I tried this, and as I expected I ran
>> out of memory.
>> One problem is duplicates. To get rid of duplicate fractions, for all y
>> <12000, I only want to check fractions with numerators x (i.e. fractions
>> xry), where x and y are coprime.(e.g so 1/2 will only be counted once, and
>> not 3/6, 4/8 etc)
>>
>> My method:
>> NB. check coprime
>>
>> coprime =. =&0@:(+/)@:(e.&q:)
>>
>> NB. get coprimes. (since index starts at zero, and we do not want to
>> compare with 0 or 1, we start at index 2
>> NB. and then +2 to get the coprime number.
>> gcp =. >:@: >:@:I.@: (coprime"(0 0))(}.@:>:@: i.)
>> e.g. gcp 12 returns 5 7 11.
>> Next,
>>
>> NB. get fractions between 1/3 and 1/2
>> fracs =. +/@:,@:(1r3&<*.1r2&>)@:(%~ func2"0)
>> If I test for 2 3 4 5 6 7 8 to see how many fractions are between 1/3 and
>> 1/2I get
>> fracs >: i.8
>> 3
>> This is the same value as shown on the question webpage for the value
>> d=8.For d = 12000if I run the verb, I get an out of memory error.Any ideas
>> how to proceed? I am wondering if my approach is salvageable, or I need to
>> go back to the drawing board.
>>
>>
>>
>>
>>
>>
>
>
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