Ooops, I meant
+/12001>".(": f) rplc ' ';',';'r';','
I always have trouble with greater-thans...
Skip
Skip Cave
Cave Consulting LLC
On Sun, Jun 15, 2014 at 4:36 PM, Skip Cave <[email protected]> wrote:
> 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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