interesting,
isP
(-: |.)@":
(* isP)@:[^:(0=])/ 0,~ /:~ ~. , */~ 100+i.900
906609
timespacex '(* (-: |.)@":)@:[^:(0=])/ 0,~ /:~ ~. , */~ 100+i.900'
0.167316 3.56726e7
btw, there is an optimization close to short circuiting built in:
({~ (-: |.)@":"0 i. 1:) \:~ ~. , */~ 100+i.900
906609
timespacex '({~ (-: |.)@":"0 i. 1:) \:~ ~. , */~ 100+i.900'
0.0977559 3.56744e7
----- Original Message -----
From: David Lambert <[email protected]>
To: programming <[email protected]>
Cc:
Sent: Sunday, June 8, 2014 8:28:47 PM
Subject: Re: [Jprogramming] Project Euler Question 4
"short circuit" tacit evaluation.
Solve PE#4 starting with the palindromes inclusively between *:100 999
Note 'Introducing "short circuit" tacit evaluation'
(* is_solution)^:unsolved/
where is_solution and unsolved are Boolean verbs,
and unsolved is inexpensive.
I hope this is a new J idea.
Please read through to the end.
Where possible I use assertions to demonstrate and test.
)
NB. factoring is expensive.
factor=: [:*/"1&>[:,[:{[:<@(^ i.@:>:)/"1[:|:2&p:
NB. tally of and factors of 12
assert (6 ; 1 2 3 4 6 12) -: (# ; /:~)@:factor 12
digits =: 10&#.^:_1
assert (i._3) -: digits 210
is_palindrome =: (-: |.)@:digits f.
assert 0 1 1 -: is_palindrome&> 48 68886 3
Filter =: (#~`)(`:6)
assert 8 9 11 -: is_palindrome&> Filter (+ i.)8
NB. A is (a vector of) the palindromes from squared 100 to 999
NB. Slow! Probably due to is_palindrome operating atomically.
NB. However, the important point is the increasing order.
A=:100 is_palindrome&> Filter@:(<. + [: i. >. >:@:- <.)&*: 999
assert 10001 997799 -: ({.,{:) A NB. true by inspection
1798 = # A NB. potentially factor this tally. I cannot assert this.
has_3_digits =: 3 = #@digits
assert 1 0 -: has_3_digits&> 234 9923
NB. or 2 = [: <. 10 ^. ]
NB. or 100&<: *. <&1000
NB. or 3 = #@":
three_digit_factors =: [: has_3_digits&> Filter factor
assert 125 625 250 100 500 200 400 -: three_digit_factors *:100
final_test =: 1 e. ((% e. ]) three_digit_factors)
assert 1 0 -: final_test&> 10000 10001
NB. this is it!
906609 = (* final_test)@:[^:(0=])/A,0
> Date: Sun, 08 Jun 2014 12:31:53 -0400
> From: Henry Rich<[email protected]>
> To:[email protected]
> Subject: Re: [Jprogramming] Project Euler Question 4
> Message-ID:<[email protected]>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> Tacit variant:
>
> break2=:1 :'(": 13!:8 (15"_))^:u"u :: (LF taketo }.@(13!:12)@(''''"_) )'
> (-: |.)@("."0@":)"0 break2 \:~ ~. , */~ 100+i.900
>
> Henry Rich
>
> On 6/8/2014 12:05 PM, 'Pascal Jasmin' via Programming wrote:
>> >an optimization to short circuit on first true, while sorting them in order.
>> >
>> > break1 =: (1 : 'for_i. y do. if. u i do. i return. end. end.')
>> >
>> > (-: |.)@": break1 \:~ ~. , */~ 100+i.900
>> >906609
>> >
>> > timespacex '(-: |.)@": break1 \:~ ~. , */~ 100+i.900'
>> >0.0350227 3.56737e7
>> > timespacex ' >./ (#~ (-: |.)@":"0) , */~ 100+i. 900'
>> >0.26438 1.78811e7
>> >
>> >----- Original Message -----
>> >From: Roger Hui<[email protected]>
>> >To: Programming forum<[email protected]>
>> >Cc:
>> >Sent: Sunday, June 8, 2014 11:10:07 AM
>> >Subject: Re: [Jprogramming] Project Euler Question 4
>> >
>> >Or apply the phrase to 900+i.100 instead of to 100+i.900.
>> >
>> >
>> >
>> >
>> >
>> >On Sun, Jun 8, 2014 at 8:07 AM, Roger Hui<[email protected]> wrote:
>> >
>>> >>Slightly shorter if the formatted results are not converted back into
>>> >>numbers:
>>> >>
>>> >> >./ (#~ (-: |.)@("."0@":)"0) , */~ 100+i. 900
>>> >>906609
>>> >> >./ (#~ (-: |.)@":"0) , */~ 100+i. 900
>>> >>906609
>>> >>
>>> >>A different and perhaps less brutish method is to start with 6-digit
>>> >>palindromes and see which ones have two 3-digits factors.
>>> >>
>>> >>
>>> >>
>>> >>
>>> >>On Sun, Jun 8, 2014 at 8:00 AM, Henry Rich<[email protected]> wrote:
>>> >>
>>>> >>>Brute force seems right for this problem. You could write your
>>>> >>>solution
>>>> >>>as
>>>> >>>
>>>> >>> >./ (#~ (-: |.)@("."0@":)"0) , */~ 100+i. 900
>>>> >>>
>>>> >>>Henry Rich
>>>> >>>
>>>> >>>On 6/8/2014 10:46 AM, Jon Hough wrote:
>>>> >>>
>>>>> >>>>I am pretty pleased that I completed Project Euler Q4.
>>>>> >>>>Question:
>>>>> >>>>
>>>>> >>>>
>>>>> >>>>A palindromic number reads the same both ways. The largest palindrome
>>>>> >>>>made from the product of two 2-digit numbers is 9009 = 91 99.
>>>>> >>>>
>>>>> >>>>Find the largest palindrome made from the product of two 3-digit
>>>>> >>>>numbers.
>>>>> >>>>
>>>>> >>>>
>>>>> >>>>https://projecteuler.net/problem=4
>>>>> >>>>
>>>>> >>>>
>>>>> >>>>My solution:
>>>>> >>>>
>>>>> >>>>base =. "."0 ":
>>>>> >>>>
>>>>> >>>>NB. Tests if y value is palindrome.
>>>>> >>>>is_palindrome =. (# =)@:(=|.)@:base
>>>>> >>>>NB. multiply all 3 digit nums (100 ~ 999)
>>>>> >>>>mult =: (* is_palindrome"0) @ (*/)
>>>>> >>>>NB. Create list of 3 digit nums.
>>>>> >>>>list =. 100+ i.900
>>>>> >>>>
>>>>> >>>>(>./)@:, list mult list
>>>>> >>>>Although I'm glad got the answer, I'm wondering if it could be made
>>>>> >>>>terser, or if there is a much terser way to solve it? I went for a
>>>>> >>>>brute
>>>>> >>>>force approach.
>>>>> >>>>For comparison here is a Haskell answer I found:
>>>>> >>>>maximum[a*b|a<-[100..999],b<-[a..999],reverse(show(a*b))==show(a*b)]
>>>>> >>>>Regards.
>>>>> >>>>----------------------------------------------------------------------
>>>>> >>>>For information about J forums seehttp://www.jsoftware.com/forums.htm
>>>>> >>>>
>>>>> >>>>
>>>>> >>>>----------------------------------------------------------------------
>>>> >>>For information about J forums seehttp://www.jsoftware.com/forums.htm
>>>> >>>
>>> >>
>>> >>
>> >----------------------------------------------------------------------
>> >For information about J forums seehttp://www.jsoftware.com/forums.htm
>> >
>> >----------------------------------------------------------------------
>> >For information about J forums seehttp://www.jsoftware.com/forums.htm
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