Yeah, it is for PE - ive done the other pandigital problems in C, but i wanted 
to try this one in J.
My biggest problem seems to be trying to flatten the pandigitals. E.g.

9 8 7 6 5 4 3 2 1

I need to turn into 987654321. Is it possible to do that or do I need to do
1+2*10+3*100+...+9*100000000?

___________________________

David Vaughan

On 8 Jul 2011, at 12:38, Ric Sherlock <[email protected]> wrote:

> Project Euler?
> 
> To get the permutations in reverse order you could reverse the string
> or reverse the list of permutation indicies. i.e.
>   0 1 2 3 A. 'cba'
>   3 2 1 0 A. 'abc'
> 
> In J you are better off testing the whole array of permutations rather
> than looping through them, so
>   (1&p: # ]) i. 50
> or
>   (#~ 1&p:) i. 50
> 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47
> 
> On Fri, Jul 8, 2011 at 10:53 PM, David Vaughan
> <[email protected]> wrote:
>> I'm trying to find the highest 1-9 pandigital prime, so I need to get the 
>> permutations in reverse order if possible. Also, how can I apply my 
>> primeTest script to the results?
>> 
>>   primeTest =: 3 : 'if. #@:q:y do. y end. '
>> This is what I have so far. The rank of y and of the list of permutations 
>> don't match up, and it feels like my 'if.' approach isnt great.
>> 
>> Can anyone offer any advice as to how achieve this?
>> 
>> Thanks.
>> 
>> ___________________________
>> 
>> David Vaughan
>> 
>> On 8 Jul 2011, at 11:29, Ric Sherlock <[email protected]> wrote:
>> 
>>> (i.@!@# A. ]) 'abcd'
>>>  or
>>> (A.~ i.@!@#) 'abcd'
>>> 
>>> see also:
>>> http://rosettacode.org/wiki/Find_the_missing_permutation#J
>>> 
>>> On Fri, Jul 8, 2011 at 9:44 PM, David Vaughan
>>> <[email protected]> wrote:
>>>> 
>>>> How would you go about getting all the permutations of a string in J?
>>>> ___________________________
>>>> 
>>>> David Vaughan
>>>> ----------------------------------------------------------------------
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>>>> 
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