One nice thing about 5 p: over totient is that it returns integer instead
of float.

   3!:0 totient 10
8
   (3!:0) 5 p: 10
4

On Wed, Mar 15, 2017 at 10:25 AM, 'Mike Day' via Programming <
[email protected]> wrote:

> Had a new look at the problem.  Reworked verbs as I'd not kept the
> old.  First pass,  pretty brutal,  took ~100 sec,  in j806beta with avx.
> Later efforts to speed it up were worse!  Evidently better to compare
> sorted character reps of numbers than sorted base 10 decodes.
>
> M
>
> Please reply to [email protected].
> Sent from my iPad
>
> > On 15 Mar 2017, at 07:48, 'Mike Day' via Programming <
> [email protected]> wrote:
> >
> > Only just up - having breakfast!  I evidently solved it,  as the
> project's discussion thread
> > is open to me,  but I don't have a record of my method.
> >
> > Early-ish PE problems were amenable to brute force,  and I expect that's
> how I would have
> > done it.
> >
> > AFAIrecall, later problems involving phi require insight how phi is
> calculated:
> >
> >   if n = */p^q,         phi(n) = */(p^(q-1)) * p-1
> > eg
> >    q:360
> > 2 2 2 3 3 5
> >    p =: 2 3 5
> >    q =: 3 2 1
> >
> >   */(p^q-1) * p-1
> > 96
> >
> >   5 p: 360
> > 96
> >
> > so n/phi(n) is
> >   */@:(%<:) p
> > 3.75
> >
> > No time for more - away from wifi for 10 hours or so!
> > Mike
> >
> >
> >
> >> On 15/03/2017 03:09, Don Guinn wrote:
> >> My second approach was brute force like you did. It gave the same
> number as
> >> my first approach. And interestingly it was faster than my first
> approach.
> >> Happens sometimes. The only thing I can think of is that I found the
> answer
> >> but I didn't supply what they wanted. I supplied the first number that
> had
> >> as a permutation of the digits given sorted by (%totient)n . I cheated.
> >> Stole totient from J phrases.
> >>
> >> The only thing I can think of is that they wanted something other than
> the
> >> n that I found.
> >>
> >> On Tue, Mar 14, 2017 at 8:12 PM, 'Jon Hough' via Programming <
> >> [email protected]> wrote:
> >>
> >>> I just tried it and got the right answer. But my approach is
> essentially
> >>> brute force:
> >>> I basically stringified  (":) the totient result, sorted it, and
> compared
> >>> to the sorted stringified original number.
> >>>
> >>> I can be more specific if you like.
> >>>
> >>> Regards,
> >>> Jon
> >>> --------------------------------------------
> >>> On Wed, 3/15/17, Don Guinn <[email protected]> wrote:
> >>>
> >>>  Subject: [Jprogramming] Project Euler
> >>>  To: "Programming forum" <[email protected]>
> >>>  Date: Wednesday, March 15, 2017, 9:37 AM
> >>>
> >>>  Has anyone out there solved problem
> >>>  70? I have worked it two ways which
> >>>  give the same answer but it is given as incorrect. I don't
> >>>  want to divulge
> >>>  what I did as that is against their rules. I must be missing
> >>>  something and
> >>>  presenting the wrong number for the result. Or is it
> >>>  possible that their
> >>>  answer is wrong?
> >>>
> >>>  Glad to discuss it in the forum, but if anyone wants to
> >>>  contact me
> >>>  privately so we don't break Project Euler rules, contact me
> >>>  at
> >>>  [email protected]
> >>>
> >>>  Thanks.
> >>>  ------------------------------------------------------------
> ----------
> >>>  For information about J forums see http://www.jsoftware.com/
> forums.htm
> >>> ----------------------------------------------------------------------
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> >> ----------------------------------------------------------------------
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> >
> >
> >
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