And... speaking of spiral neighbors -- yeah... not paying attention...
and that looks like an inductive thing...
So...
spiral =: ,~ $ [: /: }.@(2 # >:@i.@-) +/\@# <:@+: $ (, -)@(1&,)
aspiral=: (*: - spiral)@>.@%:
coord=: $@] #: I.@,@e.~
steps=:3 :0"0 NB. part 1
as=. aspiral y
+/|-/(1,y) coord as
)
fibspi=:3 :0"0 NB. part 2
as=. <:aspiral y+25
seq=. ;(],+/@:{) ::]L:0/|.1; }.(/:,as) {&, _1 _1|.3 3 <@(#~ ] <4&{
::_)@,;.3 as
seq{~(seq <:y)i.0
)
Beware email induced line wrap (if you see an explicit line which is
not the last line and which is not an assignment, it's really part of
the previous line).
Probably can be simplified... and painfully slow.
Now to go back and read whether anyone already posted this and I guess
try things and see which approaches are faster...
Sorry about the noise,
--
Raul
On Mon, Dec 4, 2017 at 7:11 AM, Raul Miller <[email protected]> wrote:
> Hmmm... actually, my spiral was oriented "wrong" - which doesn't
> matter, which means that I did not need the |."1 in it...
>
> Anyways... looks like I was not paying enough attention...
>
> --
> Raul
>
>
> On Mon, Dec 4, 2017 at 7:05 AM, Raul Miller <[email protected]> wrote:
>> I used Eugene's approach, also:
>>
>> spiral =: ,~ $ [: /: }.@(2 # >:@i.@-) +/\@# <:@+: $ (, -)@(1&,)
>> aspiral=: (*: - |."1@spiral)@>.@%:
>> steps=:3 :0"0
>> as=. aspiral y
>> +/|-/(($as)#:I.,1=as),($as)#:I.,y=as
>> )
>>
>> steps 1 12 23 1024
>> 0 3 2 31
>>
>> FYI,
>>
>> --
>> Raul
>>
>> On Mon, Dec 4, 2017 at 2:02 AM, Brian Schott <[email protected]> wrote:
>>> I have only worked on part 1 so far.
>>> All my verbs are monadic and take the number in question as its argument.
>>>
>>>
>>> evenQ =. = <.&.-:
>>> size =. (+evenQ)@>.@%: NB. edge length
>>> border =. size-&*:_2+size NB. border length (number)
>>> lowest =. 1 _3 4& p. @ -:@<:@size NB. border's number which needs
>>> lowest # of moves
>>> least4 =. lowest+ (i. 4)*4%~border NB. border's 4 lowest # of moves
>>> least =. -:@<:@size NB. least # of moves
>>> addto =: least4<./@:(|@-)] NB. additional # of moves
>>> total =. least + addto
>>> total"0]1 12 23 1024
>>> 0 3 2 31
>>>
>>> I found the polynomial in `lowest` by studying the pattern of numbers 1 2
>>> 11 28 53
>>> as follows. (The fact that the 2nd differences are constantly 8 lead me to
>>> the
>>> solution.)
>>>
>>> 2-/\(^: 0 1 2)1 _3 4 p. i. 5
>>> 1 2 11 28 53
>>> _1 _9 _17 _25 0
>>> 8 8 8 0 0
>>>
>>>
>>> I had seen Eugene's evolute and now WRAP, so maybe the second part is
>>> doable, but I'm tired.
>>>
>>> --
>>> (B=) <-----my sig
>>> Brian Schott
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
