[Jprogramming] Advent of code, Day 3 (spoilers probable)

[Raul Miller]

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 <rauldmiller at gmail.com> wrote:
> I used Eugene's approach, also:
>
> spiral =: ,~ $ [: /: }.@(2 # >:@i. at -) +/\@# <:@+: $ (, -)@(1&,)
> aspiral=: (*: - |."1 at 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 <schott.brian at gmail.com> 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
>> ----------------------------------------------------------------------
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