I need to learn how to recurse and fold in J. If a problem strikes my brain as
requiring iteration to solve, and a solution with the power conjunction (^:)
isn't obvious, I quickly resort to the explicit control structures as in my
solution below.
Mike
parse =: "."0 ;._2
freqs =: # (- ,. ]) +/
power =: {{)m
yfreqs =. freqs y
gamma =. #. </"1 yfreqs
epsilon =. #. >/"1 yfreqs
gamma ; epsilon ; gamma * epsilon
}}
search =: {{)a
mask =. = u/@:freqs
filter =. mask@:({~"1) # [
report =. y
position =. 0
while. (# report) > 1 do.
report =. report filter position
position =. >: position
end.
#. report
}}
o2 =: <: search
co2 =: > search
life =: o2 * co2
--
Michael P. Manti
[email protected]
> On Dec 24, 2021, at 05:31, Stefan Baumann <[email protected]> wrote:
>
> Hello Raul,
> Not day 2 but the day 3 puzzle lead me to practicing the fold conjunction:
>
> rd=: "."0;._2
> ]d=: rd 0 : 0
> 00100
> 11110
> 10110
> 10111
> 10101
> 01111
> 00111
> 11100
> 10000
> 11001
> 00010
> 01010
> )
> (*&#. -.) (-:@# < +/) d NB. (*) power consumption
> NB. Rating adverb r: u=.<: oxygen generator, u=.> CO2 scrubber
> r=: {{ x #.F..((] #~ [: (= -:@# u +/) {"1)^:(1<#@])) y }}
> ((<:r * >r) i.@{:@$) d NB. (**) life support rating
>
> Was only wondering if the r adverb could be written tacitly - currently
> have a hard time understanding
> these modifier trains...
>
> Thanks. Stefan.
>
>
>
>
>> On Thu, Dec 23, 2021 at 5:35 PM Raul Miller <[email protected]> wrote:
>>
>> previous message:
>> http://jsoftware.com/pipermail/programming/2021-December/059435.html
>> AoC day 3 details: https://adventofcode.com/2021/day/3
>>
>> This is a somewhat cleaned up version of my AoC day 3 implementation.
>> (This was my first "midnight attempt" which lends itself to illegible
>> code.)
>>
>> parse=: {{_".1j1#y}};._2
>> sample=: parse {{)n
>> 00100
>> 11110
>> 10110
>> 10111
>> 10101
>> 01111
>> 00111
>> 11100
>> 10000
>> 11001
>> 00010
>> 01010
>> }}
>>
>> NB. "part a"
>> a3=: powerConsumption=:{{
>> gamma=. (0.5 < +/%#) y
>> (*&#. -.) gamma
>> }}
>>
>> NB. "part b"
>> b3=: lifeSupportRating=:{{
>> (<: filter y) *&#. > filter y
>> }}
>>
>> NB. recursive
>> filter=:{{
>> if. 1=#y do. ,y return.end.
>> bits=. {."1 y
>> bit=. (0.5 u +/%#) bits
>> if. 1 < {:$y do.
>> bit,u filter (bit=bits)#}."1 y
>> else.
>> bit
>> end.
>> }}
>>
>> The part b puzzle smells an awful lot like a median or mode
>> calculation, and there might be a simpler approach.
>>
>> Nothing fancy here, but it works...
>>
>> --
>> Raul
>> ----------------------------------------------------------------------
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>>
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