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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