Many thanks, that helped a lot.
Stefan.
On Fri, Dec 24, 2021 at 1:31 PM Elijah Stone <[email protected]> wrote:
> Not having looked at the problem very closely, I notice the adverb is only
> ever passed <: and >. So, I think a potential simplification could come
> by instead conditionally negating the result of the comparison. That
> would let you replace (= -:@# u +/) with (m = ] = -:@# > +/) (where m is 0
> or 1), which is easier to express tacitly and could even be rewritten as a
> verb.
>
> (Or you could say u@:(] = ...), where u is either -. or ]. That would
> also be easy to make tacit, as it is just ]: @: (] = ...), but is
> obviously harder to make a verb out of.)
>
> -E
>
> On Fri, 24 Dec 2021, Stefan Baumann 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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