Also, after reflection, it turns out that getting rid of the if
statement was straightforward:
BLOCK=: {{
W=: x{::digits
X7=: W~:/(x{B)+26|y
((W+x{A)*X7)+(1+25*X7)*"(X7 <.&#&$ t)t=.<.y%x{C
}}
By the way, in case it was not obvious -- the 7 (in X7) refers to the
line number of that assignment to the X register. (Also the result
here and, thus, the argument, is a Z register assignment. Despite all
that register verbiage, the data flow here was quite straight
forward.)
And, ... there's probably other good simplifications, just waiting for
me to notice them.
FYI,
--
Raul
On Sat, Jan 15, 2022 at 4:18 PM 'Michael Day' via Programming
<programm...@jsoftware.com> wrote:
>
> ... and it's one of those "days" that part 2 is ridiculously easy cf
> part 1 !!!
>
> Mike
>
> On 15/01/2022 21:15, 'Michael Day' via Programming wrote:
> > Yes, that's my guess. I called them p q & r, and found that p = 1 or
> > 26 as q is positive or negative.
> >
> > So: having got one machine to run 14 times on different values of p q
> > r and the previous Z, I worked back
> > from Z = 0 output on the 14th machine, saving which W & Zs gave rise
> > to 0, then using those Zs on the
> > 13th machine ....
> >
> > But that gave me potentially a lot of Zs - I haven't counted - so I
> > then worked forward from the first machine
> > finding only 1452 14-ples giving rise to a final zero. Very messy - I
> > didn't really code it - but it worked!
> >
> > Haven't got round to part 2 yet.
> >
> > Mike
> >
> > On 15/01/2022 18:12, Raul Miller wrote:
> >> Actually, another caution here: it might be that other people's puzzle
> >> inputs used a different block structure from my block structure.
> >>
> >> Probably not -- it's probably the case that only my A, B and C values
> >> differed for different people. But .. that's just "probably" because I
> >> haven't looked at other people's puzzle input.
> >>
> >> FYI,
> >>
> >
>
>
> --
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