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 assignme
... 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
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 us
https://adventofcode.com/2021/day/25
This is the end of the advent of code series for last year. It was a
quick wrap up.
Here, the puzzle was about waiting for sea cucumbers to get out of the
way so you could land your submarine and launch santa's sleigh (which
makes about as much sense as santa,
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 oth
I've got this solved now,
Eugene Nonko put me on the right path here.
The trick, from my perspective, was to convert the machine code into
executable J. And, to ensure that I could use J's dimensionality to
manage multiple tests at the same time.
Specifically, I parameterized the three "block s
