I did solve it, but I didn't use J for that one.
I wrote a translator in Perl that converted their assembly into C. C,
because it's easier to read, plus I thought that some brute forcing might
be required and compiled C will run faster than interpreting their assembly.
While studying that generat
https://adventofcode.com/2021/day/24
I have not completed the day 24 puzzle.
The day 24 puzzle has a sequence of instructions representing a
calculation to verify a model number (and conceptually enable features
based on that model number -- though part A of the puzzle does not
provide any detail
Thanks to Raul's very accurate observations, I patched two of my previously
posted solutions (The array based tacit one and the explicit one).
The approach based on :&max <.@:* <./@sh5)^:(max +./@:<: ,@])^:_ start
was wrong because: it stops when no more maxes are present, i.e. does not
consider pa
The rank of a matrix (in the linear algebra, not J sense) is tricky to
determine numerically because of stability issues: an arbitrarily small
perturbation may increase the rank. A good practical start is the singular
value decomposition, available through the LAPACK extension. This reduces th
Not sure this is the preferable way but one solution might be:
load 'math/misc/matfacto'
matrixrank=: #@(] -. 0 #~ #)@(1 {:: lud)
matrixrank _3 ]\ 1 2 3 5 4 6 9 7 8
3
matrixrank _3 ]\ 1 2 3 2 4 6 9 7 8
2
Note that "rank" has lots of alternative meanings in J other than
calculating the
#$m
On Thu, Jan 13, 2022 at 2:51 PM Pawel Jakubas
wrote:
> Dear J enthusiasts,
>
> I am wondering what is the preferable way to determine the rank of a matrix
> in J.
> I would expect here below
>
>]m=: 3 3 $ 1 2 3 5 4 6 9 7 8
> 1 2 3
> 5 4 6
> 9 7 8
> rank m
> 3
>
>]m=: 3 3 $ 1 2 3 2
Dear J enthusiasts,
I am wondering what is the preferable way to determine the rank of a matrix
in J.
I would expect here below
]m=: 3 3 $ 1 2 3 5 4 6 9 7 8
1 2 3
5 4 6
9 7 8
rank m
3
]m=: 3 3 $ 1 2 3 2 4 6 9 7 8
1 2 3
2 4 6
9 7 8
rank m
2
Thanks and cheers
Pawel Jakubas
-
This seems not to have appeared in the forum, so I'm sending again.
Apologies if it turns out to be a duplicate.
I've now modified reboot2a - it's now reboot2b (!) - to accumulate
volumes of repeated intersection cuboids.
It performs significantly better than reboot2a, whose time & space I
r
