For simple arithmetic, you can get precedence into an expression using
stringreplace
For example, define:
add=: +
sub=: -
mult=: {{m*n}}
div=: {{m%n}}
Then string replace to use the names instead of the symbols at
evaluation time (make sure to throw in spaces on each side of the
names, just in case).
FYI,
--
Raul
On Sat, Dec 19, 2020 at 1:48 PM David Lambert <[email protected]> wrote:
>
> Otherwise, I've solved all with j.
>
> --- Day 18: Operation Order ---
>
> Part 2:
>
> Operator overloading usually doesn't include precedence alterations. I
> abandoned j and used python. Swapping operators and wrapping the
> numbers with a class that evaluates + as * (and multiplies for +), I
> evaluated strings rewritten from '1 + 2 * 3' to 'screwy(1) * screwy(2) +
> screwy(3)' .
>
> class screwy:
> def __init__(a, b): a.a = b
> __add__= lambda a,b: screwy(a.a * b.a)
> __mul__= lambda a,b: screwy(a.a + b.a)
>
> print(eval('screwy(1) * screwy(2) + screwy(3)').a)
>
>
>
> --- Day 15: Rambunctious Recitation ---
>
> Part 2:
>
> My j solution already appears in the programming forum as sparse array
> trouble. (IIRC) Some years ago Roger had asked for sparse array problem
> reports, which must be so because he responded. Anyway, my formulation
> was slow in j so I translated it to c. The sparse array needn't be
> sparse, and that code runs the thirty million test case in 37 seconds on
> the high end processor option at the time Thinkpad W540.
>
> For 2020 player turns the program with sparse array runs 5 times slower
> than a dense array version. With a sparse table I'm unwilling to wait
> for timing results of the thirty million loops.
>
>
> Today it looks like we're to construct a regular expression from an
> hideous tree.
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
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