Yes, I ended up indexing brackets with I. '()' =/ y , assuming that
the left & right brackets
were balanced!
Part 2 was long-winded, recursive, but evaluated quickly, luckily.
My slowest routine for any of these posers is for part 2 of Day 15; best
I managed was
~ 40 seconds.
Cheers,
Mike
On 18/12/2020 19:50, Joseph Novakovich wrote:
For part 1 I used pretty much the same approach, eval by way of
reversing (|.&.;:):
+/ {{".'('c}')'o}y['o c'=. I. 0 1=/'()'i.y=.|.&.;:y}}&> in
Mine differs mostly in avoiding library definitions, amending parens
by index. For part 2 I used ;: to calculate a depth vector for the
tokens and then I evaluate the deepest nested subexpressions at each
iteration reaching fixpoint.
https://github.com/jitwit/aoc/blob/a/J/20/18.ijs
On 12/18/20, 'Michael Day' via Programming <programm...@jsoftware.com> wrote:
Unlike in the K forum, the J-wires have been virtually silent on this
year's Advent of Code...
Anyway, today's problem, number 18, might interest J & APLers as part
1 is ALMOST
plug-it-in to J/APL; there's just one twist, so to speak, namely we
need to reverse an
arithmetic expression, eg
1 + (2 * 3) + (4 * (5 + 6))
needs to be evaluated as
((6 + 5) * 4) + (3 * 2) + 1
But, even though all numbers are single-digit, just reversing the
string doesn't work, fairly obviously:
|.'1 + (2 * 3) + (4 * (5 + 6))'
))6 + 5( * 4( + )3 * 2( + 1
So the brackets need swapping. The data is supplied as strings, so I
expected to use stringreplace,
as found in strings.ijs
But:
|. ( ')[' , '()' , '[(' ) stringreplace '1 + (2 * 3) + (4 * (5
+ 6))'
[[6 + 5) * 4) + [3 * 2) + 1
Why don't the intermediate '[' get changed to the final ')' ?
The second action in stringreplace is on the left hand argument:
_2 [\ ,(')[' , '()' , '[(' )
)[
()
[(
The first column should be items of old text and the second column
their new replacements. But the third pair seems to be ignored.
I did of course get round this little difficulty, but would have preferred
not to reinvent the wheel.
NB. Part 2 is a little more challenging, as + is defined there as taking
precedence over * .
Thanks,
Mike
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