Code Golf Asked on October 27, 2021
When encountering a problem whose input is meant to be read sequentially, but you may also want to push back onto the input like a stack, which languages have the optimal boilerplates for doing so, and how?
For example, from Create a Boolean Calculator, we have a string like:
1AND0OR1XOR1
We want to "pop" 5 chars 1AND0
, compute it to be 0
, then "push" it back to form:
0OR1XOR1
Then, repeat until there’s only 1 char left. In this case:
0
It’s trivial to write while
loops and such, but this seems common enough in codegolf problems that there must be canonical forms. I wasn’t able to find a tips question specifically about this setup though, and going through each tips question by language was difficult.
The approach you're asking seems like reduce/left fold in general. Many languages have this, such as Python (reduce(f,seq)
or functools.reduce(f,seq)
), APL (f⍨/⌽seq
), Jelly (f/seq
), and Haskell (foldl f start seq
).
As a Python example, let's assume we already have the input parsed as a list seq=[1, 'AND', 0, 'OR', 1, 'XOR', 1]
. Then reduce(f,seq)
is equivalent to
f(f(f(f(f(f(1, 'AND'), 0), 'OR'), 1), 'XOR'), 1)
The trouble here is that we need to take 3 arguments at a time. A way this could be done is by grouping most of the sequence into pairs seq2=[1, ['AND',0], ['OR',1], ['XOR',1]]
, so reduce(f,seq)
would be equivalent to
f(f(f(1, ['AND',0]), ['OR',1]), ['XOR',1])
This could work well in Jelly because it has a builtin s
that could help split into pairs (output looks funny strings are internally lists of chars).
However, a loop-based approach would work better in Python by assigning to a slice of an array:
seq=[1, 'AND', 0, 'OR', 1, 'XOR', 1]
while len(seq)>1:
seq[1:3] = [f(*seq[1:3])]
print(seq[0])
This would output f(f(f(1, 'AND', 0), 'OR', 1), 'XOR', 1)
.
As @AdHocGarfHunter notes in the comments, recursion is a good idea too:
# (ungolfed)
def r(s):
if len(s)>1:
return r(f(*s[:3]) + s[3:])
else:
return s[0]
APL has little boilerplate for this: {1=⍴⍵:⊃⍵⋄∇(3↓⍵),f3↑⍵}
(∇
is the recursion). Jelly does too, with 1
byte recursion.
Answered by fireflame241 on October 27, 2021
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