Caddyshack or: How to compress already-golfed code

Question

In golf, a knowledgable caddie helps an already good golfer to be even better. In this challenge, you build a "caddie" to help you code golf. In particular, your caddie will compress golfed code in such a way that another program can uncompress and then run the code. A good caddie will compress golfed code to be much smaller than the original golfed code. In fact, in caddie-assisted Python (say), you might be able to golf on par with a Jelly golfer!

Formally, a caddie for a language L is a pair (E,D) of programs. The program E takes as input any code c in L and outputs a string s. The program D takes s as input along with any input x of c and returns the output of c(x). We do not require E and D to be written in L. You could say that E is a transpiler that converts c into a version s that runs in the language D. We are interested in transpilers E for which E(c) tends to be smaller than c, even when c is already golfed.

Write a caddie (E,D) for any language L of your choosing. When you first post a caddie here, its score is infinity. Use the following method to update your score:

• Let C denote the challenge in code-golf with at least 10 points that was posted first after your submission.
• Let c denote the shortest posted solution to C that is written in L.
• Your score is infinity unless C and c both exist, at which point your score is size(E(c))/size(c).

Recompute your score whenever either C or c changes. Lowest score wins.

Answer 1

Python 3 Caddy, interprets BrainF***

Score: \$\frac{38}{224}\approx0.17\$ (with newlines removed)

Program E

from textwrap import wrap

x = input('Code: ')
i = ['<', '>', '+', '-', '[', ']', '.', ',']
o = ''
for c in x:
    if c in i:
        o += str(i.index(c) + 1)
if o:
    d = wrap(o, 6)
    d = [int(x) for x in d]
    o = ''
    for c in d:
        o += chr(int(c) + 128)
print('Compressed: ', o)

The way this works is by converting every instruction in the input to a digit, then concatenating those digits to a number. Then it splits that number into 6-digit groups and converts each to a Unicode character. (A number higher than 6 could be used, by that would cause errors for large programs). Then it outputs it.

Program D

x = input('Compressed: ')
p = input('Input: ')
p = list(p)
d = ''
for c in x:
    c = ord(c) - 128
    d += str(c)
d = d.replace('-', '')
i = ['<', '>', '+', '-', '[', ']', '.', ',']
o = ''
for n in d:
    n = int(n) - 1
    o += i[n]
tape = [0] * 30000
head = 0
m = {}
m2 = {}
stack = []
for i in range(len(o)):
    char = o[i]
    if char == '[':
        stack.append(i)
    elif char == ']':
        begin = stack.pop()
        m[begin] = i + 1
        m2[i] = begin
i = 0
while i < len(o):
    char = o[i]
    if char == '<':
        head -= 1
        i += 1
    elif char == '>':
        head += 1
        i += 1
    elif char == '+':
        tape[head] += 1
        i += 1
    elif char == '-':
        tape[head] -= 1
        i += 1
    elif char == '[':
        if tape[head] == 0:
            i = m[i]
        else:
            i += 1
    elif char == ']':
        if tape[head] != 0:
            i = m2[i]
        else:
            i += 1
    elif char == '.':
        print(chr(tape[head]), end='')
        i += 1
    elif char == ',':
        try:
            tape[head] = ord(p.pop(0))
        except:
            tape[head] = 0
        i += 1

This is pretty simple, it just reverses the compression algorithm and evaluates the result.