Edit: I will be posting a newer version of this question on meta-golf soon. Stay tooned!

Edit #2: I will no longer be updating the challenge, but will leave it open. The meta-golf version is available here: https://codegolf.stackexchange.com/questions/106509/obfuscated-number-golf


Most numbers can be written with only 6 different symbols:

  • e (Euler's Constant)
  • - (Subtraction, Not Negation)
  • ^ (Exponentiation)
  • (
  • )
  • ln (Natural Logarithm)

For example, you could convert the imaginary number i using this equation:



Given any integer k through any reasonable means, output the shortest representation possible of that number using only those 6 symbols.


0 => "e-e"
1 => "ln(e)"
2 => "ln(ee)"
// Since - cannot be used for negation, this is not a valid solution: 
// ln(e)-(-ln(e))
-1 => "e-e-ln(e)"


  • Ending parenthesis count towards the total amount of characters.
  • ln( only counts as 1 character.
  • Everything else counts as 1 character.
  • n^0=1
  • Order of operations apply
  • Parenthesis multiplying is acceptable, e.g. (2)(8)=16, 2(5)=10, and eln(e)=e.
  • ln e is not valid, you must do ln(e)
  • 3
    \$\begingroup\$ I think that formula (ln(ee...e)) is the best way to portray positives. Edit: no, its not. ln(e^(ln(eeeee)ln(eeee))) is better for 20 \$\endgroup\$ Commented Jan 10, 2017 at 17:59
  • 6
    \$\begingroup\$ @JulianLachniet love the idea, would like to see the first 10-20 terms of the sequence requested though. Maybe put up an example for -10 to 10 for clarification. WheatWizard has already poked a couple holes, with those holes the objective criteria of "shortest possible" is hard to determine without concrete examples. \$\endgroup\$ Commented Jan 10, 2017 at 18:03
  • \$\begingroup\$ Not sure about some of the higher ones, especially 20. \$\endgroup\$ Commented Jan 10, 2017 at 18:05
  • 2
    \$\begingroup\$ ln(eeee)^ln(ee) is a shorter than ln(eeeeeeeeeeeeeeee) for 16 \$\endgroup\$
    – Wheat Wizard
    Commented Jan 10, 2017 at 18:17
  • 8
    \$\begingroup\$ Just a word of suggestion. I think this might be more fun as a meta-golf challenge than as a code-golf challenge. Its really hard to demonstrate that some code always produces the optimal result so it might be better to score answers on how well they golf their output. \$\endgroup\$
    – Wheat Wizard
    Commented Jan 10, 2017 at 18:19

1 Answer 1


Python 3, 402 bytes

from itertools import*
from ast import*
from math import*
v,r=lambda x:'UnaryOp'not in dump(parse(x)),lambda s,a,b:s.replace(a,b)
def l(x,y):
    for s in product('L()e^-',repeat=x):
        while g!=f:f,g=g,r(g,'ee','e*e')
            if eval(g)==y and v(g):return g
def b(v):
    while 1:
        if r:return r

Example usage:

>>> b(1)
>>> b(0)
>>> b(-3)
>>> b(8)

Note that although the output format may not reflect it, the code properly counts all lengths according to the question's specifications.

This is a dumb bruteforce through all the possible lengths of strings. Then I use some replacements so that Python can evaluate it. If it's equal to what we want, I also check to exclude unary negative signs by checking the AST.

I'm not very good at golfing in Python, so here's the semi-ungolfed code if anybody wants to help!

from itertools import*
from ast import*
from math import*

def valid(ev):
    return 'UnaryOp' not in dump(parse(ev))

def to_eval(st):
    f = ''.join(st).replace('L', 'log(').replace(')(', ')*(').replace('^', '**')
    nf = f.replace('ee', 'e*e')
    while nf != f:
        f, nf = nf, nf.replace('ee', 'e*e')
    return nf

def try_length(length, val):
    for st in product('L()e^-', repeat=length):
        ev = to_eval(st) 
            if eval(ev) == val and valid(ev):
                return st

def bruteforce(val):
    for i in range(11):
        res = try_length(i, val)
        if res:
            print(i, res)
            return res
  • \$\begingroup\$ Instead of indenting with tabs you can indent with spaces for one level of indent and tabs for 2. \$\endgroup\$
    – Wheat Wizard
    Commented Jan 12, 2017 at 15:08

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