Python numbers obfuscation can be done the next way:
>>> {} > [] # i.e. 'd' > 'l' ([d]ict > [l]ist)
False
>>> [] > {} # i.e. 'l' > 'd' ([l]ist > [d]ict)
True
And it's possible to get values 0 and 1 from False and True, by implicit converting bool to int.
>>> False == 0
True
>>> True == 1
True
Other numbers could be received by applying logical operations (like "~" = bitwise NOT, or "<<" bitwise left shift).
For example 2:
2 = -(-2) = -(~1)
Let's take 16 or more first numbers, and all other numbers could be obtained by sum and/or multiplication of these first numbers.
nums = [
'([]<{})', # 0
'([]>{})', # 1
'-~([]>{})', # 2
'-~-~([]>{})', # 3
'-~-~-~([]>{})', # 4
'-~-~-~-~([]>{})', # 5
'-~-~-~-~-~([]>{})', # 6
'-~-~-~-~-~-~([]>{})', # 7
'-~-~-~-~-~-~-~([]>{})', # 8
'~-~([]>{})*~-~([]>{})', # 9
'~-~-~-~([]>{})*~([]>{})', # 10
'-~-~(~-~([]>{})*~-~([]>{}))', # 11
'(-~-~([]>{})<<-~([]>{}))', # 12
'-~(-~-~([]>{})<<-~([]>{}))', # 13
'-~-~-~-~-~-~([]>{})*-~([]>{})', # 14
'~-~([]>{})*~-~-~-~([]>{})', # 15
'(-~([]>{})<<-~-~([]>{}))', # 16
]
numbers_count = len(nums) - 1
def get_number(num):
r = ''
if num > numbers_count:
if num // numbers_count > 1:
if num % numbers_count == 0:
r += '(%s*%s)' % (get_number(num // numbers_count), numbers_count)
else:
r += '(%s*%s+%s)' % (get_number(num // numbers_count), numbers_count, num % numbers_count)
else:
r += '(%s+%s)' % (numbers_count, get_number(num - numbers_count))
else:
r = '%s' % num
return r
def convert_nums(s):
res = s
for n in xrange(numbers_count, -1, -1):
res = res.replace(str(n), nums[n])
res = res.replace('+-', '-')
return res
Example:
num = 1234567
num_str = get_number(num)
num_str = num_str[1:-1] if num_str[0] == '(' and num_str[-1] == ')' else num_str
print num_str
#((((16+2)*16+13)*16+6)*16+8)*16+7
obfuscated = convert_nums(num_str)
print obfuscated
#(((((-~([]>{})<<-~-~([]>{}))-~([]>{}))*(-~([]>{})<<-~-~([]>{}))-~(-~-~([]>{})<<-~([]>{})))*(-~([]>{})<<-~-~([]>{}))-~-~-~-~-~([]>{}))*(-~([]>{})<<-~-~([]>{}))-~-~-~-~-~-~-~([]>{}))*(-~([]>{})<<-~-~([]>{}))-~-~-~-~-~-~([]>{})
print eval(obfuscated)
#1234567
So here the number 1234567 was successfully obfuscated and restored using eval function.
1234567 = ((((16+2)*16+13)*16+6)*16+8)*16+7 = (((((-~([]>{})<<-~-~([]>{}))-~([]>{}))*(-~([]>{})<<-~-~([]>{}))-~(-~-~([]>{})<<-~([]>{})))*(-~([]>{})<<-~-~([]>{}))-~-~-~-~-~([]>{}))*(-~([]>{})<<-~-~([]>{}))-~-~-~-~-~-~-~([]>{}))*(-~([]>{})<<-~-~([]>{}))-~-~-~-~-~-~([]>{})
For negative numbers you only need to change the sign in front of the parentheses.
But for huge numbers like 12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789, will be an error:
_push: parser stack overflow
Traceback (most recent call last):
File "obfuscation.py", line 51, in
print eval(obfuscated)
MemoryError
So be careful and use nested parentheses carefully :)
The presentation with updated version of obfuscator (added Python 3 support):
http://speakerdeck.com/delimitry/numbers-obfuscation-in-python
