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Solution to Fog Creek's developer puzzle
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''' | |
http://www.fogcreek.com/jobs/dev | |
Find a 9 letter string of characters that contains only letters from: acdegilmnoprstuw | |
such that the hash(the_string) is: 945924806726376 | |
if hash is defined by the following pseudo-code: | |
Int64 hash (String s) { | |
Int64 h = 7 | |
String letters = "acdegilmnoprstuw" | |
for(Int32 i = 0; i < s.length; i++) { | |
h = (h * 37 + letters.indexOf(s[i])) | |
} | |
return h | |
} | |
For example, if we were trying to find the 7 letter string where hash(the_string) was 680131659347, the answer would be "leepadg". | |
''' | |
LETTERS = "acdegilmnoprstuw" | |
def hash(s): | |
h = 7 | |
for i in range(len(s)): | |
h = h * 37 + LETTERS.index(s[i]) | |
return h | |
def unhash(h, length): | |
h -= 7 * 37 ** length | |
return _unhash(h, "", length) | |
def _unhash(h, s, length): | |
if h == 0 and len(s) == length: | |
return s | |
elif h < 0 or len(s) == length: | |
return False | |
for i in reversed(range(len(LETTERS))): | |
new_h = h - (i * (37 ** (length - len(s) - 1))) | |
result = _unhash(new_h, s+LETTERS[i], length) | |
if result: return result | |
hash = 680131659347 | |
print(str(hash) + " >> " + unhash(hash, 7)) | |
hash = 945924806726376 | |
print(str(hash) + " >> " + unhash(hash, 9)) |
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