moth/puzzles/crypto/180rotate.py

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Python
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2009-10-13 16:08:11 -06:00
import crypto
import itertools
width = 7
alice = b'''The key for this one was essentially 0 1 2 3 4 5 6 7. The key for the next puzzle is much stronger. I bet they're glad we're not also applying a substitution cypher as a secondary step. '''
bob = b'''I take that to mean it uses the same basic algorithm. I guess it won't be too hard then, will it? The key for this puzzle is this sentence'''
alice = alice.lower().replace(b' ', b'_')
bob = bob.lower().replace(b' ', b'_')
def rotate(text):
out = bytearray()
assert len(text) % width == 0, 'At %d of %d.' % (len(text) % width, width)
slices = [bytearray(text[i:i+width]) for i in range(0, len(text), width)]
nextSlice = slices.pop(0)
while len(out) < len(text):
if nextSlice:
out.append(nextSlice.pop(0))
slices.append(nextSlice)
nextSlice = slices.pop(0)
return bytes(out)
def unrotate(text):
out = bytearray()
assert len(text) % width == 0
slices = []
for i in range(len(text) // width):
slices.append([])
inText = bytearray(text)
while inText:
slice = slices.pop(0)
slice.append(inText.pop(0))
slices.append(slice)
for slice in slices:
out.extend(slice)
return bytes(out)
print(rotate(alice))
crypto.mkIndex(rotate, unrotate, alice, bob, crypto.groups)