mirror of https://github.com/dirtbags/moth.git
48 lines
1.3 KiB
Python
48 lines
1.3 KiB
Python
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import crypto
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import itertools
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width = 7
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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. '''
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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'''
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alice = alice.lower().replace(b' ', b'_')
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bob = bob.lower().replace(b' ', b'_')
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def rotate(text):
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out = bytearray()
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assert len(text) % width == 0, 'At %d of %d.' % (len(text) % width, width)
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slices = [bytearray(text[i:i+width]) for i in range(0, len(text), width)]
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nextSlice = slices.pop(0)
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while len(out) < len(text):
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if nextSlice:
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out.append(nextSlice.pop(0))
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slices.append(nextSlice)
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nextSlice = slices.pop(0)
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return bytes(out)
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def unrotate(text):
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out = bytearray()
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assert len(text) % width == 0
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slices = []
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for i in range(len(text) // width):
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slices.append([])
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inText = bytearray(text)
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while inText:
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slice = slices.pop(0)
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slice.append(inText.pop(0))
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slices.append(slice)
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for slice in slices:
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out.extend(slice)
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return bytes(out)
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print(rotate(alice))
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crypto.mkIndex(rotate, unrotate, alice, bob, crypto.groups)
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