On May 14, 2015, at 6:10 PM, R. K. Belew wrote:
> i'm trying to make sure i understand just what `calcfp()` is
> doing, by trying to reconcile its two `fp` and `bits` return values.
> below is a short demo, showing i cannot. i'm sure i'm missing
> something simple, but don't know what!
In your pybelBits2binary you treat everything as little-endian, so bit 0 is the
furthest left, bit is second from the left, and so on.
>>> pybelBits2binary([1])
'1000000000000 ....
>>> pybelBits2binary([2])
'0100000000000 ....
>>> pybelBits2binary([3])
'0010000000000 ....
The format(num, '032b') formats a 32-bit integer value as big-endian, so given
a 32 bit word the value of 1 will set the rightmost bit, 2 will set the next
rightmost, etc.
>>> format(1, "032b")
'00000000000000000000000000000001'
>>> format(2, "032b")
'00000000000000000000000000000010'
>>> format(3, "032b")
'00000000000000000000000000000011'
You can see that both of the bits you generate have the same number of bits:
>>> from collections import Counter
>>> Counter(bits1)
Counter({'0': 979, '1': 45})
>>> Counter(bits2)
Counter({'0': 979, '1': 45})
The only difference is the arrangement. For each group of 32 bits, the one
fingerprint is in reverse order of the other:
>>> bits1[:32]
'00000000000000000000000000010000'
>>> bits2[:32][::-1]
'00000000000000000000000000010000'
>>> bits1[32:64]
'00010000000000000000000000000011'
>>> bits2[63:31:-1]
'00010000000000000000000000000011'
The simplest way to get the two fingerprints to match is to change how 'bits2'
is created, so that the bits for each integer are in little-endien order. The
current code is:
>>> bits2 = ''.join([format(num,'032b') for num in fp.fp])
>>> bits1 == bits2
False
and the change to make bits2 match bits1 is:
>>> bits2 = ''.join([format(num,'032b')[::-1] for num in fp.fp])
>>> bits1 == bits2
True
This will end up with a pure little-endian fingerprint.
By the way, if you want the byte-oriented version of the little-endian
fingerprint, where the first byte contains the first 8 bits (in big-endian
order), the second byte contains the next 8 bits, etc. then you can use the
struct module.
>>> byte_fp = struct.pack("<" + "I"*32, *fp.fp)
>>> byte_fp
'\x00\x00\x00\x08\x08\x00\x00\xc0\x00
\x06\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00"\x00\x00\x00\x00\x80\x00\x00\x00\x00\x04\x00@\x00\x00\x00\x00
\n\x02\x08\x00\x00\x00@\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x02\x00\x00\x00\x00\x08\x00\x08\x00@\x00\x02\x00\x00\x00\x00\x00\x10\x80\x00\x00\x00\x90\x00\xc0\x04\x00\x00\x00@\x08\x00\x00\x00\x01\x00@\x00\x00\x00\x00\x00\x00\x00\x00\x00@\x01\x00\x00\x00\x00\x00\x00\x02\x01\x00\x00\x01\x00\x00\x00\x88\x00@\x02@\x00\x00'
I'll construct a simple translation table to convert a byte into its
little-endian representation
>>> T = {chr(i): format(i, "08b")[::-1] for i in range(256)}
and use that table to show that the new byte_fp is the same as bits1.
>>> bits1 == "".join(T[byte] for byte in byte_fp)
True
Best regards,
Andrew
da...@dalkescientific.com
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