Re: Testing XTEA



Thanks Mike, Woolfgang and Tom. The program works well with your plaintexts
and cyphertexts, but I still don't get anything meaningfule with the Keith
Lockstone vectors - so the problem must be my understanding of those
vectors.

Mike

"Mike Amling" <spamonly@xxxxxxxxxxx> wrote in message
news:en9h5f$2le@xxxxxxxxxxxxxxxxxxxxxxxxxx
Wolfgang Ehrhardt wrote:
On Sun, 31 Dec 2006 09:06:28 -0000, "Mike Simpson"
<mike@xxxxxxxxxxxxxxxxxx> wrote:

I assume that you mean that the relation of the key to the data may be

endednedness dependant. The endedness of any of the functions used in

XTEA sould be immaterial between any two equal sized variables. Sadly,

Even the only vectors I have got use a key of 4 zero values will have

the same value either way round.

Mike.



No, it's not this easy. Even if the key is endian invariant, the
ciphertexts are different for non-invariant plaintexts. There must be
some error in your code. You might compare your values to those I just
calculated with my code

key = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
plaintext = (1,35,69,103,137,171,205,239) or in hex
(0x01,0x23,0x45,0x67,0x89,0xab,0xcd,0xef) ciphertext big-endian =
(126,102,199,28,136,137,114,33)
ciphertext little-endian = (29,212,116,253,36,239,28,46)

For an endian-independent key and plaintext,
key=01020201 03040403 05060605 07080807 (hex)
plaintext=11222211 33444433 (hex)
ciphertext=9F775B19 D1C7121D (big-endian hex) for XTEA
ciphertext=F383187D 4B933379 (big-endian hex) for Block TEA

The OP should get ciphertext=195B779F 1D12C7D1 if it's just an endian
issue.

--Mike Amling


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