Changes

 Steps:* Making The best way to attack the IV would be to repeat the 1-bit conceptual attack feasible** Capture a bunch for each of the bits. Try to do this yourself! (500?Really!)** Apply decryption** Look at one bit** Find means + plot** Find differences + plot* Automating the attack** Finding the attack points** Getting If you're stuck, here are a single bitfew hints to get you going:** Building One easy way of looping through the IV bytes* Full script in appendix Examplebits is by using two nested loops, like this: <pre>#Imports for IV Attackfrom Crypto.Cipher import AES def initPreprocessingbyte in range(self16): self.preProcessingResyncSAD0 = preprocessing.ResyncSAD.ResyncSADfor bit in range(self.parent8): self.preProcessingResyncSAD0.setEnabled # Attack bit number (Truebyte*8 + bit) self.preProcessingResyncSAD0.setReference(rtraceno=0, refpoints=(6300,6800), inputwindow=(6000,7200))</pre> self* The sample that you'll want to look at will depend on which byte you're attacking.preProcessingResyncSAD1 We had success when we used <code>location = preprocessing.ResyncSAD.ResyncSAD(self.parent) self.preProcessingResyncSAD1.setEnabled(True) self.preProcessingResyncSAD1.setReference(rtraceno=051 + byte*60</code>, refpoints=(4800,5100), inputwindow=(4700,5200)) selfbut your mileage will vary.preProcessingList = [self.preProcessingResyncSAD0,self.preProcessingResyncSAD1,] return self.preProcessingList* The bitshift operator and the bitwise-AND operator are useful for getting at a single bit:<pre>class AESIVAttack(object):# This will either result in a 0 or a 1 numSubKeys checkIfBitSet = 16  @staticmethod def leakage(textin, textout, guess, bnum, setting, statebyteToCheck >> bit):& 0x01 knownkey = [0x94, 0x28, 0x5D, 0x4D, 0x6D, 0xCF, 0xEC, 0x08, 0xD8, 0xAC, 0xDD, 0xF6, 0xBE, 0x25, 0xA4, 0x99,</pre> 0xC4If you're ''really, 0xD9really'' stuck, 0xD0, 0x1E, 0xC3, 0x40, 0x7E, 0xD7, 0xD5, 0x28, 0xD4, 0x09, 0xE9, 0xF0, 0x88, 0xA1] knownkey = str(bytearray(knownkey)) ct = str(bytearray(textin))  aes = AESthere's a working attack in Appendix ?.new(knownkey, AES.MODE_ECB) pt = aes.decrypt(ct) return getHW(bytearray(pt)[bnum] ^ guess)You should find that the secret IV is <code>C1 25 68 DF E7 D3 19 DA 10 E2 41 71 33 B0 EB 3C</precode>.