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sasc:laboratoare:03 [2015/03/19 15:02] marios.choudary |
sasc:laboratoare:03 [2017/03/07 15:32] (current) dan.dragan |
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- | ===== Laboratorul 03. ===== | + | ===== Lab 03 - PRGs ===== |
- | In this lab we shall do some exercises related to PRFs, PRPs and DES. | ||
- | Please check the course, available here: | ||
- | http://cs.curs.pub.ro/2014/pluginfile.php/13095/mod_resource/content/1/sasc_curs4.pdf | ||
- | ==== Exercise 1 ==== | + | ==== Exercise 1 (4p) ==== |
- | Let F : K × X → Y be a secure PRF with K = X = Y = {0, 1}<sup>n</sup>. | + | In this exercise we'll try to break a Linear Congruential Generator, that may be used to generate "poor" random numbers. |
+ | We implemented such weak RNG to generate a sequence of bytes and then encrypted a plaintext message. | ||
+ | The resulting ciphertext in hexadecimal is this: | ||
+ | <code> | ||
+ | a432109f58ff6a0f2e6cb280526708baece6680acc1f5fcdb9523129434ae9f6ae9edc2f224b73a8 | ||
+ | </code> | ||
- | * a) Show that F1(k,x) = F(k,x) || 0 is not a secure PRF. (for strings y and z we use y || z to denote the concatenation of y and z) | + | You know that the LCG uses the following formula to produce each byte: |
- | * b) Prove that F2(k, x) = F (k, x ⊕ 1<sup>n</sup>) is a secure PRF. Here x ⊕ 1<sup>n</sup> is the bit-wise complement of x. To prove security argue the contra-positive: a distinguisher A that breaks F2 implies a distinguisher B that breaks F and whose running time is about the same as A’s. | + | |
- | * c) Let K3 = {0, 1}<sup>n+1</sup>. Construct a new PRF F3 : K3 ×X → Y with the following property: the PRF F3 is secure, however if the adversary learns the last bit of the key then the PRF is no longer secure. This shows that leaking even a single bit of the secret key can completely destroy the PRF security property. | + | |
- | <note tip> | + | |
- | Hint: Let k3 = k || b where k ∈ {0,1}<sup>n</sup> and b ∈ {0,1}. Set F3(k3,x) to be the same as F (k, x) for all x != 0<sup>n</sup>. Define F3(k3, 0<sup>n</sup>) so that F3 is a secure PRF, but becomes easily distinguishable from a random function if the last bit of the secret key k3 is known to the adversary. Prove that your F3 is a secure PRF by arguing the contra-positive, as in part (b). | + | |
- | </note> | + | |
- | * d) Construct a new PRF F4 : K2 × X → Y that remains secure if the attacker learns any single bit of the key. Your function F2 may only call F once. Briefly explain why your PRF remains secure if any single bit of the key is leaked. | + | |
- | ==== Exercise 2 ==== | + | s_next = (a * s_prev + b) mod p |
- | Let's analyse some substitution-permutation networks (SPN). | + | where both s_prev and s_next are byte values (between 0 and 255) and p is 257. |
+ | Both a and b are values between 0 and 256. | ||
- | === SPN 1 === | ||
- | We have the SPN from this figure: | + | You also know that the first 16 letters of the plaintext are "Let all creation" and that the ciphertext was generated by xor-ing a string of consecutive bytes generated by the LCG with the plaintext. |
- | {{:sasc:laboratoare:spn_1r_reduced_2s.png|}} | + | |
- | where S denotes the AES S-box (we'll discuss this in some detail during the next lecture), and 'Permutation' is a simple permutation block that simply shifts the input 4 bits to the right as in a queue. Each input (x1, x2) is 8-bit (1 byte), as well as the keys k1, k2, and the outputs y1, y2. | + | Can you break the LCG and predict the RNG stream so that in the end you find the entire plaintext ? |
- | - How can you find the key ? | + | You may use this starting code: |
- | - Given the message/ciphertext pair ('Hi' - as characters, 0xba52 - as hex number), find the key bytes k1 and k2. Print them in ascii. | + | <code python 'ex1_weak_rng.py'> |
- | + | ||
- | <note tip> | + | |
- | For these exercises you can use the following helper/starter code: | + | |
- | </note> | + | |
- | + | ||
- | <code> | + | |
import sys | import sys | ||
import random | import random | ||
Line 41: | Line 30: | ||
import operator | import operator | ||
- | # Rijndael S-box | + | #Parameters for weak LC RNG |
- | sbox = [0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67, | + | class WeakRNG: |
- | 0x2b, 0xfe, 0xd7, 0xab, 0x76, 0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, | + | "Simple class for weak RNG" |
- | 0x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0, 0xb7, | + | def __init__(self): |
- | 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1, | + | self.rstate = 0 |
- | 0x71, 0xd8, 0x31, 0x15, 0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05, | + | self.maxn = 255 |
- | 0x9a, 0x07, 0x12, 0x80, 0xe2, 0xeb, 0x27, 0xb2, 0x75, 0x09, 0x83, | + | self.a = 0 #Set this to correct value |
- | 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3, 0x29, | + | self.b = 0 #Set this to correct value |
- | 0xe3, 0x2f, 0x84, 0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b, | + | self.p = 257 |
- | 0x6a, 0xcb, 0xbe, 0x39, 0x4a, 0x4c, 0x58, 0xcf, 0xd0, 0xef, 0xaa, | + | |
- | 0xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x02, 0x7f, 0x50, 0x3c, | + | |
- | 0x9f, 0xa8, 0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc, | + | |
- | 0xb6, 0xda, 0x21, 0x10, 0xff, 0xf3, 0xd2, 0xcd, 0x0c, 0x13, 0xec, | + | |
- | 0x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d, 0x64, 0x5d, 0x19, | + | |
- | 0x73, 0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee, | + | |
- | 0xb8, 0x14, 0xde, 0x5e, 0x0b, 0xdb, 0xe0, 0x32, 0x3a, 0x0a, 0x49, | + | |
- | 0x06, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62, 0x91, 0x95, 0xe4, 0x79, | + | |
- | 0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4, | + | |
- | 0xea, 0x65, 0x7a, 0xae, 0x08, 0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, | + | |
- | 0xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f, 0x4b, 0xbd, 0x8b, 0x8a, 0x70, | + | |
- | 0x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e, 0x61, 0x35, 0x57, 0xb9, | + | |
- | 0x86, 0xc1, 0x1d, 0x9e, 0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, | + | |
- | 0x94, 0x9b, 0x1e, 0x87, 0xe9, 0xce, 0x55, 0x28, 0xdf, 0x8c, 0xa1, | + | |
- | 0x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f, 0xb0, | + | |
- | 0x54, 0xbb, 0x16] | + | |
+ | def init_state(self): | ||
+ | "Initialise rstate" | ||
+ | self.rstate = 0 #Set this to some value | ||
+ | self.update_state() | ||
- | # Rijndael Inverted S-box | + | def update_state(self): |
- | rsbox = [0x52, 0x09, 0x6a, 0xd5, 0x30, 0x36, 0xa5, 0x38, 0xbf, 0x40, 0xa3, | + | "Update state" |
- | 0x9e, 0x81, 0xf3, 0xd7, 0xfb , 0x7c, 0xe3, 0x39, 0x82, 0x9b, 0x2f, | + | self.rstate = (self.a * self.rstate + self.b) % self.p |
- | 0xff, 0x87, 0x34, 0x8e, 0x43, 0x44, 0xc4, 0xde, 0xe9, 0xcb , 0x54, | + | |
- | 0x7b, 0x94, 0x32, 0xa6, 0xc2, 0x23, 0x3d, 0xee, 0x4c, 0x95, 0x0b, | + | def get_prg_byte(self): |
- | 0x42, 0xfa, 0xc3, 0x4e , 0x08, 0x2e, 0xa1, 0x66, 0x28, 0xd9, 0x24, | + | "Return a new PRG byte and update PRG state" |
- | 0xb2, 0x76, 0x5b, 0xa2, 0x49, 0x6d, 0x8b, 0xd1, 0x25 , 0x72, 0xf8, | + | b = self.rstate & 0xFF |
- | 0xf6, 0x64, 0x86, 0x68, 0x98, 0x16, 0xd4, 0xa4, 0x5c, 0xcc, 0x5d, | + | self.update_state() |
- | 0x65, 0xb6, 0x92 , 0x6c, 0x70, 0x48, 0x50, 0xfd, 0xed, 0xb9, 0xda, | + | return b |
- | 0x5e, 0x15, 0x46, 0x57, 0xa7, 0x8d, 0x9d, 0x84 , 0x90, 0xd8, 0xab, | + | |
- | 0x00, 0x8c, 0xbc, 0xd3, 0x0a, 0xf7, 0xe4, 0x58, 0x05, 0xb8, 0xb3, | + | |
- | 0x45, 0x06 , 0xd0, 0x2c, 0x1e, 0x8f, 0xca, 0x3f, 0x0f, 0x02, 0xc1, | + | |
- | 0xaf, 0xbd, 0x03, 0x01, 0x13, 0x8a, 0x6b , 0x3a, 0x91, 0x11, 0x41, | + | |
- | 0x4f, 0x67, 0xdc, 0xea, 0x97, 0xf2, 0xcf, 0xce, 0xf0, 0xb4, 0xe6, | + | |
- | 0x73 , 0x96, 0xac, 0x74, 0x22, 0xe7, 0xad, 0x35, 0x85, 0xe2, 0xf9, | + | |
- | 0x37, 0xe8, 0x1c, 0x75, 0xdf, 0x6e , 0x47, 0xf1, 0x1a, 0x71, 0x1d, | + | |
- | 0x29, 0xc5, 0x89, 0x6f, 0xb7, 0x62, 0x0e, 0xaa, 0x18, 0xbe, 0x1b , | + | |
- | 0xfc, 0x56, 0x3e, 0x4b, 0xc6, 0xd2, 0x79, 0x20, 0x9a, 0xdb, 0xc0, | + | |
- | 0xfe, 0x78, 0xcd, 0x5a, 0xf4 , 0x1f, 0xdd, 0xa8, 0x33, 0x88, 0x07, | + | |
- | 0xc7, 0x31, 0xb1, 0x12, 0x10, 0x59, 0x27, 0x80, 0xec, 0x5f , 0x60, | + | |
- | 0x51, 0x7f, 0xa9, 0x19, 0xb5, 0x4a, 0x0d, 0x2d, 0xe5, 0x7a, 0x9f, | + | |
- | 0x93, 0xc9, 0x9c, 0xef , 0xa0, 0xe0, 0x3b, 0x4d, 0xae, 0x2a, 0xf5, | + | |
- | 0xb0, 0xc8, 0xeb, 0xbb, 0x3c, 0x83, 0x53, 0x99, 0x61 , 0x17, 0x2b, | + | |
- | 0x04, 0x7e, 0xba, 0x77, 0xd6, 0x26, 0xe1, 0x69, 0x14, 0x63, 0x55, | + | |
- | 0x21, 0x0c, 0x7d] | + | |
def strxor(a, b): # xor two strings (trims the longer input) | def strxor(a, b): # xor two strings (trims the longer input) | ||
Line 101: | Line 62: | ||
hb = b.decode('hex') | hb = b.decode('hex') | ||
return "".join([chr(ord(x) ^ ord(y)).encode('hex') for (x, y) in zip(ha, hb)]) | return "".join([chr(ord(x) ^ ord(y)).encode('hex') for (x, y) in zip(ha, hb)]) | ||
- | |||
- | def bitxor(a, b): # xor two bit strings (trims the longer input) | ||
- | return "".join([str(int(x)^int(y)) for (x, y) in zip(a, b)]) | ||
| | ||
- | def str2bin(ss): | + | def main(): |
- | """ | + | |
- | Transform a string (e.g. 'Hello') into a string of bits | + | |
- | """ | + | |
- | bs = '' | + | |
- | for c in ss: | + | |
- | bs = bs + bin(ord(c))[2:].zfill(8) | + | |
- | return bs | + | |
- | def hex2bin(hs): | + | #Initialise weak rng |
- | """ | + | wr = WeakRNG() |
- | Transform a hex string (e.g. 'a2') into a string of bits (e.g.10100010) | + | wr.init_state() |
- | """ | + | |
- | bs = '' | + | |
- | for c in hs: | + | |
- | bs = bs + bin(int(c,16))[2:].zfill(4) | + | |
- | return bs | + | |
- | def bin2hex(bs): | + | #Print ciphertext |
- | """ | + | CH = 'a432109f58ff6a0f2e6cb280526708baece6680acc1f5fcdb9523129434ae9f6ae9edc2f224b73a8' |
- | Transform a bit string into a hex string | + | print "Full ciphertext in hexa: " + CH |
- | """ | + | |
- | return hex(int(bs,2))[2:] | + | |
- | def byte2bin(bval): | + | #Print known plaintext |
- | """ | + | pknown = 'Let all creation' |
- | Transform a byte (8-bit) value into a bitstring | + | nb = len(pknown) |
- | """ | + | print "Known plaintext: " + pknown |
- | return bin(bval)[2:].zfill(8) | + | pkh = pknown.encode('hex') |
+ | print "Plaintext in hexa: " + pkh | ||
+ | #Obtain first nb bytes of RNG | ||
+ | gh = hexxor(pkh, CH[0:nb*2]) | ||
+ | print gh | ||
+ | gbytes = [] | ||
+ | for i in range(nb): | ||
+ | gbytes.append(ord(gh[2*i:2*i+2].decode('hex'))) | ||
+ | print "Bytes of RNG: " | ||
+ | print gbytes | ||
- | def permute4(s): | + | #Break the LCG here: |
- | """ | + | #1. find a and b |
- | Perform a permutatation by shifting all bits 4 positions right. | + | #2. predict/generate rest of RNG bytes |
- | The input is assumed to be a 16-bit bitstring | + | #3. decrypt plaintext |
- | """ | + | |
- | ps = '' | + | |
- | ps = ps + s[12:16] | + | |
- | ps = ps + s[0:12] | + | |
- | return ps | + | |
- | def permute_inv4(s): | + | # Print full plaintext |
- | """ | + | p = '' |
- | Perform the inverse of permute4 | + | print "Full plaintext is: " + p |
- | The input is assumed to be a 16-bit bitstring | + | |
- | """ | + | |
- | ps = '' | + | |
- | ps = ps + s[4:16] | + | |
- | ps = ps + s[0:4] | + | |
- | return ps | + | |
- | def spn_1r_reduced_2s(k, x): | ||
- | """ | ||
- | Performs an encryption with a substitution-permutation network. | ||
- | Key k = {k1, k2}, total of 16 bits (2 x 8 bits) | ||
- | Input x = {x1, x2}, total of 16 bits (2 x 8 bits) | ||
- | Both k and x are assumed to be bitstrings. | ||
- | Return: | + | if __name__ == "__main__": |
- | a 16-bit bitstring containing the encryption y = {y1, y2} | + | main() |
- | """ | + | </code> |
- | # Split input and key | + | ==== Exercise 2 (3p) ==== |
- | x1 = x[0:8] | + | |
- | x2 = x[8:16] | + | |
- | k1 = k[0:8] | + | |
- | k2 = k[8:16] | + | |
- | #Apply S-box | + | Let's use the experiment defined earlier as a pseudorandom generator ($\mathsf{PRG}$) as follows: |
- | u1 = bitxor(x1, k1) | + | - Set a desired output length $n$ |
- | v1 = sbox[int(u1,2)] | + | - Obtain a random sequence $R$ of bits of length $n$ (e.g. using the Linear-congruential generator from Exercise 1) |
- | v1 = byte2bin(v1) | + | - For each bit $r$ in the random sequence $R$ generated in the previous step, output a bit $b$ as follows: |
+ | * if the bit $r$ is $0$, then output a random bit $b \in \{0, 1\}$ | ||
+ | * if the bit $r$ is $1$, then output $1$ | ||
- | u2 = bitxor(x2, k2) | + | a. Implement the frequency (monobit) test from [[http://csrc.nist.gov/publications/nistpubs/800-22-rev1a/SP800-22rev1a.pdf | NIST (see section 2.1)]] and check if a sequence generated by the above $\mathsf{PRG}$ (say $n=100$) seems random or not. |
- | v2 = sbox[int(u2,2)] | + | |
- | v2 = byte2bin(v2) | + | |
- | #Apply permutation | + | b. Run the test on a random bitstring (e.g. a string such as R used by the above $\mathsf{PRG}$), and compare the result of the test. |
- | pin = v1 + v2 | + | |
- | pout = permute4(pin) | + | |
- | return pout | + | If the two results are different across many iterations, this test already gives you an attacker that breaks the $\mathsf{PRG}$. |
- | + | ||
- | def spn_1r_full_2s(k, x): | + | |
- | """ | + | |
- | Performs an encryption with a substitution-permutation network. | + | |
- | Key k = {k1, k2, k3, k4}, total of 32 bits (4 x 8 bits) | + | |
- | Input x = {x1, x2}, total of 16 bits (2 x 8 bits) | + | |
- | Both k and x are assumed to be bitstrings. | + | |
- | Return: | + | <note tip>You may use a function like this to generate a random bitstring</note> |
- | a 16-bit bitstring containing the encryption y = {y1, y2} | + | <code python> |
- | """ | + | import random |
- | # Split input and key | + | def get_random_string(n): #generate random bit string |
- | x1 = x[0:8] | + | bstr = bin(random.getrandbits(n)).lstrip('0b').zfill(n) |
- | x2 = x[8:16] | + | return bstr |
- | k1 = k[0:8] | + | </code> |
- | k2 = k[8:16] | + | |
- | k3 = k[16:24] | + | |
- | k4 = k[24:32] | + | |
- | #Apply S-box | + | <note tip>Also, in Python you may find the functions sqrt, fabs and erfc from the module math useful</note> |
- | u1 = bitxor(x1, k1) | + | |
- | v1 = sbox[int(u1,2)] | + | |
- | v1 = byte2bin(v1) | + | |
- | u2 = bitxor(x2, k2) | + | ==== Exercise 3 - LFSR (3p) ==== |
- | v2 = sbox[int(u2,2)] | + | |
- | v2 = byte2bin(v2) | + | |
- | #Apply permutation | + | In this exercise we'll build a simple Linear Feedback Shift Register (LFSR). LFSRs produce random bit strings with good statistical properties, but are very easy to predict. |
- | pin = v1 + v2 | + | |
- | pout = permute4(pin) | + | |
- | #Apply final XOR | + | The register is a sequence of $n$ bits; a LFSR is defined by: |
- | po1 = pout[0:8] | + | * an initial state (the initial bit contents of the register) |
- | po2 = pout[8:16] | + | * a polynomial that describes how bit shifts should be performed |
- | y1 = bitxor(po1, k3) | + | |
- | y2 = bitxor(po2, k4) | + | |
- | return y1+y2 | + | For example, given an $18$ bit LFSRm the polynomial $X^{18} + X^{11} + 1$ and the initial state: |
- | + | <code> | |
- | def main(): | + | state = '001001001001001001' |
+ | * * | ||
+ | </code> | ||
- | #Run reduced 2-byte SPN | + | we generate a new bit $b$ by $\mathsf{xor}$-ing bits $11$ ($0$) and $18$ ($1$), thus obtaining $b = 1$. We then shift the whole register to the right (thus dropping the right-most bit, which is the bit we add to the generated random sequence) and insert $b$ to the left. Thus, the new state is: |
- | msg = 'Hi' | + | <code> |
- | key = '??' # Find this | + | state = '100100100100100100' |
- | xs = str2bin(msg) | + | |
- | ks = str2bin(key) | + | |
- | ys = spn_1r_reduced_2s(ks, xs) | + | |
- | print 'Two y halves of reduced SPN: ' + ys[0:8] + ' (hex: ' + bin2hex(ys[0:8]) + '), ' + ys[8:16] + ' (hex: ' + bin2hex(ys[8:16]) + ')' | + | |
- | + | ||
- | #Run full 2-byte SPN | + | |
- | msg = 'Om' | + | |
- | key = '????' # Find this | + | |
- | xs = str2bin(msg) | + | |
- | ks = str2bin(key) | + | |
- | ys = spn_1r_full_2s(ks, xs) | + | |
- | print 'Two y halves of full SPN (2 bytes): ' + ys[0:8] + ' (hex: ' + bin2hex(ys[0:8]) + '), ' + ys[8:16] + ' (hex: ' + bin2hex(ys[8:16]) + ')' | + | |
- | + | ||
- | + | ||
- | + | ||
- | if __name__ == "__main__": | + | |
- | main() | + | |
</code> | </code> | ||
- | === SPN 2 === | + | The process is repeated until the desired number of bits have been generated. |
- | + | ||
- | Now we have a better SPN, where the output of the permutation is XOR-ed with another 2 key bytes, as in the following figure: | + | |
- | {{:sasc:laboratoare:spn_1r_full_2s.png|}} | + | |
- | + | ||
- | - Try to find the key in this case, when given the following message/ciphertext pair ('Om', 0x0073). Print the key in ascii. | + | |
- | <note tip>You may try some kind of brute-force search</note> | ||
+ | Using the above starting state and polynomial, generate $100$ random bits and run the monobit statistical test from the previous exercise to see if their frequency seems random. |