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--- sasc:laboratoare:03 [2015/03/19 17:33]
marios.choudary
+++ sasc:laboratoare:03 [2017/03/07 15:32] (current)
dan.dragan
@@ Line 1: / Line 1: @@
-===== 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 random
@@ Line 41: / Line 30: @@
 import operator
-# Rijndael S-box
+#Parameters for weak LC RNG
-sbox =  [0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67,
+class WeakRNG:
-x2b, 0xfe, 0xd7, 0xab, 0x76, 0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59,
+    "Simple class for weak RNG"
-x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0, 0xb7,
+    def __init__(self):
-xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1,
+        self.rstate = 0
-x71, 0xd8, 0x31, 0x15, 0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05,
+        self.maxn = 255
-x9a, 0x07, 0x12, 0x80, 0xe2, 0xeb, 0x27, 0xb2, 0x75, 0x09, 0x83,
+        self.a = 0 #Set this to correct value
-x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3, 0x29,
+        self.b = 0 #Set this to correct value
-xe3, 0x2f, 0x84, 0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b,
+        self.p = 257
-x6a, 0xcb, 0xbe, 0x39, 0x4a, 0x4c, 0x58, 0xcf, 0xd0, 0xef, 0xaa,
-xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x02, 0x7f, 0x50, 0x3c,
-x9f, 0xa8, 0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc,
-xb6, 0xda, 0x21, 0x10, 0xff, 0xf3, 0xd2, 0xcd, 0x0c, 0x13, 0xec,
-x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d, 0x64, 0x5d, 0x19,
-x73, 0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee,
-xb8, 0x14, 0xde, 0x5e, 0x0b, 0xdb, 0xe0, 0x32, 0x3a, 0x0a, 0x49,
-x06, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62, 0x91, 0x95, 0xe4, 0x79,
-xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4,
-xea, 0x65, 0x7a, 0xae, 0x08, 0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6,
-xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f, 0x4b, 0xbd, 0x8b, 0x8a, 0x70,
-x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e, 0x61, 0x35, 0x57, 0xb9,
-x86, 0xc1, 0x1d, 0x9e, 0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e,
-x94, 0x9b, 0x1e, 0x87, 0xe9, 0xce, 0x55, 0x28, 0xdf, 0x8c, 0xa1,
-x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f, 0xb0,
-x54, 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"
-x9e, 0x81, 0xf3, 0xd7, 0xfb , 0x7c, 0xe3, 0x39, 0x82, 0x9b, 0x2f,
+        self.rstate = (self.a * self.rstate + self.b) % self.p
-xff, 0x87, 0x34, 0x8e, 0x43, 0x44, 0xc4, 0xde, 0xe9, 0xcb , 0x54,
-x7b, 0x94, 0x32, 0xa6, 0xc2, 0x23, 0x3d, 0xee, 0x4c, 0x95, 0x0b,
+    def get_prg_byte(self):
-x42, 0xfa, 0xc3, 0x4e , 0x08, 0x2e, 0xa1, 0x66, 0x28, 0xd9, 0x24,
+        "Return a new PRG byte and update PRG state"
-xb2, 0x76, 0x5b, 0xa2, 0x49, 0x6d, 0x8b, 0xd1, 0x25 , 0x72, 0xf8,
+        b = self.rstate & 0xFF
-xf6, 0x64, 0x86, 0x68, 0x98, 0x16, 0xd4, 0xa4, 0x5c, 0xcc, 0x5d,
+        self.update_state()
-x65, 0xb6, 0x92 , 0x6c, 0x70, 0x48, 0x50, 0xfd, 0xed, 0xb9, 0xda,
+        return b
-x5e, 0x15, 0x46, 0x57, 0xa7, 0x8d, 0x9d, 0x84 , 0x90, 0xd8, 0xab,
-x00, 0x8c, 0xbc, 0xd3, 0x0a, 0xf7, 0xe4, 0x58, 0x05, 0xb8, 0xb3,
-x45, 0x06 , 0xd0, 0x2c, 0x1e, 0x8f, 0xca, 0x3f, 0x0f, 0x02, 0xc1,
-xaf, 0xbd, 0x03, 0x01, 0x13, 0x8a, 0x6b , 0x3a, 0x91, 0x11, 0x41,
-x4f, 0x67, 0xdc, 0xea, 0x97, 0xf2, 0xcf, 0xce, 0xf0, 0xb4, 0xe6,
-x73 , 0x96, 0xac, 0x74, 0x22, 0xe7, 0xad, 0x35, 0x85, 0xe2, 0xf9,
-x37, 0xe8, 0x1c, 0x75, 0xdf, 0x6e , 0x47, 0xf1, 0x1a, 0x71, 0x1d,
-x29, 0xc5, 0x89, 0x6f, 0xb7, 0x62, 0x0e, 0xaa, 0x18, 0xbe, 0x1b ,
-xfc, 0x56, 0x3e, 0x4b, 0xc6, 0xd2, 0x79, 0x20, 0x9a, 0xdb, 0xc0,
-xfe, 0x78, 0xcd, 0x5a, 0xf4 , 0x1f, 0xdd, 0xa8, 0x33, 0x88, 0x07,
-xc7, 0x31, 0xb1, 0x12, 0x10, 0x59, 0x27, 0x80, 0xec, 0x5f , 0x60,
-x51, 0x7f, 0xa9, 0x19, 0xb5, 0x4a, 0x0d, 0x2d, 0xe5, 0x7a, 0x9f,
-x93, 0xc9, 0x9c, 0xef , 0xa0, 0xe0, 0x3b, 0x4d, 0xae, 0x2a, 0xf5,
-xb0, 0xc8, 0xeb, 0xbb, 0x3c, 0x83, 0x53, 0x99, 0x61 , 0x17, 0x2b,
-x04, 0x7e, 0xba, 0x77, 0xd6, 0x26, 0xe1, 0x69, 0x14, 0x63, 0x55,
-x21, 0x0c, 0x7d]
 def strxor(a, b): # xor two strings (trims the longer input)
@@ Line 101: / Line 62: @@
   hb = b.decode('hex')
   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>
-=== 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>
-=== SPN 3 ===
-As another example, which uses a larger block size, let's use an SPN that takes a 4-byte input x=[x1 || x2 || x3 || x4] and an 8-byte key k=[k1 || k2 || k3 || k4 || k5 || k6 || k7 || k8] as in this figure:
-{{:sasc:laboratoare:spn_1r_full_4s.png|}}
-  - Try to find the key in this case as well, using the following message/ciphertext pair:  . Again print the key in ascii.
-<note tip> This time you cannot (easily) do a brute-force on all the bytes of the last XOR. However, you may try to attack one S-box at a time. Think of the bits that affect one such S-box and find an efficient attack.
-</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.