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--- ic:laboratoare:04 [2018/10/06 18:07]
george.pirtoaca
+++ ic:laboratoare:04 [2020/11/02 01:08] (current)
acosmin.maria
@@ Line 1: / Line 1: @@
-<hidden>
+===== Laboratorul 04 - PRFs, PRPs, SPNs =====
-===== Laboratorul 04 - PRFs and PRPs =====
+<file python utils.py>
+import base64
+# CONVERSION FUNCTIONS
+def _chunks(string, chunk_size):
+    for i in range(0, len(string), chunk_size):
+        yield string[i:i+chunk_size]
+# THIS ONE IS NEW
+def byte_2_bin(bval):
+    """
+      Transform a byte (8-bit) value into a bitstring
+    """
+    return bin(bval)[2:].zfill(8)
+def _hex(x):
+    return format(x, '02x')
+def hex_2_bin(data):
+    return ''.join(f'{int(x, 16):08b}' for x in _chunks(data, 2))
+def str_2_bin(data):
+    return ''.join(f'{ord(c):08b}' for c in data)
+def bin_2_hex(data):
+    return ''.join(f'{int(b, 2):02x}' for b in _chunks(data, 8))
+def str_2_hex(data):
+    return ''.join(f'{ord(c):02x}' for c in data)
+def bin_2_str(data):
+    return ''.join(chr(int(b, 2)) for b in _chunks(data, 8))
+def hex_2_str(data):
+    return ''.join(chr(int(x, 16)) for x in _chunks(data, 2))
+# XOR FUNCTIONS
+def strxor(a, b):  # xor two strings, trims the longer input
+    return ''.join(chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b))
+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 hexxor(a, b):  # xor two hex-strings, trims the longer input
+    return ''.join(_hex(int(x, 16) ^ int(y, 16)) for (x, y) in zip(_chunks(a, 2), _chunks(b, 2)))
+# BASE64 FUNCTIONS
+def b64decode(data):
+    return bytes_to_string(base64.b64decode(string_to_bytes(data)))
+def b64encode(data):
+    return bytes_to_string(base64.b64encode(string_to_bytes(data)))
+# PYTHON3 'BYTES' FUNCTIONS
+def bytes_to_string(bytes_data):
+    return bytes_data.decode()  # default utf-8
+def string_to_bytes(string_data):
+    return string_data.encode()  # default utf-8
+</file>
 In this recitation, we shall do some exercises related to PRFs, PRPs and DES.
-Please check the course, available here:
+Please check the course available [[https://drive.google.com/file/d/1Fjybv6k5QudRB1bkAi5shVUGlUyTO_0U/view?usp=sharing|here]] and make sure you understand the basic definitions and principles before moving forward to solving the exercises.
-http://cs.curs.pub.ro/2014/pluginfile.php/13095/mod_resource/content/2/sasc_curs4_5.pdf
 ==== Exercise 1 (4p) ====
@@ Line 10: / Line 69: @@
 Let $F : K × X \to Y$ be a secure PRF with $K = X = Y = \{0, 1\}^{n}$.
-  * a) Show that $F_1(k,x) = F(k,x) \| 0$ is not a secure PRF. (for strings $y$ and $z$ we use $y \| z$ to denote the concatenation of $y$ and $z$)
+  * a) Show that $F_1(k,x) = F(k,x) \| 0$ is not a secure PRF. (for strings $y$ and $z$ we use $y \| z$ to denote the concatenation of $y$ and $z$);
   * b) Prove that $F_2(k, x) = F \left(k, x \oplus 1^{n}\right)$ is a secure PRF. Here $x \oplus 1^{n}$ is the bit-wise complement of $x$. To prove security argue the contra-positive: a distinguisher $A$ that breaks $F_2$ implies a distinguisher $B$ that breaks $F$ and whose running time is about the same as $A$’s.
   * c) Let $K_3 = \{0, 1\}^{n+1}$. Construct a new PRF $F_3 : K_3 \times X \to Y$ with the following property: the PRF $F_3$ 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.
@@ Line 16: / Line 75: @@
 Hint: Let $k_3 = k \| b$ where $k \in \{0,1\}^{n}$ and $b \in \{0,1\}$. Set $F_3(k_3,x)$ to be the same as $F (k, x)$ for all $x \neq 0^{n}$. Define $F_3\left(k_3, 0^{n}\right)$ so that $F_3$ is a secure PRF, but becomes easily distinguishable from a random function if the last bit of the secret key $k_3$ is known to the adversary. Prove that your $F_3$ is a secure PRF by arguing the contra-positive, as in part (b).
 </note>
-  * d) Construct a new PRF $F_4 : K_2 × X \to Y$ that remains secure if the attacker learns any single bit of the key. Your function $F_2$ may only call $F$ once. Briefly explain why your PRF remains secure if any single bit of the key is leaked.
+  * d) Construct a new PRF $F_4 : K × X \to Y$ that remains secure if the attacker learns any single bit of the key. Your function $F_4$ may only call $F$ once. Briefly explain why your PRF remains secure if any single bit of the key is leaked.
-==== Exercise 2 ====
+==== Exercise 2-3-4 ====
 Let's analyse some substitution-permutation networks (SPN).
-=== SPN 1 (3p) ===
+==== SPN 1 (3p) ====
 We have the SPN from this figure:
@@ Line 37: / Line 96: @@
 <code python spn1.py>
-import sys
+from utils import *
-import random
-import string
-import operator
 # Rijndael S-box
-sbox =  [0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67,
+sbox = [0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67,
 x2b, 0xfe, 0xd7, 0xab, 0x76, 0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59,
 x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0, 0xb7,
@@ Line 67: / Line 123: @@
 x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f, 0xb0,
 x54, 0xbb, 0x16]
 # Rijndael Inverted S-box
 rsbox = [0x52, 0x09, 0x6a, 0xd5, 0x30, 0x36, 0xa5, 0x38, 0xbf, 0x40, 0xa3,
-x9e, 0x81, 0xf3, 0xd7, 0xfb , 0x7c, 0xe3, 0x39, 0x82, 0x9b, 0x2f,
+x9e, 0x81, 0xf3, 0xd7, 0xfb, 0x7c, 0xe3, 0x39, 0x82, 0x9b, 0x2f,
-xff, 0x87, 0x34, 0x8e, 0x43, 0x44, 0xc4, 0xde, 0xe9, 0xcb , 0x54,
+xff, 0x87, 0x34, 0x8e, 0x43, 0x44, 0xc4, 0xde, 0xe9, 0xcb, 0x54,
 x7b, 0x94, 0x32, 0xa6, 0xc2, 0x23, 0x3d, 0xee, 0x4c, 0x95, 0x0b,
-x42, 0xfa, 0xc3, 0x4e , 0x08, 0x2e, 0xa1, 0x66, 0x28, 0xd9, 0x24,
+x42, 0xfa, 0xc3, 0x4e, 0x08, 0x2e, 0xa1, 0x66, 0x28, 0xd9, 0x24,
-xb2, 0x76, 0x5b, 0xa2, 0x49, 0x6d, 0x8b, 0xd1, 0x25 , 0x72, 0xf8,
+xb2, 0x76, 0x5b, 0xa2, 0x49, 0x6d, 0x8b, 0xd1, 0x25, 0x72, 0xf8,
 xf6, 0x64, 0x86, 0x68, 0x98, 0x16, 0xd4, 0xa4, 0x5c, 0xcc, 0x5d,
-x65, 0xb6, 0x92 , 0x6c, 0x70, 0x48, 0x50, 0xfd, 0xed, 0xb9, 0xda,
+x65, 0xb6, 0x92, 0x6c, 0x70, 0x48, 0x50, 0xfd, 0xed, 0xb9, 0xda,
-x5e, 0x15, 0x46, 0x57, 0xa7, 0x8d, 0x9d, 0x84 , 0x90, 0xd8, 0xab,
+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,
+x45, 0x06, 0xd0, 0x2c, 0x1e, 0x8f, 0xca, 0x3f, 0x0f, 0x02, 0xc1,
-xaf, 0xbd, 0x03, 0x01, 0x13, 0x8a, 0x6b , 0x3a, 0x91, 0x11, 0x41,
+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,
+x73, 0x96, 0xac, 0x74, 0x22, 0xe7, 0xad, 0x35, 0x85, 0xe2, 0xf9,
-x37, 0xe8, 0x1c, 0x75, 0xdf, 0x6e , 0x47, 0xf1, 0x1a, 0x71, 0x1d,
+x37, 0xe8, 0x1c, 0x75, 0xdf, 0x6e, 0x47, 0xf1, 0x1a, 0x71, 0x1d,
-x29, 0xc5, 0x89, 0x6f, 0xb7, 0x62, 0x0e, 0xaa, 0x18, 0xbe, 0x1b ,
+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,
+xfe, 0x78, 0xcd, 0x5a, 0xf4, 0x1f, 0xdd, 0xa8, 0x33, 0x88, 0x07,
-xc7, 0x31, 0xb1, 0x12, 0x10, 0x59, 0x27, 0x80, 0xec, 0x5f , 0x60,
+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,
+x93, 0xc9, 0x9c, 0xef, 0xa0, 0xe0, 0x3b, 0x4d, 0xae, 0x2a, 0xf5,
-xb0, 0xc8, 0xeb, 0xbb, 0x3c, 0x83, 0x53, 0x99, 0x61 , 0x17, 0x2b,
+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)
-  return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b)])
-def hexxor(a, b): # xor two hex strings (trims the longer input)
+def permute4(s):
-  ha = a.decode('hex')
+    """
-  hb = b.decode('hex')
+      Perform a permutatation by shifting all bits 4 positions right.
-  return "".join([chr(ord(x) ^ ord(y)).encode('hex') for (x, y) in zip(ha, hb)])
+      The input is assumed to be a 16-bit bitstring
+    """
+    ps = ''
+    ps = ps + s[12:16]
+    ps = ps + s[0:12]
+    return ps
-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):
-  """
-    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):
+def permute_inv4(s):
-  """
+    """
-    Transform a hex string (e.g. 'a2') into a string of bits (e.g.10100010)
+      Perform the inverse of permute4
-  """
+      The input is assumed to be a 16-bit bitstring
-  bs = ''
+    """
-  for c in hs:
+    ps = ''
-    bs = bs + bin(int(c,16))[2:].zfill(4)
+    ps = ps + s[4:16]
-  return bs
+    ps = ps + s[0:4]
+    return ps
-def bin2hex(bs):
-  """
-    Transform a bit string into a hex string
-  """
-  return hex(int(bs,2))[2:]
-def byte2bin(bval):
+def spn_1r_reduced_2s(k, x):
-  """
+    """
-    Transform a byte (8-bit) value into a bitstring
+      Performs an encryption with a substitution-permutation network.
-  """
+      Key k = {k1, k2}, total of 16 bits (2 x 8 bits)
-  return bin(bval)[2:].zfill(8)
+      Input x = {x1, x2}, total of 16 bits (2 x 8 bits)
+      Both k and x are assumed to be bitstrings.
+      Return:
+      a 16-bit bitstring containing the encryption y = {y1, y2}
+    """
-def permute4(s):
+    # Split input and key
-  """
+    x1 = x[0:8]
-    Perform a permutatation by shifting all bits 4 positions right.
+    x2 = x[8:16]
-    The input is assumed to be a 16-bit bitstring
+    k1 = k[0:8]
-  """
+    k2 = k[8:16]
-  ps = ''
-  ps = ps + s[12:16]
-  ps = ps + s[0:12]
-  return ps
-def permute_inv4(s):
+    # Apply S-box
-  """
+    u1 = bitxor(x1, k1)
-    Perform the inverse of permute4
+    v1 = sbox[int(u1, 2)]
-    The input is assumed to be a 16-bit bitstring
+    v1 = byte_2_bin(v1)
-  """
-  ps = ''
-  ps = ps + s[4:16]
-  ps = ps + s[0:4]
-  return ps
-def spn_1r_reduced_2s(k, x):
+    u2 = bitxor(x2, k2)
-  """
+    v2 = sbox[int(u2, 2)]
-    Performs an encryption with a substitution-permutation network.
+    v2 = byte_2_bin(v2)
-    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:
+    # Apply permutation
-    a 16-bit bitstring containing the encryption y = {y1, y2}
+    pin = v1 + v2
-  """
+    pout = permute4(pin)
-  # Split input and key
+    return pout
-  x1 = x[0:8]
-  x2 = x[8:16]
-  k1 = k[0:8]
-  k2 = k[8:16]
-  #Apply S-box
-  u1 = bitxor(x1, k1)
-  v1 = sbox[int(u1,2)]
-  v1 = byte2bin(v1)
-  u2 = bitxor(x2, k2)
+def spn_1r_full_2s(k, x):
-  v2 = sbox[int(u2,2)]
+    """
-  v2 = byte2bin(v2)
+      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.
-  #Apply permutation
+      Return:
-  pin = v1 + v2
+      a 16-bit bitstring containing the encryption y = {y1, y2}
-  pout = permute4(pin)
+    """
-  return pout
+    # Split input and key
+    x1 = x[0:8]
-def spn_1r_full_2s(k, x):
+    x2 = x[8:16]
-  """
+    k1 = k[0:8]
-    Performs an encryption with a substitution-permutation network.
+    k2 = k[8:16]
-    Key k = {k1, k2, k3, k4}, total of 32 bits (4 x 8 bits)
+    k3 = k[16:24]
-    Input x = {x1, x2}, total of 16 bits (2 x 8 bits)
+    k4 = k[24:32]
-    Both k and x are assumed to be bitstrings.
-    Return:
+    # Apply S-box
-    a 16-bit bitstring containing the encryption y = {y1, y2}
+    u1 = bitxor(x1, k1)
-  """
+    v1 = sbox[int(u1, 2)]
+    v1 = byte_2_bin(v1)
-  # Split input and key
+    u2 = bitxor(x2, k2)
-  x1 = x[0:8]
+    v2 = sbox[int(u2, 2)]
-  x2 = x[8:16]
+    v2 = byte_2_bin(v2)
-  k1 = k[0:8]
-  k2 = k[8:16]
-  k3 = k[16:24]
-  k4 = k[24:32]
-  #Apply S-box
+    # Apply permutation
-  u1 = bitxor(x1, k1)
+    pin = v1 + v2
-  v1 = sbox[int(u1,2)]
+    pout = permute4(pin)
-  v1 = byte2bin(v1)
-  u2 = bitxor(x2, k2)
+    # Apply final XOR
-  v2 = sbox[int(u2,2)]
+    po1 = pout[0:8]
-  v2 = byte2bin(v2)
+    po2 = pout[8:16]
+    y1 = bitxor(po1, k3)
+    y2 = bitxor(po2, k4)
-  #Apply permutation
+    return y1+y2
-  pin = v1 + v2
-  pout = permute4(pin)
-  #Apply final XOR
-  po1 = pout[0:8]
-  po2 = pout[8:16]
-  y1 = bitxor(po1, k3)
-  y2 = bitxor(po2, k4)
-  return y1+y2
 def main():
-  #Run reduced 2-byte SPN
+    # Run reduced 2-byte SPN
-  msg = 'Hi'
+    msg = 'Hi'
-  key = '??' # Find this
+    key = '??'  # Find this
-  xs = str2bin(msg)
+    xs = str_2_bin(msg)
-  ks = str2bin(key)
+    ks = str_2_bin(key)
-  ys = spn_1r_reduced_2s(ks, xs)
+    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]) + ')'
+    print('Two y halves of reduced SPN: ' + ys[0:8] + ' (hex: ' + bin_2_hex(
+        ys[0:8]) + '), ' + ys[8:16] + ' (hex: ' + bin_2_hex(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]) + ')'
+    # Run full 2-byte SPN
+    msg = 'Om'
+    key = '????'  # Find this
+    xs = str_2_bin(msg)
+    ks = str_2_bin(key)
+    ys = spn_1r_full_2s(ks, xs)
+    print('Two y halves of full SPN (2 bytes): ' + ys[0:8] + ' (hex: ' + bin_2_hex(
+        ys[0:8]) + '), ' + ys[8:16] + ' (hex: ' + bin_2_hex(ys[8:16]) + ')')
 if __name__ == "__main__":
-  main()
+    main()
 </code>
@@ Line 271: / Line 290: @@
   - Try to find the key in this case as well, using the following message/ciphertext pairs: ('Omul', 0xddcf7bc7), ('stea', 0x96d58b43), ('luna', 0x9c3f2303) . 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 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. Suppose you brute force k5 and k6, is it possible to find out k1?
 </note>
-<hidden>The solution is {{:ic:laboratoare:lab4_sol.zip|here}}.</hidden>
-</hidden>

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ic/laboratoare/04.1538838450.txt.gz · Last modified: 2018/10/06 18:07 by george.pirtoaca

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