1 | # $Id: p3.py,v 1.2 2003/11/18 19:04:03 phr Exp phr $ |
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2 | |
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3 | # Simple p3 encryption "algorithm": it's just SHA used as a stream |
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4 | # cipher in output feedback mode. |
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5 | |
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6 | # Author: Paul Rubin, Fort GNOX Cryptography, <phr-crypto at nightsong.com>. |
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7 | # Algorithmic advice from David Wagner, Richard Parker, Bryan |
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8 | # Olson, and Paul Crowley on sci.crypt is gratefully acknowledged. |
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9 | |
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10 | # Copyright 2002,2003 by Paul Rubin |
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11 | # Copying license: same as Python 2.3 license |
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12 | |
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13 | # Please include this revision number in any bug reports: $Revision: 1.2 $. |
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14 | |
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15 | from string import join |
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16 | from array import array |
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17 | import sha |
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18 | import getpass |
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19 | from time import time |
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20 | |
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21 | |
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22 | class CryptError(Exception): pass |
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23 | def _hash(str): return sha.new(str).digest() |
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24 | |
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25 | _ivlen = 16 |
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26 | _maclen = 8 |
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27 | _state = _hash(`time()`) |
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28 | |
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29 | try: |
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30 | import os |
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31 | _pid = `os.getpid()` |
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32 | except ImportError, AttributeError: |
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33 | _pid = '' |
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34 | |
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35 | def _expand_key(key, clen): |
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36 | blocks = (clen+19)/20 |
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37 | xkey=[] |
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38 | seed=key |
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39 | for i in xrange(blocks): |
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40 | seed=sha.new(key+seed).digest() |
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41 | xkey.append(seed) |
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42 | j = join(xkey,'') |
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43 | return array ('L', j) |
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44 | |
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45 | def p3_encrypt(plain,key): |
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46 | global _state |
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47 | H = _hash |
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48 | |
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49 | # change _state BEFORE using it to compute nonce, in case there's |
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50 | # a thread switch between computing the nonce and folding it into |
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51 | # the state. This way if two threads compute a nonce from the |
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52 | # same data, they won't both get the same nonce. (There's still |
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53 | # a small danger of a duplicate nonce--see below). |
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54 | _state = 'X'+_state |
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55 | |
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56 | # Attempt to make nlist unique for each call, so we can get a |
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57 | # unique nonce. It might be good to include a process ID or |
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58 | # something, but I don't know if that's portable between OS's. |
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59 | # Since is based partly on both the key and plaintext, in the |
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60 | # worst case (encrypting the same plaintext with the same key in |
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61 | # two separate Python instances at the same time), you might get |
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62 | # identical ciphertexts for the identical plaintexts, which would |
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63 | # be a security failure in some applications. Be careful. |
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64 | nlist = [`time()`, _pid, _state, `len(plain)`,plain, key] |
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65 | nonce = H(join(nlist,','))[:_ivlen] |
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66 | _state = H('update2'+_state+nonce) |
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67 | k_enc, k_auth = H('enc'+key+nonce), H('auth'+key+nonce) |
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68 | n=len(plain) # cipher size not counting IV |
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69 | |
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70 | stream = array('L', plain+'0000'[n&3:]) # pad to fill 32-bit words |
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71 | xkey = _expand_key(k_enc, n+4) |
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72 | for i in xrange(len(stream)): |
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73 | stream[i] = stream[i] ^ xkey[i] |
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74 | ct = nonce + stream.tostring()[:n] |
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75 | auth = _hmac(ct, k_auth) |
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76 | return ct + auth[:_maclen] |
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77 | |
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78 | def p3_decrypt(cipher,key): |
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79 | H = _hash |
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80 | n=len(cipher)-_ivlen-_maclen # length of ciphertext |
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81 | if n < 0: |
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82 | raise CryptError, "invalid ciphertext" |
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83 | nonce,stream,auth = \ |
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84 | cipher[:_ivlen], cipher[_ivlen:-_maclen]+'0000'[n&3:],cipher[-_maclen:] |
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85 | k_enc, k_auth = H('enc'+key+nonce), H('auth'+key+nonce) |
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86 | vauth = _hmac (cipher[:-_maclen], k_auth)[:_maclen] |
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87 | if auth != vauth: |
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88 | raise CryptError, "invalid key or ciphertext" |
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89 | |
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90 | stream = array('L', stream) |
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91 | xkey = _expand_key (k_enc, n+4) |
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92 | for i in xrange (len(stream)): |
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93 | stream[i] = stream[i] ^ xkey[i] |
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94 | plain = stream.tostring()[:n] |
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95 | return plain |
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96 | |
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97 | # RFC 2104 HMAC message authentication code |
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98 | # This implementation is faster than Python 2.2's hmac.py, and also works in |
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99 | # old Python versions (at least as old as 1.5.2). |
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100 | from string import translate |
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101 | def _hmac_setup(): |
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102 | global _ipad, _opad, _itrans, _otrans |
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103 | _itrans = array('B',[0]*256) |
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104 | _otrans = array('B',[0]*256) |
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105 | for i in xrange(256): |
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106 | _itrans[i] = i ^ 0x36 |
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107 | _otrans[i] = i ^ 0x5c |
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108 | _itrans = _itrans.tostring() |
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109 | _otrans = _otrans.tostring() |
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110 | |
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111 | _ipad = '\x36'*64 |
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112 | _opad = '\x5c'*64 |
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113 | |
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114 | def _hmac(msg, key): |
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115 | if len(key)>64: |
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116 | key=sha.new(key).digest() |
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117 | ki = (translate(key,_itrans)+_ipad)[:64] # inner |
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118 | ko = (translate(key,_otrans)+_opad)[:64] # outer |
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119 | return sha.new(ko+sha.new(ki+msg).digest()).digest() |
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120 | |
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121 | # |
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122 | # benchmark and unit test |
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123 | # |
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124 | |
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125 | def _time_p3(n=1000,len=20): |
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126 | plain="a"*len |
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127 | t=time() |
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128 | for i in xrange(n): |
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129 | p3_encrypt(plain,"abcdefgh") |
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130 | dt=time()-t |
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131 | print "plain p3:", n,len,dt,"sec =",n*len/dt,"bytes/sec" |
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132 | |
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133 | def _speed(): |
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134 | _time_p3(len=5) |
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135 | _time_p3() |
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136 | _time_p3(len=200) |
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137 | _time_p3(len=2000,n=100) |
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138 | |
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139 | def _test(): |
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140 | e=p3_encrypt |
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141 | d=p3_decrypt |
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142 | |
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143 | plain="test plaintext" |
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144 | key = "test key" |
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145 | c1 = e(plain,key) |
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146 | c2 = e(plain,key) |
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147 | assert c1!=c2 |
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148 | assert d(c2,key)==plain |
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149 | assert d(c1,key)==plain |
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150 | c3 = c2[:20]+chr(1+ord(c2[20]))+c2[21:] # change one ciphertext character |
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151 | |
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152 | try: |
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153 | print d(c3,key) # should throw exception |
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154 | print "auth verification failure" |
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155 | except CryptError: |
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156 | pass |
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157 | |
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158 | try: |
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159 | print d(c2,'wrong key') # should throw exception |
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160 | print "test failure" |
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161 | except CryptError: |
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162 | pass |
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163 | |
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164 | |
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165 | def encrypt(instring,key=None): |
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166 | """Encrypt the input string |
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167 | |
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168 | Encrypts the input string or the string contained in the file |
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169 | with name <instring>. Uses encryption key <key> |
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170 | """ |
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171 | |
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172 | try: |
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173 | if os.path.exists(instring): |
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174 | instring = open(instring,'r').read().strip() |
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175 | except: |
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176 | pass |
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177 | |
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178 | if key is None: |
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179 | key = getpass.getuser() |
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180 | return 'cry'+p3_encrypt(instring,key)+'pto' |
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181 | |
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182 | def decrypt(instring,key=None): |
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183 | """Decrypt the input string |
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184 | |
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185 | Decrypts the input string or the string contained in the file |
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186 | with name <instring>. Uses encryption key <key>. If the input |
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187 | string is not encrypted, just returns it. |
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188 | """ |
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189 | |
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190 | try: |
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191 | if os.path.exists(instring): |
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192 | instring = open(instring,'r').read().strip() |
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193 | except: |
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194 | pass |
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195 | |
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196 | ins = instring.strip() |
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197 | #if re.search('^cry.*pto$',ins): |
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198 | if ins[:3]+ins[-3:] == 'crypto': |
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199 | if key is None: |
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200 | key = getpass.getuser() |
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201 | ins = ins[3:-3] |
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202 | return p3_decrypt(ins,key) |
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203 | else: |
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204 | return instring |
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205 | |
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206 | |
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207 | _hmac_setup() |
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208 | _test() |
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209 | # _speed() # uncomment to run speed test |
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