Re: Open Cryptographic Challenge

From: Brian Gladman (brg_at_nowhere.at.all)
Date: 04/09/04 

Date: Fri, 9 Apr 2004 09:25:31 +0100

"Felino" <lapazdelmundo2003@geocities.com-dot-mx.no-spam.invalid> wrote in
message news:4074b4b6_1@127.0.0.1...
> >
> > There, that should make me look really dumb. LOL....
>
> Well, up to now, I am not looking dumb since nobody has been able to
> post the correct answer. :wink:
>
> And no, it is not a joke. It is very serious. And although some
> "patterns" were shown, what is the sequence? No one has shown it
> yet!

Since you claim to be serious about demonstrating a very secure pseudo
random function, here is one that I have 'invented' for you to crack:

f( 0) = 8402301
f( 1) = 4095870
f( 2) = 2614815
f( 3) = 2015042
f( 4) = 4414669
f( 5) = 937417
f( 6) = 6683438
f( 7) = 3580876
f( 8) = 584423
f( 9) = 7718867
f(10) = 9871829
f(11) = 4032076
f(12) = 5111536
f(13) = 3472733
f(14) = 5920997
f(15) = 1864555
f(16) = 6269705
f(17) = 3827574
f(18) = 5244252
f(19) = 6979976
f(20) = 3056631
f(21) = 718413
f(22) = 7066490
f(23) = 4464291
f(24) = 436778
f(25) = 1041598
f(26) = 990689
f(27) = 222181
f(28) = 735947
f(29) = 9506158
f(30) = 4297449
f(31) = 4392765
f(32) = 8758877
f(33) = 8283250
f(34) = 7102034
f(35) = 3838757
f(36) = 1940005
f(37) = 7323071
f(38) = 5999312
f(39) = 5310180
f(40) = 4507225
f(41) = 5231859
f(42) = 1109220
f(43) = 3066164
f(44) = 8175964
f(45) = 2110227
f(46) = 4126799
f(47) = 5705155
f(48) = 2675578
f(49) = 5125016
f(50) = 4565190
f(51) = 5306079
f(52) = 1412716
f(53) = 8274751
f(54) = 679781
f(55) = 4588169
f(56) = 9045735
f(57) = 1060974
f(58) = 8849119
f(59) = 6644132
f(60) = 1522602
f(61) = 5272203
f(62) = 9069545
f(63) = 3332600
f(64) = 7795419
f(65) = 5578397
f(66) = 4809775
f(67) = 9656449
f(68) = 6883264
f(69) = 200404
f(70) = 3083466
f(71) = 9889856
f(72) = 8896733
f(73) = 2837863
f(74) = 7274380
f(75) = 280219
f(76) = 5763438
f(77) = 7893468
f(78) = 2438071
f(79) = 9949776
f(80) = 2734044
f(81) = 4669809
f(82) = 6349496
f(83) = 5449080
f(84) = 8728239
f(85) = 1058885
f(86) = 5459207
f(87) = 2563777
f(88) = 6177890
f(89) = 2676421
f(90) = 6427928
f(91) = 3149583
f(92) = 9024499
f(93) = 3833950
f(94) = 117328
f(95) = 1329678
f(96) = 763391
f(97) = 6438160
f(98) = 7641339
f(99) = 7895424

I will even tell you that these values are generated with the following
pseudo code (where 'buf' is a 16 byte buffer and 'val' is a 32-bit unsigned
long variable):

AES_set_128bit_key(key)
for i = 0 to 99 step 1
  buf[0] = i
  for j = 1 to 15 step 1
     b[j] = 0
  end for
  AES_encrypt(buf)
  val = buf[0] + 256 *(buf[1] + 256 * (buf[2] + 256 * buf[3]))
  print "f(", i, ") = ", val % 10000000
end for

All you have to do is to tell us the value of f(100).

In fact I have given you a chance of cracking this by using a particular key
for the AES encryption. But once you choose your own secret AES key you
will have a pseudo random function that is far more secure than anything you
have shown us so far.

   Brian Gladman