Irdial-Discs

Sorry for this delayed email but it took me 
              some time to think about this
problem and a good proof.  (I'm always refering to G2text 
              here.)
Let's assume the text was really encrypted using 
              a OTP.  It's most secure
to use a pad that is as long as the original text.  The 
              message of the
G2text consists of 100 numbers so let's assume, they have chosen 
              an OTP
that is 100 numbers long.
It's obvious that the result of the plain-text 
              depends on the OTP.  Even
better, you could try a first OTP that decrypts your cipher 
              into a serious
English sentence.  But the problem is, you choose a second, 
              different OTP
that translates the message into yet another English sentence 
              but is
totally different from the first result.  How can you decide 
              now what
result is better?  You have a lot of possibilities to choose 
              your OTP that
all translate into proper English.
In fact, every message that consists of 100 
              numbers is a possible
plain-text.  You can choose one of them and apply that 
              OTP-algorithm
backwards to calculate the one OTP that is needed to transform 
              your
cipher-text into your desired plain-text.  In other words, 
              for any message
that is 100 numbers long, you can calculate an OTP.
But this means that every message that consists 
              of 100 number is an
equally possible candidate for a plain-text.  And we have 
              to emphasize
"equally" here because that's the power of the OTP method.  
              If you don't
know the OTP, you simply cannot crack it because all results 
              are equally
possible and you can't decide which one is the right.
Here's a small programme written in C that applies 
              the OTP algorithm
backwards to calculate two different OTPs.  The first one 
              decrypts the
cipher-text "ABCDE" into "THILO", and the second one into "MUSIC".  
              But I
could find many more OTPs to decrypt it to _any_ words.
OTP          
              plain-text
------------+-----
0x150a0a080a THILO
0x0c17100d06 MUSIC
I hope this illustrates why I think you can't 
              crack these numbers.
> ill have to get some more cyphertext together; 
              hopefully there are some people
> who have bothered to transcribe G2, because I only have three 
              examples at the
> most.
Well, but if it's really OTP, it's useless to 
              have more G2texts - the OT
stands for "one-time"...
> >I've factorized all of the SwedRhap numbers 
              and
>
> Ahhhh at last, someone doing something for real!!!! so in 
              the example given,
> out of each five figure group, there were how many prime numbers 
              out of the
> 100? let me have some exact details please!!! <<<<excitement
See the results attached to this mail.
Anyway, please tell me what you think about 
              this proof.
Ende.
Thilo
                                    
              _________ ___ ___ ___ ___      ___
Thilo.Mezger@stud.uni-karlsruhe.de  |___ ___| | |_| | | 
              | | |     / _ \
http://www.uni-karlsruhe.de/~ua5m/     | 
              |    |  _  | | | | |___ ( (_) )
Stop Pollution - Create Eevolution     |_|    
              |_| |_| |_| |____|  \___/
/*
 *    otp1.c - small programme that proves 
              why you can't "crack"
 *             
              OTP encrypted messages if you don't know anything
 *             
              about the plain-text.
 *             
              Donation to Irdial~Discs (leaders in _any_ field)
 *             
              Thilo Mezger, 27-Nov-97
 */
#include <stdio.h>
int main(void)
{
 char cipher[] = "ABCDE";
 char otp1[6], otp2[6];
 char plain1[] = "THILO", plain2[] = "MUSIC";
 int i;
 /***  calculate OTP depending on 
              desired decryption result  ***/
 for (i=0; i<5; i++)
 {
  otp1[i] = (char)(cipher[i] ^ plain1[i]);
  otp2[i] = (char)(cipher[i] ^ plain2[i]);
 }
 otp1[5] = '\0';
 otp2[5] = '\0';
 /***  output/decrypt first example  
              ***/
 printf("0x");
 for (i=0; i<5; i++)
  printf("%02x", otp1[i]);
 printf(" ");
 for (i=0; i<5; i++)
  putchar(cipher[i] ^ otp1[i]);
 putchar('\n');
 /***  output/decrypt second example  
              ***/
 printf("0x");
 for (i=0; i<5; i++)
  printf("%02x", otp2[i]);
 printf(" ");
 for (i=0; i<5; i++)
  putchar(cipher[i] ^ otp2[i]);
 putchar('\n');
 return 0;
}41643 =  3 3 7 661
78876 =  2 2 3 3 7 313
59277 =  3 19759
54663 =  3 7 19 137
07540 =  2 2 5 13 29
19417 =  19417
31570 =  2 5 7 11 41
57034 =  2 28517
63797 =  131 487
17854 =  2 79 113
96337 =  96337
05083 =  13 17 23
43211 =  7 6173
89133 =  3 11 37 73
71676 =  2 2 3 3 11 181
17950 =  2 5 5 359
52524 =  2 2 3 3 1459
85755 =  3 5 5717
66506 =  2 11 3023
75889 =  11 6899
90350 =  2 5 5 13 139
81102 =  2 3 7 1931
71299 =  37 41 47
20404 =  2 2 5101
50074 =  2 25037
24104 =  2 2 2 23 131
62215 =  5 23 541
81516 =  2 2 3 6793
30755 =  5 6151
14501 =  17 853
27876 =  2 2 3 23 101
21790 =  2 5 2179
55166 =  2 27583
88133 =  31 2843
37523 =  157 239
32497 =  32497
73511 =  19 53 73
02010 =  2 3 5 67
78983 =  19 4157
83605 =  5 23 727
78828 =  2 2 3 6569
06841 =  6841
72071 =  97 743
50139 =  3 3 3 3 619
36680 =  2 2 2 5 7 131
83260 =  2 2 5 23 181
50776 =  2 2 2 11 577
76762 =  2 7 5483
40723 =  193 211
10197 =  3 3 11 103
47518 =  2 23 1033
03752 =  2 2 2 7 67
08360 =  2 2 2 5 11 19
13907 =  13907
97104 =  2 2 2 2 3 7 17 17
83424 =  2 2 2 2 2 3 11 79
11062 =  2 5531
09003 =  3 3001
97425 =  3 3 5 5 433
06868 =  2 2 17 101
69668 =  2 2 17417
69206 =  2 34603
53582 =  2 73 367
76594 =  2 7 5471
34026 =  2 3 53 107
55913 =  11 13 17 23
12475 =  5 5 499
73644 =  2 2 3 17 19 19
08582 =  2 7 613
20609 =  37 557
56297 =  19 2963
09330 =  2 3 5 311
73318 =  2 7 5237
82309 =  53 1553
44604 =  2 2 3 3 3 7 59
94584 =  2 2 2 3 7 563
08306 =  2 4153
65209 =  61 1069
74282 =  2 13 2857
39222 =  2 3 3 2179
94676 =  2 2 23669
94507 =  7 23 587
45127 =  45127
19858 =  2 9929
68940 =  2 2 3 3 5 383
33344 =  2 2 2 2 2 2 521
13298 =  2 61 109
36137 =  36137
77043 =  3 61 421
60423 =  3 11 1831
18019 =  37 487
23199 =  3 11 19 37
98824 =  2 2 2 11 1123
46681 =  46681
28922 =  2 14461
13373 =  43 311
70086 =  2 3 11681
22260 =  2 2 3 5 7 53
71915 =  5 19 757
91901 =  29 3169
***********************************************************
Well, I wish we could crack it (G2) but I guess 
              those 100 numbers of the G2text
are simply not enough to crack it.  You need a lot more 
              data and then try
to find similarities.  I've factorized 
              all of the SwedRhap numbers and
there were quite a lot of prime numbers 
              in it (10 I think), much more than
you would expect in a totally random 
              set of 100 numbers.  But this might
be a coincidence - there's simply not enough data to prove it.
                                    
              _________ ___ ___ ___ ___      ___
Thilo.Mezger@stud.uni-karlsruhe.de  |___ ___| | |_| | | 
              | | |     / _ \
http://www.uni-karlsruhe.de/~ua5m/     | 
              |    |  _  | | | | |___ ( (_) )
Stop Pollution - Create Eevolution     |_|    
              |_| |_| |_| |____|  \___/
On January 20, 1992 at 2100 UTC, on 9092 khz, 
              was the following E5
message:
         458,458,458,1-2-3-4-5-6-7-8-9-0 
              etc.
         Count 64, Count 
              64
         110-18   
              999-99   930-35   048-94   664-13
         570-50   
              914-64   198-78   973-60   168-29
         729-28   
              013-82   504-23   029-03   140-16
         127-72   
              524-59   464-43   666-14   541-02
         369-24   
              999-99   011-66   714-55   121-96
         579-38   
              950-01   373-39   560-02   818-31
         455-38   
              727-07   812-83   101-71   612-26
         574-31   
              660-67   216-13   418-92   138-20
         785-65   
              951-46   520-33   426-03   108-65
         276-68   
              855-36   992-67   681-78   964-75
         021-72   
              672-36   611-99   127-06   517-43
         104-01   
              117-78   606-76   525-59   050-76
         862-44   
              739-73   999-99   110-18
         Repeat, etc, 
              End
Note that the odds of the first group being 
              the same as the last are
1 in 100,000. Also, even if you 
              accept that the second group by
accident was a "funny" one like 999-99, the odds of the penultimate
group matching it are also 1 in 100,000. 
              For both these "coincidences"
to show up in the same message 
              is 1 in 10,000,000,000; not 
              even
considering the *third* 999-99 (group 22)