The Hunt For Theta Seven - You Could Win 750,000,000 Cr!
- Thread starter Zac Cocken
- Start date
Well, how'd you manage it?
Good job CMDR!
Well, that seems convincing. Might wanna click "reply" on the original post with that, I'd hope it doesn't matter but I think it's bolded. Maybe you'll trigger some fancy automatic detection that, I dunno, blows up some more bombs or something.
Anyway, can't wait to see how I was one step away from the solution or something
congratulations on figuring it out!
Anyway, can't wait to see how I was one step away from the solution or something
congratulations on figuring it out!Can you please write down how you came to that message?
Panjabell system has 12 million population.
Not really nessecary.
If he can provide a methodical overview of how he went from the encrypted message to the supposedly decrypted message, then others can mimic his methods and see if results are equal. If it is, we got ourselves a winner, and theta better be packin'
Definitely, but until then, nothing is certainNot really nessecary.
If he can provide a methodical overview of how he went from the encrypted message to the supposedly decrypted message, then others can mimic his methods and see if results are equal. If it is, we got ourselves a winner, and theta better be packin'
Thank you! I should have thought of that. My kids keep distracting me.
The new sequences are:
Zeros:
61 76 99 71 113 41 96 91 69 117 141 108 102 104 92 96 83 80 91 150 114 144 163 83 123 94 162 131 150 98 74 98 75 111 95 155 131 97 151 105 144 155 64 123 89 127 52 103 51 88 147 114 121 90 153 104 84 151 113 117 123 73 126 106 104 75 90
Ones:
117 55 103 134 90 79 115 95 98 76 91 57 99 106 81 128 147 99 125 55 123 107 108 82 91 124 109 149 96 59 116 47 61 112 112 111 124 153 134 109 80 131 97 83 123
Twos:
126 74 81 103 51 84 114 90 102 78 79 132 115 105 93 138 108 115 129 29 59 112 129 89 135 98 152 128 81 90 87 51 89 98 129 113 111 131 84 129 105 99 118 144 49
Threes:
67 104 83 148 140 71 152 94 87 130 78
Overlying the data with duplicit log(grid 7x8):
Will yield these groups though:
Also worth noting is how each row is divisible by 7, same as the traffic log rows.
Using the duplicit row, it divides data into 3 parts of 56 values. So the table can be perfectly overlaid.
Will yield these groups though:
| 0 | 61 76 113 96 117 103 91 117 141 134 90 95 102 98 76 57 92 78 132 106 150 163 83 138 128 131 129 98 99 125 107 112 82 71 75 109 95 89 98 97 152 128 64 96 59 116 89 98 112 111 129 153 87 114 90 153 131 104 131 151 130 123 126 144 83 49 |
| 1 | 126 99 74 41 103 69 90 99 96 115 80 91 114 148 144 123 108 115 162 140 59 98 131 149 152 144 155 81 89 90 127 94 103 47 88 147 121 117 105 106 118 97 123 75 78 |
| 2 | 67 71 81 55 104 84 114 108 79 102 91 83 79 83 105 93 94 81 147 29 74 55 123 108 91 111 129 155 151 105 123 87 52 51 51 112 124 134 109 80 113 84 84 113 73 99 104 90 |
| 3 | 51 115 104 150 124 135 61 111 129 |
Also worth noting is how each row is divisible by 7, same as the traffic log rows.
Using the duplicit row, it divides data into 3 parts of 56 values. So the table can be perfectly overlaid.
Last edited:
The character count is correct. I'm very curious what the decryption method was. As the same characters are using different numbers in the encrypted message, so he figured out the key.
Good job if this is it! 
CMDRs,
Apologies for the delayed explanation of how I deciphered this. I had to high-tail it to Panjabell to look for anything unusual. There is nothing apparent there, but I haven't searched exhaustively yet. Please feel free to join in the hunt at Panjabell.
Like many, I tried a number of techniques to no avail until I stumbled upon the right one. The key was indeed the fake traffic logs. Using the incrementing Type-7 numbers to determine order, the other ship types spelled out a 7x7 matrix, namely:
0 | 0 | 1 | 2 | 1 | 1 | 2 |
0 | 2 | 1 | 0 | 1 | 3 | 0 |
2 | 0 | 0 | 1 | 2 | 0 | 2 |
3 | 0 | 0 | 2 | 0 | 0 | 2 |
0 | 2 | 3 | 0 | 1 | 1 | 0 |
2 | 2 | 0 | 1 | 1 | 0 | 1 |
0 | 1 | 2 | 1 | 1 | 2 | 0 |
Assuming this to be the key, I arranged the numerical ciphertext provided in the Galnet article into a matrix 7 columns by 24 rows. I then multiplied it by the inverse of the key above. The result yielded a reasonable number of unique values, 27 (with 30 being an outlier). To me, this appeared to be the alphabet (single case) plus a space. I started with 1=A, 2=B, etc. I set the remaining values after Z to space. This deciphered to the message I posted, to my surprise.
Good luck recreating my results. I used Excel for all of this. Good luck to us all in tracking down Theta 7 from here, and I'll see you in the black!
o7,
CMDR Therion Cygni
Awesome!!!
And nuts! I was just working on that!!
Attachments
congrats Therion
That's awesome, this has been fun to follow along!
OK, could you please explain this a bit simpler to me? I understood maybe half of that... to be more precise: What do you mean by "multiplied it by the inverse of the key above"?
I feel dumb now. Well done!
Run an internet search for matrix inversion. There are a number of educational sources, and even some handy tools to help with the calculations.