Search content
Sort by

Showing 2 of 2 results by Samsondenali
Post
Topic
Board Bitcoin Discussion
Re: New Bitcoin Puzzle
by
Samsondenali
on 17/06/2016, 14:43:18 UTC
Stuck and unsure which direction to go at this point and what parts of the available information are relevant for this stage (no luck with any of the approaches so far):

- use content of the qr codes?
- anagrams?
- use the markings?
- use the pipes?
- decode the aztec as something else? (since no proper orientation markers, and apparent improbable symmetry suggesting it might not be an aztec code)
- select appropriate pixels from the aztec using markings or pipes, and decode entire code? (the aztec core does not contain proper orientation markers though)
- select appropriate pixels from the aztec using markings or pipes, and decode part of the code? (the aztec core mode message indicates 2 layers and 7 data codewords, so presumably the keypad code is <= 7 characters)

Any opinions or hints to narrow it down?

A phone pad?
1=0
2= ABC
3= DEF
4=GHI
5= JKL
6= MNO
7 = PQRS
8 = TUV
9= WXYZ

Triangle orientation could determine if it's A, B, or C. So, 2a would be A, 2b would be B, and 2c would be C?

Was just looking at my phone and thought of this. Not sure how it would apply to the squares that have four letters.

I haven't had much time to spend on this, but isn't this a modified pigpen cipher? Everyone has exhausted this possibility? Could we all share the different variations of pigpen cipher we've tried? 

I like the sudoku suggestion. That does trigger some ideas. But before I go down a sudoku rabbit hole... I think we should consolidate the most obvious choices for a basic substitution cipher. And make a list of all the strings that have been checked to date.
Post
Topic
Board Bitcoin Discussion
Re: New Bitcoin Puzzle
by
Samsondenali
on 06/06/2016, 18:35:41 UTC
4Pollo, would you mind making those corrections on your grid and reposting. I'm not adept with the style and formatting on bitcointalk.

Thanks for corrections! Here is the updated grid (with the 6 corrections so far indicated by underscores):
Code:
6b 7a 8b             7a 2b 2b    1d             6a    9b
8a    1d    1d    3d 7a 1a       1d    2a 3b 1a 6d 5a   
        _8a_      6b 1a      _7d_5d 5b 4a    7b 2d     
      9b 2d            _8d_3b 5a 3d 8b          2d     
         6a 7b 9a    8d          7b 7b       8d    4d   
4d    6b    3d 2b 7b 2b 2b 7b 2a 9b 5b 4d    8d 1b 4b   
   1b 8a    2a 4b                      2a 1b 8b         
6d             3d    6b 9a 4a 8b 3d    7d 8a           
   1a    9a 4a 6d    7d          1a    5a               
3a    6a_3b_   7d    4a    7d    6b    3d    3b 7a   _4d_
               3d    3d          9b    4b 3b 4b    1b   
            9a 5a    9a 5a 3a 6a 4b   _7a_            9b
         3b 1a 9b                      2b 8a    8d 6b   
   8a 2a 6a    8a 1b 1b 8a 7a 1a 7a 3b 9a 7a    6b    4d
   7d    9b       5b 2b    6b    6d    1b 1b 1b         
      #b          3b 1b 6b 3b 3a             3b 7d     
      1b 6a    3a 8b 5b 9a       7a 2a       8d         
   2d 6b *a 6d 5d    5d    1a 1b 7a 5b    6d    7b    5d
7b    2d             3d    8d 3d 9b             7d 8b 6a
Code:
# = 6 (or 2 or 3)
* = 2 (or 1 or 4)

I disagree. I think the newly changed 3b is 2b. Imo