New Bitcoin Puzzle

[mi1994]

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.

Doesn't change much really, you're still just assigning a letter to a symbol, the main issue is that it's unlikely to be plaintext English as the frequencies of each symbol are so off what would be expected.
Looking at frequencies of other languages it looks closer to Italian, at least at the high end, which isn't impossible given the Rome connection.

[Pompobit]

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.

Doesn't change much really, you're still just assigning a letter to a symbol, the main issue is that it's unlikely to be plaintext English as the frequencies of each symbol are so off what would be expected.
Looking at frequencies of other languages it looks closer to Italian, at least at the high end, which isn't impossible given the Rome connection.

I'm Italian, if you have something to do a frequency attack in Italian, I could try to give it a sense

[sirbamf]

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.

Doesn't change much really, you're still just assigning a letter to a symbol, the main issue is that it's unlikely to be plaintext English as the frequencies of each symbol are so off what would be expected.
Looking at frequencies of other languages it looks closer to Italian, at least at the high end, which isn't impossible given the Rome connection.

I'm Italian, if you have something to do a frequency attack in Italian, I could try to give it a sense

Am I the only one confused? What are we supposed to be looking for?

[Coinnoiseur]

There is also some symmetry in the way the 'dark' pipes connect to the blocks.

On the left and right they connect to the same 6 'mirrored' blocks, tough one pipe connection has a different orientation.
On the top and bottom 3 out of 4 blocks are the same, one is different.

Also taking the 24 blocks that those pipes connect to, there is an abundance of 'bottom-right' triangles: 13 of the 24.

[Edit: Color blindness yay... checked the sheet and looked again... the pipe at the top only connects to 3 blocks)

[mephala]

Everything else failing, I too think we need to make some sense out of the green piping in the background. Especially the dark green piping doesn't seem random at all

[4pollo]

Everything else failing, I too think we need to make some sense out of the green piping in the background. Especially the dark green piping doesn't seem random at all

Here is a possible segmentation of the image by content:
http://imgur.com/iylMHIc

This animation illustrates possible depths of the image content and might help to see which pipe goes where. (here it assumes QR on top of AZTEC CODE AND DARK PIPES on top of YELLOWISH PIPES on top of LIGHT PIPES on top of RED GRID)
http://imgur.com/V5Gfo3I

GIMP file with image content split out to different layers:
https://mega.nz/#!6gYB2K7D!7WD3zHx0RfQ0y3po6Ij3SYSAWV7ATDkJkX9AerybMOI

Disclaimer: another red-green colorblind person here, so do not blindly trust this segmentation. It probably needs corrections and needs to be split out further still.

[franky1]

ill just leave this here as a random thing.. it may be nothing. im only mentioning the "cream" coloured square symmetry. although there is more alterations with the dark green pipes too. but ill leave you to think about it

[coin@coin]

Just throwing down assumptions from what has been found so far:

So the back of the map, the first layer, the dark and light red squares as shown in the animation here: http://imgur.com/V5Gfo3I (Thanks to 4pollo)
look like a 12 x 12 chess board.

Google search for 12x12 chessboard shows Millenial Chess

http://www.chessvariants.com/large.dir/contest/milchess.html

I wonder if what has been found so far could be used on the chess board to pinpoint specific squares or the order to look at them.

Like these below from 4pollo?

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[2b]   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)
[2b] = might be 3b

When looking at the map on the game online the message that comes up is:

"The journey inward begins here" is scrawled on the back.

Maybe back means scrawled on the first layer of the map picture?

Scrawled means: write (something) in a hurried, careless way.

Also why is the game picture on TextAdventures faded out, discoloured, is that trying to tell us to ignore certain colours on the map?



I'm only assuming just thought I throw some ideas for people smarter than me to figure it out.

[Prompyboo]

Im so confused.....

[mirth23]

all,

I've been lurking in the background this time around but have worked through some of the same paths that have already been posted on here. At any rate, I tend to like to facilitate rubber duck debugging on here by producing summaries of what we know when we're stuck, so here goes:

* There are 15 QR codes. They appear to be text adventure room descriptions. Speculation is that they will be clues to other rooms in the text game. Many of them correspond to components of the Porto Alchemica (which I suspect will be important later). There is a reference to Mechanica Mathematica that doesn't seem to lead anywhere.

* One QR is a URL to a text adventure game that wants a keypad entry. Other items in the game don't seem relevant at this point (shirt, shoes, jacket, etc). A small amount of interaction is possible but we are pretty much stuck in a cell with a keypad.

* The QR codes have three different patterns of 4 or 6 gold dots around them. It's not clear what, if anything, this means.

* The tan squares make a non-compliant Aztec code (specifically, it is missing targeting pixels in the upper right of the interior square). Some attempts have been made to use the Aztec decoding algorithm by hand since the upper left portion of the code is all that should matter (the rest is CRC), but it didn't lead to a meaningful outcome. That said, Aztec decoding is voodoo magic so someone else might have better luck.

* Each Aztec pixel has a triangle and an orange dot. There are 26 unique combinations of dots + triangles (9*3 - 1 combo which never used), which seems to imply a basic alphabetical substitution cipher.
** Straightforward mappings combining dots and triangles (e.g., upper left dot + upper left triangle position == A, etc) don't cause anything to jump out for any of several 'normal' orderings of dots and triangles. If this is the mapping I would strongly suggest that the bottom right triangle is T-Z since that triangle is missing a dot in the bottom right position. Performing this mapping results in poor frequency analysis results. My suspicion is that a mapping like this is correct and that we need to understand which letters we need to extract from the grid, and in which order. "The journey inward begins here" may imply reading inwards, and the use of a 2 layer Aztec code might imply in a counterclockwise spiral.
** Other approaches to substitute letters have been promising but have gotten nowhere. Pigpen does not map well. The phone keypad is an interesting approach and results in slightly nicer frequency analysis but you can only do it if you do things like arbitrarily throw out Q and Z, and we haven't had any hints to try to map this to anything having to do with a phone. Also, I find it very hard to ignore the fact that there are 26 combinations of dot + triangle, which keeps me not wanting to jump away from a more straightforward substitution approach.

* Brown/green areas outside the Aztec code area. Unclear if they mean anything or are just art. (There was a LOT of 'just art' in earlier OP games.)

* The image is hosted on an image zoom service. Component image files do not have any EXIF clues. Neither does the map that you can download from the text adventure game.

* In the past there have been two things in the first puzzle step - proof of prize in the form of a public bitcoin address, and a hint leading to the next step. So it's useful to keep in mind that at any point we might encounter both, and that may be leading to confusion now.

[4pollo]

all,

I've been lurking in the background this time around but have worked through some of the same paths that have already been posted on here. At any rate, I tend to like to facilitate rubber duck debugging on here by producing summaries of what we know when we're stuck, so here goes:

* There are 15 QR codes. They appear to be text adventure room descriptions. Speculation is that they will be clues to other rooms in the text game. Many of them correspond to components of the Porto Alchemica (which I suspect will be important later). There is a reference to Mechanica Mathematica that doesn't seem to lead anywhere.

* One QR is a URL to a text adventure game that wants a keypad entry. Other items in the game don't seem relevant at this point (shirt, shoes, jacket, etc). A small amount of interaction is possible but we are pretty much stuck in a cell with a keypad.

* The QR codes have three different patterns of 4 or 6 gold dots around them. It's not clear what, if anything, this means.

* The tan squares make a non-compliant Aztec code (specifically, it is missing targeting pixels in the upper right of the interior square). Some attempts have been made to use the Aztec decoding algorithm by hand since the upper left portion of the code is all that should matter (the rest is CRC), but it didn't lead to a meaningful outcome. That said, Aztec decoding is voodoo magic so someone else might have better luck.

* Each Aztec pixel has a triangle and an orange dot. There are 26 unique combinations of dots + triangles (9*3 - 1 combo which never used), which seems to imply a basic alphabetical substitution cipher.
** Straightforward mappings combining dots and triangles (e.g., upper left dot + upper left triangle position == A, etc) don't cause anything to jump out for any of several 'normal' orderings of dots and triangles. If this is the mapping I would strongly suggest that the bottom right triangle is T-Z since that triangle is missing a dot in the bottom right position. Performing this mapping results in poor frequency analysis results. My suspicion is that a mapping like this is correct and that we need to understand which letters we need to extract from the grid, and in which order. "The journey inward begins here" may imply reading inwards, and the use of a 2 layer Aztec code might imply in a counterclockwise spiral.
** Other approaches to substitute letters have been promising but have gotten nowhere. Pigpen does not map well. The phone keypad is an interesting approach and results in slightly nicer frequency analysis but you can only do it if you do things like arbitrarily throw out Q and Z, and we haven't had any hints to try to map this to anything having to do with a phone. Also, I find it very hard to ignore the fact that there are 26 combinations of dot + triangle, which keeps me not wanting to jump away from a more straightforward substitution approach.

* Brown/green areas outside the Aztec code area. Unclear if they mean anything or are just art. (There was a LOT of 'just art' in earlier OP games.)

* The image is hosted on an image zoom service. Component image files do not have any EXIF clues. Neither does the map that you can download from the text adventure game.

* In the past there have been two things in the first puzzle step - proof of prize in the form of a public bitcoin address, and a hint leading to the next step. So it's useful to keep in mind that at any point we might encounter both, and that may be leading to confusion now.

Great summary mirth23! A few things to add:

* The phone substitution made me realize that 'space' could be a possible substitution as well.

* A correction to QR code 11: http://textadventures.co.uk/games/view/5craldfdmkkzngf1davsna/generation-a-days-destroyed
'Generation A' is a book by Douglas Coupland (https://en.wikipedia.org/wiki/Generation_A) and according to that page, the name 'Generation A' is based on a commencement speech by Kurt Vonnegut (http://versailles1.tripod.com/syracuse.html). No idea yet how this relates to 'days destroyed' or 'bitcoin days destroyed'.

* note the different quotes around "Mechanica Mathematica". This indicates that in case we try doing something with the quoted parts of the text, this should not be treated in the same way as the actual quotes.

* 181 aztec pixels corresponds roughly to the length of a bitcoin address in binary (?)

* The hi-res image seems to be a scan of the artwork. The low-res map image from the text adventure seems to be a photo of the artwork hanging from a wall (see the shadows). The different lighting makes it easier to distinguish some of the red squares in the low-res version than in the hi-res version.

Also why is the game picture on TextAdventures faded out, discoloured, is that trying to tell us to ignore certain colours on the map?

* Good point! I'd agree that it means something, and would guess that it indicates that for level 1, we need to use the QR codes.

* Besides 4 or 6 gold dots, there are the 4 possible orientations (=possible encoding of 2 bits) for the QR codes, and potential mirroring (= possible encoding of 1 bit). Together with the dots, that gives 4 potential bits encoded per QR code.

* I've been trying to recreate the QR-codes after reading them out, with varying success: some are identical, some are partly identical, some are different altogether. My thinking was that with the error correction level L in the QR codes it would be possible to flip up to 7% of the bits and still read it out successfully. Those flipped bits could potentially be used to encode letters or something else directly into the pixels of the QR code. These could then be recovered by taking the difference of the pristine and the altered QR code.

[coin@coin]

all,

* Each Aztec pixel has a triangle and an orange dot. There are 26 unique combinations of dots + triangles (9*3 - 1 combo which never used), which seems to imply a basic alphabetical substitution cipher.

Guess what? The keypad in the cell is a QWERTY Keyboard. A QWERTY Keyboard has exactly 26 letter keys, only letters not including numbers nor special characters or other keys. Counting just the letters.

Can that lead somewhere? "The journey inward begins here", so if G (in the middle of the keyboard) is the middle square can the other letters be mapped easily?
Or maybe decide where to start from with the squares and replacing the first 6 with QWERTY (1 per square) in a spiral way as mirth23 mentioned?

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[2b]   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)
[2b] = might be 3b

Unfortunately I don't know how the mapping works... But I'm sure others here can have a go.

[Coinnoiseur]

'Generation A' is a book by Douglas Coupland (https://en.wikipedia.org/wiki/Generation_A)

It might also be 'Generation Algorithm', which also could be a hint as it seems we are looking for some algorithm to generate a key from the blocks in the artwork.

Guess what? The keypad in the cell is a QWERTY Keyboard. A QWERTY Keyboard has exactly 26 letter keys, only letters not including numbers nor special characters or other keys. Counting just the letters.

I totally overlooked that as a possible hint (while actually in the room/game atmosphere it stands out like a sore thumb).

According to the game "It's a full QWERTY keypad.", which is usually used referring to phones with keypads, so maybe something like this:

https://i.imgur.com/aJtoJDa.jpg

But that would still leave the too even distribution.

[alphabetacanary]

Okay, I have a theory.

Suppose the puzzle creator generated an aztec code with the password which is 19x19 pixels.

Take a copy of the original code and rotate it 180 degrees.

Now cut out a triangle from the lower right of the new rotated image (defined by top right, bottom right and bottom left corners).  Paste this over the original image.

Now, the pixels along the line defined by top right and bottom left would have been swapped.  But it looks like the pixels in this image are mirrored.  So that's 9 pixels that were left untouched (if you don't count the very center pixel).  That would mean exactly 181 pixels were 'lost' in this process.

19+18+17+16+...2+1 - 9 = 181

Now suppose you wanted to encode the 'missing' lower right corner in the pixels of the newly created 'invalid' aztec image.  You would need 181 pixels to encode on/off, right?

Suppose you went through this process and discovered that the now invalid aztec code you created (original + pasted part) only had 179 pixels set (just because that's the way the encoding went for the top left part).   What would you do?  Well, just add 2 pixels to the lower right corner since that's the section that's going to be restored anyway.  (That would explain why there are 2 extra pixels in the bottom mirror part.)

So, in short, I believe the goal of this first step is to restore the original aztec code. There are 181 pixels that need to be restored in the lower right corner and I believe that information is given in the 181 pixels of this image.

The only part that is troubling is that you only need on/off information for those 181 pixels so the 3x3 grid and triangles are still a mystery.  This theory might be totally wrong but it would explain the 181 count and 'mirrored' aztec image.



(181 pixels to reconstruct = # of pixels in mirrored invalid aztec)

[coin@coin]

That sounds really interesting!
If I just knew how to do it...

But others on here I'm sure they'll have a shot.

[4pollo]

Another observation about the QR codes, after going through http://www.thonky.com/qr-code-tutorial/data-masking

Code:

All codes use mask pattern #2, except for:

- QR code 03 (The corridor is lined with pulsating light bulbs in iron sconces. In an alcove there is a fallen statue with an inscription “When azoth and fire whiten Latona, Diana comes unclothed.”) 
- QR code 10 (There is an iron passageway, it is very dark, you walk slowly with your hand against the wall guiding you... just in case. You come across a fortified iron door guarded by two mechanical gargoyles. There is an inscription on the door “The dragon guards the key.”)

which use mask pattern #6.

That said, it could just be a coincidence of course.

[mi1994]

Okay, I have a theory.

Suppose the puzzle creator generated an aztec code with the password which is 19x19 pixels.

Take a copy of the original code and rotate it 180 degrees.

Now cut out a triangle from the lower right of the new rotated image (defined by top right, bottom right and bottom left corners).  Paste this over the original image.

Now, the pixels along the line defined by top right and bottom left would have been swapped.  But it looks like the pixels in this image are mirrored.  So that's 9 pixels that were left untouched (if you don't count the very center pixel).  That would mean exactly 181 pixels were 'lost' in this process.

19+18+17+16+...2+1 - 9 = 181

Now suppose you wanted to encode the 'missing' lower right corner in the pixels of the newly created 'invalid' aztec image.  You would need 181 pixels to encode on/off, right?

Suppose you went through this process and discovered that the now invalid aztec code you created (original + pasted part) only had 179 pixels set (just because that's the way the encoding went for the top left part).   What would you do?  Well, just add 2 pixels to the lower right corner since that's the section that's going to be restored anyway.  (That would explain why there are 2 extra pixels in the bottom mirror part.)

So, in short, I believe the goal of this first step is to restore the original aztec code. There are 181 pixels that need to be restored in the lower right corner and I believe that information is given in the 181 pixels of this image.

The only part that is troubling is that you only need on/off information for those 181 pixels so the 3x3 grid and triangles are still a mystery.  This theory might be totally wrong but it would explain the 181 count and 'mirrored' aztec image.

https://i.imgur.com/z68P9Um.png

(181 pixels to reconstruct = # of pixels in mirrored invalid aztec)

The way you have split it it is still invalid as the top right orientation marker is missing, and the first 5 codewords are still readable giving ".e: " (ignore the quotes) which doesn't work, and doesn't resemble the start of anything meaningful.

[alphabetacanary]

The way you have split it it is still invalid as the top right orientation marker is missing, and the first 5 codewords are still readable giving ".e: " (ignore the quotes) which doesn't work, and doesn't resemble the start of anything meaningful.

1. What about this configuration?  The top right marker could be set.
2. Wouldn't it be possible for the error correction code words to correct bits that were intentionally flipped in the data code words?

[4pollo]

TL;DR: me rambling on possible directions for the Aztec code. Conclusion: try a lot of things randomly and hope the ECC saves us.

HOW WAS THIS AZTEC CODE POTENTIALLY GENERATED?

Suppose that the resulting code can be generated either by
A) rotating the Aztec code 180 degrees, and then overlaying it on the original;
B) rotating the Aztec code 180 degrees, and setting the result using any commutative logical operation (i.e. f(x,y)=f(y,x)) to get a symmetric result (there are 8 possible truth tables that satisfy that requirement);
C) taking half the original Aztec code and adding it's mirror to get a complete one.
D) other?

option A
based on the missing orientation marks, we can cross off option A, or we would have seen them in the result.

option B
the 8 possible truth tables are

Code:

xy   a b c d e f g h
00 | 0 0 1 1 0 0 1 1
01 | 0 0 0 0 1 1 1 1
10 | 0 0 0 0 1 1 1 1
11 | 0 1 0 1 0 1 0 1

* the resulting code is not fully white or fully black codes, so possibilities a and h can be ignored.
* overlaying the bottom right orientation marker on the top left orientation marker (https://en.wikipedia.org/wiki/Aztec_Code#/media/File:Aztec_Code_with_desc.png) means 11->1, given the result (same for the bulls-eye on top of the bulls-eye), hence possibilities c, e and g can be ignored.

This leaves

Code:

xy   b d f 
00 | 0 1 0
01 | 0 0 1
10 | 0 0 1
11 | 1 1 1

* the top right orientation marker on top of the bottom left orientation marker means 10->0, 01->0, 00->0, given the resulting code. Therefore, possibilities d and f can be dismissed.

This leaves only

Code:

xy   b
00 | 0
01 | 0
10 | 0
11 | 1

* This means that in the resulting code only those pixels that have a mirrored black pixel remain visible. However, that would mean that the top left orientation code would disappear, except for one pixel....which it did not. Hence possibility b can also be dismissed.

Conclusion: no commutative function was used to generate a symmetric Aztec code from the original Aztec code and the original aztec code rotated 180 degrees.

option C
This option is possible if you take a subset of the pixels (not necessarily half) of the original Aztec code, and then overlay a 180 degree rotated version of that. This is the option analyzed further below.

option D
Not sure what other ways there would be to generate a symmetric code. Or perhaps it's not an Aztec code in the first place... 


REMOVE THE SYMMETRY, BY REMOVING ONE HALF?

Ignoring the 2 extra pixels and the bulls-eye, there is 65 pairs of mirrored pixels and therefore 2^65 ways of removing the symmetry and removing 1 pixel from each pair, and 3^65 possibilities if we also consider the option of keeping both pixels of the pair. Therefore, it's not feasible to go through all possibilities unless a big percentage of those possibilities decode successfully due to the error correction capability of the Aztec code.

Therefore, if removing the symmetry and reading out the result is the right approach, we try assuming a few things to limit the possibilities:

1) that the bulls-eye remains intact

2) a more structured way to leave out half the pixels by requiring that the halves are contiguous areas of pixels (with one of the possibilities given above). Any path from the bulls-eye to the edge of the code, together with its mirrored path defines two such halves that are mirrors of each other.

3) we can further try assuming that the mirrored half contains two correct orientation markers (as shown in https://en.wikipedia.org/wiki/Aztec_Code#/media/File:Aztec_Code_with_desc.png). Therefore, it must be a half that contains the top-left and bottom-left orientation marker (i.e. the top-right and bottom-right orientation markers are wrong in the Aztec code) that is picked and rotated.

4) the orientation marks of the missing half can be filled in correctly

However, even with these assumptions there are still many ways to do split up the code in two halves and remove one of the halves. We might get lucky though if the error correction level of the original Aztec code is high enough and the exact path does not really matter. Either way, it is probably something we would want to do programmatically.

Alternatively to removing one half completely (which corresponds to flipping all the 1s to 0s in that half), we could randomly guess the missing half of the code and hope the ECC will allow us to read the code .


REMOVE THE SYMMETRY, BY EXPLORING CODES WITHIN A CERTAIN EDIT DISTANCE?

Finally, in case we don't make the assumption of the halves being contiguous areas, we could still explore randomly the codes within a certain edit distance of the given code, and hope the ECC will allow us to read the code successfully. So, assume a correct bulls-eye with correct positioning markers, randomly flip up to n pixels in the given code, and attempt to read the result.


ARE WE MISSING SOMETHING?

Still:

- the 2 non-symmetric pixels are really bothering me
- we have not used the markings in any way yet.

So, are we missing something here? Did we miss a clue? Or are we just failing to see the forest for the trees?

[franky1]

i think we are definetly missing something. even on the text game i thought we would have had some clue within the game to get the passcode to get throw the door. afterall the game inventor gave us the map before getting into the game so obviously there must be something else