Sorry, pardon? Can you expound on this a bit?
It says “Connection detected : Access denied” 
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I was looking at the relationship between the two sibling images connection/CDAD(reddit) and fragment/C3q5NYF(freeimage) due to all the unrest about them and @pr1sm checked to confirm findings.
They are the “same image”:
Fragment: The fragment has perfectly quantized palette colors (exactly 7), clean 20×20 dot bit blocks (that might be 10x10 nibble packs), and a well-defined 5×8×100 grid structure (this we know already).
Connection: Has interpolated/mixed RGB values (e.g. 127 unique B-values vs 4 in the fragment). Blurred residuals and shearing are present in two directions.
By splitting the quantized palettes off of the image and rolling magenta +3 right and blue +1 right (as blocks), we place the blocks into their position as displayed in the CDAD image.
You can then observe the quant/shear aberration occuring in the two directions, displacing the colours out (magenta shears into R/B via near flip and cyan shears out to green/blue via near-low near-high flip also). You can observe the output here:
Imgur: Imgur: The magic of the Internet
(Edit: slightly updated palette order for a lower mismatch delta).
This topic of “is it, or isn’t it” can be put to rest with confidence.
Maybe in a lore context CDAD is an imperfect read from the memory bloc, and here we have a correct fragmented bloc. Guess we’ll know soon.
@the_architect - operators requesting file integrity check
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Nicely done. Pretty conclusive.
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Wow you reverse-ingeneered PM’s work!
Nicely done folks
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I guess we are just waiting on @the_architect to boot up this morning and see if @7101334’s post periods results!
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Hoping they have given us a day off 
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Next scheduled social media post is tomorrow, so anything that comes out before then is likely triggered.
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Just done catching up, you guys are doing incredible work, happy to sit back and watch the pro’s at play 
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Nice work!
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Just curious, but would we get anything in binary if all the different colors were separated, and then condensed into a single “ERROR”-sized cell?
I’m doubtful, as it looks as if cyan would be mostly 1s.
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Very clever : )
It’s a slightly complex question and depends on perspective of how to read/decode the permutes.
If you split all 7 out and treat them as equally possible positions in the layer stack, you have thousands of possible solves - realistically we don’t want that and want to keep the carrier (2 layers) intact and at the rear, so we just shuffle the 5 resulting in a much more reasonable a little over a hundred possible solves.
With those, you can either sort/treat them as different bit orders, or treat groups as the same, or treat all content layers as a single flattened value.
I grabbed the wrong line as an explainer before, so edited:
As an example, if you flatten all to 1 except carrier base 0 you get:
- Binary:
11111111 11111111 11111111 11111111 11111111 - Bytes:
0xFF 0xFF 0xFF 0xFF 0xFF - ASCII:
ÿÿÿÿÿ
Because all 5 end up replacing the carrier base like you mentioned. There are a bunch of other options, but exhausting most of them has resulted in garbage output.
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After tinkering with the spectrogram previously when it was first dropped I got ; 71, 37, 32, 67, 31, (inaudible but sounded like) (51/57/35/53 or something with a 5 or 3 in it, hard to say, somebody check me on it? if you need the audio file i used let me know)
my first thought was cords but it could be possible that it was a input to something? idk how useful it is or i am just stating old news, but it is worth the shot
Adding to this, I’ve given a few of these (the most obvious attempts) a try, and the result is always jumbled exotic or missing characters that make no sense.
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Yeah, all 120 of those permutes produce garbage.
All 5040 also do
The 5060 line report was a blast to read.
The most unique hits were simply:
[255][255][255][255][255] (2100 perms)
ARR[3]R (720 perms)
[251][255][255][179][255] (72 perms)
[255][255][255][179][255] (588 perms)
USR_V (168 perms)
A[223]V[3]V (120 perms)
ERR[3]R (120 perms)
CRR[3]S (120 perms)
U[223]V_V (84 perms)
W[223]V_W (84 perms)
WSR_W (84 perms)
E[223]V[3]V (48 perms)
C[223]V[3]W (48 perms)
GRR[3]S (48 perms)
G[223]V[3]W (36 perms)
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Yellow (or green, depends CMYN/K or RGB) dots are too sparse to be random.
Picture is translated few pixel towards the right (first left vertical pixels are in fact far right)
Some dark spots (but not black) stand out when normally true black patterns are painted in white (for visibility contrast). They’re painted in red to make them more visible.
Some spots clearly appear in straight vertical lines
Not all have been painted in white here, but these spots are even in some “error cell” (made a few to demonstrate)
On a sidenote,
@the_architect seems to ‘rush’ solutions on mondays when operators stalled over the weekend (when most are available). Obviously, there has to be a timetable up to 10th anniversary and it seems reasonable to think each week will have a puzzle with last clue to solve it being given during the weekend. So Monday interactions with PM may be considered help and hints before next puzzle.
I was specifically talking about the combined block question, not the whole available list of bitplanes.
But, sparseness is the opposite of what you want for a binary decode.
The transform space needs to be significant in variety to produce meaningful decode in this way.
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New hint ?
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Y’all, the password is EMILY.
The anomalous grid resulting in QSRSV was intended to mark a new offset of +1 to the right. This results in:
01000101 01001101 01001001 01001100 01011010
or, EMILZ. Looked suspiciously close to EMILY, so I gave that a shot. Turns out, that’s the password.
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