[ARG] The Pizza Code Mystery

Black Mesa Archive The Cafeteria
#4561

I need some help here. I don’t understand the big words and math stuff…

the side scrolling signal, next to the angel spit song. I need some help decoding it please. I found these
Quadrature phase, oscillations that are said to be in quadrature if they are separated in phase by 90° (π/2, or λ/4)

Quadrature phase-shift keying (QPSK), a phase-shift keying of using four quadrate points on the constellation diagram, equispaced around a circle

In Optical phase space, quadratures are operators which represent the real and imaginary parts of the complex amplitude; see also in-phase and quadrature components
marking down 1’s and 0’s with the top edges I got this for all 4 lines.
10101010101010101010101
10011001100110011001100
11100001111000011110000
11111110000000011111111

when I run it through a translator, I got this for oct. 52525253 14631463 41703607 74007760
for hex I got aa aa ab 33 33 33 87 87 87 f0 0f f0
for ascii85 I got this. WiE*U1GafAL[YiD

can anyone tell me what any of it might mean? I has the dumb.
youtu.be/g9M9G4Wqc1U

#4562

[attach=4567,none,256][/attach]

Look at how each line is half the frequency of the previous line, so it looks like it could be the outputs of a binary counter.

Instead of writing the bits out like this:

10101010101010101010101
10011001100110011001100
11100001111000011110000
11111110000000011111111

…write them out like this:

1111
1100
1101
1010
1011
1000
1001
0110
0111
0100
0101
0010
0011
0000
0001
1110
1111
1100
1101
1010
1011
1000
1001

where the least significant bits are from the first line, and most significant bits are from the 4th line.

If we compare this sequence to that of a binary counter, we see that the 2nd and 3rd lines (1st and 2nd bits, respectively) have been inverted. It is also not starting at 0, and there is some overlap.

[code]Binary Counter | Binary signals on monitor

Binary | Decimal | Binary | Decimal
-----------------|--------------------------
1001 | 9 | 1111 | 15
1010 | 10 | 1100 | 12
1011 | 11 | 1101 | 13
1100 | 12 | 1010 | 10
1101 | 13 | 1011 | 11
1110 | 14 | 1000 | 8
1111 | 15 | 1001 | 9
0000 | 0 | 0110 | 6
0001 | 1 | 0111 | 7
0010 | 2 | 0100 | 4
0011 | 3 | 0101 | 5
0100 | 4 | 0010 | 2
0101 | 5 | 0011 | 3
0110 | 6 | 0000 | 0
0111 | 7 | 0001 | 1
1000 | 8 | 1110 | 14
1001 | 9 | 1111 | 15
1010 | 10 | 1100 | 12
1011 | 11 | 1101 | 13
1100 | 12 | 1010 | 10
1101 | 13 | 1011 | 11
1110 | 14 | 1000 | 8
1111 | 15 | 1001 | 9[/code]

I’ve identified the sound file where that sound is coming from: [tt]bms_sounds_misc_dir.vpk -> sound\bms_ambience\machine_ambience\m_ambience_comp2.wav[/tt]

The spectrogram looks interesting (but I’m too lazy to upload a screenshot right now). There are some interesting pulses in the upper end of the spectrum. It would be interesting to know what’s making this sound, but I going to bet that this is just a stock sound that has nothing to do with the ARG.

Another thing, those consoles with the array of red (or green) lights are the ones that were “upgraded” from the consoles with the databank server write displays.
qe_console_widea_2.png

#4563

[align=center][/align]

@Flavrans who wrote:[/size]
[/size]
Brainfuck Source Code[/size]

I had forgotten all about the Brainfuck code so, I went looking back and found a post by pointless talking about a js code on the KXBM site that was composed of a sequence of hex digits where “a sound is played when the digits is at least 4 and the effect is long and when the digit is odd. The digits are chosen so that it looks random at the first sight.”

Is there a way to substitute our HEX code in the script and see what happens?

#4564

FYI, it wasn’t me who wrote the “Brainfuck Source Code” label in the source code box, it’s the stupid forum software. The source code detection code of the syntax highlighter seems to think it’s code written in the Brainfuck programming language, apparently because it sees a number of Brainfuck characters in a row and concludes it must be Brainfuck source code.

Yes, but @faed has already tried:

The format of the hex encoded morse code was explained by @1942rob:

#4565

///B/AI5BM/AI5BM/AI5BM/BLACK/MESA/RESEARCH/FACILITY/BIODOME/LABS./TRAPPED,/NEED/RESCUE./EVAC/ZONE/NOT/SAFE./MILITARY/HOSTILE//K//////////////////////////////

guess we already know that part…

#4566

this is something that has not been tried yet, and i think it relates really well to what we are trying to solve. i linked BENALOHPAILLIER, alice and bob, how deep the rabbit hole goes, and time reveals all. someone has access to running this in a script hopefully, i did all the leg work, just need someone to solve. divided up into a few parts. He said he would give us a REALLY HARD ONE, that can be solved, i think this would qualify.
its based on
A common use of elliptic curve cryptography is to generate a shared secret key for use in secret key systems. The Diffie-Hellman protocol can be applied to elliptic curves as follows:

  • Alice generates a secret key r[sub]A[/sub]
  • Alice calculates P[sub]A[/sub] = r[sub]A[/sub]*e[sub]1[/sub] and sends it to Bob.
  • Bob generates a secret key r[sub]B[/sub]
  • Bob calculates P[sub]B[/sub] = r[sub]B[/sub]*e[sub]1[/sub] and sends it to Alice.
  • Alice receives P[sub]B[/sub] and calculates P = r[sub]A[/sub]*P[sub]B[/sub]
  • Bob receives P[sub]A[/sub] and calculates P = r[sub]B[/sub]*P[sub]A[/sub]
  • Alice and Bob can use P to generate a shared secret key suitable for AES or some other system. Some protocols use the x coordinate of P as the shared key, or a hash can be applied to P to generate the key. Note that if Alice already has published an elliptic curve public key, then Bob and Alice can agree to use Q in place of P[sub]A[/sub].
  • I’m leaning more towards part 3 , is the first few numbers are 3b, 28 or 2b, 6d and 3a. all of which are in the first blocks on the halos file.
    part 1 ,

A=2101 B=2103

A’s secret number = 2101
Scalar Multiply: 2101*(2,9)
Calculating 1*(2,9) = 0*(2,9) + 1*(2,9) = (Zero) + (2,9)
Adding (Zero) and (2,9)
Sum is (2,9)
Working result: 1*(2,9) = (2,9)
Doubling 1*(2,9) = (2,9)
x1^2 = 4
3x1^2 = 12
3x1^2+a = 16
2y1 = 18
lambda = (3x1^2+a)/(2y1) = 64
lambda^2 = 49
x3 = lambda^2-x1-x1 = 45
x1-x3 = 28
lambda(x1-x3) = 17
y3 = lambda*(x-x3)-x1 = 8
Double value: 2*(2,9) = (45,8)
Doubling 2*(2,9) = (45,8)
x1^2 = 37
3x1^2 = 40
3x1^2+a = 44
2y1 = 16
lambda = (3x1^2+a)/(2y1) = 56
lambda^2 = 12
x3 = lambda^2-x1-x1 = 64
x1-x3 = 52
lambda(x1-x3) = 1
y3 = lambda*(x-x3)-x1 = 64
Double value: 4*(2,9) = (64,64)
Calculating 5*(2,9) = 1*(2,9) + 4*(2,9) = (2,9) + (64,64)
Adding (2,9) and (64,64)
y2-y1 = 55
x2-x1 = 62
lambda = (y2-y1)/(x2-x1) = 57
lambda^2 = 54
x3 = lambda - x1 - x2 = 59
x1-x3 = 14
lambda*(x1-x3) = 17
y3 = lambda*(x1-x3)-y1 = 8
Sum is (59,8)
Working result: 5*(2,9) = (59,8)
Doubling 4*(2,9) = (64,64)
x1^2 = 49
3x1^2 = 5
3x1^2+a = 9
2y1 = 57
lambda = (3x1^2+a)/(2y1) = 45
lambda^2 = 37
x3 = lambda^2-x1-x1 = 51
x1-x3 = 13
lambda(x1-x3) = 17
y3 = lambda*(x-x3)-x1 = 24
Double value: 8*(2,9) = (51,24)
Doubling 8*(2,9) = (51,24)
x1^2 = 45
3x1^2 = 64
3x1^2+a = 68
2y1 = 48
lambda = (3x1^2+a)/(2y1) = 31
lambda^2 = 38
x3 = lambda^2-x1-x1 = 7
x1-x3 = 44
lambda(x1-x3) = 15
y3 = lambda*(x-x3)-x1 = 62
Double value: 16*(2,9) = (7,62)
Calculating 21*(2,9) = 5*(2,9) + 16*(2,9) = (59,8) + (7,62)
Adding (59,8) and (7,62)
y2-y1 = 54
x2-x1 = 19
lambda = (y2-y1)/(x2-x1) = 29
lambda^2 = 60
x3 = lambda - x1 - x2 = 65
x1-x3 = 65
lambda*(x1-x3) = 39
y3 = lambda*(x1-x3)-y1 = 31
Sum is (65,31)
Working result: 21*(2,9) = (65,31)
Doubling 16*(2,9) = (7,62)
x1^2 = 49
3x1^2 = 5
3x1^2+a = 9
2y1 = 53
lambda = (3x1^2+a)/(2y1) = 35
lambda^2 = 18
x3 = lambda^2-x1-x1 = 4
x1-x3 = 3
lambda(x1-x3) = 34
y3 = lambda*(x-x3)-x1 = 43
Double value: 32*(2,9) = (4,43)
Calculating 53*(2,9) = 21*(2,9) + 32*(2,9) = (65,31) + (4,43)
Adding (65,31) and (4,43)
y2-y1 = 12
x2-x1 = 10
lambda = (y2-y1)/(x2-x1) = 58
lambda^2 = 27
x3 = lambda - x1 - x2 = 29
x1-x3 = 36
lambda*(x1-x3) = 29
y3 = lambda*(x1-x3)-y1 = 69
Sum is (29,69)
Working result: 53*(2,9) = (29,69)
Doubling 32*(2,9) = (4,43)
x1^2 = 16
3x1^2 = 48
3x1^2+a = 52
2y1 = 15
lambda = (3x1^2+a)/(2y1) = 65
lambda^2 = 36
x3 = lambda^2-x1-x1 = 28
x1-x3 = 47
lambda(x1-x3) = 2
y3 = lambda*(x-x3)-x1 = 30
Double value: 64*(2,9) = (28,30)
Doubling 64*(2,9) = (28,30)
x1^2 = 3
3x1^2 = 9
3x1^2+a = 13
2y1 = 60
lambda = (3x1^2+a)/(2y1) = 44
lambda^2 = 19
x3 = lambda^2-x1-x1 = 34
x1-x3 = 65
lambda(x1-x3) = 20
y3 = lambda*(x-x3)-x1 = 61
Double value: 128*(2,9) = (34,61)
Doubling 128*(2,9) = (34,61)
x1^2 = 20
3x1^2 = 60
3x1^2+a = 64
2y1 = 51
lambda = (3x1^2+a)/(2y1) = 11
lambda^2 = 50
x3 = lambda^2-x1-x1 = 53
x1-x3 = 52
lambda(x1-x3) = 4
y3 = lambda*(x-x3)-x1 = 14
Double value: 256*(2,9) = (53,14)
Doubling 256*(2,9) = (53,14)
x1^2 = 40
3x1^2 = 49
3x1^2+a = 53
2y1 = 28
lambda = (3x1^2+a)/(2y1) = 45
lambda^2 = 37
x3 = lambda^2-x1-x1 = 2
x1-x3 = 51
lambda(x1-x3) = 23
y3 = lambda*(x-x3)-x1 = 9
Double value: 512*(2,9) = (2,9)
Doubling 512*(2,9) = (2,9)
x1^2 = 4
3x1^2 = 12
3x1^2+a = 16
2y1 = 18
lambda = (3x1^2+a)/(2y1) = 64
lambda^2 = 49
x3 = lambda^2-x1-x1 = 45
x1-x3 = 28
lambda(x1-x3) = 17
y3 = lambda*(x-x3)-x1 = 8
Double value: 1024*(2,9) = (45,8)
Doubling 1024*(2,9) = (45,8)
x1^2 = 37
3x1^2 = 40
3x1^2+a = 44
2y1 = 16
lambda = (3x1^2+a)/(2y1) = 56
lambda^2 = 12
x3 = lambda^2-x1-x1 = 64
x1-x3 = 52
lambda(x1-x3) = 1
y3 = lambda*(x-x3)-x1 = 64
Double value: 2048*(2,9) = (64,64)
Calculating 2101*(2,9) = 53*(2,9) + 2048*(2,9) = (29,69) + (64,64)
Adding (29,69) and (64,64)
y2-y1 = 66
x2-x1 = 35
lambda = (y2-y1)/(x2-x1) = 10
lambda^2 = 29
x3 = lambda - x1 - x2 = 7
x1-x3 = 22
lambda*(x1-x3) = 7
y3 = lambda*(x1-x3)-y1 = 9
Sum is (7,9)
Working result: 2101*(2,9) = (7,9)
Product is (7,9)
A sends (7,9)

B’s secret number = 2103
Scalar Multiply: 2103*(2,9)
Calculating 1*(2,9) = 0*(2,9) + 1*(2,9) = (Zero) + (2,9)
Adding (Zero) and (2,9)
Sum is (2,9)
Working result: 1*(2,9) = (2,9)
Doubling 1*(2,9) = (2,9)
x1^2 = 4
3x1^2 = 12
3x1^2+a = 16
2y1 = 18
lambda = (3x1^2+a)/(2y1) = 64
lambda^2 = 49
x3 = lambda^2-x1-x1 = 45
x1-x3 = 28
lambda(x1-x3) = 17
y3 = lambda*(x-x3)-x1 = 8
Double value: 2*(2,9) = (45,8)
Calculating 3*(2,9) = 1*(2,9) + 2*(2,9) = (2,9) + (45,8)
Adding (2,9) and (45,8)
y2-y1 = 70
x2-x1 = 43
lambda = (y2-y1)/(x2-x1) = 33
lambda^2 = 24
x3 = lambda - x1 - x2 = 48
x1-x3 = 25
lambda*(x1-x3) = 44
y3 = lambda*(x1-x3)-y1 = 35
Sum is (48,35)
Working result: 3*(2,9) = (48,35)
Doubling 2*(2,9) = (45,8)
x1^2 = 37
3x1^2 = 40
3x1^2+a = 44
2y1 = 16
lambda = (3x1^2+a)/(2y1) = 56
lambda^2 = 12
x3 = lambda^2-x1-x1 = 64
x1-x3 = 52
lambda(x1-x3) = 1
y3 = lambda*(x-x3)-x1 = 64
Double value: 4*(2,9) = (64,64)
Calculating 7*(2,9) = 3*(2,9) + 4*(2,9) = (48,35) + (64,64)
Adding (48,35) and (64,64)
y2-y1 = 29
x2-x1 = 16
lambda = (y2-y1)/(x2-x1) = 24
lambda^2 = 8
x3 = lambda - x1 - x2 = 38
x1-x3 = 10
lambda*(x1-x3) = 27
y3 = lambda*(x1-x3)-y1 = 63
Sum is (38,63)
Working result: 7*(2,9) = (38,63)
Doubling 4*(2,9) = (64,64)
x1^2 = 49
3x1^2 = 5
3x1^2+a = 9
2y1 = 57
lambda = (3x1^2+a)/(2y1) = 45
lambda^2 = 37
x3 = lambda^2-x1-x1 = 51
x1-x3 = 13
lambda(x1-x3) = 17
y3 = lambda*(x-x3)-x1 = 24
Double value: 8*(2,9) = (51,24)
Doubling 8*(2,9) = (51,24)
x1^2 = 45
3x1^2 = 64
3x1^2+a = 68
2y1 = 48
lambda = (3x1^2+a)/(2y1) = 31
lambda^2 = 38
x3 = lambda^2-x1-x1 = 7
x1-x3 = 44
lambda(x1-x3) = 15
y3 = lambda*(x-x3)-x1 = 62
Double value: 16*(2,9) = (7,62)
Calculating 23*(2,9) = 7*(2,9) + 16*(2,9) = (38,63) + (7,62)
Adding (38,63) and (7,62)
y2-y1 = 70
x2-x1 = 40
lambda = (y2-y1)/(x2-x1) = 55
lambda^2 = 43
x3 = lambda - x1 - x2 = 69
x1-x3 = 40
lambda*(x1-x3) = 70
y3 = lambda*(x1-x3)-y1 = 7
Sum is (69,7)
Working result: 23*(2,9) = (69,7)
Doubling 16*(2,9) = (7,62)
x1^2 = 49
3x1^2 = 5
3x1^2+a = 9
2y1 = 53
lambda = (3x1^2+a)/(2y1) = 35
lambda^2 = 18
x3 = lambda^2-x1-x1 = 4
x1-x3 = 3
lambda(x1-x3) = 34
y3 = lambda*(x-x3)-x1 = 43
Double value: 32*(2,9) = (4,43)
Calculating 55*(2,9) = 23*(2,9) + 32*(2,9) = (69,7) + (4,43)
Adding (69,7) and (4,43)
y2-y1 = 36
x2-x1 = 6
lambda = (y2-y1)/(x2-x1) = 6
lambda^2 = 36
x3 = lambda - x1 - x2 = 34
x1-x3 = 35
lambda*(x1-x3) = 68
y3 = lambda*(x1-x3)-y1 = 61
Sum is (34,61)
Working result: 55*(2,9) = (34,61)
Doubling 32*(2,9) = (4,43)
x1^2 = 16
3x1^2 = 48
3x1^2+a = 52
2y1 = 15
lambda = (3x1^2+a)/(2y1) = 65
lambda^2 = 36
x3 = lambda^2-x1-x1 = 28
x1-x3 = 47
lambda(x1-x3) = 2
y3 = lambda*(x-x3)-x1 = 30
Double value: 64*(2,9) = (28,30)
Doubling 64*(2,9) = (28,30)
x1^2 = 3
3x1^2 = 9
3x1^2+a = 13
2y1 = 60
lambda = (3x1^2+a)/(2y1) = 44
lambda^2 = 19
x3 = lambda^2-x1-x1 = 34
x1-x3 = 65
lambda(x1-x3) = 20
y3 = lambda*(x-x3)-x1 = 61
Double value: 128*(2,9) = (34,61)
Doubling 128*(2,9) = (34,61)
x1^2 = 20
3x1^2 = 60
3x1^2+a = 64
2y1 = 51
lambda = (3x1^2+a)/(2y1) = 11
lambda^2 = 50
x3 = lambda^2-x1-x1 = 53
x1-x3 = 52
lambda(x1-x3) = 4
y3 = lambda*(x-x3)-x1 = 14
Double value: 256*(2,9) = (53,14)
Doubling 256*(2,9) = (53,14)
x1^2 = 40
3x1^2 = 49
3x1^2+a = 53
2y1 = 28
lambda = (3x1^2+a)/(2y1) = 45
lambda^2 = 37
x3 = lambda^2-x1-x1 = 2
x1-x3 = 51
lambda(x1-x3) = 23
y3 = lambda*(x-x3)-x1 = 9
Double value: 512*(2,9) = (2,9)
Doubling 512*(2,9) = (2,9)
x1^2 = 4
3x1^2 = 12
3x1^2+a = 16
2y1 = 18
lambda = (3x1^2+a)/(2y1) = 64
lambda^2 = 49
x3 = lambda^2-x1-x1 = 45
x1-x3 = 28
lambda(x1-x3) = 17
y3 = lambda*(x-x3)-x1 = 8
Double value: 1024*(2,9) = (45,8)
Doubling 1024*(2,9) = (45,8)
x1^2 = 37
3x1^2 = 40
3x1^2+a = 44
2y1 = 16
lambda = (3x1^2+a)/(2y1) = 56
lambda^2 = 12
x3 = lambda^2-x1-x1 = 64
x1-x3 = 52
lambda(x1-x3) = 1
y3 = lambda*(x-x3)-x1 = 64
Double value: 2048*(2,9) = (64,64)
Calculating 2103*(2,9) = 55*(2,9) + 2048*(2,9) = (34,61) + (64,64)
Adding (34,61) and (64,64)
y2-y1 = 3
x2-x1 = 30
lambda = (y2-y1)/(x2-x1) = 64
lambda^2 = 49
x3 = lambda - x1 - x2 = 22
x1-x3 = 12
lambda*(x1-x3) = 58
y3 = lambda*(x1-x3)-y1 = 68
Sum is (22,68)
Working result: 2103*(2,9) = (22,68)
Product is (22,68)
B sends (22,68)

#4567

Endless stream of nonsense.

edit: the stream is gone, but I think everyone can guess who it was

#4568

What stream are you talking about?

#4569

I had posted a fairly long list of equations, 4 of them, two different variations. I do not possess anything to solve something so long so I posted it asking if someone has the ability to run it. it was based on elliptic curve cryptography and the Diffie-Hellman protocol.

with A and B values set as the “clockers” clock times in game. then trying with the A and B values reversed. Pointless was kind enough to point out it was a stream of nonsense rather than read and understand what I put up. So i’ll figure out how to run it in an app and post what I find.

#4570

I think i found what is hidden behind the X01 and X02 doors!

It’s either this:

youtube.com/watch?v=hyCIpKAIFyo

This:

youtube.com/watch?v=M6r4pAqOBDY

Or this:

youtube.com/watch?v=j-ATEH4wzrQ

I’m obviously joking here but these videos are still nice.

#4571

In a PM entitled “i did it,” @drgibbles claims that he has found a pattern in the Hex Code:

I’m asking you to post it here if you have, man. Share it with everyone–no need to be secretive and send it only to me or a select few. We’re all working together on this.

#4572

I remember when my dad showed me that first video in your post.
It was amazing when i saw it, but i laughed when my dad told me that it was a real recording of a machine.

#4573

Someone suggested to me in a PM that we might have missed something in the original “21 goes into 1” clue:

/*! *Oh hai there, someones bein a nosy bounder aren't they... *Looking for ARG related stuffs? *I have a hint for you *21 goes into 1 *Don't trust anyone, it's not safe. */It’s in the letters that are extraneous or missing in the misspelled words:

The highlighted letters spell out AGES.

However, as I said in my reply to the other party, nosy isn’t technically a misspelling, as both nosy and nosey are valid spellings of the word. But who knows? Maybe it could be a clue. Could it be related to Tempus omnia revelant (Time reveals all things) ? Or is it an acronym? Or anagram?

#4574

#4575

haha! That is fantastic! WITH A RESOUNDING. … YES!!!

#4576

So that’s a no to the zirconium then?

#4577

no good…

#4578

Have we ever considered that the Hexcode has something to do with the Tempus omnia revelant page? I was talking to drgibbles about this; On the Tempus omnia revelant there are phrases like “Tempus omnia revelant” and “Vox populi vox Dei” which translates to “Time reveals all” and “Voice of the people, Voice of God” respectively. We know this already obviously. So if you translate the page (using chrome: right click a blank spot on the page, click translate, and then change ‘translate from’ to Latin) it changes certain words. The top of the page does not read “Time reveals all” like it should. Instead it said “convergence” Possibly a reference to “21 into 1”? it also completely changes the layout of the long niobium codes. It changes a lot of the values. Also the letters that are bolded and Italicized are separated from the words they were in. (I know that part has been solved but it’s interesting how they separated). I have other possible leads that aren’t developed enough to post about yet, but I thought that could lead to something.

#4579

New development: Ok so I did a bit more digging and found a note from Dr. Horn on a white board that I’m sure you’ve all seen:casquade ciphers..JPG “Use cascade ciphers only”

A cascade cipher is the process of encrypting an already encrypted message one or more times, either using the same or a different algorithm. Each time you encrypt a code, it adds a layer. So what if the Hex code is in layers? I would try different methods myself but you guys have already tried so many that I think you’d be better suited for this.

#4580

I don’t know if anyone else feels this way but the fact that Storm the calls the 752 hex code HALOS file bothers me for some reason.

Founded in 2004, Leakfree.org became one of the first online communities dedicated to Valve’s Source engine development. It is more famously known for the formation of Black Mesa: Source under the 'Leakfree Modification Team' handle in September 2004.