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Laughingstok335 karma

Hey guys, I responded to the author but no response yet. I got a hit. But a little background. I edited the script to include the zodiacAlphabet into the regular alphabet and also added slices of the original encryption (5 characters long each) as words to look for as hits.

I received the following hit.. please help me understand.. did I just find 3 instances of 3 slices of its own "encryption?"

Processing: Pass #1136403 [ha>MznYtzynSczaGHGaxGMSUMSxYy>ccHeYYMHHc>OSGnMSicGXGaxT9t1qzYGT>zhtaUHhxMtSoxYMc]

MATCH FOUND: yBulhSE7hkSeX3BF4FBVq7FsleVnk4XXxLn}C4dXubebSIeTXFfVBLxa73:hKqxuSy7BiduVC7FfLE7X (Count: 3, X3BF4 3BF4F BF4FB ), cycle 1136408

    a equals (L)    b equals (#_)   c equals ()     d equals (-)    e equals (]G)
    f equals (:A)   g equals ()     h equals (T)    i equals (})
    j equals ()     k equals (2)    l equals (1)    m equals ()
    n equals (V)    o equals ()     p equals ()     q equals (6)
    r equals ()     s equals (R)    t equals ()     u equals (C9)
    v equals ()     w equals ()     x equals (;E)   y equals (Y)
    z equals ()     = equals ()     > equals (I)    _ equals ()
    - equals ()     ; equals ()     : equals (H)    ' equals ()
    ] equals ()     { equals ()     } equals (U)    & equals ()
    # equals ()     % equals ()     1 equals ()     2 equals ()
    3 equals (S3)   4 equals (4=)   5 equals ()     6 equals ()
    7 equals (7O)   8 equals (F)    9 equals ()     A equals ()
    B equals (B)    C equals (Z)    E equals (K)    F equals (8)
    G equals ()     H equals ()     I equals (%)    K equals (')
    L equals (>)    M equals ()     O equals ()     R equals ()
    S equals (M{)   T equals (&)    U equals ()     V equals (W5)
    W equals ()     X equals (X)    Y equals ()     Z equals ()

Laughingstok112 karma

I just hit it again.. same exact thing.

Processing: Pass #4060692 [65VrrUAVrUUX}S5CLC5ShZZSrXS}UV}}VA}LXLW}VlXhUqBS}ClA5G5gVgArZh5VC6V5wWfSXVZ#GAZ}]

MATCH FOUND: BBUAXUdxX8UlX3BF4FBB}B{kAlBV8yXXAWV%U4UXU8lkU8LeXF64BUszxyUX%}sUdBxB}UeBUx{]UdBX (Count: 3, X3BF4 3BF4F BF4FB ), cycle 4060699

    a equals ()     b equals ()     c equals ()     d equals ({K)   e equals (&C)
    f equals ()     g equals ()     h equals ()     i equals ()
    j equals ()     k equals (R#)   l equals (])    m equals ()
    n equals ()     o equals ()     p equals ()     q equals ()
    r equals ()     s equals (;)    t equals ()     u equals ()
    v equals ()     w equals ()     x equals (O)    y equals (S=)
    z equals (L)    = equals ()     > equals ()     _ equals (2%)
    - equals ()     ; equals ()     : equals ()     ' equals (U)
    ] equals (A)    { equals (8)    } equals (}6)   & equals ()
    # equals ()     % equals (')    1 equals ()     2 equals ()
    3 equals (3)    4 equals (W4)   5 equals ()     6 equals (:)
    7 equals (B5)   8 equals (_)    9 equals (-)    A equals (E1)
    B equals (Y7)   C equals ()     E equals ()     F equals (F)
    G equals ()     H equals ()     I equals (Z9)   K equals ()
    L equals (G)    M equals (HI)   O equals ()     R equals ()
    S equals ()     T equals (X)    U equals (M)    V equals (V)
    W equals (>)    X equals (T)    Y equals ()     Z equals ()

Laughingstok76 karma

So I added the zodiac alphabet into the $normalAlphabet scalar, and then added the cipherText in blocks of 5 characters to the @signalWords array. My theory is that this guy may have encrypted his message twice, so basically I am also hunting for the following "words" that will trigger a hit:

$normalAlphabet="abcdefghijklmnopqrstuvwxyz=>_-;:']{}&#%123456789ABCEFGHIKLMORSTUVWXYZ";

@signalWords=("children","others","around","shall","about","people","would","missed","police","because","thing","there","could","rather","light","school","vallejo","useing","using","triger","trigger","howers","cerous","serious","meannie","meany","killed","victom","victim","speaking","lyeing","lying","slaves","afterlife","zodiac","intersting","interesting","idenity","identity","woeman","women","woman","untill","until","YB91T","B91TM","91TMK","1TMKO","TMKOT","MKOT2","KOT2M","OT2M]","T2M]X","2M]X3","M]X3B","]X3BF","X3BF4","3BF4F","BF4FB","F4FB5","4FB56","FB567","B5678","5678R","678R1","78R1]","8R1]5","R1]5V","1]5V2","]5V2=","5V2=X","V2=XX","2=XXE","=XXE>","XXE>V","XE>VU","E>VUZ",">VUZ4","VUZ4-","UZ4-X","Z4-X9","4-X9","-X9]","X9]#","9]#M","_]#M%","]#M%G","#M%G&","M%G&X","%G&XF","G&XF:","&XF:W","XF:WB","F:WBI",":WBI;","WBI;L","BI;LO","I;LOS",";LOSH","LOSHT","OSHT'","SHT'6","HT'6;","T'6;9","'6;9{","6;9{Y",";9{YO","9{YOB","{YOB}","YOB}-","OB}-C","B}-C5","}-C5Z","-C5ZO","C5ZO8","5ZO8A","ZO8AI","O8AIK","8AIK7","AIK7X");

Laughingstok15 karma

"They don't make 'em like they used to."

Laughingstok5 karma

It's not that a particle "knows" it is being observed, it's more like we're observing the Universe through photographs. I.E. - What you see right now is more like a film (yes, from a movie) in the idea that you are essentially collecting photons on your retina to "observe" them at any particular moment. This holds true to anything with mass. Anytime any piece of matter impacts another piece of matter, you could say it has been "observed." I.E. - It's natural state has been changed. It was moving, it collided with something, now it has been altered and thus takes a "final form."

With me so far? For example, how can you ever truly know the temperature of something? Because as soon as you interact with it to take the temperature, you have already, in some fundamentally small way, altered it's temperature by trying to take it's measurement.

In this way, it's not to say the electrons are aware of their own observation, but it's more like saying you take a picture of your friend sprinting from one place to another. Depending on how fast the shutter is on your camera, you can either see your friend very clearly, but lose any idea on how fast he is going, or you get a blur (and can get an idea of speed) but you lose clarity on your friend. You can't have it both ways. In this way, an electron is observed like a photograph of a moving object. The moment you take the measurement (the moment you "snap the picture to observe/measure/be aware of it), you have essentially caused that particle to reveal it's absolute nature.

So thus, "observing" a particle travelling through a slit is essentially determining where it was going, and thus is ends up going where you think it was going. The film paper (on the back wall in the double-slit experiment) is also an observer. So if you don't observe while the electrons are passing through the slits, they ultimately get observed on the film paper. Thus they take their "final form" on the wall and show a wave pattern. But if you try to sneak and watch them earlier, then you "collapse the wave pattern" into it's "final form" and the electron, from that point on, goes exactly where you expect it to go.

That's the best ELI5 I can do on it. It'd liken to your friend running from one end of the football field to the other at incredible speed, so fast you can't actually see him with your normal eye, but at any one moment you take a pic, and you see exactly where he is. To make it crazier though, Feynman essentially says that your friend is going 'everywhere at once' and it's not until you look for him, does he actually show up in his true spot/state.

I personally liken the "everywhere at once" thing to a mathematical idea, rather than reality. Though someone may be able to explain it better. In other words, I tend towards the idea that you can calculate the probability of a particle veering off into another galaxy and coming back, but in actuality it's really just going straight ahead by all mathematical odds. I.E. - It could maybe, possibly, in some far off branch of chance, actually go to another galaxy, but it's far more likely to just go straight ahead. This is essentially how you can predict the odds of where a particle will be, but you can "never know for sure" without observing.