I was forced to write a blog post. So I made a code breaking contest. | Hacker News Reader

I was forced to write a blog post. So I made a code breaking contest.

carwoo.com

If you base64 decode this

IThGMj4wS0gvR0M9PE5QQEEnMyxNIiFHJE1NVTdYSS5PTTdFT1FCVUtMLVBW Qj4nNkBUIDsqLUNTLjVCJS87IkwnT1E/LiZRI1E4ITlBLzBXNU5SNEgmVDM8 QFM0Ty4wLFAvL0ohNTUxQzU2Jy82LVEpN0gpSjs3NCE5Q1lXQlUwUVc1MUFX NDI7LUQnO09BSEI9WVdNRktOTTdLNkI+Pj5HIDtKS1YkSVY0I0pRPTtYVSY9 KyhZVEAqJykrJSIpIjcsI0ckWD4ySlRIIkAhUyw+VStUQVY/NiUpNFZVMis8 IylXMiU9VVFKSFI3WkI3S0A/QTMhIkkuP0pFMVJKQixYJSFGJz8+TD89WFgu JyRSJ1hYIlIoVEFVLlUwOjc0Oi4iQDlNOUVPSUdHOkM5Mj9LWilCMlkmPj1X M0otSiwmKFU4Jy9IJzc5WkFVQ1VFPFFAQ0tHOClYLDklPDs/T1BPS0hTPDAi WkRBQVVSOTIhKjxBR0ZNNypRNzwsMVI=

The first 5 characters are "!8F2>"

Those 5 characters represent the first "T". Another "T" won't necessarily result in the same code. Actually, the odds of ever seeing that code again for a "T" is low.

The message is:

DON’T FORGET TO DRINK YOUR OVALTINE

Seriously, don't downvote me. That's the decrypted message.

You are right. Congrats!! There is actually an exclamation point at the end, but close enough! send an email to erik@carwoo.com and we can figure out how to get you the $500. Also, if you would detail your solution...that would be awesome!

Here's a Ruby version of the solution that was posted to your blog: http://pastie.org/1390793

Here's a blog post that details my solution somewhat. Fun challenge for a lazy Sunday. :)

http://grack.com/blog/2010/12/19/solving-the-carwoo-code-bre...

Well.. I guess I'm too late, but as long as the solution isn't revealed, I'm still going to play just for the fun of it.

I have reason to believe that: "0KH/GC=" encodes "H" "<NP@A'3,M"!" encodes "E" ...

Finally...

[Do not read further if you want to solve it yourself]

1. The encoded string can be divided into groups of chars, each representing one (decoded) char. These groups occur every 5, 7, 11, and 13 chars. That is, chars 0-4 correspond to group 1, chars 5-11 correspond to group 2, etc. After the first 4 groups, the lengths repeat. Thus group 5 is chars 36-40.

2. As hinted in this thread, the space is of 59 chars, and there's an offset of 32. So, for each char in each group, subtract 32, sum them up, mod by 59. Add 32 to this result, and that's your decoded character ascii code.

Example: "!8F2>" corresponds to [33, 56, 70, 50, 62]. Subtracting 32 from each of those is [1, 24, 38, 18, 30]. The sum of those is 111. 111 % 59 = 52. Adding back the 32 offset, yields ascii code 84, corresponding to "T".

I had most of the pieces to this, but didn't make the leap to the variance in encoding length being a series of primes. Ah well, maybe next time.

Here are four different, but all valid encodings of THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG

Each one can be decoded with the same decoding method.

JkpXLlVZIz5MLTxKJj1PUFA2R09BPUtTLSI1Oi9DMEwzM0VYSiE5S1gzVTA4 WksxS0VSVS0pQzlCLTg1VDpATlUpWihIVVIgIDE/V1o2JlA+RkNZJ1BLIFY6 Q1QoWCA2KzlPKlBVTS8mKihNKU9JM1EmK1kkRSxOSD43O1JEWTApOD1JRk48 Ozk1UzJUTkZJOSY2V1AqVUwwQCogVUk6UiIvTDtDSjBGSlNQNCxTNCNIL0VE IC1QLDwnRFE8ISpITSRNOE8iMzBSQlYtQT8mWEFOIkVPSSJXMCkxRT8tPFEj QFlWN1JBIEosWjAmODNCJzM3QCUmTCgwKyowJ1NWWkwyJ0ZFTDggRlMxIUJY JD88NS48SDQpJUNTQitFVS4qOCk8KztaQEI6TysmUC9XNiotNSEtIEYmKS4h QVZXNlI/TEpFPihDITIxIFcrKlQ/UUdWOEk+TFNZOldDQEtWN0c+QzgrQCFW KT82U0k9QVlMQFhAOiw/OT9LICkjQSM=

NCs6SixWVishJy4xTSw0SlQsM041SzRFKDIvSVIoPDZaQyY8ODwzMjMqRlNB Uzo1QT0tLyRONyFHOVEhV0A0QzNIOzVOPkg3Kk4tVEhPRCspPkQoQFQnVjw5 IkVFWVkyLEk/PEIuUVFKRyJNUllOVTI/SFNENDw6JDQxLkUxTVo7RCcrQFQ/ LUs4KyUuSSQuSUo7IU0kVU1PQihBMSkxV0I0PFVCR00nJTxKPTMvUiJCI0A9 IjYiNFcuOjdSQUJSNlkpKj05SSFSLThGRDdUTy03PUwxPkIwLFJQRzRBSCdN OTEqU0A2NEJCIkgxOyQzUUYgNTxHUDM2JCdXRE1XTz9aJUAqVlkrRDErIDoh PixMO0A5OCEuMCQwQjk6NFg7QVpQRUk2SkAsKFpINVc+WjouPSUjLkg7WCpZ LU8zSi1ULVlTNic/VUYjLFk9MCNYQz8yIzcvIlk4IzdVMkdVTyNGI1QhPjAh UTdWT0gxRUVULkY9SUI+N1lVPSMvVlY=

MzgsTStTREc4JClWNyswNiUxLkUgWitJQjVUSzYxVk07ODwlIExYM1AsM0o4 PSZHQzY6LUorQVVQPzY3V1g1S1FZRUtQWjJaJU9AKSlGMkpLMUtDLEgyQkxL PFYgRzFJNSg0VTsnWTkiVSc4Ni8iQUJHJUxGSS5TNjxPLlZDJzZHLCJKQDxN RywwPDpHK0lWQ00tPEMlRD1GU000TE4gIi9OKEhINT4+VUgzMEkhNDEkIFUh IzYtQD9SS0U1J1kvPzImN1pLWVE1JT1IICInIFVTUjBGJT5TUkM0JlojLzw2 UlgjSz9ATDo8NiM7RlAuP0wwIykxKyhRKCYvR1QoSS8xLUVRTjlIO0UnNixI J1ZLI0tWICcyRCIxT1NCUDkoVCtGPzojUihGUS9TMkA/Kk4/RC9JVkg4WUgk QDsjPlhMRDE8ViUoSkpDSC4pRU9VLDo4LEUmNztMO1c3PEwrM1okTS1GJ1Qs R0YrUjBBIFoqMi5AWkdTPFBONDRHJ0U=

It never encodes things the same each time your run it, but they can all be decoded the same way. There are no strange initial conditions you need to know about like time or some other kind of token.

each decodes to exactly 383 chars.

Unencoded:

    !8F2>0KH/GC=<NP@A'3,M"!G$MMU7XI.OM7EOQBUKL-PVB>'6@T ;*-CS.5B%/;"L'OQ?.&Q#Q8!9A/0W5NR4H&T3<@S4O.0,P//J!551C56'/6-Q)7H)J;74!9CYWBU0QW51AW42;-D';OAHB=YWMFKNM7K6B>>>G ;JKV$IV4#JQ=;XU&=+(YT@*')+%")"7,#G$X>2JTH"@!S,>U+TAV?6%)4VU2+<#)W2%=UQJHR7ZB7K@?A3!"I.?JE1RJB,X%!F'?>L?=XX.'$R'XX"R(TAU.U0:74:."@9M9EOIGG:C92?KZ)B2Y&>=W3J-J,&(U8'/H'79ZAUCUE<Q@CKG8)X,9%<;?OPOKHS<0"ZDAAUR92!*<AGFM7*Q7<,1R

    &JW.UY#>L-<J&=OPP6GOA=KS-"5:/C0L33EXJ!9KX3U08ZK1KERU-)C9B-85T:@NU)Z(HUR 1?WZ6&P>FCY'PK V:CT(X 6+9O*PUM/&*(M)OI3Q&+Y$E,NH>7;RDY0)8=IFN<;95S2TNFI9&6WP*UL0@* UI:R"/L;CJ0FJSP4,S4#H/ED -P,<'DQ<!*HM$M8O"30RBV-A?&XAN"EOI"W0)1E?-<Q#@YV7RA J,Z0&83B'37@%&L(0+*0'SVZL2'FEL8 FS1!BX$?<5.<H4)%CSB+EU.*8)<+;Z@B:O+&P/W6*-5!- F&).!AVW6R?LJE>(C!21 W+*T?QGV8I>LSY:WC@KV7G>C8+@!V)?6SI=AYL@X@:,?9?K )#A#

    4+:J,VV+!'.1M,4JT,3N5K4E(2/IR(<6ZC&<8<323*FSAS:5A=-/$N7!G9Q!W@4C3H;5N>H7*N-THOD+)>D(@T'V<9"EEYY2,I?<B.QQJG"MRYNU2?HSD4<:$41.E1MZ;D'+@T?-K8+%.I$.IJ;!M$UMOB(A1)1WB4<UBGM'%<J=3/R"B#@="6"4W.:7RABR6Y)*=9I!R-8FD7TO-7=L1>B0,RPG4AH'M91*S@64BB"H1;$3QF 5<GP36$'WDMWO?Z%@*VY+D1+ :!>,L;@98!.0$0B9:4X;AZPEI6J@,(ZH5W>Z:.=%#.H;X*Y-O3J-T-YS6'?UF#,Y=0#XC?2#7/"Y8#7U2GUO#F#T!>0!Q7VOH1EET.F=IB>7YU=#/VV

    38,M+SDG8$)V7+06%1.E Z+IB5TK61VM;8<% LX3P,3J8=&GC6:-J+AUP?67WX5KQYEKPZ2Z%O@))F2JK1KC,H2BLK<V G1I5(4U;'Y9"U'86/"ABG%LFI.S6<O.VC'6G,"J@<MG,0<:G+IVCM-<C%D=FSM4LN "/N(HH5>>UH30I!41$ U!#6-@?RKE5'Y/?2&7ZKYQ5%=H "' USR0F%>SRC4&Z#/<6RX#K?@L:<6#;FP.?L0#)1+(Q(&/GT(I/1-EQN9H;E'6,H'VK#KV '2D"1OSBP9(T+F?:#R(FQ/S2@?*N?D/IVH8YH$@;#>XLD1<V%(JJCH.)EOU,:8,E&7;L;W7<L+3Z$M-F'T,GF+R0A Z*2.@ZGS<PN44G'E

All 4 strings share 15 characters at the same position, not sure if it's relevant though. (I doubt it since the clear string is 43 characters long)

The position/character couples are: 14: P, 20: M, 36: O, 92: 4, 137: ;, 158: >, 171: #, 177: U, 204: ", 207: S, 211: +, 276: ", 281: U, 364: U, 366: 9

Huh?

    >> a.index("P")
    => 14
    >> b.index("P")
    => 15
    >> c.index("P")
    => 218
    >> d.index("P")
    => 40

Sry, bug in my code, I only tested the first 2 strings.

Here are some integer values for the characters of two of the QBFs

[72, 55, 81, 36, 86, 49, 49, 37, 38, 61, 62, 86, 40, 63, 57, 43, 57, 62, 50, 47, 36, 43, 68, 38, 36, 54, 54, 86, 67, 46, 69, 52, 78, 37, 61, 33, 46, 77, 55, 41, 49, 70, 70, 53, 46, 74, 32, 50, 63, 40, 81, 82, 51, 90, 58, 82, 69, 77, 54, 42, 73, 67, 74, 80, 84, 85, 48, 86, 75, 71, 77, 61, 32, 58, 74, 44, 54, 53, 33, 78, 73, 41, 54, 69, 51, 44, 39, 32, 38, 60, 85, 68, 78, 80, 47, 33, 33, 32, 65, 35, 51, 34, 44, 35, 65, 39, 45, 73, 63, 34, 63, 79, 86, 63, 71, 80, 34, 45, 52, 52, 89, 61, 41, 60, 83, 35, 46, 39, 75, 87, 77, 50, 72, 58, 77, 53, 48, 48, 38, 70, 73, 71, 35, 77, 49, 85, 79, 73, 89, 46, 44, 47, 70, 35, 38, 50, 44, 49, 77, 74, 54, 67, 54, 45, 40, 35, 46, 80, 59, 33, 55, 58, 33, 75, 35, 53, 34, 42, 87, 67, 51, 77, 84, 67, 41, 83, 75, 72, 85, 45, 39, 55, 56, 81, 67, 39, 36, 80, 33, 52, 81, 88, 79, 57, 58, 88, 45, 87, 51, 59, 50, 54, 70, 40, 73, 86, 43, 71, 71, 79, 65, 67, 61, 40, 39, 76, 85, 33, 82, 59, 87, 72, 54, 74, 67, 87, 35, 82, 54, 50, 60, 85, 59, 57, 86, 79, 44, 70, 47, 62, 72, 53, 76, 68, 87, 35, 49, 41, 65, 58, 75, 48, 84, 80, 37, 63, 82, 50, 51, 65, 88, 78, 52, 55, 85, 33, 44, 33, 35, 89, 42, 32, 71, 61, 67, 48, 35, 55, 88, 41, 64, 58, 67, 74, 80, 62, 80, 63, 41, 38, 48, 43, 64, 33, 67, 71, 35, 55, 87, 44, 41, 61, 36, 37, 88, 39, 75, 70, 84, 88, 49, 73, 44, 70, 43, 33, 43, 37, 37, 71, 78, 82, 85, 85, 68, 49, 57, 66, 89, 44, 57, 37, 40, 62, 50, 44, 40, 72, 48, 39, 53, 35, 68, 55, 73, 70, 33, 49, 55, 61, 58, 39, 50, 40, 68, 82, 52, 83, 78, 78, 34, 41, 71, 61, 58, 81, 39, 68, 53, 69, 85, 68, 33]

[33, 56, 70, 50, 62, 48, 75, 72, 47, 71, 67, 61, 60, 78, 80, 64, 65, 39, 51, 44, 77, 34, 33, 71, 36, 77, 77, 85, 55, 88, 73, 46, 79, 77, 55, 69, 79, 81, 66, 85, 75, 76, 45, 80, 86, 66, 62, 39, 54, 64, 84, 32, 59, 42, 45, 67, 83, 46, 53, 66, 37, 47, 59, 34, 76, 39, 79, 81, 63, 46, 38, 81, 35, 81, 56, 33, 57, 65, 47, 48, 87, 53, 78, 82, 52, 72, 38, 84, 51, 60, 64, 83, 52, 79, 46, 48, 44, 80, 47, 47, 74, 33, 53, 53, 49, 67, 53, 54, 39, 47, 54, 45, 81, 41, 55, 72, 41, 74, 59, 55, 52, 33, 57, 67, 89, 87, 66, 85, 48, 81, 87, 53, 49, 65, 87, 52, 50, 59, 45, 68, 39, 59, 79, 65, 72, 66, 61, 89, 87, 77, 70, 75, 78, 77, 55, 75, 54, 66, 62, 62, 62, 71, 32, 59, 74, 75, 86, 36, 73, 86, 52, 35, 74, 81, 61, 59, 88, 85, 38, 61, 43, 40, 89, 84, 64, 42, 39, 41, 43, 37, 34, 41, 34, 55, 44, 35, 71, 36, 88, 62, 50, 74, 84, 72, 34, 64, 33, 83, 44, 62, 85, 43, 84, 65, 86, 63, 54, 37, 41, 52, 86, 85, 50, 43, 60, 35, 41, 87, 50, 37, 61, 85, 81, 74, 72, 82, 55, 90, 66, 55, 75, 64, 63, 65, 51, 33, 34, 73, 46, 63, 74, 69, 49, 82, 74, 66, 44, 88, 37, 33, 70, 39, 63, 62, 76, 63, 61, 88, 88, 46, 39, 36, 82, 39, 88, 88, 34, 82, 40, 84, 65, 85, 46, 85, 48, 58, 55, 52, 58, 46, 34, 64, 57, 77, 57, 69, 79, 73, 71, 71, 58, 67, 57, 50, 63, 75, 90, 41, 66, 50, 89, 38, 62, 61, 87, 51, 74, 45, 74, 44, 38, 40, 85, 56, 39, 47, 72, 39, 55, 57, 90, 65, 85, 67, 85, 69, 60, 81, 64, 67, 75, 71, 56, 41, 88, 44, 57, 37, 60, 59, 63, 79, 80, 79, 75, 72, 83, 60, 48, 34, 90, 68, 65, 65, 85, 82, 57, 50, 33, 42, 60, 65, 71, 70, 77, 55, 42, 81, 55, 60, 44, 49, 82]

Erik says that all of the strings can be decoded with the same method. Since the input string can result in multiple different valid encoded forms that implies to me that the method for decoding these strings is also encoded in the string (it seems that otherwise we would have to guess magic inputs).

However, assuming that approach, it is interesting to me that each resulting string ended up the same length. Trying to tease out some encoding code from the outputs now...

The character distribution of the encoded message for the example ("THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG"):

    8: " " (note that the original also has 8 spaces)
    4: "O"
    3: "E"
    2: "H", "R", "T", "U"
    1: "Z", "B", "Y", "N", "C", "D", "X", "M", "P",
       "W", "F", "Q", "G", "L", "S", "A", "I", "K",
       "J", "V"

The character distribution of the encoded message (to be solved):

    12: ";", " ", "I"
    11: "H"
    10: "*", "-", "F", "2"
     9: "?", "S", "L", "O", "E", "T"
     8: "N", "4", "W"
     7: "3", "(", "J", "#", "/", "U"
     6: "C", ">", "<", """, "D", "A", ".", "%", "0"
     5: "9", "V", "@", "+", "$", "Y", "G", "6"
     4: "=", ":", "Z", "P", "R", "&", "7", "1", "B",
        ",", "!"
     3: "8", ")", "K", "X", "'", "5"
     2: "Q", "M"

383 is prime. So it can't be a substitution cipher.

Well, not a substitution cipher in which each substitution is of equal length. :-)

Given that sample (see blog comments), it is a little trickier than a simple substitution.

Edit: I stand corrected on the original statement about lengths - wasn't paying attention and was performing integer division.

Are you sure it's exactly 8? I get ~8.9 times its own length.

    var fox = "THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG";
    var decodedfox = atob("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");

    console.log(decodedfox.length/fox.length);

26/3 is ~8.67, I wonder if that is relevant.

> encodes to exactly 8 times it's own length

really? I have strlen(decoded-text) at 383 chars, and qbf as 43 chars, giving the encoded as 8.906976744186 times the size of the original...

You did the same thing I did which is to forget to force floating point division.

I am working on the assumption that the first few chars could be a randomly generated salt and the rest is equal sized blocks of encrypted content. The random salt results in a different encoding each time.

This is what I have so far and it seems to fit the original encoded message too.

The salt is the first 29 chars.

Each encrypted block is 59 chars.

So for the QBFs : 29 + 6x59

For the original message : 29 + 5x59

Note: I have never come across such sizes for salt and block though. I may be totally off.

Here are some numbers for this approach. Here are another two QBFs

SALT: [52, 43, 58, 74, 44, 86, 86, 43, 33, 39, 46, 49, 77, 44, 52, 74, 84, 44, 51, 78, 53, 75, 52, 69, 40, 50, 47, 73, 82] BLOCKS: [40, 60, 54, 90, 67, 38, 60, 56, 60, 51, 50, 51, 42, 70, 83, 65, 83, 58, 53, 65, 61, 45, 47, 36, 78, 55, 33, 71, 57, 81, 33, 87, 64, 52, 67, 51, 72, 59, 53, 78, 62, 72, 55, 42, 78, 45, 84, 72, 79, 68, 43, 41, 62, 68, 40, 64, 84, 39, 86] [60, 57, 34, 69, 69, 89, 89, 50, 44, 73, 63, 60, 66, 46, 81, 81, 74, 71, 34, 77, 82, 89, 78, 85, 50, 63, 72, 83, 68, 52, 60, 58, 36, 52, 49, 46, 69, 49, 77, 90, 59, 68, 39, 43, 64, 84, 63, 45, 75, 56, 43, 37, 46, 73, 36, 46, 73, 74, 59] [33, 77, 36, 85, 77, 79, 66, 40, 65, 49, 41, 49, 87, 66, 52, 60, 85, 66, 71, 77, 39, 37, 60, 74, 61, 51, 47, 82, 34, 66, 35, 64, 61, 34, 54, 34, 52, 87, 46, 58, 55, 82, 65, 66, 82, 54, 89, 41, 42, 61, 57, 73, 33, 82, 45, 56, 70, 68, 55] [84, 79, 45, 55, 61, 76, 49, 62, 66, 48, 44, 82, 80, 71, 52, 65, 72, 39, 77, 57, 49, 42, 83, 64, 54, 52, 66, 66, 34, 72, 49, 59, 36, 51, 81, 70, 32, 53, 60, 71, 80, 51, 54, 36, 39, 87, 68, 77, 87, 79, 63, 90, 37, 64, 42, 86, 89, 43, 68] [49, 43, 32, 58, 33, 62, 44, 76, 59, 64, 57, 56, 33, 46, 48, 36, 48, 66, 57, 58, 52, 88, 59, 65, 90, 80, 69, 73, 54, 74, 64, 44, 40, 90, 72, 53, 87, 62, 90, 58, 46, 61, 37, 35, 46, 72, 59, 88, 42, 89, 45, 79, 51, 74, 45, 84, 45, 89, 83] [54, 39, 63, 85, 70, 35, 44, 89, 61, 48, 35, 88, 67, 63, 50, 35, 55, 47, 34, 89, 56, 35, 55, 85, 50, 71, 85, 79, 35, 70, 35, 84, 33, 62, 48, 33, 81, 55, 86, 79, 72, 49, 69, 69, 84, 46, 70, 61, 73, 66, 62, 55, 89, 85, 61, 35, 47, 86, 86]

SALT: [51, 56, 44, 77, 43, 83, 68, 71, 56, 36, 41, 86, 55, 43, 48, 54, 37, 49, 46, 69, 32, 90, 43, 73, 66, 53, 84, 75, 54] BLOCKS: [49, 86, 77, 59, 56, 60, 37, 32, 76, 88, 51, 80, 44, 51, 74, 56, 61, 38, 71, 67, 54, 58, 45, 74, 43, 65, 85, 80, 63, 54, 55, 87, 88, 53, 75, 81, 89, 69, 75, 80, 90, 50, 90, 37, 79, 64, 41, 41, 70, 50, 74, 75, 49, 75, 67, 44, 72, 50, 66] [76, 75, 60, 86, 32, 71, 49, 73, 53, 40, 52, 85, 59, 39, 89, 57, 34, 85, 39, 56, 54, 47, 34, 65, 66, 71, 37, 76, 70, 73, 46, 83, 54, 60, 79, 46, 86, 67, 39, 54, 71, 44, 34, 74, 64, 60, 77, 71, 44, 48, 60, 58, 71, 43, 73, 86, 67, 77, 45] [60, 67, 37, 68, 61, 70, 83, 77, 52, 76, 78, 32, 34, 47, 78, 40, 72, 72, 53, 62, 62, 85, 72, 51, 48, 73, 33, 52, 49, 36, 32, 85, 33, 35, 54, 45, 64, 63, 82, 75, 69, 53, 39, 89, 47, 63, 50, 38, 55, 90, 75, 89, 81, 53, 37, 61, 72, 32, 34] [39, 32, 85, 83, 82, 48, 70, 37, 62, 83, 82, 67, 52, 38, 90, 35, 47, 60, 54, 82, 88, 35, 75, 63, 64, 76, 58, 60, 54, 35, 59, 70, 80, 46, 63, 76, 48, 35, 41, 49, 43, 40, 81, 40, 38, 47, 71, 84, 40, 73, 47, 49, 45, 69, 81, 78, 57, 72, 59] [69, 39, 54, 44, 72, 39, 86, 75, 35, 75, 86, 32, 39, 50, 68, 34, 49, 79, 83, 66, 80, 57, 40, 84, 43, 70, 63, 58, 35, 82, 40, 70, 81, 47, 83, 50, 64, 63, 42, 78, 63, 68, 47, 73, 86, 72, 56, 89, 72, 36, 64, 59, 35, 62, 88, 76, 68, 49, 60] [86, 37, 40, 74, 74, 67, 72, 46, 41, 69, 79, 85, 44, 58, 56, 44, 69, 38, 55, 59, 76, 59, 87, 55, 60, 76, 43, 51, 90, 36, 77, 45, 70, 39, 84, 44, 71, 70, 43, 82, 48, 65, 32, 90, 42, 50, 46, 64, 90, 71, 83, 60, 80, 78, 52, 52, 71, 39, 69]

Quick thought: each byte is a printable ascii character between space (0x20) and 'Z' (0x5a) inclusive. All these bytes appear in all the encoded files. Does anything interesting happen if you interpret the files as encoded in base 59?

(Edit: obviously I'm talking about the files after they have been base64 decoded…)

getting warmer. the range of ascii characters is correct. The code isn't as simple though as just base59ing it.

It seems more like the string decodes to multiple base 59 numbers. The cipher is another layer on top of that.

The strange properties of the encryption lead me to believe it was some binary modification (just a pattern), but that may have been looking too far into the problem. I started to see a pattern by deleting bits initially but it disappeared as I progressed.

I'm not even sure where to start with using the unencoded string.

"A Christmas Story" is pretty great christmas movie. One of the greatest scenes is when Ralphie decodes the secret orphan annie message. I changed it a bit so that someone wouldn't just guess it.

it seemed fitting as it is the only encoded message in any holiday movie I could think of! :)

I have no idea where to begin with the decryption, but I like this guy's style.

I would start here: http://www.opinionatedgeek.com/dotnet/tools/base64decode/

22 subscribers to their blog's RSS feed. The web has changed.

"I" (11 times) followed closely by "F" and "H" (both 9 times) seem to the most common letters.

I doubt this information will be helpful since for every character there are ~9 characters after encryption.

you are on the right track.

Other observation: the encoded string's characters seem to range from ASCII 32 to 90.

That range is 58 characters which (perhaps suspiciously) is size of alphabet + 2^5. (Although a whitespace character is also encoded, making that suspicion a little less "round".)

From comparing the multiple encodings of Quick Brown Fox, the value distribution across that range is fairly even and every value is used at least once.

I have the top five as:

; (12 times) I (11 times) * (10 times) - (10 times) 2 (10 times)

and for the quick-brown-fox sample he posted in the comments, I have:

! (14 times) # (13 times) C (11 times) , (11 times) U (10 times)

So I'm not convinced this is a fruitful path of inquiry.

Erik, is there some kind of offset within the ciphertext?

erik, maybe you should give them a clue where to start...

I gave some clues of sorts in the comments.

it's 500 bones brosuf...i can't just give it away. I will add some more clues next week.

Given that people on HM should be fine with decoding from base64, we're trying to break this code:

    @(2=(!+Y0NTI;C+U;Y-LV,(73*DP#P&3FNS.#%<O<)H3O4=-DX;FE(&:-'40
    #?)%AB8;P#@2<!ZNCSS$?=H*7EJ,!"ULVF:6IF'J "2GFT*FH,>40*W(T @-
    Y5LF$32/#"#.?!OWI429U;$2LS8VH/A+QZF7HI0(;*#9EJ<G%->?LS<6HE&V
    5Y/;I.3TD%I*2O?K>-?-IGN?Z38WH T?-SBTD1A0*/2WN35UJ1CN64/=4:79
    RFAJC2S,@;-90 9+"W%T:)6ILRI/JB'"$D%".HNI6I+$C(.*WA.;>*MX?;*L
    S;@11K/2<EO>UW-;4>4GHEL&

I'm trying to make sure I'm seeing the correct things.

the reason i base64 encoded it is that wordpress was trying to be smart and changing characters for me...which in a coded message is...not a good idea.

I'm aware of those problems, however I'm trying to confirm if the above is the same as what I should be trying to decode. It's a waste of time to decode the wrong string.

I get the same thing FWIW (using python's base64.decodestring.)

Just a note, this is not correct. It is the result of having spaces in the string from the line breaks.

See https://gist.github.com/747869

apparently it's ruby..

Maybe some kind of crazy ruby unpack function? It looks alot like a pack string from perl, but thats a guess.

I wish I knew. I'm a python guy.

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