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Zack
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« on: December 12, 2003, 11:56:07 AM » |
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Write an algo to decrypt this: J#hmlt#tkbw#zlv$qf#wkjmhjmd-#Gjg#kf#ejqf#pj{#pklwp#lq#lmoz#ejuf<#Tfoo/#wl#wfoo#zlv#wkf#wqvwk/#jm#boo#wkjp#f{`jwfnfmw/#J$uf#hjmgb#olpw#wqb`h#nzpfoe-#Avw#afjmd#bp#wkjp#jp#b#-77#Nbdmvn#---#zlv$uf#dlw#wl#bph#zlvqpfoe#lmf#rvfpwjlm9#$Gl#J#effo#ov`hz<$#Tfoo/#gl#zb#svmh< Your score will increase if: 1) You actually manage to decrypt it 2) How readable the code is. [EDIT]Actually, that isn't going to work since there might be some unprintable chars there...here's a text file with the same thing, pure ASCII: http://www.thegeekery.org/ciph.txt [/EDIT] |
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f only life let you press CTRL-Z.
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Freebasic is like QB, except it doesn't suck.
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na_th_an
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« Reply #1 on: December 12, 2003, 12:32:22 PM » |
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' I saw the # that looked spaces. ASC("#")=35, so it was SPACE+3
' I figured out that you XORed something to encode this, as I cheated
' seeing your latest subroutines ;) So I tried...
OPEN "ciph.txt" FOR INPUT AS #1
LINE INPUT #1, ciphered$
CLOSE #1
deciphered$ = ""
FOR i% = 1 TO LEN(ciphered$)
current$ = MID$(ciphered$, i%, 1)
dec% = ASC(current$) XOR 3
deciphered$ = deciphered$ + CHR$(dec%)
NEXT i%
PRINT deciphered$ |
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Zack
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« Reply #2 on: December 12, 2003, 01:57:04 PM » |
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Very good, Nath!
5 points for code readability, and 3 for just figuring it out.
You should have PMed me the result, now nobody else can try.
I might make another challenge, more secure algo.
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f only life let you press CTRL-Z.
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Freebasic is like QB, except it doesn't suck.
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whitetiger0990
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« Reply #3 on: December 12, 2003, 02:12:00 PM » |
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yes, please do. i wanna try
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 [size=10]Back by popular demand! I will byte and nibble you bit by bit until nothing remains but crumbs.[/size] |
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na_th_an
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« Reply #4 on: December 12, 2003, 05:05:42 PM » |
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Aw sorry about that - Now I realize I should've PMed you...
My apollogies. :oops:
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Neo
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« Reply #5 on: December 15, 2003, 05:58:51 AM » |
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Very good, Nath!
5 points for code readability, and 3 for just figuring it out.
You should have PMed me the result, now nobody else can try.
I might make another challenge, more secure algo. Hehe. lol. I already saw it was some kind of practical example of your new-discovered encryption... where the decryption and encryption routines are the same  I don't mind that I can't enter anything. :roll: |
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Anonymous
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« Reply #6 on: December 15, 2003, 05:10:39 PM » |
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in the spirit of password-less encryption, i give you this. This is my feeble newbie attempt at encryption from 3 yrs ago. The algo is something pretty neat though... i dont know if anyone has ever done anything like this before. Mad props if you can do it! EDIT: Note: You must open this in qbasic to see all the characters properly... i think |
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whitetiger0990
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« Reply #7 on: December 15, 2003, 06:12:38 PM » |
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dead link.
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[size=10]Back by popular demand! I will byte and nibble you bit by bit until nothing remains but crumbs.[/size] |
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Anonymous
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« Reply #8 on: December 15, 2003, 06:21:14 PM » |
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my apologies... it should be thisEDIT:This one should work... finally lol sorry my bad. |
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Lachie Dazdarian
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« Reply #9 on: December 17, 2003, 07:19:12 AM » |
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And how do you decryipt in the first place? I mean, the stuff Nathan just did is like magic to me.
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Neo
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« Reply #10 on: December 17, 2003, 10:04:53 AM » |
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When using a XOR-encryption (only), the encryption is the same as the decryption |
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Lachie Dazdarian
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« Reply #11 on: December 17, 2003, 10:30:14 AM » |
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Right...How did he knew it was a XOR encryption? And the exact code used?
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na_th_an
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« Reply #12 on: December 17, 2003, 03:30:45 PM » |
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Well, first of all I examined the stream. I saw that the symbol # repeated all over the place, in a pseudocyclic fashion, so I guessed it was the encoded "SPACE" character. As it looked so much as the "word delimiter" I guessed that a straightforward method was used, and that each symbol represented one only character. I noticed that ASC("#") = ASC(" ") XOR 3, partially guessing 'cause this guy had posted two XOR encryption/decryption algos earlier that week. I cheated  |
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