HTTPCOLONSLASHSLASHRJBDOTSOCDOTPORTDOTACDOTUKSLASHPUNKSLASH RIGHTRIGHTRIGHTRIGHTDOTZIP

Transcription

1 Portsmouth Cipher Step (morse code) HTTPCOLONSLASHSLASHRJBDOTSOCDOTPORTDOTACDOTUKSLASHPUNKSLASH RIGHTRIGHTRIGHTRIGHTDOTZIP Step 2 ALMPINYPMYWQECLEZIWLMJXIHFCXLVIIMXMWKSSHXSWIIXLEXCSYHMHRSXK MZIRXLEXGEIWEVAVSXILM WQIWWEKIWSZIVXASXLSYWERHCIEVWEKSMXMWKSSHXSWIIXLEXTVSKVIWW LEWFIIRQEHIMRXLIJMIPHSJ GVCTXSKVETLCWMRGIXLIJEPPSJXLIVISERIQTMVIXLMWRIBXGLETLEHELEHMR XLEXLXXTGSPSRWPEWLW PEWLVNFHSXWSGHSXTSVXHSXEGHSXYOWPEWLTYROWPEWLUAIVXCYMSTHS XDMT (caesar) WHILE JULIUS MAY HAVE SHIFTED BY THREE IT IS GOOD TO SEE THAT YOU DID NOT GIVEN THAT CAESAR WROTE HIS MESSAGES OVER TWO THOUSAND YEARS AGO IT IS GOOD TO SEE THA T PROGRESS HAS BEEN MADE IN THE FIELD OF CRYPTOGRAPHY SINCE THE FALL OF THE REOA N EMPIRE THIS NEXT CHAP HAD A HAD IN THAT HTTP COLON SLASH SLASH RJB DOT SOC DOT PORT DOT AC DOT UK SLASH PUNKSLASHQWERTYU

2 IOPDOTZIP Step 3 WKZRGZUBHXIPESXJFIGDCQWKUCLWZAYJKZGFCAUGQRITHLMQRTHERXACWI ATKLAZMFCBTXHOXREYMMJTHUSEXIHWKHJDMJVGJPSGQOXYTRINPAQEWVW CPQUCUNIJGMNIHIHEFLMGHOWIJKXYXGNIZXQJGIRNNWUGQLLVKUBWQPCAGV GROZWTIXIWHPYPWBLEWZMGHOVIJLGFEMHAZPIDWCTQBZXQTKXJHAXWHEEN XUHRUKRDJQOJEYMPDJDGWCTQBEWABHIRIWICTJHPLVKDIZKPHZSEMFCAXDK NRVRRBZCJWDXZFCUAGJCERXRMOIRSQYLRKYWBSGIQJWBVINPVMGDHILXAW DQTIWVJOUHWORVWHJFWKXDCXYXDIZKPHZWCTQB (Vigenere) GOVANBATTISTABELLASOMUSTBEROLLINGINHISGRAVECONSIDERINGHECAME UPWITHITFIRSTALASEVERYONEKNOWSTHISCIPHERASTHATOFBLAISEDEVIGNER ESNOTATTRIBUTINGITTOTHEITALIANCRYPTOGRAPHERWHOCAMECENTURIESBE FOREHIMVISITINGHTTPCOLONSLASHSLASHRJBDOTSOCDOTPORTDOTACDOTUKS LASHPUNKSLASHWILLLEAPYOUFURTHERFORWARDONTHISJOURNEYTHROUGHT IMEASSUMINGYOUADDACHARACTERSUFFIXOFBGWKZQPNDSIOAXEFCLUMTHY VRFOLLOWINGTHEFORWARDSLASH Step 4 OPAPNMRORYFRXEYLNCYNSYPEYMXYXHELDUNQWRLLOXOYUEYMOMXEXVR SIYSEYNGLROWTORXACXMHYZORYLYROGXLYSMYLSOLYROAHYWYFXCADHY TRWVLERAPEOASXSXCWONRDNLLRYUVYYLFEYLSEHUWHMXSYQYOYWYXHY MDTTQSDPBRGTWATPNGZNTQOMGMEONNUWNBQYKMATYEGTRONYZMSOGSO SRYNISQYSFISBYCGYQPYPIZTQQXFBRYWIPNBXTQSEFTOYDIAEVNPRAWDWCGR TCGINOMPLURPSZRITSGTFMRRRZTPNMTXTRTHRZWQQIHR (Bifid cipher, using the picture provided)

3 AND SO ENDS OUR TRIP THROUGH TIME THE FINAL STOP ON THIS PART OF YOUR JOURNEY CONCLUDING WITH FELIX DELASTELLE THE FRENCH AMATEUR CRYPTOGRAPHER RENOWN FOR HIS WORK IN THE FIELD OF CRYPTOGRAPHY TIME TO THINK DIFFERENTLY TO CONTINUE TO THE NEXT STAGE YOU WILL NEED CYPHER WYTWOVFLXOTHREEQ DOT ONION YOU SHOULD KEEP PORTSM SAFE FOR THE FUTURE REMEMBER TO CONVERT NUMBERS FROM STRING TO INTEGER CYPHER WYTWOVFLXOTHREEQ DOT ONION cypherwy2vflxo3q.onion Clue: portsm ('keep portsm safe for the future...') Step 5 If we go on the onion site that we've just found, we get a zip file containing a.ppm image. After looking through it, it was discovered that there was a pattern of binary for the RGB values of '2x 20x 2x' which were on top of each other. It was possible to convert the binary from left to right, top to bottom, i.e.: C o

4 n Until we get the whole message, repeated: C o n g r a t u l a t i o n s. N o w t h a t w a s n ' t s o d i f f i c u l t w a s i t? N o w o n w a r d t o t h e f i n a l s t a g e, y o u ' l l n e e d t h e s e c o n d p a r t t o t h e f i n a l p u z z l e w h i c h i s t h o o 5 a n d t h e n e x t s t a g e i s f o u n d i t p u n k s 6 a y s w 6 u j p o l. o n i o n Clue: thoo5 Step 6 The code generated from the previous step was: import binascii import struct import sys val="4cc2ae9a45702a881e2ef1a7573d0ca57c02002c19e5567f7b0d05205caa30821efbeec75a2d4c e0116b86f2a12187f6a195b3e7ac4d36b6b ebfe5e2874ca127ea1014e4a88fe2f7175c6b 216c069f af11176c043bfe64564fab10c34a37b d3aa4f0e46f078ab02bf550e21e25593e 1c34aae278c57c9ba c8f01a834b35c5fbb392a800d726edd162262f e544cc0370e75b264 fd0c3385f4a6765f8cf3e f415ed2614de0d25e59f5f60d0a5de65a326eeb0537f5ec390b92b3fe1 5b9f0e91f7db8bc6c9b0b4a bb9d93bf8ed4f308108aa1c0416ed54a4cdfa5070cbfd7987a78 a64052a8e2b48b2885fd2e69d3a9bb7a956bbae b4f26716e1e7f6f892a0f77df356b5c5f706ec 9cb4f de27b472fd32e3a5e7d6f26b63e27059afc5e6a3ba85a a4f98ffab73f 992ee631637ad f07f3f65c42a9a2ce4572fa4b6646f0aa5992acbe36e038f677f177e5783d37c27 272ea98756d b8b229f48a548994e3623a9df6f78cd612b e48371fdb10eb0b4f1de915a 32d61c86460ecc36e0ec5ca3f5a1bcf6474ac a91b01edfd0e3c92b d65a2bab10ef03 ef85b0881a1706ddf0b feeb4d814569b782bf733c346fad1" val="4cc2ae9a45702a881e2ef1a7573d0ca57c0200" # xor two messages def xor(a, b): return ''.join([chr(ord(a[i]) ^ ord(b[i])) for i in range(len(a))]) #stream cipher constants a = m = 2**31 # run the stream cipher! def generatestream(key, length): x = key stream = struct.pack(">l", x) while len(stream) < len(msg): x = (a*x) % m stream += struct.pack(">l", x)

5 return stream msg = "Test message " key = 0x # key - you need to crack this! stream = generatestream(key, len(msg)) ciphertext = xor(msg, stream) print "ciphertext: ", binascii.hexlify(ciphertext) print len(stream) print len(msg) #To decrypt: for key in range(0, 0xfffff): stream = generatestream(key, len(val)) plaintext = xor(val, stream) print plaintext The cipher is intentionally weak with the first four bytes of the key stream X-OR ing the message stream: def generatestream(key, length): x = key stream = struct.pack(">l", x) while len(stream) < len(msg): x = (a*x) % m stream += struct.pack(">l", x) return stream Thus we just created a list of common words (with upper and lower case starting letters), and generated keys for these for the first four bytes: Method: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace gencode class Program static void Main(string[] args) string str = "4cc2ae9a"; string [] commonwords = "and", "look","this","the","for","you","with ","have","with", "from","she","they","will","there","who ","make","like","which","when", "can","make","good","your","than","into","just";

6 foreach (string s1 in commonwords) for (int i = 0; i < 2; i++) string s = Truncate(s1, 4); string instr = s; if (i == 0) else instr = Convert.ToString(char.ToUpper(instr[0]))+Convert.ToString(instr[1]) + Convert.ToString(instr[2]) + instr[3]; s); string hex = ConvertAsciiToHex(instr); Console.WriteLine("print g(0x" + xor(hex, str) + ") # " + public static string Truncate( string value, int maximumlength) if (value.length == 3) return (value + " "); if (string.isnullorempty(value) == true) return value; if (value.length > maximumlength) return value.substring(0, maximumlength); return value; public static string xor(string hex1, string hex2) int dec1 = Convert.ToInt32(hex1, 16); int dec2 = Convert.ToInt32(hex2, 16); int result = dec1 ^ dec2; string hexresult = result.tostring("x"); return hexresult; public static string ConvertAsciiToHex(string asciistring) string hex = ""; foreach (char c in asciistring) int tmp = c; hex += String.Format("0:X2", (uint)system.convert.touint32(tmp.tostring())); return hex; This generated all the required key that would generate the first four letters of the cipher: print g(0x2daccaba) # and print g(0xdaccaba) # And print g(0x20adc1f1) # look print g(0xadc1f1) # Look print g(0x38aac7e9) # this print g(0x18aac7e9) # This print g(0x38aacbba) # the print g(0x18aacbba) # The

7 print g(0x2aaddcba) # for print g(0xaaddcba) # For print g(0x35addbba) # you print g(0x15addbba) # You print g(0x3babdaf2) # with print g(0x1babdaf2) # With print g(0x24a3d8ff) # have print g(0x4a3d8ff) # Have print g(0x3babdaf2) # with print g(0x1babdaf2) # With print g(0x2ab0c1f7) # from print g(0xab0c1f7) # From print g(0x3faacbba) # she print g(0x1faacbba) # She print g(0x38aacbe3) # they print g(0x18aacbe3) # They print g(0x3babc2f6) # will print g(0x1babc2f6) # Will print g(0x38aacbe8) # ther print g(0x18aacbe8) # Ther print g(0x3baac1ba) # who print g(0x1baac1ba) # Who print g(0x21a3c5ff) # make print g(0x1a3c5ff) # Make print g(0x20abc5ff) # like print g(0xabc5ff) # Like print g(0x3baac7f9) # whic print g(0x1baac7f9) # Whic print g(0x3baacbf4) # when print g(0x1baacbf4) # When print g(0x2fa3c0ba) # can print g(0xfa3c0ba) # Can print g(0x21a3c5ff) # make print g(0x1a3c5ff) # Make print g(0x2badc1fe) # good print g(0xbadc1fe) # Good print g(0x35addbe8) # your print g(0x15addbe8) # Your print g(0x38aacff4) # than print g(0x18aacff4) # Than print g(0x25acdaf5) # into print g(0x5acdaf5) # Into print g(0x26b7ddee) # just print g(0x6b7ddee) # Just Next we imported this into the Python code: import binascii import struct import sys import string val="4cc2ae9a45702a881e2ef1a7573d0ca57c02002c19e5567f7b0d05205caa30821efbeec7 5a2d4c e0116b86f2a12187f6a195b3e7ac4d36b6b ebfe5e2874ca127ea10 14e4a88fe2f7175c6b216c069f af11176c043bfe64564fab10c34a37b d3aa4f 0e46f078ab02bf550e21e25593e1c34aae278c57c9ba c8f01a834b35c5fbb392a800d7 26edd162262f e544cc0370e75b264fd0c3385f4a6765f8cf3e f415ed2614de0 d25e59f5f60d0a5de65a326eeb0537f5ec390b92b3fe15b9f0e91f7db8bc6c9b0b4a bb9d93bf8ed4f308108aa1c0416ed54a4cdfa5070cbfd7987a78a64052a8e2b48b2885fd2e6 9d3a9bb7a956bbae b4f26716e1e7f6f892a0f77df356b5c5f706ec9cb4f de27b472fd32e3a5e7d6f26b63e27059afc5e6a3ba85a a4f98ffab73f992ee ad f07f3f65c42a9a2ce4572fa4b6646f0aa5992acbe36e038f677f177e5783d3 7c27272ea98756d b8b229f48a548994e3623a9df6f78cd612b e48371fdb10

8 eb0b4f1de915a32d61c86460ecc36e0ec5ca3f5a1bcf6474ac a91b01edfd0e3c92b d65a2bab10ef03ef85b0881a1706ddf0b feeb4d814569b782bf733c346fa d1" # xor two messages def xor(a, b): return ''.join([chr(ord(a[i]) ^ ord(b[i])) for i in range(len(a))]) #stream cipher constants a = m = 2**31 # run the stream cipher! def generatestream(key, length): x = key stream = struct.pack(">l", x) while len(stream) < len(msg): x = (a*x) % m stream += struct.pack(">l", x) return stream def g( key1 ): stream = generatestream(key1, len(cipher)) plaintext = xor(cipher, stream) return plaintext #To decrypt: msg="" cipher = binascii.unhexlify(val) for x in range(1, len(cipher)): msg+="a" print g(0x2daccaba) # and print g(0xdaccaba) # And print g(0x20adc1f1) # look print g(0xadc1f1) # Look print g(0x38aac7e9) # this print g(0x18aac7e9) # This print g(0x38aacbba) # the print g(0x18aacbba) # The print g(0x2aaddcba) # for print g(0xaaddcba) # For... print g(0x21a3c5ff) # make print g(0x1a3c5ff) # Make print g(0x2badc1fe) # good print g(0xbadc1fe) # Good print g(0x35addbe8) # your print g(0x15addbe8) # Your print g(0x38aacff4) # than print g(0x18aacff4) # Than print g(0x25acdaf5) # into print g(0x5acdaf5) # Into print g(0x26b7ddee) # just print g(0x6b7ddee) # Just Then when run gives garbled text, apart from "Good", which gives:

9 Good work! If you are reading this you must have cracked the stream cipher. Hopefully in cracking this you have seen the value in creating a cryptographically secure random number generator, the same weakness in this stream cipher can also be seen in some computer-based casino games which can be used to predict play. The end is just around the corner, but to reach the end, take this final piece of the puzzle and the clue: Final piece: 3jhmu Clue: Final step If we look at the previous clues given in steps 4, 5, and 6, we get the words: portsm, thoo5, and 3jhmu. The video of Shrek basically tells us to take the 3 clues, put them together, and go to a third onion website on the tor network: portsmthoo53jhmu.onion

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