How to write a program to decode a message?

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  • anderson1373
    New Member
    • Nov 2009
    • 23
    #1

    How to write a program to decode a message?

    My program needs to fix an encrypted text file. The person who wrote it put it in “leet speak”, with special characters representing various letters. For example, consider the sample below:
    1 4|-| 50 |_33+.
    is really
    I am so leet.
    Here is the conversion table I need to use.
    A 4
    B 8
    C [
    D |)
    E 3
    F |#
    G 6
    H #
    I 1
    J ]
    K |\
    L |_
    M |-|
    N |\|
    O 0
    P |*
    Q 0\
    R 2
    S 5
    T +
    U |_|
    V \/
    W \/\/
    X ><
    Y 7
    Z 7_
    Your program should read leet speak sentences from a file called "leet.txt" (one sentence per line). It should covert each sentence to normal English and print it to the screen.
    here is what I have:
    Code:
    d= {"4":"A", "8":"B", "[":"C", "|)":"D", "3":"E", "|#":"F", "6":"G", "#":"H", "1":"I", "]":"J", "|\\":"K", "|_":"L", "|-|":"M", "|\\|":"N", "0":"O", "|*":"P", "0\\":"Q", "2":"R", "5":"S", "+":"T", "|_|":"U", "\\/":"V", "\\/\\/":"W", "><":"X", "7":"Y", "7_":"Z"}
infile = open("leet.txt", "r")
for line in infile:
	line.split
	window = 4
	i = 0
	while i < len(line):
		t = line[i:i+window]
		i += window
	print (line)
    It is printing in leet but where do I put in the line to map it to the dictionary? And How do I move down?
    Tags: None
    • dwblas
      Recognized Expert Contributor
      • May 2008
      • 626
      #2
      You look up each character in the dictionary http://www.java2s.com/Code/Python/Di...lternative.htm http://books.google.com/books?id=1Hx...ership&f=false. Of course, the character "|" and others signal a multiple character conversion, so for those characters you would have to test for 2 characters, then 3, etc. until a match is found.
      Code:
      d= {"4":"A", "8":"B", "[":"C", "|)":"D", "3":"E", "|#":"F", "6":"G", "#":"H", "1":"I", "]":"J", "|\\":"K", "|_":"L", "|-|":"M", "|\\|":"N", "0":"O", "|*":"P", "0\\":"Q", "2":"R", "5":"S", "+":"T", "|_|":"U", "\\/":"V", "\\/\\/":"W", "><":"X", "7":"Y", "7_":"Z"}

ch ='4'
print "%s translates to %s" % (ch, d[ch])
ch =']'
print "%s translates to %s" % (ch, d[ch])

        Comment

        • anderson1373
          New Member
          • Nov 2009
          • 23
          #3
          We are supposed to start with 4 which is the longest character and work back to 1.

            Comment

            • dwblas
              Recognized Expert Contributor
              • May 2008
              • 626
              #4
              You will have to start with 4 characters and remove the right-most under a while() loop until it is found. Ask you instructor what happens if you have "O" followed by "V" or "W". How can you tell that it is not a "Q":
              "OV" = 0\/ ## will find "Q" = "0\" going from 4 chars to one
              "OW" = 0\/\/ ## will find "Q" = "0\"
              "Q" = 0\
              "O" = 0
              Unless you copied the dictionary wrong. Changing "Q" to a forward slash, 0/ would fix it.

                Comment

                • dwblas
                  Recognized Expert Contributor
                  • May 2008
                  • 626
                  #5
                  A very quick and dirty solution that hopefully will help you solve this for yourself. It does not allow for a malformed leet that is not found in the lookup dictionary, and probably other things as well, but that is up to you to do.
                  Code:
                  def from_leet(d, phrase, ctr):
    to_add = 4
    if ctr+to_add >= len(phrase):
        to_add = len(phrase) - ctr    

    for x in range(to_add, 0, -1):
        test_this = phrase_leet[ctr:ctr+x]
        if test_this in d:
            return(ctr+x, d[test_this])
    return ctr, ""

def to_leet(d, phrase_in):
    d_reverse = {}
    for key in d:
        d_reverse[d[key]] = key
    phrase_in = phrase_in.upper()
    output_list = []
    for ch in phrase_in:
        output_list.append(d_reverse[ch])
    return("".join(output_list))


d= {"4":"A", "8":"B", "[":"C", "|)":"D", "3":"E", "|#":"F", "6":"G", 
    "#":"H", "1":"I", "]":"J", "|\\":"K", "|_":"L", "|-|":"M", "|\\|":"N", 
    "0":"O", "|*":"P", "0\\":"Q", "2":"R", "5":"S", "+":"T", "|_|":"U", 
    "\\/":"V", "\\/\\/":"W", "><":"X", "7":"Y", "7_":"Z"}

##phrase = "quickbrownfox"     ## "ow" problem
phrase = "whydoesthecagedbirdsing"
phrase_leet = to_leet(d, phrase)
print phrase_leet, "\n"

ctr = 0
output_list = []
while ctr < len(phrase_leet):
    ctr, letter = from_leet(d, phrase_leet, ctr)
    output_list.append(letter)

print "".join(output_list)

                    Comment

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