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DESCRIPTION OF THE ALGORITHM The compressor algorithm builds a String translation Table from the text being compressed. The string translation table maps fixed-length codes (usually 12- Bit ) to strings. The string table is initialized with all single- Character strings (256 entries in the case of 8-bit characters). As the compressor character-serially examines the text, it stores every unique two-character string into the table as a code/character Concatenation , with the code mapping to the corresponding first character. As each two-character string is stored, the first character is outputted. Whenever a previously-encountered string is read from the input, the longest such previously-encountered string is determined, and then the code for this string concatenated with the extension character (the next character in the input) is stored in the table. The code for this longest previously-encountered string is outputted and the extension character is used as the beginning of the next string. The decompressor algorithm only requires the compressed text as an input, since it can build an identical string table from the compressed text as it is recreating the original text. However, an abnormal case shows up whenever the sequence ''character''/''string''/''character''/''string''/''character'' (with the same character for each ''character'' and string for each ''string'') is encountered in the input and ''character''/''string'' is already stored in the string table. When the decompressor reads the code for ''character''/''string''/''character'' in the input, it cannot resolve it because it has not yet stored this code in its table. This special case can be dealt with because the decompressor knows that the extension character is the previously-encountered ''character''.Welch, T. A. (June 1984). " A technique for high-performance data compression ." ''Computer''. Vol. 17, pp. 8-19. ALGORITHM Compressor algorithm: Decompressor algorithm: Example == Compression The following table shows the result of executing compressor algorithm on this input: TOBEORNOTTOBEORTOBEORNOT
After compression we have this sequence of 9-bits codes in output: TOBEORNOT<256><258><260><265><259><261><263>
== Decompression The result of applying the decompression algorithm to the compressed 9-bit sequence is presented in this table:
USES The method became widely used in the program Compress , which became a more or less standard utility in Unix systems circa 1986. (It has since disappeared from many for both legal and technical reasons.) Several other popular compression utilities also used the method, or closely related ones. It became very widely used after it became part of the GIF image format in 1987 . It may also (optionally) be used in TIFF files. LZW compression provided a better compression ratio, in most applications, than any well-known method available up to that time. It became the first widely used universal data compression method on computers. It would typically compress large English texts to about half of their original sizes. Today, an implementation of the algorithm is contained within the popular Adobe Acrobat software program. EXAMPLE This example shows the LZW algorithm in action, showing the status of the output and the Dictionary at every stage, both in encoding and decoding the message. In order to keep things clear, let us assume that we're dealing with a simple alphabet - capital letters only, and no punctuation or spaces. This example has been constructed to give reasonable compression on a very short message; when used on real data, repetition is generally less pronounced, and so the initial parts of a message will see little compression. As the message grows, however, the Compression Ratio tends asymptotically to the maximum.Jacob Ziv and Abraham Lempel; ''Compression of Individual Sequences Via Variable-Rate Coding'' , IEEE Transactions on Information Theory, September 1978. A message to be sent might then look like the following: TOBEORNOTTOBEORTOBEORNOT# The # is a marker used to show that the end of the message has been reached. Clearly, then, we have 27 symbols in our alphabet (the 26 capital letters ''A'' through ''Z'', plus the ''#'' character). A computer will render these as strings of Bit s; 5-bit strings are needed to give sufficient combinations to encompass the entire dictionary. As the dictionary grows, the strings will need to grow in length to accommodate the additional entries. A 5-bit string gives 25 = 32 possible combinations of bits, and so when the 33rd dictionary word is created, the algorithm will have to start using 6-bit strings. Note that since the all-zero string 00000 is used, and is labeled "0", the 33rd dictionary entry will be labeled '''32'''. The initial dictionary, then, will consist of the following: # = 00000 A = 00001 B = 00010 C = 00011 . . . Z = 11010 Encoding If we weren't using LZW, and just sent the message as it stands (25 symbols at 5 bits each), it would require 125 bits. We will be able to compare this figure to the LZW output later. We are now in a position to apply LZW to the message. Symbol: Bit Code: New Dictionary Entry: (= output) T 20 = 10100 O 15 = 01111 28: TO <--- Don't forget, we originally had 27 symbols, so the next one is 28th. B 2 = 00010 29: OB E 5 = 00101 30: BE O 15 = 01111 31: EO <--- start using 6-bit strings R 18 = 010010 32: OR N 14 = 001110 33: RN O 15 = 001111 34: NO T 20 = 010100 35: OT TO 28 = 011100 36: TT BE 30 = 011110 37: TOB OR 32 = 100000 38: BEO TOB 37 = 100101 39: ORT EO 31 = 011111 40: TOBE RN 33 = 100001 41: EOR OT 35 = 100011 42: RNO # 0 = 000000 43: OT# This is somewhat clearer: Current Next Output Value Extended Sequence Char (# of bits) Dictionary NULL T T O 20 = 5 bits 27: TO <-- This IS the 28th entry, but the initial entries are numbered 0-26 so this is #27. O B 15 = 5 bits 28: OB B E 2 = 5 bits 29: BE E O 5 = 5 bits 30: EO O R 15 = 5 bits 31: OR R N 18 = 6 bits 32: RN <-- Starting at R, 6 bits are used {floor(lg2(init_dict_size + num_chars_output)) + 1} N O 14 = 6 bits 33: NO i.e. O: floor(lg2(27 + 4)) + 1 = 5 bits -> 01111 O T 15 = 6 bits 34: OT R: floor(lg2(27 + 5)) + 1 = 6 bits -> 010010 T T 20 = 6 bits 35: TT TO B 27 = 6 bits 36: TOB BE O 29 = 6 bits 37: BEO OR T 31 = 6 bits 38: ORT TOB E 36 = 6 bits 39: TOBE EO R 30 = 6 bits 40: EOR RN O 32 = 6 bits 41: RNO OT # 34 = 6 bits 42: OT# # 0 = 6 bits
In using LZW we have made a saving of 28 bits out of 125 -- we have reduced the message by almost 22%. If the message were longer, then the dictionary words would begin to represent longer and longer sections of text, allowing repeated words to be sent very compactly. Decoding Imagine now that we have received the message produced above, and wish to decode it. We need to know in advance the initial dictionary used, but we can reconstruct the additional entries as we go, since they are always simply Concatenation s of previous entries. Bits: Output: New Entry: Full: Partial: 10100 = 20 T 28: T? 01111 = 15 O 28: TO 29: O? 00010 = 2 B 29: OB 30: B? 00101 = 5 E 30: BE 31: E? 01111 = 15 O 31: EO 32: O? <--- start using 6-bit strings 010010 = 18 R 32: OR 33: R? 001110 = 14 N 33: RN 34: N? 001111 = 15 O 34: NO 35: O? 010100 = 20 T 35: OT 36: T? 011100 = 28 TO 36: TT 37: TO? <--- for 36, only add 1st element 011110 = 30 BE 37: TOB 38: BE? of next dictionary word 100000 = 32 OR 38: BEO 39: OR? 100101 = 37 TOB 39: ORT 40: TOB? 011111 = 31 EO 40: TOBE 41: EO? 100001 = 33 RN 41: EOR 42: RN? 100011 = 35 OT 42: RNO 43: OT? 000000 = 0 # The only slight complication comes if the newly-created dictionary word is sent immediately. In the decoding example above, when the decoder receives the first symbol, T, it knows that symbol 28 begins with a T, but what does it end with? The problem is illustrated below. We are decoding part of a message that reads '''ABABA''': Bits: Output: New Entry: Full: Partial: . . . 011101 = 29 AB 46: (word) 47: AB? 101111 = 47 AB? <--- what do we do here? At first glance, this may appear to be asking the impossible of the decoder. We know ahead of time that entry 47 should be ABA, but how can the decoder work this out? The critical step is to note that 47 is built out of 29 plus whatever comes next. 47, therefore, ends with "whatever comes next". But, since it was sent immediately, it must also start with "whatever comes next", and so must end with the same symbol it starts with, namely '''A'''. This trick allows the decoder to see that 47 must be ABA. More generally the situation occurs whenever the encoder encounters the input of the form ''cScSc'', where ''c'' is a single character, ''S'' is a string and ''cS'' is already in the dictionary. The encoder outputs the symbol for ''cS'' putting new symbol for ''cSc'' in the dictionary. Next it sees the ''cSc'' in the input and sends the new symbol it just inserted into the dictionary. By the reasoning presented in the above example this is the only case where the newly-created symbol is sent immediately. PYTHON EXAMPLE #Lempel-Ziv-Welch compression algorithm #Translated to python by Thomas Van Durme #Last edited: September 20, 2006 class LZW: """Returns the compressed string in utf8 format""" def compress(self,uncompressed): if isinstance(uncompressed, str): chars = int(256) mydict = dict() buffer = list() result = str() for i in range(chars): mydict {Link without Title} = i for i in uncompressed: if len(buffer) == 0: xstr = str(ord(i)) else: xstr = self.__join(buffer,"-")+"-"+str(ord(i)) if mydict.has_key(xstr): buffer.append(ord(i)) else: result += unichr(mydict {Link without Title} ).encode('utf8') mydict {Link without Title} = chars chars += 1 del buffer buffer = list() buffer.append(ord(i)) if len(buffer) != 0: result += unichr(mydict {Link without Title} ).encode('utf8') return result else: raise TypeError """Returns the decompressed string, input is a utf8 compressed string""" def decompress(self,compressed): if isinstance(compressed, str): chars = int(256) mydict = dict() for i in range(chars): mydict {Link without Title} = unichr(i).encode('utf8') decoded = compressed.decode('string_escape').decode('utf8') buffer = str() chain = str() result = str() for i in decoded: code = ord(i) if buffer == "": current = mydict {Link without Title} buffer = current result += current else: if code<=255: current = mydict {Link without Title} result += current chain = buffer+current mydict {Link without Title} = chain chars += 1 buffer = current else: if mydict.has_key(code): chain = mydict {Link without Title} else: chain = buffer+buffer {Link without Title} result += chain mydict = buffer+chain[0 chars += 1 buffer = chain return result else: raise TypeError def __join(self,mylist,delimiter): if isinstance(mylist,list) and isinstance(delimiter,str): result = str() for i in range(0,len(mylist)): try: if i+1 == len(mylist): result += str(mylist {Link without Title} ) else: result += str(mylist {Link without Title} )+delimiter except TypeError: pass return result else: raise TypeError How to use: lzw = LZW() a = lzw.compress(" print a b = lzw.decompress(a) print b PATENT ISSUES Various and Mark N. Wegman and assigned to IBM , originally filed on June 1 , 1983 , and by Welch, assigned to Sperry Corporation, later Unisys Corporation, filed on June 20 , 1983. On June 20 , 2003 , this patent on the LZW algorithm expired {Link without Title} . US Patent 4,558,302 is the one that has caused the most controversy (See Graphics Interchange Format#Unisys And LZW Patent Enforcement ). LEMPEL-ZIV-WELCH VS. ZIV-LEMPEL-WELCH Although the name of the algorithm refers to the inventors as Lempel, Ziv and Welch, some people claim that the intellectual property rightly goes to Ziv first, so the method should be called the ''Ziv-Lempel-Welch algorithm'', and not the ''Lempel-Ziv-Welch algorithm''. Others who distinguish between the algorithm and the code prefer calling the algorithm ''LZ'' and the code written by Welch as ''LZW''. SEE ALSO
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