This document discusses various data compression techniques including Huffman coding, adaptive Huffman coding, arithmetic coding, statistical and adaptive algorithms, and dictionary modeling. It provides details on how each technique works, such as assigning codes based on symbol frequencies for Huffman coding, adapting the code tree for adaptive Huffman coding, and dividing the numeric range based on symbol probabilities for arithmetic coding. It also discusses calculating entropy, the adaptive character word length algorithm, overhead considerations, and how dictionary-based algorithms encode variable-length strings as tokens.

1. DATA COMPRESSION
PRESENTED BY,
R.RAMADEVI,
II – M. SC(CS&IT).

2. HUFFMAN CODE
• THE ALGORITHM AS DESCRIBED BY DAVID HUFFMAN ASSIGNS EVERY SYMBOL TO A LEAF NODE
OF A BINARY CODE TREE.
• THESE NODES ARE WEIGHTED BY THE NUMBER OF OCCURRENCES OF THE CORRESPONDING
SYMBOL CALLED FREQUENCY OR COST.
• THE TREE STRUCTURE RESULTS FROM COMBINING THE NODES STEP-BY-STEP UNTILALL OF
THEM ARE EMBEDDED IN A ROOT TREE.
• THE ALGORITHM ALWAYS COMBINES THE TWO NODES PROVIDING THE LOWEST FREQUENCY IN
A BOTTOM UP PROCEDURE. THE NEW INTERIOR NODES GETS THE SUM OF FREQUENCIES OF
BOTH CHILD NODES.

3. CODE TREE ACCORDING TO HUFFMAN
Symbol Frequency
A 24
B 12
C 10
D 8
E 8
----> total 186 bit
(With 3 bit per code word)
The branches of the tree represent the binary values 0 and 1 according to the rules
for common prefix-free code trees. The path from the root tree to the corresponding
leaf node defines the particular code word.
The following example bases on a data source using a set of five different
symbols. The symbol's frequencies are:

4. CODE TREE ACCORDING TO HUFFMAN
Length Length
-------------------------------------
-------------------------------------
-
A 24 0 1
24
B 12 100 3
36
C 10 101 3
30
D 8 110 3
24
E 8 111 3
24
-------------------------------------
----------------------------------
186 bit tot.138 bit
(3 bit code)

5. ADAPTIVE HUFFMAN CODE
• THE ADAPTIVE HUFFMAN CODE WILL BE DEVELOPED IN THE COURSE OF THE CODING
STEP BY STEP.
• IN CONTRAST TO THE STATIC OR DYNAMIC CODING THE SYMBOL DISTRIBUTION
WILL NOT BE DETERMINED PREVIOUSLY.
• IT WILL BE GENERATED IN PARALLEL BY USING ALREADY PROCESSED SYMBOLS.
• PRINCIPLE:
• THE WEIGHT OF EACH NODE REPRESENTING THE LAST ENCODED SYMBOL WILL
BE INCREASED.
• AFTERWARDS THE CORRESPONDING PART OF THE TREE WILL BE ADAPTED.
• THUS THE TREE APPROACHES GRADUALLY THE CURRENT DISTRIBUTION OF THE
SYMBOLS. HOWEVER, THE CODE TREE ALWAYS REFLECTS THE "PAST" AND NOT THE
REAL DISTRIBUTION.

6. • INITIALIZATION:
• BECAUSE THE ADAPTIVE HUFFMAN CODE USES PREVIOUSLY ENCODED
SYMBOLS, AN INITIALIZATION PROBLEM ARISES AT THE BEGINNING OF THE
CODING.
• AT FIRST THE CODE TREE IS EMPTY AND DOES NOT CONTAIN SYMBOLS FROM
ALREADY ENCODED DATA. TO SOLVE THIS, A SUITABLE INITIALIZATION HAS TO
BE UTILIZED. AMONG OTHERS THE FOLLOWING OPTIONS ARE AVAILABLE:
• A STANDARD DISTRIBUTION WILL BE USED THAT IS AVAILABLE AT THE ENCODER
AS WELL AS THE DECODER.
• AN INITIAL CODE TREE WILL BE GENERATED WITH A FREQUENCY OF 1 FOR EACH
SYMBOL.
• A SPECIAL CONTROL CHARACTER WILL BE ADDED IDENTIFYING NEW SYMBOLS
FOLLOWING.

7. • ALGORITHM ADAPTIVE HUFFMAN CODE
• A NUMBER OF RULES AND CONVENTIONS IS REQUIRED FOR THE ALGORITHM
DESCRIBED HEREINAFTER.
• IT BASES ON A PROCEDURE STARTING WITH AN EMPTY TREE AND INTRODUCING
A SPECIAL CONTROL CHARACTER.
• NO PREVENTIVE ASSUMPTIONS ABOUT THE CODE DISTRIBUTION WILL BE MADE.
• CONTROL CHARACTER FOR SYMBOLS THAT ARE NOT PART OF THE TREE:
• A CONTROL CHARACTER IS DEFINED IN ADDITION TO THE ORIGINAL SET OF
SYMBOLS.
• THIS CONTROL CHARACTER IDENTIFIES SYMBOLS THAT, AT RUN-TIME, DO NOT
BELONG TO THE CURRENT CODE TREE. IN THE LITERATURE THIS CONTROL
CHARACTER IS DENOTED AS A NYA OR NYT (NOT YET AVAILABLE OR
TRANSMITTED).

8. WEIGHT OF A NODE:
• ANY NODE HAS AN ATTRIBUTE CALLED THE WEIGHT OF THE NODE.
• THE WEIGHT OF A LEAF NODE IS EQUAL TO THE FREQUENCY WITH WHICH THE
CORRESPONDING SYMBOL HAD BEEN CODED SO FAR.
• THE WEIGHT OF THE INTERIOR NODES IS EQUAL TO THE SUM OF THE TWO
SUBORDINATED NODES. THE CONTROL CHARACTER NYA GETS THE WEIGHT 0.
STRUCTURE OF THE CODE TABLE:
• ALL NODES ARE SORTED ACCORDING TO THEIR WEIGHT IN ASCENDING ORDER.
THE NYA NODE ALWAYS PROVIDES THE LOWEST NODE IN THE HIERARCHY.

9. SIBLING PROPERTY:
• ANY PAIR OF NODES THAT ARE NEIGHBORS IN THE HIERARCHY REFER TO THE
SAME PARENT NODE (NODE 2N-1 AND 2N -> 1 & 2, 3 & 4, 5 & 6, ETC.).
• THE PARENT NODE IS ALWAYS ORDERED AT A HIGHER LEVEL IN THE HIERARCHY
BECAUSE ITS WEIGHT RESULTS FROM THE SUM OF THE SUBORDINATED NODES.
THIS IS DENOTED AS SIBLING PROPERTY.
ROOT
• THE NODE WITH THE HIGHEST WEIGHT IS PLACED AT THE HIGHEST LEVEL IN THE
HIERARCHY AND FORMS THE CURRENT ROOT OF THE TREE.
INITIAL TREE
• THE INITIAL TREE ONLY CONTAINS THE NYA NODE WITH THE WEIGHT 0.

10. • DATA FORMAT FOR UNCODED SYMBOLS
• IN THE SIMPLEST CASE SYMBOLS WHICH ARE NOT CONTAINED IN THE CODE TREE ARE
ENCODED LINEARLY (E.G. WITH 8 BIT FROM A SET OF 256 SYMBOLS).
• BETTER COMPRESSION EFFICIENCY COULD BE ACHIEVED, IF THE DATA FORMAT WILL BE
ADAPTED TO THE NUMBER OF REMAINING SYMBOLS.
• A VARIETY OF OPTIONS ARE AVAILABLE TO OPTIMIZE THE CODE LENGTH. A SUITABLE
PROCEDURE WILL BE PRESENTED GUARANTEEING FULL UTILIZATION OF THE RANGE OF
VALUES.
• MAXIMUM NUMBER OF NODES
• ASSUMING A SET OF N SYMBOLS THE TOTAL NUMBER OF NODES RESULTS IN 2N-1 (LEAF AND
INTERIOR NODES). ACCORDINGLY THE MAXIMUM NUMBER OF NODES IS 511 IF THE STANDARD
UNIT IS THE BYTE.
• BLOCK
• ALL NODES OF IDENTICAL WEIGHT WILL BE SUMMARIZED LOGICALLY TO A BLOCK.
NODES WHICH ARE PART OF A BLOCK ARE ALWAYS NEIGHBORING IN THE CODE TABLE
REPRESENTING THE CODE TREE.

11. ARITHMETIC CODING:
• HUFFMAN CODING ACHIEVES THE SHANNON VALUE ONLY IF THE
CHARACTER/SYMBOL PROBABILITIES ARE ALL INTEGER POWERS OF ½.
• THE CODEWORDS PRODUCED USING ARITHMETIC CODING ALWAYS ACHIEVE THE
SHANNON VALUE. ARITHMETIC CODING, HOWEVER, IS MORE COMPLICATED
THAN HUFFMAN CODING.
• CONSIDER THE TRANSMISSION OF A MESSAGE COMPRISING A STRING OF
CHARACTERS WITH PROBABILITIES OF:
• E=0.3, N=0.3, T=0.2, W=0.1, . =0.1

12. • AT THE END OF EACH CHARACTER STRING MAKING UP A MESSAGE, A KNOWN
CHARACTER IS SENT WHICH, IN THIS EXAMPLE, IS A PERIOD..
• WHEN THIS IS DECODED AT THE RECEIVING SIDE, THE DECODER INTERPRETS
THIS AS THE END OF THE STRING/MESSAGE.
• ARITHMETIC CODING YIELDS A SINGLE CODEWORD FOR EACH CHARACTERS.
THE FIRST STEP IS DIVIDE THE NUMERIC RANGE FROM 0 TO 1 INTO A NUMBER OF
DIFFERENT CHARACTERS PRESENT IN THE MESSAGE TO BE SENT-INCLUDING THE
TERMINATION CHARACTER-AND THE SIZE OF EACH SEGMENT BY THE
PROBABILITY OF THE RELATED CHARACTER.
• THERE ARE ONLY FIVE DIFFERENT CHARACTERS, THERE ARE FIVE SEGMENTS,
THE WIDTH OF EACH SEGMENT BEING DETERMINED BY THE PROBABILITY OF
THE RELATED CHARACTER.

13. • FOR EXAMPLE, THE CHARACTER E HAS A PROBABILITY OF 0.3 AND HENCE IS ASSIGNED THE
RANGE FROM 0.3 TO 0.6, AND SO ON.
• HOWEVER AN ASSIGNMENT IS THE RAGE, SAY, 0.8 TO 0.9, MEANS THAT THE PROBABILITY IN THE
CUMULATIVE RANGE IS FROM 0.8 TO 0.8999…
• ONCE THIS HAS BEEN DONE , WE ARE READY TO START THE ENCODING PROCESS. WE ASSUME
THE CHARACTER STRING/MESSAGE TO BE ENCODED IS THE SINGLE WORD WENT.
• THE FIRST CHARACTER TO BE ENCODED W IS IN THE RANGE 0.8 TO 0.9.THE FINAL (NUMERIC)
CODEWORD IS A NUMBER IN THE RANGE 0.8 TO0.8999...
• SINCE EACH SUBSEQUENT CHARACTER IN THE STRING SUBDIVIDES THE RANGE 0.8 TO 0.9 INTO
PROGRESSIVELY SMALLER SEGMENTS EACH DETERMINED BY THE PROBABILITIES OF THE
CHARACTERS IN THE STRING.
EXAMPLE CHARACTER SET AND THEIR PROBABILITIES:
• CHARACTER SET |--------E----------|--------N--------|-----------T--------|-------W-------|-.--| CUMULATIVE
0 0.3 0.6 0.8 0.9 PROBABILITIES

14. • THE NEXT CHARACTER IN THE STRING IS E AND HENCE ITS RANGE (0.8 TO 0.83) IS
AGAIN SUBDIVIDED INTO FIVE SEGMENTS.
• WITH THE NEW ASSIGNMENTS, THEREFORE, THE CHARACTER E HAS A RANGE
FROM 0.8TO 0.809(0.809+0.3*0.03), AND SO ON.
• THIS PROCEDURE CONTINUES UNTIL THE TERMINATION CHARACTER (.) IS
ENCODED. AT THIS POINT, THE SEGMENT RANGE OF (.) FROM 0.81602 TO 0.8162
AND HENCE THE CODEWORD FOR THE COMPLETE STRING IS ANY NUMBER
WITHIN THE RANGE:
• IN THE STATIC MODE, THE DECODER KNOWS THE SET OF CHARACTERS THAT ARE
PRESENT IN THE ENCODED MESSAGES IT RECEIVES AS WELL AS THE SEGMENT TO
WHICH EACH CHARACTER HAS BEEN ASSIGNED AND ITS RELATED RANGE.

15. • FOR EXAMPLE, IF THE RECEIVED CODEWORD IS DAY, 0.8161, THEN THE
DECODER CAN READILY DETERMINE FROM THIS THAT THE FIRST CHARACTER IS
W SINCE IT IS THE ONLY CHARACTER WITHIN THE RANGE 0.8 TO 0.9.
• THE SECOND CHARACTER MUST BE E SINCE 0.8161 IS WITHIN THE RANGE 0.8
TO 0.83.THIS PROCEDURE THEM REPEATS UNTIL IT DECODES THE KNOWN
TERMINATION CHARACTER (.)
• AT WHICH POINT IT HAS RECREATED THE, SAY, ASCII STRING RELATING TO
WENT AND PASSES THIS ON FOR PROCESSING.

16. STATISTICAL AND ADAPTIVE
• THE ADAPTIVE CHARACTER WORD LENGTH ALGORITHM THAT HAS BEEN
PROPOSED CAN BE USED TO ENHANCE THE EFFECTIVENESS OF STATISTICAL
DATA COMPRESSION TECHNIQUES.
• IN THIS SECTION, WE GIVE A BRIEF DESCRIPTION TO ONE OF THE MOST
SOPHISTICATED AND EFFICIENT STATISTICAL DATA COMPRESSION TECHNIQUES,
NAMELY, HUFFMAN CODING

17. HUFFMAN CODING:
• HUFFMAN CODING IS A SOPHISTICATED AND EFFICIENT STATISTICAL LOSSLESS DATA
COMPRESSION TECHNIQUE, IN WHICH THE CHARACTERS IN A DATA FILE ARE
CONVERTED TO A BINARY CODE, WHERE THE MOST COMMON CHARACTERS IN THE
FILE HAVE THE SHORTEST BINARY CODES, AND THE LEAST COMMON HAVE THE
LONGEST.
• IN HUFFMAN CODING, FOR EXAMPLE, THE TEXT FILE TO BE COMPRESSED IS READ TO
CALCULATE THE FREQUENCIES FOR ALL THE CHARACTERS USED IN THE TEXT,
INCLUDING ALL LETTERS, DIGITS, AND
• PUNCTUATION.
• THE FIRST STEP IN BUILDING A HUFFMAN CODE IS TO ORDER THE CHARACTERS FROM
HIGHEST TO LOWEST FREQUENCY OF OCCURRENCE.

18. • THE SECOND STEP IS TO CONSTRUCT A BINARY TREE STRUCTURE. THIS BEGINS
BY SELECTING THE TWO LEAST-FREQUENT CHARACTERS: LOGICALLY GROUP
THEM TOGETHER, AND THEIR FREQUENCIES ARE ADDED.
• THEN, SELECT THE TWO ELEMENTS THAT HAVE THE LOWEST FREQUENCIES,
REGARDING THE PREVIOUS COMBINATION AS A SINGLE ELEMENT, GROUP THEM
TOGETHER AND ADD THEIR FREQUENCIES.
• CONTINUE IN THE SAME WAY TO SELECT THE TWO ELEMENTS WITH THE LOWEST
FREQUENCY, GROUP THEM TOGETHER, AND ADD THEIR FREQUENCIES, UNTIL
ONE ELEMENT REMAINS

19. • IN STATISTICAL COMPRESSION ALGORITHMS, IT IS REQUIRED FIRST TO FIND THE
PROBABILITIES (I.E., FREQUENCIES, F) FOR ALL CHARACTERS USED IN THE DATA FILE. FOR
TEXT FILES, CHARACTERS INCLUDE ALL LETTERS, DIGITS, AND PUNCTUATIONS.
• THESE CHARACTERS ARE TO BE ORDERED IN THE SEQUENCE (S) FROM HIGHEST
FREQUENCIES TO THE LOWEST.
• THE SECOND STEP IS TO FIND THE EQUIVALENT BINARY CODE FOR EACH CHARACTER
ACCORDING TO THE STATISTICAL DATA COMPRESSION TECHNIQUE THAT HAS BEEN USED,
E.G., HUFFMAN CODING, WHERE THE MOST COMMON CHARACTERS IN THE FILE HAVE THE
SHORTEST BINARY CODES, AND THE LEAST COMMON HAVE THE LONGEST.
• THEN THESE BINARY CODES ARE USED TO CONVERT ALL CHARACTERS IN THE DATA FILE TO
A BINARY CODE. TYPICALLY, IN ALL STATISTICALALGORITHMS, A CHARACTER WORD
LENGTH OF 8 BITS IS USED, WHERE THE EQUIVALENT DECIMAL VALUE FOR EACH 8 BITS (0–
255) IS CALCULATED AND CONVERTED TO A CHARACTER THAT IS WRITTEN TO THE OUTPUT
FILE.

20. • THE SIZE OF THE ORIGINAL DATA FILE (S DATA) IN BYTES IS GIVEN BY
• N
• S DATA= FI
• I=1
• ENTROPY IS A MEASURE OF THE INFORMATION CONTENT OF THE DATA FILE AND THE
SMALLEST NUMBER OF BITS PER CHARACTER NEEDED, ON AVERAGE, TO REPRESENT THE
COMPRESSED FILE. THEREFORE, THE ENTROPY OF A COMPLETE DATA FILE WOULD BE THE SUM
OF THE INDIVIDUAL CHARACTERS’ ENTROPY. THE ENTROPY OF A CHARACTER IS REPRESENTED
AS THE NEGATIVE LOGARITHM OF ITS FREQUENCY AND EXPRESSED USING BASE TWO. WHERE
THE FREQUENCY OF EACH CHARACTER OF THE ALPHABET IS CONSTANT, THE ENTROPY IS
CALCULATED AS
• N
• E = - - I LOG 2( I)
• I=1
•

21. • THE ACW ALGORITHM WILLACHIEVE A CODING RATE OF (2.94 × 8)/B OR A
COMPRESSION RATIO OF B/2.94, WHERE B IS THE OPTIMUM CHARACTER WORD
LENGTH. THUS, FOR 9 AND 10 CHARACTER WORD LENGTHS, FOR EXAMPLE, THE
CODING RATES ARE 2.16 (C = 3.06) AND 2.35 (C = 3.40), RESPECTIVELY. THE
OPTIMUM VALUE OF B DEPENDS ON A NUMBER OF FACTORS:
• 1. THE SIZE AND THE TYPE OF THE DATA FILE,
• 2. THE CHARACTERS FREQUENCIES WITHIN THE DATA FILE,
• 3. THE DISTRIBUTION OF CHARACTERS WITHIN THE FILE, AND
• 4. THE EQUIVALENT BINARY CODE USED FOR EACH CHARACTER.

22. • IT IS CLEAR THAT USING THE ACW ALGORITHM, THE COMPRESSION RATIO, FOR
ANY STATISTICAL COMPRESSION ALGORITHM, INCREASES LINEARLY WITH B,
AND IT CAN BE 4 TIMES ITS ORIGINAL VALUE IF B IS EQUAL TO 32 BITS.
• IN ORDER TO DECOMPRESS THE FILE EFFICIENTLY, THE NUMBER OF
POSSIBILITIES AND THEIR DECREMENTALLY ORDERED ACTUAL DECIMAL VALUES
MUST BE STORED INTO THE HEADER OF THE COMPRESSED FILE.
• THE MAXIMUM ADDED OVERHEAD OCCURS WHEN THE VALUE OF B IS 32 BITS
AND ALL THE POSSIBILITIES (0–255) ARE USED.
• THE MAXIMUM OVERHEAD IS EQUAL TO 2563 B. THIS CAN BE REDUCED TO
2050 B, IF THESE VALUES ARE STORED IN HEXADECIMAL NOTATION.

23. DICTIONARY MODELING
• THE COMPRESSION ALGORITHMS WE STUDIED SO FAR USE A STATISTICAL MODEL TO
ENCODE SINGLE SYMBOL.
• 1. COMPRESSION: ENCODE SYMBOLS INTO BIT STRINGS THAT USE FEWER BITS.
• DICTIONARY-BASED ALGORITHMS DO NOT ENCODE SINGLE SYMBOLS AS
VARIABLE-LENGTH BIT STRINGS; THEY ENCODE VARIABLE-LENGTH STRINGS OF
SYMBOLS AS SINGLE TOKENS
• 2. THE TOKENS FORM AN INDEX INTO A PHRASE DICTIONARY
• 3. IF THE TOKENS ARE SMALLER THAN THE PHRASES THEY REPLACE,
COMPRESSION OCCURS.
• DICTIONARY-BASED COMPRESSION IS EASIER TO UNDERSTAND BECAUSE IT
USES A STRATEGY THAT PROGRAMMERS ARE FAMILIAR WITH-> USING INDEXES INTO
DATABASES TO RETRIEVE INFORMATION FROM LARGE AMOUNTS OF STORAGE.

24. • 4. TELEPHONE NUMBERS
• 5. POSTAL CODES
•
• DICTIONARY-BASED COMPRESSION: EXAMPLE
• CONSIDER THE RANDOM HOUSE DICTIONARY OF THE ENGLISH LANGUAGE,
SECOND EDITION, UNABRIDGED. USING THIS DICTIONARY, THE STRING:
• A GOOD EXAMPLE OF HOW DICTIONARY BASED COMPRESSION WORKS CAN
BE CODED AS:
• 1/1 822/3 674/4 1343/60 928/75 550/32 173/46 421/2

25. • CODING:
• USES THE DICTIONARYAS A SIMPLE LOOKUP TABLE
• EACH WORD IS CODED AS X/Y, WHERE, X GIVES THE PAGE IN THE
DICTIONARY AND Y GIVES THE NUMBER OF THE WORD ON THAT PAGE.
• THE DICTIONARY HAS 2,200 PAGES WITH LESS THAN 256 ENTRIES PER PAGE:
THEREFORE X REQUIRES 12 BITS AND Y REQUIRES 8 BITS, I.E., 20 BITS PER
WORD (2.5 BYTES PER WORD).
• USING ASCII CODING THE ABOVE STRING REQUIRES 48 BYTES,
• WHEREAS OUR ENCODING REQUIRES ONLY 20 (<-2.5 * 8) BYTES: 50%
COMPRESSION.
• ADAPTIVE DICTIONARY-BASED COMPRESSION
• BUILD THE DICTIONARYADAPTIVELY,

26. • NECESSARY WHEN THE SOURCE DATA IS NOT PLAIN TEXT, SAY AUDIO OR VIDEO DATA.
• IS BETTER TAILORED TO THE SPECIFIC SOURCE.
• ORIGINAL METHODS DUE TO ZIP AND LEMPEL IN 1977 (LZ77) AND 1978 (LZ78). TERRY
WELCH IMPROVED THE SCHEME IN 1984 (CALLED LZW COMPRESSION). IT IS USED IN,
UNIX COMPRESS,
• AND , GIF.
• LZ77: A SLIDING WINDOW TECHNIQUE IN WHICH THE DICTIONARY CONSISTS OF A SET
OF FIXED LENGTH PHRASES FOUND IN A WINDOW INTO THE PREVIOUSLY PROCESSED TEXT
• LZ78: INSTEAD OF USING FIXED-LENGTH PHRASES FROM A WINDOW INTO THE TEXT, IT
BUILDS PHRASES UP ONE SYMBOL AT A TIME, ADDING A NEW SYMBOL TO AN EXISTING PHRASE
WHEN A
• MATCH OCCURS.

27. • LZW ALGORITHM
• PRELIMINARIES:
• A DICTIONARY THAT IS INDEXED BY CODES IS USED.
• THE DICTIONARY IS ASSUMED TO BE INITIALIZED WITH 256 ENTRIES
(INDEXED WITH ASCII CODES THROUGH 255) REPRESENTING THE ASCII TABLE.
• THE COMPRESSION ALGORITHM ASSUMES THAT THE OUTPUT IS EITHER A
FILE OR A COMMUNICATION CHANNEL. THE INPUT BEING A FILE OR BUFFER.
• CONVERSELY, THE DECOMPRESSION ALGORITHM ASSUMES THAT THE
INPUT IS A FILE OR A COMMUNICATION CHANNEL AND THE OUTPUT IS A FILE OR
A BUFFER.

28. • LZW ALGORITHM
• LZW COMPRESSION:
• SET W = NIL
• LOOP
• READ A CHARACTER K
• IF WK EXISTS IN THE DICTIONARY
• W = WK
• ELSE
• OUTPUT THE CODE FOR W
• ADD WK TO THE DICTIONARY
• W = K
• END LOOP

29. THANKYOU