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Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
Huffman Student
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Huffman Student

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  • 1. CSE 326 Huffman coding Richard Anderson
  • 2. Coding theory <ul><li>Conversion, Encryption, Compression </li></ul><ul><li>Binary coding </li></ul><ul><li>Variable length coding </li></ul>Code examples 000,001,010,011,100,101 1,01,001,0001,00001,000001 00,010,011,100,11,101 A B C D E F
  • 3. Decode the following 11010010010101011 100100101010 Ambiguous Prefix code E 0 T 11 N 100 I 1010 S 1011 E 0 T 10 N 100 I 0111 S 1010
  • 4. Prefix code <ul><li>No prefix of a codeword is a codeword </li></ul><ul><li>Uniquely decodable </li></ul>A 00 1 00 B 010 01 10 C 011 001 11 D 100 0001 0001 E 11 00001 11000 F 101 000001 101
  • 5. Prefix codes and binary trees <ul><li>Tree representation of prefix codes </li></ul>A 00 B 010 C 0110 D 0111 E 10 F 11
  • 6. Construct the tree for the following code E 0 T 11 N 100 I 1010 S 1011
  • 7. Minimum length code <ul><li>Average cost </li></ul><ul><li>Average leaf depth </li></ul><ul><li>Huffman tree – tree with minimum weighted path length </li></ul><ul><li>C(T) – weighted path length </li></ul>
  • 8. Compute average leaf depth A 00 1/4 B 010 1/8 C 0110 1/16 D 0111 1/16 E 1 1/2
  • 9. Huffman code algorithm <ul><li>Derivation </li></ul><ul><ul><li>Two rarest items will have the longest codewords </li></ul></ul><ul><ul><li>Codewords for rarest items differ only in the last bit </li></ul></ul><ul><li>Idea: suppose the weights are with and the smallest weights </li></ul><ul><ul><li>Start with an optimal code for and </li></ul></ul><ul><ul><li>Extend the codeword for to get codewords for and </li></ul></ul>
  • 10. Huffman code H = new Heap() for each w i T = new Tree(w i ) H.Insert(T) while H.Size() &gt; 1 T 1 = H.DeleteMin() T 2 = H.DeleteMin() T 3 = Merge(T 1 , T 2 ) H.Insert(T 3 )
  • 11. Example: Weights 4, 5, 6, 7, 11, 14, 21 4 5 6 7 11 14 21
  • 12. Draw a Huffman tree for the following data values and show internal weights: 3, 5, 9, 14, 16, 35
  • 13. Correctness proof <ul><li>The most amazing induction proof </li></ul><ul><li>Induction on the number of code words </li></ul><ul><li>The Huffman algorithm finds an optimal code for n = 1 </li></ul><ul><li>Suppose that the Huffman algorithm finds an optimal code for codes size n, now consider a code of size n + 1 . . . </li></ul>
  • 14. Key lemma <ul><li>Given a tree T, we can find a tree T’, with the two minimum cost leaves as siblings, and C(T’) &lt;= C(T) </li></ul>
  • 15. Modify the following tree to reduce the WPL 29 10 19 6 4 13 6 10 3 5 5
  • 16. Finish the induction proof <ul><li>T – Tree constructed by Huffman </li></ul><ul><li>X – Any code tree </li></ul><ul><li>Show C(T) &lt;= C(X) </li></ul><ul><li>T’ and X’ – Trees from the lemma </li></ul><ul><li>C(T’) = C(T) </li></ul><ul><li>C(X’) &lt;= C(X) </li></ul><ul><li>T’’ and X’’ – Trees with minimum cost leaves x and y removed </li></ul>
  • 17. X : Any tree, X’: – modified, X’’ : Two smallest leaves removed <ul><li>C(X’’) = C(X’) – x – y </li></ul><ul><li>C(T’’) = C(T’) – x – y </li></ul><ul><li>C(T’’) &lt;= C(X’’) </li></ul><ul><li>C(T) = C(T’) = C(T’’) + x + y &lt;= C(X’’) + x + y = C(X’) &lt;= C(X) </li></ul>

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