1. Huffman Compression

2. Compress + Expand

3. Compress Expand A Tabulate frequency counts Build trie
Build codebook (write trie)
Encode input
Build codebook (read trie)
Decode compressed message
Compress: compute freq,
buildTrie() A
Trie
Compress: writeTrie()
Expand: readTrie()
Compress: write length
Expand: read length
Compress: encode
Expand: decode
Compressed file format:
Codebook
Input length
Compressed message

4. Compute frequency counts
Eerie eyes seen near lake.
Contain letters:
‘E’ ‘e’ ‘r’ ‘i’ ‘y’ ‘s’ ‘n’ ‘a’ ‘r’ ‘l’ ‘k’ ‘.’ space
Generate frequency counts:
E: 1, e: 8, r: 2, i: 1, y: 1, s: 2, n: 2, a: 2, l: 1, k: 1, .: 1, space: 4

5. Build trie (buildTrie())
Merge nodes based on frequency until there is only one left
Special case: only one node
Priority queue
delMin: Removes and returns a smallest key on this priority queue.
insert: adds a new key to the priority queue.
Node{
char ch
int freq
Node left, right }

6. Build trie
private static Node buildTrie(int[] freq) { MinPQ<Node> pq = new MinPQ<Node>(); for (char ch = 0; ch < R; ++ch if (freq[ch] > 0) pq.insert(new Node(c, freq[c], null, null)); while (pq.size() > 1) { Node x=pq.delMin(); Node y=pq.delMin(); Node parent = new Node(‘\0’, x.freq + y.freq, x, y); pq.insert(parent); } return pq.delMin();

7. E 1 i y l k . r 2 s n a sp 4 e 8
private static Node buildTrie(int[] freq) {
MinPQ<Node> pq = new MinPQ<Node>();
for (char ch = 0; ch < R; ++ch
if (freq[ch] > 0)
pq.insert(new Node(c, freq[c], null, null));
while (pq.size() > 1) {
Node x=pq.delMin();
Node y=pq.delMin();
Node parent = new Node(‘\0’, x.freq + y.freq, x, y);
pq.insert(parent); }
return pq.delMin();

8. y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2 sp 4 e 8
private static Node buildTrie(int[] freq) {
MinPQ<Node> pq = new MinPQ<Node>();
for (char ch = 0; ch < R; ++ch
if (freq[ch] > 0)
pq.insert(new Node(c, freq[c], null, null));
while (pq.size() > 1) {
Node x=pq.delMin();
Node y=pq.delMin();
Node parent = new Node(‘\0’, x.freq + y.freq, x, y);
pq.insert(parent); }
return pq.delMin(); 2 E 1 i 1

9. An Introduction to Huffman Coding
March 21, 2000 y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2 2 sp 4 e 8 E 1 i 1
private static Node buildTrie(int[] freq) {
MinPQ<Node> pq = new MinPQ<Node>();
for (char ch = 0; ch < R; ++ch
if (freq[ch] > 0)
pq.insert(new Node(c, freq[c], null, null));
while (pq.size() > 1) {
Node x=pq.delMin();
Node y=pq.delMin();
Node parent = new Node(‘\0’, x.freq + y.freq, x, y);
pq.insert(parent); }
return pq.delMin();
Mike Scott

10. An Introduction to Huffman Coding
March 21, 2000 k 1 . 1 r 2 s 2 n 2 a 2 2 sp 4 e 8
private static Node buildTrie(int[] freq) {
MinPQ<Node> pq = new MinPQ<Node>();
for (char ch = 0; ch < R; ++ch
if (freq[ch] > 0)
pq.insert(new Node(c, freq[c], null, null));
while (pq.size() > 1) {
Node x=pq.delMin();
Node y=pq.delMin();
Node parent = new Node(‘\0’, x.freq + y.freq, x, y);
pq.insert(parent); }
return pq.delMin(); E 1 i 1 2 y 1 l 1
Mike Scott

11. An Introduction to Huffman Coding
March 21, 2000 2 k 1 . 1 r 2 s 2 n 2 a 2 2 sp 4 e 8 y 1 l 1 E 1 i 1
private static Node buildTrie(int[] freq) {
MinPQ<Node> pq = new MinPQ<Node>();
for (char ch = 0; ch < R; ++ch
if (freq[ch] > 0)
pq.insert(new Node(c, freq[c], null, null));
while (pq.size() > 1) {
Node x=pq.delMin();
Node y=pq.delMin();
Node parent = new Node(‘\0’, x.freq + y.freq, x, y);
pq.insert(parent); }
return pq.delMin();
Mike Scott

12. An Introduction to Huffman Coding
March 21, 2000 r 2 s 2 n 2 a 2 2 2 sp 4 e 8 y 1 l 1 E 1 i 1
private static Node buildTrie(int[] freq) {
MinPQ<Node> pq = new MinPQ<Node>();
for (char ch = 0; ch < R; ++ch
if (freq[ch] > 0)
pq.insert(new Node(c, freq[c], null, null));
while (pq.size() > 1) {
Node x=pq.delMin();
Node y=pq.delMin();
Node parent = new Node(‘\0’, x.freq + y.freq, x, y);
pq.insert(parent); }
return pq.delMin(); 2 k 1 . 1
Mike Scott

13. An Introduction to Huffman Coding
March 21, 2000 r 2 s 2 n 2 a 2 2 sp 4 e 8 2 2 k 1 . 1 E 1 i 1 y 1 l 1
private static Node buildTrie(int[] freq) {
MinPQ<Node> pq = new MinPQ<Node>();
for (char ch = 0; ch < R; ++ch
if (freq[ch] > 0)
pq.insert(new Node(c, freq[c], null, null));
while (pq.size() > 1) {
Node x=pq.delMin();
Node y=pq.delMin();
Node parent = new Node(‘\0’, x.freq + y.freq, x, y);
pq.insert(parent); }
return pq.delMin();
Mike Scott

14. An Introduction to Huffman Coding
March 21, 2000 n 2 a 2 2 sp 4 e 8 2 2 E 1 i 1 y 1 l 1 k 1 . 1 4 r 2 s 2
Mike Scott

15. An Introduction to Huffman Coding
March 21, 2000 4 4 4 2 sp 4 e 8 2 2 r 2 s 2 n 2 a 2 k 1 . 1 E 1 i 1 y 1 l 1
Mike Scott

16. An Introduction to Huffman Coding
March 21, 2000 4 4 4 e 8 2 2 r 2 s 2 n 2 a 2 E 1 i 1 y 1 l 1 6 2 sp 4 k 1 . 1
Mike Scott

17. An Introduction to Huffman Coding
March 21, 2000 4 4 6 4 e 8 2 sp 4 2 2 r 2 s 2 n 2 a 2 k 1 . 1 E 1 i 1 y 1 l 1
Mike Scott

18. An Introduction to Huffman Coding
March 21, 2000 4 6 e 8 2 2 2 sp 4 k 1 . 1 E 1 i 1 y 1 l 1 8 4 4 r 2 s 2 n 2 a 2
Mike Scott

19. An Introduction to Huffman Coding
March 21, 2000 4 6 e 8 8 2 2 2 sp 4 4 4 k 1 . 1 E 1 i 1 y 1 l 1 r 2 s 2 n 2 a 2
Mike Scott

20. An Introduction to Huffman Coding
March 21, 2000 8 e 8 4 4 10 r 2 s 2 n 2 a 2 4 6 2 2 2 sp 4 E 1 i 1 y 1 l 1 k 1 . 1
Mike Scott

21. An Introduction to Huffman Coding
March 21, 2000 8 e 8 10 4 4 4 6 2 2 r 2 s 2 n 2 a 2 2 sp 4 E 1 i 1 y 1 l 1 k 1 . 1
Mike Scott

22. An Introduction to Huffman Coding
March 21, 2000 10 16 4 6 2 2 e 8 8 2 sp 4 E 1 i 1 y 1 l 1 k 1 . 1 4 4 r 2 s 2 n 2 a 2
Mike Scott

23. An Introduction to Huffman Coding
March 21, 2000 10 16 4 6 e 8 8 2 2 2 sp 4 4 4 E 1 i 1 y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2
Mike Scott

24. An Introduction to Huffman Coding
March 21, 2000 26 16 10 4 e 8 8 6 2 2 2 sp 4 4 4 E 1 i 1 y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2
Mike Scott

25. An Introduction to Huffman Coding
March 21, 2000 26 16 10 4 e 8 8 6 2 2 2 sp 4 4 4 E 1 i 1 y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2
Mike Scott

26. Generate codebook (buildCode())
Perform a traversal of the tree to obtain new code words
Going left is a 0 going right is a 1
code word is only completed when a leaf node is reached
private static void buildCode(String[] st, Node x, String s) {
if (!x.isLeaf()) {
buildCode(st, x.left, s + '0');
buildCode(st, x.right, s + '1'); }
else {
st[x.ch] = s;
st is the code book, in which st[ch] stores the code for character ch, for example:
st[‘E’]=“0000“
st[‘s’]=“1101”
String s records the path to the current node x, for example, “left, left, left, right” is encoded as “0001”

27. An Introduction to Huffman Coding
March 21, 2000
Generate codebook
Char Code E i y l k
space 011
e 10 r s n a E 1 i sp 4 e 8 2 y l k . r s n a 6 10 16 26
Mike Scott

28. Encode message Eerie eyes seen near lake.
Char Code E i y l k
space 011
e 10 r s n a
Eerie eyes seen near lake.
Codebook
Input length
Compressed message

29. Expand Compress A Tabulate frequency counts Build trie
Build codebook (write trie)
Encode input
Build codebook (read trie)
Decode compressed message
Compress: compute freq,
buildTrie() A
Trie
Compress: writeTrie()
Expand: readTrie()
Compress: write length
Expand: read length
Compress: encode
Expand: decode
Compressed file format:
Codebook
Input length
Compressed message

30. Build codebook (readTrie())
readTrie / writeTrie
How to deserialize (read) trie ? Boolean flag + data
If flag == 1 (leaf), then the next 8bits represent a character
If flag == 0 (internal node), then recursively desterilize left and right children
Codebook
Input length
Compressed message

31. Build codebook readTrie / writeTriesp e - y l k . r s n a
readTrie / writeTrie
00001E(8bits)1i(8bits)01y(8bits) 1l(8bits)...
// write bitstring-encoded trie to standard output
private static void writeTrie(Node x) {
if (x.isLeaf()) {
BinaryStdOut.write(true);
BinaryStdOut.write(x.ch, 8);
return; }
BinaryStdOut.write(false);
writeTrie(x.left);
writeTrie(x.right);

32. Build codebook readTrie / writeTriesp e - y l k . r s n a
readTrie / writeTrie
00001E(8bits)1i(8bits)01y(8bits) 1l(8bits)...
private static Node readTrie() {
boolean isLeaf = BinaryStdIn.readBoolean();
if (isLeaf) {
return new Node(BinaryStdIn.readChar(), -1, null, null); }
else {
return new Node('\0', -1, readTrie(), readTrie());

33. Decode message E i sp e - y l k . r s n a
Scan the bit string while traversal the tree: 0 go left, 1 go right
end at leaf node
Eerie eyes seen near lake.