channel: 41 session: cs2420 HUFFMAN CODING
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1 channel: 41 session: cs2420 HUFFMAN CODING cs2420 Introduction to Algorithms and Data Structures Fall
2 channel: 41 session: cs2420 administrivia 2
3 channel: 41 session: cs2420 -assignment 11 due tonight -assignment 12 out -assignment 13 (final project) requires a partner 3
4 last time 4
5 min-max heap 5
6
7 1 min
8 1 min max
9 1 min max min
10 1 min max min max 6
11 -a min-max heap further extends the heap order property -for any node E at even depth, E is the minimum element in its subtree -for any node O at odd depth, O is the maximum element in its subtree -the root is considered to be at even depth (zero) 7
12 today 8
13 -file compression -data compression -data decompression -Huffman s algorithm 9
14 file compression 10
15 -why do we care about file compression? -reducing traffic on networks and the internet -reducing disk space requirements -various media formats -MP3 -MPEG4 -JPEG - -YouTube, Netflix, GoogleMaps, etc. would not be possible without compression 11
16 exercise -suppose we the following string stored in a text file: ddddddddddddabc -how many bytes of disk space does it take to store these 15 characters using ASCII? 12
17 binary -each bit represents power of on off off on off off on on -sum up all the bits that are on = 147 -how can we convert the other way? 13
18 what is the binary representation of the number 39? A) B) C) D)
19 There are 10 types of people in the world: those who understand binary, and those who don t. 15
20 -all data in a computer is stored in binary -a bit is a binary digit -a byte is an 8-bit value -common unit of data in a computer -bytes
21 how many different values can 4 bits hold? A) 7 B) 8 C) 15 D) 16 E) 31 F) 32 17
22 ASCII -each character corresponds to one byte -remember, a byte is just an 8-bit number! (0-255) -for example: = 32 = (blank space) = 59 = ; = 65 = A = 66 = B 18
23 -a simple file containing the text Hello is stored on the disk as: the text editor knows to treat each byte as an ASCII value -in reality, they are just bytes 19
24 exercise -suppose we the following string stored in a text file: ddddddddddddabc -how many bytes of disk space does it take to store these 15 characters using ASCII? -is there any way to represent this file in fewer bytes? -hint: take advantage of repeated characters 20
25 data compression 21
26 -allow number of bits for each character to vary, instead of using the full 8 for every character -represent common characters with fewer bits, represent rare characters like q and z with more bits -with ASCII we know each char is 8 bits -how do we reproduce (decompress) original file if characters are of varying lengths? 22
27 -for example: - e may be the two-bit value 11 - a may be the two-bit value 00 - q may be the four-bit value z may be the four-bit value what characters does the following file contain? is there any way to know? -we must include the secret decoder ring with the compressed file 23
28 binary trie -no, I did not misspell it! -a binary trie is a binary tree in which a left branch represents a 0 and a right branch represents a 1 c d a: 000 -the path to a node represents its encoding a b b: 001 c: 01 d: 1 24
29 ddddddddddddabc d takes 15 bytes (120 bits) in ASCII c a: 000 b: is a b c: 01 d: 1 less than 3 bytes (20 bits) why is the d near the top of the tree? 25
30 data decompression 26
31 -read the bits of the compressed file one at a time on 0 go left, on 1 go right d -when a leaf node is reached, print char contained in node -start back at the root for the next bit a b c a: 000 b: 001 c: 01 d: 1 27
32 -what does the following bitstring encode? d -how many bytes is this encoded in? what if it was in ASCII? a b c a: 000 b: 001 c: 01 d: 1 28
33 what string do these bits encode? A) low B) wow C) wool D) were o l r w H d e 29
34 (re)building the compression tree 30
35 -the compressed file now contains non-ascii, varying length bytes -decompression needs the same tree in order to decode -must add header information that describes the tree -this header should not be compressed! -we will get there in a bit 31
36 building the compression tree 32
37 hello world -count the frequency of each different character h = 1 e = 1 l = 3 o = 2 = 1 w = 1 r = 1 d = 1 -construct a binary trie with highest frequency characters near the top, lowest near the bottom 33
38 start with a separate tree for each character each tree is a single root node, holding the character and its frequency :1 h:1 d:1 e:1 r:1 w:1 o:2 l:3 hello world 34
39 merge the two lowest weight trees together into one new tree make a new parent node with their combined weight; smaller node on the left, larger on the right lots of ties here! h, e, w, r, d, and need a tie-breaker... more on this later... :1 h:1 d:1 e:1 r:1 w:1 o:2 l:3 hello world 35
40 remove and h trees from the list add new merged tree to the list note: non-leaf nodes don t have a character, only a weight 2 :1 h:1 d:1 e:1 r:1 w:1 o:2 l:3 hello world 36
41 continue process of merging two smallest trees which is next? 2 2 :1 h:1 d:1 e:1 r:1 w:1 o:2 l:3 hello world 37
42 continue process of merging two smallest trees which is next? :1 h:1 d:1 e:1 r:1 w:1 o:2 l:3 hello world 38
43 continue process of merging two smallest trees which is next? :1 h:1 d:1 e:1 r:1 w:1 o:2 l:3 hello world 39
44 continue process of merging two smallest trees which is next? o:2 2 :1 h:1 d:1 e:1 r:1 w:1 l:3 hello world 40
45 continue process of merging two smallest trees which is next? 7 l: o:2 2 :1 h:1 d:1 e:1 r:1 w:1 hello world 41
46 done! o:2 2 l:3 4 r:1 w:1 2 2 :1 h:1 d:1 e:1 hello world 42
47 new character encoding h: 1101 e: l : 10 o: :1100 w: 011 r : 010 o:2 2 l:3 4 d: 1110 r:1 w:1 2 2 :1 h:1 d:1 e:1 hello world
48 Huffman s algorithm 44
49 -idea is to encode most frequent characters with fewest bits -use a binary trie -common characters will be near the top of the tree -uncommon near the bottom -every file has a different frequency of characters, and therefore a different compression tree 45
50 1. count occurrences of each character in a string 2. for each character, create a leaf node to store the character and count (ie. weight) 3. place each leaf node into a priority queue 4. construct the binary trie 5. write header with binary trie information 6. compress string using character codes 46
51 1. count occurrences of each character in a string 2. for each character, create a leaf node to store the character and count (ie. weight) 3. place each leaf node into a priority queue 4. construct the binary trie 5. write header with binary trie information 6. compress string using character codes 46
52 I heart data. 47
53 I heart data. character frequency I 1 2 h 1 e 1 a 3 r 1 t 2 d
54 1. count occurrences of each character in a string 2. for each character, create a leaf node to store the character and count (ie. weight) 3. place each leaf node into a priority queue 4. construct the binary trie 5. write header with binary trie information 6. compress string using character codes 49
55 I heart data. character frequency I 1 2 h 1 I:1 :2 h:1 e:1 e 1 a 3 r 1 a:3 r:1 t:2 d:1.:1 t 2 d
56 -but, let s think about what happens when we write our compressed codes -files must be a whole number of bytes -but, the length of our compressed string will not necessarily be a factor of 8 -what do we do? 51
57 -to solve this, we add a special EOF leaf node EOF:1 -then, if we encounter this during decompression, we know we are done -and we ignore the remaining bits 52
58 I heart data. character frequency I 1 2 h 1 I:1 :2 h:1 e:1 e 1 a 3 a:3 r:1 t:2 d:1 r 1 t 2 d 1. 1 EOF 1.:1 EOF:1 53
59 1. count occurrences of each character in a string 2. for each character, create a leaf node to store the character and count (ie. weight) 3. place each leaf node into a priority queue 4. construct the binary trie 5. write header with binary trie information 6. compress string using character codes 54
60 -how do we determine the highest priority? -what happens if two nodes have the same priority? -need a tie-breaker! -use ASCII value for character to break tie -lowest values have highest priority 55
61 I heart data. I:1 :2 h:1 e:1 a:3 r:1 t:2 d:1.:1 EOF:1 PQ: 56
62 I heart data. :2 a:3 t:2 I:1 h:1 e:1 r:1 d:1.:1 EOF:1 PQ: 57
63 I heart data. character frequency ASCII I h 1 72 e a 3 97 r t d EOF
64 I heart data. :2 a:3 t:2 I:1 h:1 e:1 r:1 d:1.:1 EOF: PQ: 59
65 I heart data. :2 a:3 t:2 I:1 h:1 e:1 r:1 d:1.:1 EOF: PQ: 60
66 I heart data. :2 a:3 t:2 I:1 h:1 e:1 r:1 d:1.:1 EOF: PQ: 61
67 I heart data. :2 a:3 t:2 I:1 h:1 e:1 r:1 d:1.: PQ: EOF:1 62
68 I heart data. :2 a:3 t:2 I:1 h:1 e:1 r:1 d: PQ: EOF:1.:1 63
69 I heart data. :2 a:3 t:2 h:1 e:1 r:1 d: PQ: EOF:1.:1 I:1 64
70 I heart data. :2 a:3 t:2 h:1 e:1 r: PQ: EOF:1.:1 I:1 d:1 65
71 I heart data. :2 a:3 t:2 h:1 r: PQ: EOF:1.:1 I:1 d:1 e:1 66
72 I heart data. :2 a:3 t:2 r:1 114 PQ: EOF:1.:1 I:1 d:1 e:1 h:1 67
73 I heart data. :2 a:3 t:2 PQ: EOF:1.:1 I:1 d:1 e:1 h:1 r:1 68
74 I heart data. a:3 :2 t: PQ: EOF:1.:1 I:1 d:1 e:1 h:1 r:1 69
75 I heart data. a:3 t:2 116 PQ: EOF:1.:1 I:1 d:1 e:1 h:1 r:1 :2 70
76 I heart data. a:3 PQ: EOF:1.:1 I:1 d:1 e:1 h:1 r:1 :2 t:2 71
77 I heart data. PQ: EOF:1.:1 I:1 d:1 e:1 h:1 r:1 :2 t:2 a:3 72
78 1. count occurrences of each character in a string 2. for each character, create a leaf node to store the character and count (ie. weight) 3. place each leaf node into a priority queue 4. construct the binary trie 5. write header with binary trie information 6. compress string using character codes 73
79 -merge the two lowest weight trees together -make a new parent node with their combined weight -reinsert new tree back into the queue -but, what about ties? -when trees have more than one node, break the tie with the ASCII value of the leftmost character 74
80 I heart data. PQ: EOF:1.:1 I:1 d:1 e:1 h:1 r:1 :2 t:2 a:3 where are the two lowest weight trees? 75
81 I heart data. PQ: EOF:1.:1 I:1 d:1 e:1 h:1 r:1 :2 t:2 a:3 where are the two lowest weight trees? 76
82 I heart data. PQ: I:1 d:1 e:1 h:1 r:1 :2 t:2 a:3 2 EOF:1.:1 77
83 I heart data. PQ: I:1 d:1 e:1 h:1 r:1 :2 t:2 a:3 2 EOF:1.:1 where do we insert this new tree? 78
84 I heart data. PQ: I:1 d:1 e:1 h:1 r:1 :2 t:2 a: EOF:1 0.:1 where do we insert this new tree? 79
85 I heart data. PQ: I:1 d:1 e:1 h:1 r:1 2 :2 t:2 a:3 EOF:1.:1 80
86 I heart data. PQ: I:1 d:1 e:1 h:1 r:1 2 :2 t:2 a:3 EOF:1.:1 AND repeat 81
87 I heart data. PQ: e:1 h:1 r:1 2 :2 t:2 a:3 2 EOF:1.:1 I:1 d:1 82
88 I heart data. PQ: e:1 h:1 r:1 2 :2 2 t:2 a:3 EOF:1.:1 I:1 d:1 83
89 I heart data. PQ: r:1 2 :2 2 t:2 a:3 EOF:1.:1 I:1 d:1 2 e:1 h:1 84
90 I heart data. PQ: r:1 2 :2 2 2 t:2 a:3 EOF:1.:1 I:1 d:1 e:1 h:1 85
91 I heart data. PQ: :2 2 2 t:2 a:3 I:1 d:1 e:1 h:1 r:1 2 EOF:1.:1 86
92 I heart data. PQ: :2 2 2 t:2 a:3 I:1 d:1 e:1 h:1 3 r:1 2 EOF:1.:1 87
93 I heart data. PQ: :2 2 2 t:2 a:3 3 I:1 d:1 e:1 h:1 r:1 2 EOF:1.:1 88
94 I heart data. PQ: 2 t:2 a:3 3 e:1 h:1 r:1 2 EOF:1.:1 :2 2 I:1 d:1 89
95 I heart data. PQ: 2 t:2 a:3 3 e:1 h:1 r:1 2 EOF:1.:1 4 :2 2 I:1 d:1 90
96 I heart data. PQ: 2 t:2 a:3 3 4 e:1 h:1 r:1 2 :2 2 EOF:1.:1 I:1 d:1 91
97 I heart data. PQ: a:3 3 4 r:1 2 :2 2 EOF:1.:1 I:1 d:1 4 2 t:2 e:1 h:1 92
98 I heart data. PQ: a: r:1 2 :2 2 2 t:2 EOF:1.:1 I:1 d:1 e:1 h:1 93
99 I heart data. PQ: 4 4 :2 2 2 t:2 I:1 d:1 e:1 h:1 6 a:3 3 r:1 2 EOF:1.:1 94
100 I heart data. PQ: :2 2 2 t:2 a:3 3 I:1 d:1 e:1 h:1 r:1 2 EOF:1.:1 95
101 I heart data. PQ: 6 a:3 3 r:1 2 EOF:1.: :2 2 2 t:2 I:1 d:1 e:1 h:1 96
102 I heart data. PQ: 6 8 a: r:1 2 :2 2 2 t:2 EOF:1.:1 I:1 d:1 e:1 h:1 97
103 I heart data a: r:1 2 :2 2 2 t:2 EOF:1.:1 I:1 d:1 e:1 h:1 98
104 while PQ contains at least two nodes { dequeue two trees,l and R node I = new internal node I.left = L I.right = R I.weight = L.weight + R.weight } PQ.enqueue(I) //PQ contains only one node root = PQ.dequeue() 99
105 1. count occurrences of each character in a string 2. for each character, create a leaf node to store the character and count (ie. weight) 3. place each leaf node into a priority queue 4. construct the binary trie 5. write header with binary trie information 6. compress string using character codes 100
106 -the compressed files needs info to be able to rebuild the same compression tree -we call this the header -what information do we need to store to reconstruct the compression tree? 101
107 1. store character frequencies -for every character -for only non-zero characters must store pair 2. use a standardized letter frequency -use the same standard for compression and decompression 3. store the tree using a pre-order traversal -more complex, but smallest 102
108 -storing character frequencies -write each character, followed by the frequency -characters are 1 byte, integers are 4 bytes -so, each character will require 5 bytes to write -how do we mark the end of the header (and beginning of the compressed string)? -write out null and 0 -NOTE: we are going to write out our file as hex 103
109 hexadecimal -hexadecimal is the base-16 number system -we only have 10 digits (0-9), so to use a number base greater than 10 we need more symbols -in hex, we use the letters A through F -A represents the value ten -F represents the value fifteen 104
110 counting -in binary, we reset and add a digit at 1 -in decimal, we reset and add a digit at 9 -in hex, we reset and add a digit at F (ie. 15) 8 9 A B C D E F 10 (= sixteen) 11 (= seventeen) 105
111 -each decimal place represents a power of 16 1AF = (1 * 16 2 ) + (A * 16 1 ) + (F * 16 0 ) = =
112 -hex is useful because each hex digit corresponds to one half-byte (ie. 4 bits) -reading raw byte data is almost always done in hex -two hex digits makes up a single byte -bytes in hex: 1A FF D5 107
113 hex to binary -each hex digit is a specific 4-bit sequence 0 = = 0001 E = 1110 F = converting from hex to binary is as simple as representing each digit with its bit-sequence 12 EF = a single byte is two hex digits -the bytes in the above are 12 and EF 108
114 what is the hex value of these 8 bits? A) B2 B) A2 C) 12 D)
115 I heart data. character hex value frequency I h 68 1 e 65 1 a 61 3 r E t 74 2 d E 1 EOF
116 1. count occurrences of each character in a string 2. for each character, create a leaf node to store the character and count (ie. weight) 3. place each leaf node into a priority queue 4. construct the binary trie 5. write header with binary trie information 6. compress string using character codes 111
117 I heart data a: r:1 2 :2 2 2 t:2 EOF:1.:1 I:1 d:1 e:1 h:1 112
118 -want to create a look-up table to quickly convert from character to bit code -what data structure can we use for this? 113
119 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
120 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
121 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
122 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
123 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
124 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
125 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
126 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
127 I heart data. character hex value bit-code I h e a r now what? t d E 0111 EOF
128 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
129 I heart data. character hex value bit-code I h e a r are we done? t d E 0111 EOF
130 I heart data. character hex value bit-code I h e a r t d E 0111 EOF
131 I heart data. character hex value bit-code I h e a r now, convert to hex t d E 0111 EOF
132 I heart data. character hex value bit-code I h e a r t d now, convert to hex A9 B8 2F 29 E3 B0. 2E 0111 EOF
133 I heart data. character hex value bit-code I h e E A9 B8 2F 29 E3 B0 a r t d E 0111 EOF
134 I heart data. character hex value bit-code I h e a r t d E E A9 B8 2F 29 E3 B0 we just went from 13 bytes to 61 bytes. why should we care about this???? EOF
135 decompression 130
136 what steps do I need to take to decompress this file? E A9 B8 2F 29 E3 B0 131
137 next time 132
138 -reading -chapter 12 in book -homework -assignment 11 due tonight 133
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