exercise in the previous class sunny rain 45 15 12 28 Y1 X sunny rain 43 57 Y2 Q1: Compute P(X=Y1) and P(X=Y2): P(X=Y1) = 0.73 and P(X=Y2) = 0 Q2:Compute I(X; Y1) and I(X; Y2): H(X) =ℋ 0.57 =0.986 bit to compute H(X|Y1), determine some probabilities; X sunny rain 0.75 0.25 0.30 0.70 Y1 P(X|Y1) P(Y1=sunny)= 0.6, P(Y1=rain)=0.4 H(X|Y1=sunny) = ℋ 0.75 =0.811 H(X|Y1=rain) = ℋ 0.30 =0.881 H(X|Y1) = 0.6×0.811+0.4×0.881=0.839 I(X; Y1) = H(X) – H(X|Y1) = 0.986 – 0.839= 0.147 bit

exercise in the previous class (cnt’d) sunny rain 45 15 12 28 Y1 X sunny rain 43 57 Y2 Q2:Compute I(X; Y1) and I(X; Y2): H(X) =ℋ 0.57 =0.986 bit to compute H(X|Y2), determine some probabilities; X sunny rain 1 Y2 P(X|Y2) P(Y2=sunny)= 0.43, P(Y2=rain)=0.57 H(X|Y2=sunny) = ℋ 0 =0 H(X|Y2=rain) = ℋ 1 =0 H(X|Y2) = 0.43×0+0.57×0=0 I(X; Y2) = H(X) – H(X|Y2) = 0.986 – 0= 0.986 bit Q3: Which is the better forecasting? Y2 gives more information.

chapter 2: compact representation of information

the purpose of chapter 2 We learn how to encode symbols from information source. source coding data compression the purpose of source encoding: to give representations which are good for communication to discard (捨てる) redundancy (冗長性) We want a source coding scheme which gives ... as precise (正確) encoding as possible as compact encoding as possible source encoder 0101101

plan of the chapter basic properties needed for source coding uniquely decodable immediately decodable Huffman code construction of Huffman code extensions of Huffman code theoretical limit of the “compression” related topics today

words and terms Meanwhile, we consider symbol-by-symbol encodings only. M...the set of symbols generated by an information source. For each symbol in M, associate a sequence (系列) over {0, 1}. codewords (符号語): sequences associated to symbols in M code (符号): the set of codewords alphabet: {0, 1} in this case... binary code M sunny cloudy rainy C 00 010 101 three codewords; 00, 010 and 101 code C = {00, 010 and 101} 011 is NOT a codeword, for example

encoding and decoding encode ... to determine the codeword for a given symbol decode ... to determine the symbol for a given codeword sunny cloudy rainy 00 010 101 encode decode encode = 符号化 decode = 復号（化） NO separation symbols between codewords; 010 00 101 101 ... NG, 01000101101 ... OK Why? {0, 1 and “space”} ... the alphabet have three symbols, not two

uniquely decodable codes A code must be uniquely decodable (一意復号可能). Different symbol sequences are encoded to different 0-1 sequences. uniquely decodable  codewords are all different, but the converse (逆) does not hold in general. a1 a2 a3 a4 C1 00 10 01 11 C2 011 111 C3 C4 with the code C3... a1 a3 a1 a4 a2 0110 yes yes no no

more than uniqueness a1 a2 a3 a4 C1 00 10 01 11 C2 011 111 011 111 consider a scenario of using C2... a1, a4, a4, a1 is encoded to 01111110. The 0-1 sequence is transmitted by 1 bit/sec. When does the receiver find that the first symbol is a1? seven seconds later, the receiver obtains 0111111: if 0 comes next, then 0 - 111 - 111 - 0  a1, a4, a4, a1 if 1 comes next, then 01 - 111 - 111  a2, a4, a4 We cannot finalize the first symbol even in the seven seconds later.  buffer to save data, latency (遅延) of decoding...

immediately decodable codes A code must be uniquely decodable, and if possible, it should be immediately decodable (瞬時復号可能). Decoding is possible without looking ahead the sequence. If you find a codeword pattern, then decode it immediately.  important property from an engineering viewpoint. formally writing... If 𝒔∈ 0,1 ∗ is written as 𝑐 1 𝒔 𝟏 =𝒔 with 𝑐 1 ∈𝐶 and 𝒔 1 ∈ 0,1 ∗ , then there is no 𝑐 2 ∈𝐶 and 𝒔 2 ∈ 0,1 ∗ such that 𝑐 2 𝒔 𝟐 =𝒔. c1 s1 c2 s2 ≠

prefix condition If a code is NOT immediately decodable, then there is 𝒔∈ 0,1 ∗ such that 𝒔= 𝑐 1 𝒔 𝟏 = 𝑐 2 𝒔 𝟐 with different 𝑐 1 and 𝑐 2 . c1 s1 c2 s2 = the codeword 𝑐 1 is a prefix (語頭) of 𝑐 2 (c1 is the same as the beginning part of c2) Lemma: A code C is immediately decodable if and only if no codeword in C is a prefix of other codewords. (prefix condition, 語頭条件) a1 a2 a3 a4 C2 01 011 111 “0” is a prefix of “01” and “011” “01” is a prefix of “011”

break: prefix condition and user interface The prefix condition is important in engineering design. bad example: strokes for character writing on Palm PDA graffiti (ver. 2) some needs of two strokes prefix condition violated “– –”, and “=”, “– 1” and “+” graffiti (ver. 1) basically one stroke only

how to achieve the prefix condition easy ways to construct codes with the prefix condition: let all codewords have the same length put a special pattern at the end of each codeword C = {011, 1011, 01011, 10011} ... “comma code” ... too straightforward select codewords by using a tree structure (code tree) for binary codes, we use binary trees for k-ary codes, we use trees with degree k a code tree with degree 3

construction of codes (k-ary case) how to construct a k-ary code with M codewords construct a k-ary tree T with M leaf nodes for each branch (枝) of T, assign a label in {0, ..., k – 1} sibling (兄弟) branches cannot have the same label for each of leaf nodes of T, traverse T from the root to the leaf, with concatenating (連接する) labels on branches  the obtained sequence is the codeword of the node

example construct a binary code with four codewords Step 1 Step 2 1 1 00 01 10 11 the constructed code is {00, 01, 10, 11}

example (cnt’d) other constructions; we can choose different trees, different labeling... 1 1 C1={0, 10, 110, 111} C2={0, 11, 101, 100} 1 C3={01, 000, 1011, 1010} The prefix condition is always guaranteed.  Immediately decodable codes are constructed.

the “best” among immediately decodable codes 1 1 C1={0, 10, 110, 111} C3={01, 000, 1011, 1010} C1 seems to give more compact representation than C3. codeword length = [1, 2, 3, 3] Can we construct more compact immediately decodable codes? codeword length = [1, 1, 1, 1]? codeword length = [1, 2, 2, 3]? codeword length = [2, 2, 2, 3]? ? What is the criteria (基準) ?

Kraft’s inequality Theorem: A) If a k-ary code {c1, ..., cM} with |ci| = li is immediately decodable, then 𝑘 − 𝑙 1 +…+ 𝑘 − 𝑙 𝑀 ≤1 (Kraft’s inequality) holds. B) If 𝑘 − 𝑙 1 +…+ 𝑘 − 𝑙 𝑀 ≤1, then we can construct a k-ary immediately decodable code {c1, ..., cM} with |ci| = li. proof omitted in this class ... use results of graph theory [trivia] The result is given in the Master’s thesis of L. Kraft.

back to the examples Can we construct more compact immediately decodable codes? codeword length = [1, 2, 2, 3]? … 2 −1 + 2 −2 + 2 −2 + 2 −3 = 9 8 >1 We cannot construct an immediately decodable code. codeword length = [2, 2, 2, 3]? … 2 −2 + 2 −2 + 2 −2 + 2 −3 = 7 8 <1 We can construct an immediately decodable code, by simply constructing a code tree....

to the next step basic properties needed for source coding uniquely decodable immediately decodable Huffman code construction of Huffman code extensions of Huffman code theoretical limit of the “compression” related topics today

the measure of efficiency We want to construct a good source coding scheme. easy to use ... immediately decodable efficient ... what is the efficiency? We try to minimize... the expected length of a codeword for representing one symbol: 𝑖=1 𝑀 𝑝 𝑖 𝑙 𝑖 symbol a1 a2 : aM probability p1 p2 pM codeword c1 c2 cM length l1 l2 lM average codeword length

computing the average codeword length symbol a1 a2 a3 a4 probability 0.4 0.3 0.2 0.1 C1 10 110 111 C2 C3 00 01 11 C1: 0.4×1+ 0.3×2+ 0.2×3+ 0.1×3 = 1.9 C2: 0.4×3+ 0.3×3+ 0.2×2+ 0.1×1 = 2.6 C3: 0.4×2+ 0.3×2+ 0.2×2+ 0.1×2 = 2.0 It is expected that... C1 gives the most compact representation in typical cases.

Huffman code Huffman algorithm gives a clever way to construct a code with small average codeword length. prepare isolated M nodes, each attached with a probability of a symbol (node = size-one tree) repeat the following operation until all trees are joined to one select two trees T1 and T2 having the smallest probabilities join T1 and T2 by introducing a new parent node the sum of probabilities of T1 and T2 is given to the new tree David Huffman 1925-1999

“merger of small companies” example “merger of small companies” 0.05 D 0.1 C 0.25 B 0.6 A 0.05 D 0.1 C 0.15 0.25 B 0.6 A 0.05 D 0.1 C 0.15 0.25 B 0.4 0.6 A 0.05 D 0.1 C 0.15 0.25 B 0.4 0.6 A 1.0 1

exercise A B C D E prob. 0.2 0.1 0.3 codewords compare the average length with the equal-length codes...

exercise A B C D E F prob. 0.3 0.2 0.1 codewords compare the average length with the equal-length codes...

different construction, same efficiency We may have multiple options on the code construction: several nodes have the same small probabilities labels can be assigned differently to branches Different option results in a different Huffman code, but... the average length does not depend on the chosen option. 0.4 a1 0.2 a2 a3 0.1 a4 a5 0.4 a1 0.2 a2 a3 0.1 a4 a5

summary of today’s class basic properties needed for source coding uniquely decodable immediately decodable Huffman code construction of Huffman code extensions of Huffman code theoretical limit of the “compression” related topics today

exercise Construct a binary Huffman code for the information source given in the table. Compute the average codeword length of the constructed code. Can you construct a 4-ary Huffman code for the source? A B C D E F G H prob. 0.363 0.174 0.143 0.098 0.087 0.069 0.045 0.021