General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

Example 210 The Morse Code in the table above induces a character code µ: R → {., −}.
Codes on Strings
c○: Michael Kohlhase 121
Definition 211 A function c ′ : A ∗ → B ∗ is called a code on strings or short string code if
c ′ is an injective function.
Theorem 212 ( ) There are character codes whose extensions are not string codes.
Proof: we give an example
P.1 Let A = {a, b, c}, B = {0, 1}, c(a) = 0, c(b) = 1, and c(c) = 01.
P.2 The function c is injective, hence it is a character code.
P.3 But its extension c ′ is not injective as c ′ (ab) = 01 = c ′ (c).
Question: When is the extension of a character code a string code?(so we can encode strings)
Definition 213 A (character) code c: A → B + is a prefix code iff none of the codewords
is a proper prefix to an other codeword, i.e.,
∀x, y ∈ A.x = y ⇒ (c(x) ⊳ c(y) ∧ c(y) ⊳ c(x))
c○: Michael Kohlhase 122
We will answer the question above by proving one of the central results of elementary coding
theory: prefix codes induce string codes. This plays back the infinite task of checking that a string
code is injective to a finite task (checking whether a character code is a prefix code).
Prefix Codes induce Codes on Strings
Theorem 214 The extension c ′ : A ∗ → B ∗ of a prefix code c: A → B + is a string code.
Proof: We will prove this theorem via induction over the string length n
P.1 We show that c ′ is injective (decodable) on strings of length n ∈ N.
P.1.1 n = 0 (base case):
If |s| = 0 then c ′ (ɛ) = ɛ, hence c ′ is injective.
P.1.2 n = 1 (another):
If |s| = 1 then c ′ = c thus injective, as c is char. code.
P.1.3 Induction step (n to n + 1):
P.1.3.1 Let a = a0, . . ., an, And we only know c ′ (a) = c(a0), . . ., c(an).
P.1.3.2 It is easy to find c(a0) in c ′ (a): It is the prefix of c ′ (a) that is in c(A). This is uniquely
determined, since c is a prefix code. If there were two distinct ones, one would have to be a
prefix of the other, which contradicts our assumption that c is a prefix code.
P.1.3.3 If we remove c(a0) from c(a), we only have to decode c(a1), . . ., c(an), which we can do by
inductive hypothesis.
P.2 Thus we have considered all the cases, and proven the assertion.
c○: Michael Kohlhase 123
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