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

Definition 274 Given a cost c let Λ(e) be the length of e considering variables as single
characters. We define
Lemma 275 σ(n) ≤ 5n for n > 0.
Proof: by induction on n
P.1.1 base case:
σ(c) := max({Λ(e) | e ∈ Ebool ∧ (C(e) ≤ c)})
The cost 1 expressions are of the form (v◦w) and (−v), where v and w are variables. So the
length is at most 5.
P.1.2 step case:
σ(n) = Λ((e1◦e2)) = Λ(e1) + Λ(e2) + 3, where C(e1) + C(e2) ≤ n − 1.
so σ(n) ≤ σ(i) + σ(j) + 3 ≤ 5 · C(e1) + 5 · C(e2) + 3 ≤ 5 · n − 1 + 5 = 5n
Corollary 276 max({Λ(e) | e ∈ En}) ≤ 5 · κ(n)
c○: Michael Kohlhase 152
An Upper Bound For κ(n)-cost Expressions
Idea: e ∈ En has at most n variables by definition.
Let An := {x1, . . ., xn, 0, 1, ∗, +, −, (, )}, then #(An) = n + 7
Corollary 277 En ⊆ 5κ(n)
i=0 An i and #(En) ≤ n+75κ(n)+1 −1
n+7
Proof Sketch: Note that the Aj are disjoint for distinct n, so
⎛
5κ(n)
#( ⎝ An i
⎞
5κ(n)
⎠) = #(An i 5κ(n)
) = #(An i 5κ(n)
) = n + 7 i = n + 75κ(n)+1 − 1
n + 7
i=0
Solving for κ(n)
n+75κ(n)+1 −1
n+7
≥ 2 2n
n + 7 5κ(n)+1 ≥ 2 2n
i=0
i=0
i=0
c○: Michael Kohlhase 153
(as n + 75κ(n)+1 ≥ n+75κ(n)+1−1 n+7 )
5κ(n) + 1 · log 2(n + 7) ≥ 2 n (as log a(x) = log b(x) · log a(b))
5κ(n) + 1 ≥
κ(n) ≥ 1/5 ·
2 n
log 2 (n+7)
κ(n) ∈ Ω( 2 n
log 2 (n) )
2 n
log2 (n+7) − 1
c○: Michael Kohlhase 154
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