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

P.1.2.3 then there are ei ∈ Ebool, such that fi = fei and C(ei) = 2 n−1 + d. (IH)
P.1.2.4 thus f = fe, where e := xn ∗ e0 + xn ∗ e1 and κ(n) = 2 · 2 n−1 + 2d + 4.
c○: Michael Kohlhase 149
The next proof is quite a lot of work, so we will first sketch the overall structure of the proof,
before we look into the details. The main idea is to estimate a cleverly chosen quantity from
above and below, to get an inequality between the lower and upper bounds (the quantity itself is
irrelevant except to make the proof work).
A Lower Bound for the Cost of BF with n Variables
Theorem 270 κ ∈ Ω( 2 n
log 2 (n) )
Proof: Sketch (counting again!)
P.1 the cost of a function is based on the cost of expressions.
P.2 consider the set En of expressions with n variables of cost no more than κ(n).
P.3 find an upper and lower bound for #(En): (Φ(n) ≤ #(En) ≤ Ψ(κ(n)))
P.4 in particular: Φ(n) ≤ Ψ(κ(n))
P.5 solving for κ(n) yields κ(n) ≥ Ξ(n) so κ ∈ Ω( 2 n
log 2 (n) )
We will expand P.3 and P.5 in the next slides
c○: Michael Kohlhase 150
A Lower Bound For κ(n)-Cost Expressions
Definition 271 En := {e ∈ Ebool | e has n variables and C(e) ≤ κ(n)}
Lemma 272 #(En) ≥ #(B n → B)
Proof:
P.1 For all fn ∈ B n → B we have C(fn) ≤ κ(n)
P.2 C(fn) = min({C(e) | fe = fn}) choose efn with C(efn ) = C(fn)
P.3 all distinct: if eg ≡ eh, then feg = feh and thus g = h.
Corollary 273 #(En) ≥ 2 2n
Proof: consider the n dimensional truth tables
P.1 2n entries that can be either 0 or 1, so 22n possibilities
so #(B n → B) = 2 2n
c○: Michael Kohlhase 151
An Upper Bound For κ(n)-cost Expressions
P.2 Idea: Estimate the number of Ebool strings that can be formed at a given cost by looking at
the length and alphabet size.
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