Question

7. Prove the following assertions by using the definitions of the notations in- volved, or disprove them by giving a specific counterexample. a. If t(n) e (g(n)), then g(n) E S2(t(n I. Θ(gg(n))-e(g(n)), where α > 0. c. Θ(g(n))-: 0(g(n))n Ω (g(n)). d. For any two nonnegative functions t (n) and g(n) defined on the set of nonnegative integers, either t (n) e 0(g(n)), or t (n) e Ω(g(n)), or both

Answer 1

a.

If f(n) is O(g(n)) then
   there exist a constant c > 0, and a constant n₀ such that for all n ≥ n₀: f(n) ≤ c*g(n)

hence,
   there exist a constant c > 0, and a constant n₀ such that for all n ≥ n₀: g(n) ≥ (1/c)*f(n)
     note that since c > 0, then the constant (1/c) > 0

hence,
   there exist a constant k > 0, namely k = (1/c), and a constant n₀ such that for all n ≥ n₀: g(n) ≥ k*f(n)

which is the definition of g(n) = Ω (f(n)).

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b:

First of all we recall the definition of the big-Θ-notation:

Definition : f(n) ∈ O(g(n)) if ∀n ≥ n₀, ∃0 < c₁ ≤ c₂ such that c₁g(n) ≤ f(n) ≤ c₂g(n)

If we state now the definition for both cases:

c₁g(n) ≤ f(n) ≤ c₂g(n) for Θ(g(n)) -------------------- (1)

c₁αg(n) ≤ f(n) ≤ c₂αg(n) for Θ(αg(n)) -------------------- (2)

As α > 0, redefining c₁ and c₂ as c˜_1/2 = αc_1/2 the we can rewrite (2) as:

c˜₁g(n) ≤ f(n) ≤ c˜₂g(n)

which complies with the inequality (1)

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c:

We are required to prove that Θ(g(n)) = O(g(n)) ∩ Ω(g(n)). The proof is in two parts.

Part-1: First we prove that Θ(g(n)) ⊆ O(g(n)) ∩ Ω(g(n)):

• We start by considering an arbitrary function f(n) that is in Θ(g(n))

f(n) ∈ Θ(g(n))   -------------------- (1)

• From the definition of Θ we know that this means:

∃c₁, c₂, n₀ : ∀n > n₀ : c₁ g(n) ≤ f(n) ≤ c₂ g(n)    -------------------- (2)

• If we elide the first inequality, and substitute c for c2, this gives us

∃c, n₀ : ∀n > n₀ : f(n) ≤ c g(n)    -------------------- (3)

which, from the definition of O, tells us

f(n) ∈ O(g(n))      -------------------- (4)

• Similarly, if we elide the second inequality, and substitute c for c₁, this gives us

∃c, n₀ : ∀n > n₀ : c g(n) ≤ f(n) -------------------- (5)

which, from the definition of Ω, tells us

f(n) ∈ Ω(g(n)) -------------------- (6)

• From the definition of ∩ for sets, we know that line 4 and line 6 together imply that

f(n) ∈ O(g(n)) ∩ Ω(g(n)) -------------------- (7)

• Recall the assumption (line 1) that f(n) was an arbitrary function in Θ(g(n)). So we have, using line 1, the rule of universal generalization, and line 7

∀f : f(n) ∈ Θ(g(n)) =⇒ f(n) ∈ O(g(n)) ∩ Ω(g(n)) -------------------- (8)

This implies that Θ(g(n)) ⊆ O(g(n)) ∩ Ω(g(n)) -------------------- (9)

Part-2: Now you would need to do the same for the second (superset) part of the proof i.e., Θ(g(n)) ⊇ O(g(n)) ∩ Ω(g(n))

   f(n) ∈ O(g(n))    [assumption] -------------------- (10)

f(n) ∈ Ω(g(n))    [assumption] -------------------- (11)

∃c₁, n₁ : ∀n > n₁ : f(n) ≥ c₁ g(n) [def. of Ω] -------------------- (12)

∃c₂, n₀ : ∀n > n₀ : f(n) ≤ c₂ g(n)   [def. of O] -------------------- (13)

∃c₁, c₂, n₀ : ∀n > max(n₀, n₁) : c₁ g(n) ≤ f(n) ≤ c₂ g(n) [combining (12) and (13) above] -------------------- (14)

f(n) ∈ Θ(g(n)) [def. of Θ and (10)] -------------------- (15)

Θ(g(n)) ⊇ O(g(n)) ∩ Ω(g(n)) [∀f : f ∈ O ∧ f ∈ Ω ⇒ f ∈ Θ] -------------------- (16)

Combining these subset (9) and superset (16) results, we get Θ(g(n)) = O(g(n)) ∩ Ω(g(n))

[ because A ⊆ B ∧ A ⊇ B ⇒ A = B ]