COMPS-2-9. [3] For each of the following pairs of

To determine the relationships between the pairs of functions, we will analyze each pair based on the definitions of Big O, Omega, and Theta notation.

Analysis of Each Pair: 1. (a)

f(n) = \log n^2

and

g(n) = \log n + 5

-

f(n) = 2 \log n

and

g(n)

grows as

\log n

. - Thus,

f(n) \in O(g(n))

is false. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is false. 2. (b)

f(n) = \sqrt{n}

and

g(n) = \log n^2

-

g(n) = 2 \log n

grows slower than

f(n)

. - Thus,

f(n) \in O(g(n))

is false. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is false. 3. (c)

f(n) = \log^2 n

and

g(n) = \log n

-

f(n)

grows faster than

g(n)

. - Thus,

f(n) \in O(g(n))

is false. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is false. 4. (d)

f(n) = n

and

g(n) = \log^2 n

-

f(n)

grows much faster than

g(n)

. - Thus,

f(n) \in O(g(n))

is false. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is false. 5. (e)

f(n) = n \log n + n

and

g(n) = \log n

-

f(n)

grows faster than

g(n)

. - Thus,

f(n) \in O(g(n))

is false. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is false. 6. (f)

f(n) = 10

and

g(n) = \log 10

- Both are constants. - Thus,

f(n) \in O(g(n))

is true. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is true. 7. (g)

f(n) = 2^n

and

g(n) = 10n^2

-

f(n)

grows exponentially while

g(n)

is polynomial. - Thus,

f(n) \in O(g(n))

is false. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is false. 8. (h)

f(n) = 2^n

and

g(n) = 3n

- Again,

f(n)

grows exponentially while

g(n)

is linear. - Thus,

f(n) \in O(g(n))

is false. -

f(n) \in \Omega(g(n))

is true. -

f(n) \in \Theta(g(n))

is false. Summary of Relationships: - (a)

O: \text{false}, \Omega: \text{true}, \Theta: \text{false}

- (b)

O: \text{false}, \Omega: \text{true}, \Theta: \text{false}

- (c)

O: \text{false}, \Omega: \text{true}, \Theta: \text{false}

- (d)

O: \text{false}, \Omega: \text{true}, \Theta: \text{false}

- (e)

O: \text{false}, \Omega: \text{true}, \Theta: \text{false}

- (f)

O: \text{true}, \Omega: \text{true}, \Theta: \text{true}

- (g)

O: \text{false}, \Omega: \text{true}, \Theta: \text{false}

- (h)

O: \text{false}, \Omega: \text{true}, \Theta: \text{false}

This concludes the analysis of the function pairs.

For the given pairs of functions, let's analyze each one based on the definitions of Big O, Omega, and Theta notation. (a)

f(n) = 3n^2, g(n) = n^2

- Here,

f(n)

is a constant multiple of

g(n)

. Thus,

f(n)

is in

\Theta(g(n))

.

Key Concept

Big Theta notation indicates that two functions grow at the same rate.

Explanation

Since

f(n)

is a constant multiple of

g(n)

, they are asymptotically equivalent.

(b)

f(n) = 2n^4 - 3n^2 + 7, g(n) = n^5

- The leading term of

f(n)

is

2n^4

, which grows slower than

n^5

. Thus,

f(n)

is in

O(g(n))

.

Key Concept

Big O notation indicates an upper bound on the growth rate of a function.

Explanation

Since

f(n)

grows slower than

g(n)

, it is bounded above by

g(n)

.

(c)

f(n) = \log n, g(n) = \log n + \frac{1}{n}

- As

n

grows large,

\frac{1}{n}

becomes negligible, so

f(n)

is in

\Theta(g(n))

.

Key Concept

Big Theta notation indicates that two functions grow at the same rate.

Explanation

The additional term

\frac{1}{n}

does not affect the asymptotic growth of

g(n)

.

(d)

f(n) = 2^k \log n, g(n) = n^k

- For large

n

,

n^k

grows faster than

2^k \log n

. Thus,

f(n)

is in

O(g(n))

.

Key Concept

Big O notation indicates an upper bound on the growth rate of a function.

Explanation

The polynomial term

n^k

dominates the logarithmic term

2^k \log n

as

n

increases.

(e)

f(n) = 2^n, g(n) = 2^{2n}

- Here,

g(n)

grows exponentially faster than

f(n)

. Thus,

f(n)

is in

O(g(n))

.

Key Concept

Big O notation indicates an upper bound on the growth rate of a function.

Explanation

Since

g(n) = 2^{2n}

grows much faster than

f(n) = 2^n

,

f(n)

is bounded above by

g(n)

.

Final Answers: - (a)

D

- (b)

A

- (c)

C

- (d)

D

-

To prove that

n^2 = O(2^n)

, we need to show that there exist constants C > 0 and

n_0

such that for all

n \geq n_0

, the inequality

n^2 \leq C \cdot 2^n

holds.

Let's choose

C = 1

and find

n_0

.

For

n \geq 5

, we can check the inequality:

-

n^2 = 25

when

n = 5

and

2^n = 32

when

n = 5

.

-

25 \leq 32

holds true.

Now, let's check for larger values of

n

:

- For

n = 6

,

n^2 = 36

and

2^n = 64

.

- For

n = 7

,

n^2 = 49

and

2^n = 128

.

- For

n = 8

,

n^2 = 64

and

2^n = 256

.

As

n

increases,

2^n

grows much faster than

n^2

.

Therefore, we can conclude that

n^2 = O(2^n)

.

To analyze how much slower each algorithm gets when the input size is doubled or increased by one, we will evaluate the running times for both scenarios.

(a) Doubling the Input Size - For \( n^2 \): - New running time:

(2n)^2 = 4n^2

- Slower by a factor of 4. - For \( n^3 \): - New running time:

(2n)^3 = 8n^3

- Slower by a factor of 8. - For \( 100n^2 \): - New running time:

100(2n)^2 = 400n^2

- Slower by a factor of 4. - For \( n \log n \): - New running time:

2n \log(2n) = 2n(\log n + 1) = 2n \log n + 2n

- Slower by a factor of approximately 2 (dominant term). - For \( 2^n \): - New running time:

2^{2n} = (2^n)^2

- Slower by a factor of

2^n

. (b) Increasing the Input Size by One - For \( n^2 \): - New running time:

(n+1)^2 = n^2 + 2n + 1

- Slower by a factor of approximately 1 (dominant term is

2n

). - For \( n^3 \): - New running time:

(n+1)^3 = n^3 + 3n^2 + 3n + 1

- Slower by a factor of approximately 1 (dominant term is

3n^2

). - For \( 100n^2 \): - New running time:

100(n+1)^2 = 100(n^2 + 2n + 1) = 100n^2 + 200n + 100

- Slower by a factor of approximately 1 (dominant term is

200n

). - For \( n \log n \): - New running time:

(n+1) \log(n+1)

- Slower by a factor of approximately 1 (dominant term is

\log n + 1

). - For \( 2^n \): - New running time:

2^{n+1} = 2 \cdot 2^n

- Slower by a factor of 2.

In conclusion, the algorithms slow down differently based on their running times when the input size is doubled or increased by one.

To find a positive constant

C

for each pair of functions

f(n)

and

g(n)

such that

f(n) \leq C \cdot g(n)

for all n > 1 , we will analyze each pair:

(a) For

f(n) = n^2 + n + 1

and

g(n) = 2n^3

: - As

n

grows,

n^2 + n + 1

is dominated by

n^2

. - We can find

C

such that

n^2 + n + 1 \leq C \cdot 2n^3

. - For n > 1 , we can choose

C = \frac{1}{2}

. - Thus,

f(n) \leq \frac{1}{2} \cdot 2n^3

holds true. (b) For

f(n) = n\sqrt{n} + n^2

and

g(n) = n^2

: - Here,

n\sqrt{n}

is

n^{3/2}

, which grows slower than

n^2

. - We can find

C

such that

n\sqrt{n} + n^2 \leq C \cdot n^2

. - For n > 1 , we can choose

C = 2

. - Thus,

f(n) \leq 2n^2

holds true. (c) For

f(n) = n^2 - n + 1

and

g(n) = \frac{n^2}{2}

: - We need

n^2 - n + 1 \leq C \cdot \frac{n^2}{2}

. - For n > 1 , we can choose

C = 2

. - Thus,

f(n) \leq 2 \cdot \frac{n^2}{2}

holds true.

Therefore, the constants are:

(a)

C = \frac{1}{2}

, (b)

C = 2

, (c)

C = 2

To prove that if

f_1(n) = \Omega(g_1(n))

and

f_2(n) = \Omega(g_2(n))

, then

f_1(n) + f_2(n) = \Omega(g_1(n) + g_2(n))

, we start by using the definitions of the asymptotic notations.

By definition,

f_1(n) = \Omega(g_1(n))

means there exist constants C_1 > 0 and

n_1

such that for all

n \geq n_1

,

f_1(n) \geq C_1 \cdot g_1(n)

.

Similarly,

f_2(n) = \Omega(g_2(n))

implies there exist constants C_2 > 0 and

n_2

such that for all

n \geq n_2

,

f_2(n) \geq C_2 \cdot g_2(n)

.

Let

n_0 = \max(n_1, n_2)

.

For all

n \geq n_0

, we have:

-

f_1(n) \geq C_1 \cdot g_1(n)

-

f_2(n) \geq C_2 \cdot g_2(n)

Therefore, adding these inequalities gives:

f_1(n) + f_2(n) \geq C_1 \cdot g_1(n) + C_2 \cdot g_2(n)

This implies that

f_1(n) + f_2(n)

is bounded below by a linear combination of

g_1(n)

and

g_2(n)

.

Thus, we can choose

C = \min(C_1, C_2)

and conclude that

f_1(n) + f_2(n) = \Omega(g_1(n) + g_2(n))

.

Therefore, the statement is proven.

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