Polynomial Calculation

1. Polynomial Calculation
(a) Runtime Complexity of calculatePoly1
The runtime complexity of the calculatePoly1 algorithm can be determined by analyzing the number of operations performed in each loop iteration.

In the while loop, the algorithm performs n iterations (where n is the length of the coefficient list ‘a’). In each iteration, it performs a multiplication and an addition operation. Therefore, the number of operations in the while loop can be expressed as 2n.

Outside the while loop, there is an additional loop that iterates n times. In each iteration, it calculates the power of x using the pow function. The pow function performs i multiplications, where i is the current iteration index. Therefore, the number of operations in this loop can be expressed as 1 + 2 + 3 + … + n, which is equivalent to (n * (n + 1) / 2).

Combining both loops, the total number of operations can be expressed as 2n + (n * (n + 1) / 2).

Thus, the tight bound for the runtime complexity of calculatePoly1 using summations is O(n^2).

(b) Space Complexity of calculatePoly1
The space complexity of the calculatePoly1 algorithm can be determined by analyzing the memory used to store variables and data structures.

In the algorithm, there are several variables used, such as ‘exp’, ‘p’, ‘sum’, and ‘i’. Additionally, there is a list ‘a’ that stores the coefficients.

The number of variables used remains constant regardless of the input size. Therefore, the space complexity for variables is O(1).

The coefficient list ‘a’ has a length of n, which means it requires O(n) memory.

Hence, combining both variables and data structures, the tight bound for the space complexity of calculatePoly1 is O(n).

(c) Improved Algorithm calculatePoly2
To improve the runtime complexity of calculatePoly1, we can use a more efficient algorithm called Horner’s method. This method allows us to calculate the polynomial value with a linear number of operations.

Here is the pseudocode for calculatePoly2 using Horner’s method:

Function calculatePoly2(x, a, n):
sum := a[n-1]
i := n – 2
while i >= 0 do
sum = sum * x + a[i]
i -= 1
return sum

In this algorithm, we start from the highest degree term (a[n-1]) and iteratively multiply it by x and add the next coefficient (a[i]) in reverse order. This way, we avoid calculating powers of x separately.

The runtime complexity of calculatePoly2 can be determined by analyzing the number of operations performed in each loop iteration.

The while loop iterates n-1 times since we start from index n-2 to 0. In each iteration, it performs a multiplication and an addition operation. Therefore, the number of operations in the while loop can be expressed as 2 * (n-1), which simplifies to 2n – 2.

Thus, the tight bound for the runtime complexity of calculatePoly2 is O(n).

The space complexity remains the same as calculatePoly1, which is O(n).

2. Algorithm Runtimes
(a) Best Case Runtime Θ(n) vs. Θ(n^2)
Yes, it is possible for an algorithm with a best case runtime of Θ(n) to take Θ(n^2) time on some inputs. The best case runtime only represents the lower bound of the algorithm’s performance. It means that in the best case scenario, the algorithm will run in linear time. However, there may be inputs where the algorithm exhibits a higher time complexity due to specific conditions or data characteristics.

(b) Best Case Runtime Θ(n) vs. Θ(n^2)
No, it is not possible for an algorithm with a best case runtime of Θ(n) to take Θ(n^2) time on all inputs. The best case runtime represents the lower bound on the algorithm’s performance. It means that in any scenario, the algorithm will run in linear time. If an algorithm consistently takes quadratic time on all inputs, it cannot have a best case runtime of Θ(n).

(c) Worst Case Runtime O(n^2) vs. o(n^2)
Yes, it is possible for an algorithm with a worst case runtime of O(n^2) to take o(n^2) time on some inputs. The big O notation represents an upper bound on the algorithm’s performance. It means that in the worst case scenario, the algorithm will run in quadratic time or less. However, there may be inputs where the algorithm exhibits a lower time complexity due to specific conditions or data characteristics.

(d) Worst Case Runtime O(n^2) vs. o(n^2)
No, it is not possible for an algorithm with a worst case runtime of O(n^2) to take o(n^2) time on all inputs. The big O notation represents an upper bound on the algorithm’s performance. It means that in any scenario, the algorithm will run in quadratic time or less. If an algorithm consistently takes subquadratic time on all inputs, it cannot have a worst case runtime of O(n^2).

3. Proof of Bound: O(18n lg(n) + 2n^2 = O(n^2))
To prove that 18n lg(n) + 2n^2 = O(n^2), we need to show that there exists a positive constant c and an input size n0 such that for all n >= n0, 18n lg(n) + 2n^2 <= c * n^2.

Let’s simplify and analyze each term separately:

For n >= 1 and lg(n) >= 0, we have: 18n lg(n) <= 18n * lg(n).

Since lg(n) <= n for all n >= 1, we can further simplify: 18n lg(n) <= 18n * n = 18n^2.

Therefore, for all n >= 1, 18n lg(n) <= 18n^2.

For n >= 1, we have: 2n^2 <= 2n^2.

Therefore, for all n >= 1, 2n^2 <= 2n^2.

Now let’s combine both terms:

For all n >= 1: 18n lg(n) + 2n^2 <= 18n^2 + 2n^2 = 20n^2.

Now we can choose c = 20 and n0 = 1:

For all n >= n0 = 1: 18n lg(n) + 2n^2 <= c * n^2.

Therefore, we have proven that 18n lg(n) + 2n^2 = O(n^2).

4. Exponential Functions
For exponential functions f(n) = kn and g(n) = mn, where k and m are positive constants with k != m:

To determine if their growth rates are asymptotically equivalent:

We need to compare their ratios f(n)/g(n) and g(n)/f(n).

Let’s calculate these ratios:

f(n)/g(n) = (kn)/(mn) = k/m g(n)/f(n) = (mn)/(kn) = m/k

Since k and m are positive constants with k != m, their ratios are not equal.

Therefore, their growth rates are not asymptotically equivalent.

5. Property of Transitivity for O
To prove the property of transitivity for O:

Let’s assume we have three asymptotically non-negative functions f(n), g(n), and h(n):

Given f(n) = O(g(n)) and g(n) = O(h(n)), we want to prove f(n) = O(h(n)).

By definition, f(n) = O(g(n)) implies that there exist positive constants c1 and n1 such that for all n >= n1, f(n) <= c1 * g(n).

Similarly, g(n) = O(h(n)) implies that there exist positive constants c2 and n2 such that for all n >= n2, g(n) <= c2 * h(n).

Combining both inequalities:

For all n >= max(n1, n2): f(n) <= c1 * g(n) <= c1 * (c2 * h(n)) = (c1 * c2) * h(n)

Let c = c1 * c2 be a positive constant.

For all n >= max(n1, n2), we have: f(n) <= c * h(n)

Therefore, f(n) = O(h(n)).

Hence, we have proven the property of transitivity for O.

6. Ranking Functions by Growth Rate
To rank the provided functions from least to greatest growth rate:

We need to compare their growth rates and prove the ordering.

Let’s analyze each function:

f1(n) = √n

This function has a square root growth rate.

To prove f1(n) = O(f5(n)), we need to show that there exists a positive constant c and an input size n0 such that for all n >= n0, √n <= c * lg(n).

Let’s choose c = √2 and n0 = 4:

For all n >= n0 = 4: √n <= √(n/4)*√4 <= √(n/4)*lg(4) <= √(n/4)*lg(√16) <= (√(n/4)*lg(√16))/lg(16) <= (√(n/4)*lg(√16))/4 <= (√(n/4)lg(√16))/4 <= (√(n/4)(4/4))/4 <= (√(n/4)4)/(44) <= √(n/4)/4 <= (√(n/4)/√16)/4 <= (√(n/64))/4 <= √(n/64)/√64 <= √(n)/64 Therefore, f1(n) = O(f5/n)), which implies f1(n) ≤ f5/n.

Hence, f1(n) ≤ f5/n.