Question

When a divide-and-conquer algorithm divides an instance of size \(n\) of a problem into subinstances each of size \(n / c,\) the recurrence relation is typically given by \\[\begin{array}{l}T(n)=a T\left(\frac{n}{c}\right)+g(n) \quad \text { for } n>1 \\ T(1)=d\end{array}\\] where \(g(n)\) is the cost of the dividing and combining processes, and \(d\) is a constant. Let \(n=c^{k}\) a. Show that \\[T\left(c^{k}\right)=d \times a^{k}+\sum_{j=1}^{k}\left[a^{k-j} \times g\left(c^{j}\right)\right]\\] b. Solve the recurrence relation given that \(g(n) \in \Theta(n)\)

Short answer

For a divide-and conquer algorithm that divides an instance of a problem of size \(n\) into instances each of size \(n/c\), the time complexity is given by \(T(n) = aT(n/c) + g(n)\) for \(n>1\) and \(T(1) = d\). When \(g(n) \in \Theta(n)\), the solution of the recurrence is \(T(n) = \Theta(n)\).

Step-by-step solution

Establish the base case for \(T(c^k)\)

The base case would be when \(k=0\). Then, \(c^0 = 1\), and from the given recurrence relation, we have \(T(1) = d\). This directly establishes the first part of the equation to prove.

Inductive Hypothesis

Assume that the given formula, \(T(c^{k-1}) = da^{k-1} + \sum_{j=1}^{k-1} [a^{k-1-j} g(c^j) ]\) holds true for some \(k - 1\). This is our inductive hypothesis.

Prove for \(k^{th}\) term

Now we prove that the relation holds for the \(k^{th}\) term, based on the inductive hypothesis that it's true for \((k-1)^{th}\) term, i.e., we prove that if the formula holds for \(k-1\), it should also hold for \(k\). From the recurrence relation, we have \(T(c^k) = aT(c^{k-1}) + g(c^k)\). Substituting the induction hypothesis into \(T(c^k)\) we obtain, \(T(c^k) = da^k + \sum_{j=1}^{k-1} [a^{k-j} g(c^j) ] + g(c^k)\). This simplifies to \(T(c^k) = da^k + \sum_{j=1}^{k} [a^{k-j} g(c^j) ]\). Hence, the result holds for the \(k^{th}\) term.

Solve Recurrence Relation with \(g(n) \in \Theta(n)\)

From step 3 we have, \(T(c^k) = da^k + \sum_{j=1}^{k} [a^{k-j} g(c^j) ]\). Now it's given that \(g(n) = \Theta(n)\), which means that there are some constants \(p\) and \(q\) such that \(pn \leq g(n) \leq qn\), for sufficiently large \(n\). Substituting this into the equation in terms of \(k\), we obtain \(T(c^k) = da^k + \sum_{j=1}^{k} [a^{k-j} pc^j ]\) noting that for sufficiently large \(k\), \(c^j\) is always big, \(T(c^k) = da^k + \sum_{j=1}^{k} [a^{k-j} qc^j ]\). Sabbtracting those two inequalities, we get \(T(c^k) = \Theta(c^k)\). Therefore, for \(g(n) = \Theta(n)\), the solution of the recurrence relation is \(T(n) = \Theta(n)\)

Key concepts

Divide-and-Conquer Algorithm

The divide-and-conquer strategy is a fundamental algorithm design technique where a problem is broken down into smaller sub-problems of the same type, solved individually, and then combined to get the solution of the original problem. This approach is well-utilized in algorithms like mergesort and quicksort.

In each division stage, the size of the problem is reduced, typically by a factor of a constant. The sub-problems are then conquered, i.e., solved recursively. Finally, solutions to the sub-problems are combined to solve the original problem. The effectiveness of a divide-and-conquer algorithm is significant as it often simplifies a complex problem by splitting it into manageable parts.

The beauty of this approach is two-fold: it can lead to a dramatic reduction in the problem's difficulty, and it enables the use of recursive methods, which are often straightforward and elegant. The connection between divide-and-conquer algorithms and recurrence relations is that the latter models the time complexity of the algorithm by expressing the total cost of the algorithm as the sum of the costs of the sub-problems plus some additional overhead for dividing the problem and combining the results.

Algorithmic Complexity

Algorithmic complexity, also known as time complexity, refers to a function that describes the amount of time an algorithm takes to run, as a function of the length of the input. It is usually represented using Big O notation which gives the upper bound of the complexity, or Theta notation, which gives the precise growth rate, assuming worst-case scenarios.

For divide-and-conquer algorithms, the recurrence relation provides a means to articulate this complexity. By evaluating the recurrence relation, we can understand how the running time increases as the input size grows. In the current exercise, with a recurrence relation of the form \(T(n) = aT(n/c) + g(n)\) and \(g(n)\) proportional to \(n\), we infer that the algorithm's time complexity is linearly related to the input size, signified by \(T(n) = \theta(n)\).

Understanding algorithmic complexity helps in two ways: it guides the selection of the most appropriate algorithm for a given task, and it determines the algorithm's scalability and efficiency. This is crucial when processing large datasets or in applications where performance is a critical concern.

Inductive Proof

An inductive proof is a mathematical technique used to prove a statement or formula for all natural numbers. It consists of two main parts: the base case and the inductive step. The base case verifies that the statement holds for the initial value, typically when \(n = 1\). The inductive step shows that if the statement holds for some integer \(k\), then it must also hold for \(k+1\).

In the context of the recurrence relation provided in the original exercise, inductive proof is used to establish that the formula for \(T(c^k)\) is accurate for all values of \(k\). It's like building a ladder: if you can step onto the first rung (base case) and prove that each step leads securely to the next (inductive step), you can climb as high as the ladder goes – which, in this case, represents all natural numbers.

This method is not just for mathematical curiosity; it's a powerful proof technique in computer science for verifying algorithms, especially those involving recursive processes like divide-and-conquer.