Question

Given the recurrence relation \\[\begin{array}{l} T(n)=7 T\left(\frac{n}{5}\right)+10 n \quad \text { for } n>1 \\\T(1)=1\end{array}\\] find \(T(625)\)

Short answer

The solution to the recurrence relation \(T(625) = T(7)T\left(\frac{n}{5}\right) + 10 n\) given \(T(1) = 1\) is \(T(625) = 65151\).

Step-by-step solution

Start with the given \(n\)

We are asked to find \(T(625)\). To calculate this, we will first substitute \(n=625\) into the recurrence relation.

Apply Recurrence Relation

Substitute \(n=625\) into the recurrence relation to get \(T(625) = 7T(125) + 6250\). Now, we need to calculate \(T(125)\). Repeat the same process i.e., substitute \(n=125\) into the recurrence relation to get \(T(125) = 7T(25) + 1250\). Continue this iterative process until we reach \(T(1)\).

Final Calculation

Here we will repeatedly substitute \(n\) in the relation till we reach \(T(1)\): \(T(625) = 7T(125) + 6250 = 7[7T(25) + 1250] + 6250 = 49T(25) + 13750 = 49[7T(5) + 250] + 13750 = 343T(5) + 24350 = 343[7T(1) + 50] + 24350 = 2401T(1) + 41150 = 2401*1 + 41150 = 65151\).

Key concepts

Time Complexity

Time complexity is a way to measure how the runtime of an algorithm increases as the input size increases. It tells us how efficient an algorithm is when scaling up to larger inputs. Time complexity is often expressed using Big O notation, which describes the worst-case scenario in terms of performance.

For example, if an algorithm has a time complexity of \(O(n^2)\), it means that as the input size \(n\) doubles, the time it takes to complete will roughly quadruple. When examining a recurrence relation like \(T(n)=7 T\left(\frac{n}{5}\right)+10 n\), understanding its time complexity helps us predict how much effort is needed to solve problems given in larger sizes.

An algorithm with a lower time complexity is more efficient and typically preferable unless other factors, like space complexity or maintainability, are more important in specific contexts.

Divide and Conquer Algorithms

Divide and conquer algorithms are a popular strategy for solving complex problems by breaking them into simpler sub-problems. The process involves three main steps: dividing the problem into smaller parts, solving each part, and then combining the results.

Each step helps make the original problem easier to manage and solve. The recurrence relation \(T(n)=7 T\left(\frac{n}{5}\right)+10 n\) is a fantastic illustration of this technique. Here's how it works:

• Divide: Split the problem into smaller sub-problems of size \(\frac{n}{5}\).
• Conquer: Solve each sub-problem recursively, assuming smaller versions of the problem have a known solution (this is where the recurrence relation becomes handy).
• Combine: Use the solutions from the sub-problems to construct a solution to the equivalent larger problem.

This technique is widely used because it can drastically reduce complexity, making otherwise daunting tasks like sorting large datasets manageable.

Master Theorem

The Master Theorem is a handy tool for solving recurrence relations, especially those that arise in divide and conquer algorithms. It provides a way to determine the asymptotic behavior of these recurrences without repeatedly substituting and solving them manually.

The theorem applies to recurrence relations of the form:

\[T(n) = aT\left(\frac{n}{b}\right) + f(n)\]

where:

• \(a\) is the number of subproblems in the recursion,
• \(b\) is the factor by which the subproblem size is reduced, and
• \(f(n)\) represents the cost of dividing the problem and combining the results.

Understanding this theorem helps simplify complex recurrence relations into a form that can be more easily analyzed, allowing for quick determination of time complexity. For the exercise given, the relationship \(T(n)=7 T\left(\frac{n}{5}\right)+10 n\) can be analyzed using the Master Theorem, revealing useful insights about how efficiently the algorithm performs as \(n\) grows.