Question

Find a big \(O\) upper bound (the best you know how to get) on solutions to the recurrence \(T(n)=T(n / 4)+T(n / 2)+n^{2}\), with \(T(n)=1\) if \(n<4 .\)

Short answer

The best big \( O \) upper bound for the recurrence is \( O(n^2) \).

Step-by-step solution

Determine Recurrence Structure

The recurrence relation given is \( T(n) = T(n/4) + T(n/2) + n^2 \) with the base case \( T(n) = 1 \) for \( n < 4 \). This means for any problem of size \( n \), it splits into two sub-problems of sizes \( n/4 \) and \( n/2 \) and performs \( n^2 \) additional work.

Identify Recurrence Type

This recurrence does not fit the standard forms for binary or balanced recurrences like the Master Theorem due to the differing denominators (4 and 2). The Master Theorem applies in cases where the subproblems are equally divided. Therefore, alternative approaches such as making substitutions or the recursion tree method are needed.

Use Recursion Tree Method

Construct a recursion tree for the recurrence to identify the total work done. The root's cost is \( n^2 \), and splits into two children each costing \((n/4)^2\) and \((n/2)^2\). This process will continue until the subproblems reach base case size \( n < 4 \).

Analyze Each Level of the Tree

At each level of the tree, calculate the total cost:- Level 0: \( n^2 \)- Level 1: \( (n/4)^2 + (n/2)^2 = n^2 / 16 + n^2 / 4 = 5n^2/16 \)- Level 2: Each subproblem further breaks into new subproblems calculated similarly. Sum their work.The general pattern for each level can be observed as a geometric series.

Aggregate Total Work

The total work is given by the sum of work at every level of the tree. The sum of the geometric series can be approximated by:\[ T(n) = \sum_{i=0}^{\log_4 n} a^i k \quad \text{where} \quad a = \frac{5}{16} \text{and} \quad k = n^2 \]As \( a < 1 \), apply the geometric series sum formula.\[ T(n) = n^2 \cdot \frac{1-(5/16)^{\log_4 n}}{1-5/16} \approx O(n^2) \] when simplifying.

Simplification and Conclusion

The dominant term in the geometric sum results from the root cost, which indicates that the total work done per division and conquering is bounded by \( O(n^2) \). Therefore, the observational bound of our recursion tree's summation is \( O(n^2) \).

Key concepts

Recurrence Relations

Recurrence relations are equations or inequalities that describe a function in terms of its value at smaller inputs. They are frequently used to characterize the running time of recursive algorithms. If an algorithm divides a problem into smaller subproblems, solves them recursively, and then combines their solutions, recurrence relations are often used to model its time complexity. In the given exercise, the recurrence relation given is:\[ T(n) = T(n/4) + T(n/2) + n^2 \]with the base condition:\[ T(n) = 1 \quad \text{for} \quad n < 4 \]This equation indicates that for an input size of \( n \), the algorithm splits into two problems of sizes \( n/4 \) and \( n/2 \), and performs an additional \( n^2 \) amount of work.When approaching such problems, the recurrence relation helps in breaking down the total work done into recognizable patterns, which can then be solved using various methods like the recursion tree method or the substitution method.

Recursion Tree Method

The recursion tree method is a visual tool used to estimate the time complexity of a recurrence relation by drawing a tree. Each node of the tree represents a recursive call and its associated cost, providing valuable insight into how the recursive process unfolds.A recursion tree starts with the initial problem at the root node. For our recurrence \[ T(n) = T(n/4) + T(n/2) + n^2 \]the root has a cost of \( n^2 \). This splits into two children nodes with costs \( (n/4)^2 \) and \( (n/2)^2 \), representing the subproblems' costs. Subsequent levels of the tree represent further subdivisions until the base condition \( n < 4 \) is met.As we add up the costs at each level of the tree, we notice a repeated pattern. These patterns often reveal themselves as geometric series, which are easier to sum. An essential concept is that the total work of the tree is dictated by the sum of work at all levels, often dominated by the level with the most significant contributions. In this exercise, simplifying the series showed that the dominant work translates to an \( O(n^2) \) complexity, suggesting the bulk of work occurs near the root node.

Big O Notation

Big O notation is a mathematical concept used to describe the upper bound of an algorithm's running time. It characterizes how the runtime grows with input size, focusing on the term that increases most rapidly as \( n \), the number of elements, becomes very large.In the provided exercise, the goal was to determine the Big O upper bound for the recurrence relation:\[ T(n) = T(n/4) + T(n/2) + n^2 \]After evaluating the recurrence using the recursion tree method, it was found that each level's cumulative work follows a geometric series. This insight leads us to \[ T(n) \approx O(n^2) \],which means that the highest order of growth for this algorithm is \( n^2 \).Big O notation allows us to strip away lower-order terms and constant factors, highlighting the most significant term that affects an algorithm's efficiency in terms of time. This makes it instrumental in analyzing algorithms, helping developers understand the efficiency and scalability of their code.