Solve T(n) = 25T(n/5) + N²(log N)²: A Step-by-Step Guide

Aug 15, 2025 by Sebastian Müller 57 views

Cracking the Code: Solving the Recurrence Relation T(n) = 25T(n/5) + n²(log n)²

Hey everyone! Today, we're diving deep into the fascinating world of recurrence relations, and we're tackling a particularly juicy one: T(n) = 25T(n/5) + n²(log n)². This bad boy looks intimidating at first glance, but don't worry, we'll break it down step by step and conquer it together. You mentioned you've tried the iteration method and the Master Theorem, but hit a wall. That's perfectly alright! Sometimes, these methods need a little extra finesse, or perhaps a different approach is more suitable. So, buckle up, grab your thinking caps, and let's get started!

Understanding the Challenge: Why the Usual Suspects Aren't Working

Before we jump into solutions, let's quickly address why the methods you initially tried might not have worked as expected. This understanding is crucial for choosing the right strategy. Let's explore why the iteration method and the Master Theorem might be facing some challenges here.

The Iteration Method: A Tangled Web

The iteration method, in essence, involves repeatedly expanding the recurrence relation until a pattern emerges. We substitute T(n/5) in terms of T(n/25), then T(n/25) in terms of T(n/125), and so on. While this method can be powerful, it can also lead to a tangled web of terms, especially when the non-recursive term (in our case, n²(log n)²) is a bit complex. The logarithmic term really throws a wrench in the gears here. When you repeatedly substitute, the log terms start nesting within each other, making it difficult to identify a clean, summable pattern. Imagine expanding log(n/5), then log(n/25), and so on – you'll end up with a series of logarithmic expressions that are tough to simplify directly. That's why, although iteration is a valid approach in principle, the algebra can quickly become overwhelming in this particular case. So, if you felt like you were drowning in logarithms, you're not alone!

The Master Theorem: A Close but No Cigar

The Master Theorem is a fantastic tool for solving many recurrence relations of the form T(n) = aT(n/b) + f(n). It provides a cookbook-like approach, allowing you to determine the asymptotic complexity of T(n) by comparing f(n) with n^(log_b a). However, there are some limitations to its applicability. In our case, we have a = 25, b = 5, and f(n) = n²(log n)². This means n^(log_b a) = n^(log_5 25) = n². Now, here's the tricky part: we need to compare f(n) = n²(log n)² with n². The Master Theorem has three cases, each dealing with a different relationship between f(n) and n^(log_b a) (up to polynomial factors). The crucial cases involve checking if f(n) = O(n^(log_b a - ε)) or f(n) = Ω(n^(log_b a + ε)) for some constant ε > 0, or if f(n) = Θ(n^(log_b a) * log^k n) for some non-negative integer k. In our scenario, f(n) grows slightly faster than n² due to the (log n)² factor. While it seems like it might fit into one of the Master Theorem cases, the logarithmic difference throws a curveball. The standard Master Theorem doesn't directly handle cases where f(n) has a logarithmic factor on top of the polynomial factor. The theorem is designed for polynomial differences, not logarithmic ones. So, while you're on the right track trying the Master Theorem, the (log n)² term makes it a bit too complex for the basic version. Don't feel discouraged – recognizing this limitation is a key step in problem-solving!

The Substitution Method: Our Hero Emerges

So, if iteration is messy and the Master Theorem isn't a perfect fit, what's our next move? Enter the substitution method! This technique is all about transforming the recurrence into a form we can solve, often by introducing a new variable. The logarithmic term n²(log n)² hints that a smart substitution involving logarithms might be the key. Let's see how this works in practice.

The Magic Substitution

The core idea is to get rid of that pesky log n term. We can do this by making a substitution that transforms the logarithmic part into a polynomial. Let's try substituting n = 5^k. Why 5? Because our recurrence involves dividing n by 5, so powers of 5 seem like a natural fit. This substitution implies that k = log_5 n. Now, let's define a new function, S(k) = T(5^k). This might seem abstract, but it's a crucial step in simplifying our recurrence. We're essentially rewriting T(n) in terms of k, where k represents the exponent of 5 that gives us n. Let's see how this substitution transforms our original recurrence:

T(n) = 25T(n/5) + n²(log n)²
Substitute n = 5^k: T(5^k) = 25T(5^k / 5) + (5^k)²(log (5^k))²
Simplify: T(5^k) = 25T(5^(k-1)) + (5^(2k))(k log(5))²
Substitute S(k) = T(5^k): S(k) = 25S(k - 1) + 5^(2k) * k² * (log 5)²

Notice what happened? The log n term has magically transformed into k, a much simpler variable. The n² term became 5^(2k), which is still exponential, but that's okay – we've dealt with exponentials before. The key is that we've replaced the log n with a polynomial term (k²), making the recurrence much more manageable. We can even get rid of the constant factor (log 5)² since we care about asymptotic behavior; constants don't change the growth rate. We can simply absorb it into the constants we'll deal with later. So, our simplified recurrence looks like this:

S(k) = 25S(k - 1) + 5^(2k) * k²

This looks much better, doesn't it? We've successfully transformed a complex recurrence into a more standard form. Now, the question is: how do we solve this new recurrence for S(k)?

Tackling the Transformed Recurrence: Another Substitution (or the Master Theorem, Revisited!)

We have S(k) = 25S(k - 1) + 5^(2k) * k². This recurrence still looks a bit intimidating, but we're closer than we think. There are a couple of ways we can approach this:

Another Substitution: We could try another substitution to further simplify the recurrence. Notice the 5^(2k) term. Since 5² = 25, we could try substituting R(k) = S(k) / 25^k to potentially eliminate the exponential term. This is a common trick when dealing with recurrences that have both recursive and exponential components.
Master Theorem (Again!): Remember how the regular Master Theorem didn't work initially? Well, with our new recurrence, it might be applicable, or at least a modified version of it. The key is to express the recurrence in the form the Master Theorem expects: T(n) = aT(n/b) + f(n). Our recurrence is currently in a slightly different form (in terms of S(k) and S(k-1)), but we can adapt the Master Theorem's principles.

Let's explore the second option first because it's often the quicker route if it works.

Master Theorem, the Remix!

Even though our recurrence isn't exactly in the Master Theorem's standard form, the underlying principles still apply. Let's think about how the Master Theorem works. It compares the cost of the recursive calls with the cost of the work done outside the recursive calls. In our case:

S(k) = 25S(k - 1) + 5^(2k) * k²

We can think of this as:

a = 25 (each call makes 25 recursive calls)
The "size" reduction is by 1 (k becomes k-1, which is like dividing by a constant in the original Master Theorem)
f(k) = 5^(2k) * k² = 25^k * k² (the cost of the work outside the recursive calls)

Now, we need to compare f(k) with the "size" of the problem raised to the power of something related to a. In the original Master Theorem, it's n^(log_b a). Here, since we're reducing the "size" by 1 in each recursive call, we're dealing with an additive reduction rather than a multiplicative one. So, instead of a logarithm, we're looking at the base of the recursion, which is 25. So, we compare f(k) = 25^k * k² with 25^k. Notice that f(k) grows faster than 25^k due to the k² term. This suggests that the cost of the work done outside the recursive calls dominates the overall complexity. In Master Theorem terms, this often corresponds to the case where T(n) = Θ(f(n)). However, we need to be a bit careful because our recurrence isn't exactly in the standard Master Theorem form. But the intuition is strong.

The Grand Finale: Unraveling the Solution

Based on our intuition from the Master Theorem (and to confirm it rigorously, we could do an inductive proof, but we'll skip that for brevity here), we strongly suspect that S(k) = Θ(25^k * k²). Remember, we made the substitution S(k) = T(5^k) and k = log_5 n. Let's substitute back to get T(n):

S(k) = Θ(25^k * k²)
T(5^k) = Θ(25^k * k²)
T(n) = Θ(25^(log_5 n) * (log_5 n)²)

Now, we need to simplify this expression. Remember that a^(log_b c) = c^(log_b a). Applying this to 25^(log_5 n), we get:

25^(log_5 n) = n^(log_5 25) = n²

So, our solution becomes:

T(n) = Θ(n² * (log_5 n)²)

Since log_5 n is just a constant multiple of log n, we can finally write the solution in the more familiar form:

T(n) = Θ(n² (log n)²)

The Takeaway: A Symphony of Techniques

Wow! We made it! That was a journey, but we successfully solved the recurrence relation T(n) = 25T(n/5) + n²(log n)². The key wasn't just one method, but a combination of techniques:

Recognizing the Limitations: We understood why the direct iteration method and the basic Master Theorem weren't directly applicable.
The Power of Substitution: We used the substitution n = 5^k to transform the recurrence into a more manageable form.
Master Theorem Intuition: We adapted the principles of the Master Theorem to guide us towards the solution.
Back-Substitution: We carefully substituted back to express the solution in terms of the original variable, n.

This problem highlights the importance of having a flexible toolkit and knowing when to apply different techniques. Recurrence relations can be challenging, but with practice and a good understanding of the methods, you can conquer them! So, keep practicing, keep exploring, and keep those brains buzzing! You got this!

If you guys have any questions, feel free to ask! We can dive deeper into any specific step or explore other recurrence-solving techniques. Keep up the great work!