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

Hey everyone! Ever stared down a recurrence relation that just seems…unsolvable? Well, you're not alone! Today, we're diving deep into a fascinating recurrence relation: T(n) = 25T(n/5) + n²(log n)². Our mission? To break it down, understand the challenges, and emerge victorious with a solution. Let's get started, guys!

The Challenge: Why This Recurrence Relation is Tricky

So, what makes this particular recurrence relation so challenging? At first glance, it might seem like a candidate for the Master Theorem, a powerful tool for solving many recurrence relations of the form T(n) = aT(n/b) + f(n). However, as our friend who initially posed the question pointed out, the Master Theorem might not be directly applicable here. Why is that? Let's break it down:

The Master Theorem has specific conditions that need to be met. It works best when the function f(n), which represents the non-recursive part of the relation, has a polynomial relationship with n. In simpler terms, f(n) should be something like n, n², n³, or even n^k for some constant k. But in our case, we have f(n) = n²(log n)². The presence of the (log n)² term throws a wrench in the works because it's not a simple polynomial.

This logarithmic component introduces a level of complexity that the basic Master Theorem can't handle directly. Think of it like trying to fit a square peg into a round hole. The theorem is a great tool, but it's not a universal solution for every recurrence relation out there. We need to explore alternative methods to tackle this beast. The iteration method, also known as the substitution method, is another common technique for solving recurrences. It involves repeatedly expanding the recurrence relation until a pattern emerges. However, in this case, the n²(log n)² term makes the iteration process quite cumbersome and difficult to manage. You'd end up with a tangled mess of terms, and spotting a clear pattern would be like finding a needle in a haystack.

So, we've established that the usual suspects – the Master Theorem and the iteration method – might not be the most straightforward paths to a solution here. But don't worry, guys! This is where the fun begins. We need to get creative and think outside the box. We need a strategy that can effectively deal with the pesky (log n)² term. This is where a clever substitution comes into play. By introducing a new variable, we can transform the recurrence into a more manageable form. This is a common trick in problem-solving – when faced with a complex problem, try to simplify it by changing your perspective or representation. It’s like putting on a different pair of glasses to see things more clearly.

The Substitution Game: Taming the Logarithm

Okay, so we need a substitution. But what should we substitute? The key lies in that (log n)² term. It's the source of our troubles, so let's try to tame it. A common strategy when dealing with logarithms in recurrence relations is to make a substitution that transforms the logarithmic term into a polynomial one. This often involves substituting n with an exponential function.

Let's try the substitution n = 5^k. Why 5? Because our recurrence relation involves dividing n by 5 in the recursive term T(n/5). This choice will simplify things nicely. If n = 5^k, then log₅(n) = k. Notice how we've effectively replaced the logarithm with a simple variable k. This is a huge step forward!

Now, let's rewrite our recurrence relation in terms of k. We define a new function S(k) = T(5^k). This substitution is crucial because it allows us to work with a recurrence relation in terms of k instead of n, which will be much easier to handle. Substituting n = 5^k into our original recurrence, we get:

S(k) = T(5^k) = 25T(5^k / 5) + (5^k)²(log (5^k))²

Now, let's simplify this expression. First, 5^k / 5 = 5^(k-1). Also, remember that log (5^k) = k log(5). So, we have:

S(k) = 25T(5^(k-1)) + (5^(2k))(k log(5))²

Since T(5^(k-1)) = S(k-1), we can substitute that in. Also, let's pull out the (log(5))² term, which is just a constant, and absorb it into a new constant c. This will make our equation cleaner and easier to work with. We get:

S(k) = 25S(k-1) + c * (5^(2k)) * k²

This looks much better! We've transformed the original recurrence relation into a new one that's arguably more manageable. The logarithmic term is gone, replaced by a polynomial term in k. This is a significant victory in our quest to solve the recurrence.

Applying the Master Theorem (Finally!)

Now, look at what we've got! The transformed recurrence relation, S(k) = 25S(k-1) + c * (5^(2k)) * k², looks much more amenable to the Master Theorem. Remember, the Master Theorem is a powerful tool for solving recurrences of the form T(n) = aT(n/b) + f(n). Our transformed recurrence has a similar structure, but instead of n, we have k, and instead of dividing by b, we're subtracting 1 from k in the recursive term.

To apply the Master Theorem, we need to identify the values of a, b, and f(n) in our transformed recurrence. In this case, a = 25. The term S(k-1) implies a division by an implicit b value. To see this more clearly, imagine rewriting S(k-1) as S(k / (5/5)) – this helps visualize the division by a constant factor, even though it's a subtraction in this case. However, for the Master Theorem in the form we usually see it, we need a division by a constant within the function argument. Because of this slight difference, we'll use an adapted approach that's conceptually similar to the Master Theorem but tailored for this specific form. The key idea remains the same: we'll compare the growth rate of the recursive part (25S(k-1)) with the non-recursive part (c * (5^(2k)) * k²).

The function f(k) = c * (5^(2k)) * k² represents the non-recursive work done at each level of the recursion. We need to compare the growth rate of f(k) with the growth rate of the recursive part, which is determined by the 25S(k-1) term. The 25 suggests an exponential growth in the recursive calls, while the f(k) term also has an exponential component (5^(2k)), but with an additional polynomial factor k². The exponential part 5^(2k) can be rewritten as (25^k), which clearly shows a strong exponential growth.

Now, we need to figure out which case of the Master Theorem (or its conceptual equivalent in this adapted approach) applies. Let's compare the growth rates. The recursive part grows roughly as 25^k, while the non-recursive part grows as 25^k * k². Since the non-recursive part has an additional polynomial factor k², it grows faster than the recursive part. This suggests that the solution will be dominated by the non-recursive part. In the context of the Master Theorem, this corresponds to Case 3, where f(n) grows faster than n^(log_b a). In our adapted context, it means that the c * (5^(2k)) * k² term will determine the overall growth of S(k).

Based on this analysis, we can conjecture that the solution to the recurrence S(k) = 25S(k-1) + c * (5^(2k)) * k² will be of the form S(k) = O(5^(2k) * k²). This is a crucial step. We've used the intuition from the Master Theorem (or its adapted form) to make an educated guess about the solution's asymptotic behavior. Now, we need to rigorously verify this guess using another powerful technique: the substitution method (also known as induction).

Proving Our Guess: Substitution (Induction) to the Rescue

We've made a conjecture: S(k) = O(5^(2k) * k²). This means we believe that there exists a constant C such that S(k) ≤ C * 5^(2k) * k² for all sufficiently large k. To prove this, we'll use induction, a fundamental proof technique in computer science and mathematics.

The base case is the starting point of our induction. We need to show that our inequality holds for some initial value of k. Let's choose k = 1. We need to ensure that S(1) ≤ C * 5^(21) * 1² = 25C* for a sufficiently large C. The actual value of S(1) depends on the base case of the original recurrence T(n), but we can always choose a C large enough to satisfy this condition. This is the beauty of asymptotic analysis – we're interested in the behavior for large inputs, so we can always adjust the constant C to make the base case work.

Now comes the inductive step. This is the heart of the proof. We assume that our inequality holds for some arbitrary k, and we need to show that it also holds for k + 1. This is like climbing a ladder: if you can get on the first rung (base case) and you know you can always reach the next rung (inductive step), then you can climb the entire ladder.

Our inductive hypothesis is: S(k) ≤ C * 5^(2k) * k². We want to prove that S(k + 1) ≤ C * 5^(2(k+1)) * (k+1)². Let's start with the recurrence relation for S(k + 1):

S(k + 1) = 25S(k) + c * 5^(2(k+1)) * (k+1)²

Now, we apply our inductive hypothesis. We replace S(k) with its upper bound:

S(k + 1) ≤ 25 * C * 5^(2k) * k² + c * 5^(2(k+1)) * (k+1)²

Our goal is to show that this expression is less than or equal to C * 5^(2(k+1)) * (k+1)². Let's simplify and manipulate the inequality:

S(k + 1) ≤ C * 25 * 5^(2k) * k² + c * 5^(2k+2) * (k+1)² S(k + 1) ≤ C * 5^(2k+2) * k² + c * 5^(2k+2) * (k+1)² S(k + 1) ≤ 5^(2k+2) * [C * k² + c * (k+1)²]

Now, we want to show that this is less than or equal to C * 5^(2(k+1)) * (k+1)² = C * 5^(2k+2) * (k+1)². So, we need to prove:

5^(2k+2) * [C * k² + c * (k+1)²] ≤ C * 5^(2k+2) * (k+1)²

Dividing both sides by 5^(2k+2), we get:

C * k² + c * (k+1)² ≤ C * (k+1)²

Rearranging the terms, we have:

c * (k+1)² ≤ C * (k+1)² - C * k² c * (k+1)² ≤ C * [(k+1)² - k²] c * (k+1)² ≤ C * [k² + 2k + 1 - k²] c * (k+1)² ≤ C * (2k + 1)

Now, we need to find a value of C that satisfies this inequality for all sufficiently large k. Notice that the left side is a quadratic function of k, while the right side is a linear function of k. For sufficiently large k, the linear function will always be smaller than the quadratic function. Therefore, we can always choose a C large enough to make this inequality hold.

To make this more concrete, let's expand the left side: c * (k² + 2k + 1) ≤ C * (2k + 1). We can choose a large enough C such that ck² + 2ck + c ≤ 2Ck + C holds for all k greater than some constant. This confirms that our inductive step holds.

We've successfully proven both the base case and the inductive step. Therefore, by the principle of mathematical induction, our conjecture S(k) = O(5^(2k) * k²) is correct! This is a major accomplishment. We've rigorously shown the asymptotic behavior of S(k).

The Grand Finale: Back to T(n)

We've conquered the transformed recurrence S(k) = O(5^(2k) * k²). But remember, our original goal was to solve T(n) = 25T(n/5) + n²(log n)². We made the substitution n = 5^k and S(k) = T(5^k). Now, it's time to reverse the substitution and express our solution in terms of n.

Since S(k) = O(5^(2k) * k²), we have T(5^k) = O(5^(2k) * k²). We know that n = 5^k, so 5^(2k) = (5^k)² = n². Also, k = log₅(n), so k² = (log₅(n))². Substituting these back into our solution, we get:

T(n) = O(n² * (log₅(n))²)

Since the base of the logarithm only affects the constant factor, we can write this more generally as:

T(n) = O(n²(log n)²)

And there you have it, guys! We've successfully solved the recurrence relation. It was a challenging journey, but we tackled it step by step, using a combination of substitution, the Master Theorem (in spirit, if not in direct application), and mathematical induction. This problem highlights the importance of having a diverse toolkit of problem-solving techniques and the ability to adapt them to new situations.

Key Takeaways and Pro Tips

• Substitution is your friend: When faced with a complex recurrence, don't be afraid to try substitutions to simplify the problem. Look for terms that are causing difficulty, like logarithms, and try to eliminate them with a clever substitution.
• Master the Master Theorem (but know its limits): The Master Theorem is a powerful tool, but it's not a one-size-fits-all solution. Understand its conditions and when it applies. If it doesn't directly apply, try to adapt its concepts or use other methods.
• Induction is your proof superpower: If you make a conjecture about the solution, use mathematical induction to rigorously prove it. This gives you confidence in your answer and ensures that it's correct.
• Think outside the box: Recurrence relations can be tricky, and sometimes you need to get creative. Don't be afraid to try different approaches and combine techniques.

Solving recurrence relations is a fundamental skill in computer science, especially in the analysis of algorithms. By mastering these techniques, you'll be well-equipped to tackle a wide range of problems. So keep practicing, keep exploring, and never stop learning!

I hope this breakdown was helpful, guys! If you have any questions or want to discuss other recurrence relations, feel free to leave a comment below. Let's keep the learning going! This will help to improve your understanding and skills in algorithms and data structures. Remember, the more you practice, the better you become at solving these types of problems. Happy coding!