How do you solve recurrence relations with the master theorem?

1 What is the master theorem?

The master theorem is a formula that provides the asymptotic behavior of a recurrence relation of the form T(n) = aT(n/b) + f(n), where a and b are positive constants, and f(n) is a non-negative function. The master theorem states that the asymptotic behavior depends on the relationship between f(n) and n^(log_b a). Specifically, if f(n) = O(n^(log_b a - e)) for some constant e > 0, then T(n) = Theta(n^(log_b a)); if f(n) = Theta(n^(log_b a)), then T(n) = Theta(n^(log_b a) log n); and if f(n) = Omega(n^(log_b a + e)) for some constant e > 0, and if af(n/b) <= cf(n) for some constant c < 1 and all sufficiently large n, then T (n)= Theta (f (n)). Intuitively, the master theorem compares the cost of recursive calls with the cost of non-recursive work. If recursive calls dominate, then the running time is proportional to the number of leaves in the recursion tree; if non-recursive work dominates, then it is proportional to the cost of the root; and if both are balanced, it is proportional to the product of levels and cost per level in the recursion tree.

2 How to apply the master theorem?

To apply the master theorem, you must first identify the values of a, b, and f(n) from the recurrence relation. Afterwards, compare f(n) with n^(log_b a) to determine which case of the master theorem applies. Then, plug in the values of a, b, and f(n) into the corresponding formula for the case and simplify if possible. Finally, write down the asymptotic bound of T(n) using Theta, O, or Omega notation.

3 What are some examples of the master theorem?

Let's see some examples of how to use the master theorem to solve recurrence relations.

Example 1: T(n) = 2T(n/4) + sqrt(n)

In this example, a = 2, b = 4, and f(n) = sqrt(n). Comparing f(n) with n^(log_b a), we get:

sqrt(n) = O(n^(log_4 2 - e)) for any e < 1/2

Therefore, this falls under case 1 of the master theorem, and we can use the formula:

T(n) = Theta(n^(log_b a)) = Theta(n^(log_4 2)) = Theta(sqrt(n))

Example 2: T(n) = 3T(n/3) + n log n

In this example, a = 3, b = 3, and f(n) = n log n. Comparing f(n) with n^(log_b a), we get:

n log n = Theta(n^(log_3 3)) = Theta(n)

Therefore, this falls under case 2 of the master theorem, and we can use the formula:

T(n) = Theta(n^(log_b a) log n) = Theta(n log n log n)

Example 3: T(n) = T(n/2) + n^2

In this example, a = 1, b = 2, and f(n) = n^2. Comparing f(n) with n^(log_b a), we get:

n^2 = Omega(n^(log_2 1 + e)) for any e > 0

Therefore, this falls under case 3 of the master theorem, and we need to check the regularity condition:

af(n/b) <= cf(n) for some constant c < 1 and all sufficiently large n

Plugging in the values, we get:

f(n/2) <= cf(n)

n^2 / 4 <= cn^2

c >= 1/4

Since we can find a constant c < 1 that satisfies the condition, the regularity condition holds, and we can use the formula:

T(n) = Theta(f(n)) = Theta(n^2)

4 Here’s what else to consider

This is a space to share examples, stories, or insights that don’t fit into any of the previous sections. What else would you like to add?