Solving the Recurrence Relation Tn 2Tn-1 n Using Master Theorem and Other Techniques

Introduction: In computer science, recurrence relations are crucial in analyzing the time complexity of algorithms. One common method to solve recurrence relations is by using the Master Theorem, but this theorem applies to specific forms of recurrence. Let's explore the solution to the recurrence relation Tn 2Tn-1 n using both the Master Theorem and another method—the method of iteration or expansion.

Understanding the Master Theorem

The Master Theorem is a powerful tool for solving recurrence relations of the form:

[ T(n) aTleft(frac{n}{b}right) f(n) ]

where:

( a geq 1 ) ( b geq 1 ) ( f(n) ) is asymptotically positive (( f(n) > 0 ) for large ( n )).

Applying the Master Theorem to the recurrence relation Tn 2Tn-1 n involves checking if the relation fits into the standard form. Here, we have:

( a 2 ) ( b 1 ) (since we are reducing ( n ) by 1) ( f(n) n )

Let's analyze the conditions:

( b 1 ) which is not greater than 1.

Since the condition ( b > 1 ) is not satisfied, the Master Theorem cannot be directly applied to this recurrence relation.

Solving the Recurrence Relation Using the Method of Iteration

A suitable alternative to the Master Theorem is the method of iteration, also known as the expansion method. Let's start by expanding the recurrence relation:

[ T(n) 2T(n-1) n ]

Expanding further:

[ T(n-1) 2T(n-2) (n-1) ]

Substituting this into the original equation:

[ T(n) 2left(2T(n-2) (n-1)right) n 2^2T(n-2) 2(n-1) n ]

Continuing the expansion:

[ T(n-2) 2T(n-3) (n-2) ]

Substituting again:

[ T(n) 2^2left(2T(n-3) (n-2)right) 2(n-1) n 2^3T(n-3) 2^2(n-2) 2(n-1) n ]

Following this pattern, after ( k ) expansions:

[ T(n) 2^kT(n-k) 2^0n 2^1(n-1) 2^2(n-2) ldots 2^{k-1}(n-(k-1)) ]

Eventually, this leads to the base case when ( n k ):

[ T(1) C ] (some constant)

So, when ( k n-1 ):

[ T(n) 2^{n-1}T(1) sum_{i0}^{n-1} 2^i(n-i) ]

Calculating the sum:

[ T(n) 2^{n-1}C 2(n2^{n-1}-(n2^{n-1}-2^{n}-n 1)) ]

By simplifying:

[ T(n) 2^nT(1) 2^n - n - 2 ]

So, the final solution is:

[ T(n) 2^nT(1) 2^n - n - 2 ]

Conclusion: The method of iteration provides a systematic way to solve the recurrence relation ( Tn 2Tn-1 n ), revealing its complexity as ( 2^n 2^n - n - 2 ).