# How to solve Recurrence Relation for Complexity Analysis of Algorithms

In the previous post, we discussed the analysis of loops. Many algorithms are recursive. When we analyze them, we get a recurrence relation for time complexity. We get running time on an input of size n as a function of n and the running time on inputs of smaller sizes. For example in Merge Sort, to sort a given array, we divide it into two halves and recursively repeat the process for the two halves. Finally, we merge the results. Time complexity of Merge Sort can be written as T(n) = 2T(n/2) + cn. There are many other algorithms like Binary Search, Tower of Hanoi, etc.

There are mainly three ways of solving recurrences:

__Substitution Method:__

We make a guess for the solution and then we use mathematical induction to prove the guess is correct or incorrect.

For example consider the recurrence T(n) = 2T(n/2) + n

We guess the solution as T(n) = O(nLogn). Now we use induction to prove our guess.

We need to prove that T(n) <= cnLogn. We can assume that it is true for values smaller than n.

T(n) = 2T(n/2) + n

<= 2cn/2Log(n/2) + n

= cnLogn – cnLog2 + n

= cnLogn – cn + n

<= cnLogn

__Recurrence Tree Method:__

__Recurrence Tree Method:__

In this method, we draw a recurrence tree and calculate the time taken by every level of the tree. Finally, we sum the work done at all levels. To draw the recurrence tree, we start from the given recurrence and keep drawing till we find a pattern among levels. The pattern is typically arithmetic or geometric series.

For example, consider the recurrence relation

T(n) = T(n/4) + T(n/2) + cn

^{2}cn

^{2}^{ }/ \

T(n/4) T(n/2)If we further break down the expression T(n/4) and T(n/2),

we get the following recursion tree.cn

^{2}

/ \

c(n^{2})/16 c(n^{2})/4

/ \ / \

T(n/16) T(n/8) T(n/8) T(n/4)Breaking down further gives us following

cn

^{2}

/ \

c(n^{2})/16 c(n^{2})/4

/ \ / \

c(n^{2})/256 c(n^{2})/64 c(n^{2})/64 c(n^{2})/16

/ \ / \ / \ / \To know the value of T(n), we need to calculate the sum of tree

nodes level by level. If we sum the above tree level by level,we get the following series T(n) = c(n^2 + 5(n^2)/16 + 25(n^2)/256) + ….

The above series is a geometrical progression with a ratio of 5/16.To get an upper bound, we can sum the infinite series. We get the sum as (n

^{2})/(1 – 5/16) which is O(n^{2})

__Master Method:__

__Master Method:__

Master Method is a direct way to get the solution. The master method works only for the following type of recurrences or for recurrences that can be transformed into the following type.

T(n) = aT(n/b) + f(n) where a >= 1 and b > 1

There are the following three cases:

- If f(n) = O(n
^{c}) where c < Log_{b}a then T(n) = Θ(n^{Log}_{b}^{a}) - If f(n) = Θ(n
^{c}) where c = Log_{b}a then T(n) = Θ(n^{c}Log n) - If f(n) = Ω(n
^{c}) where c > Log_{b}a then T(n) = Θ(f(n))

**How does this work?**

The master method is mainly derived from the recurrence tree method. If we draw the recurrence tree of T(n) = aT(n/b) + f(n), we can see that the work done at the root is f(n), and work done at all leaves is Θ(n^{c}) where c is Log_{b}a. And the height of the recurrence tree is Log_{b}n

In the recurrence tree method, we calculate the total work done. If the work done at leaves is polynomially more, then leaves are the dominant part, and our result becomes the work done at leaves (Case 1). If work done at leaves and root is asymptotically the same, then our result becomes height multiplied by work done at any level (Case 2). If work done at the root is asymptotically more, then our result becomes work done at the root (Case 3).

**Examples of some standard algorithms whose time complexity can be evaluated using the Master Method **

- Merge Sort: T(n) = 2T(n/2) + Θ(n). It falls in case 2 as c is 1 and Log
_{b}a] is also 1. So the solution is Θ(n Logn) - Binary Search: T(n) = T(n/2) + Θ(1). It also falls in case 2 as c is 0 and Log
_{b}a is also 0. So the solution is Θ(Logn)

**Notes: **

- It is not necessary that a recurrence of the form T(n) = aT(n/b) + f(n) can be solved using Master Theorem. The given three cases have some gaps between them. For example, the recurrence T(n) = 2T(n/2) + n/Logn cannot be solved using master method.
- Case 2 can be extended for f(n) = Θ(n
^{c}Log^{k}n)

If f(n) = Θ(n^{c}Log^{k}n) for some constant k >= 0 and c = Log_{b}a, then T(n) = Θ(n^{c}Log^{k+1}n)

For more details, please refer: Design and Analysis of Algorithms.

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