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Akra-Bazzi method for finding the time complexities
  • Last Updated : 04 May, 2021

Master’s theorem is a popular method to solve time complexity recurrences of the form:

T(n) = aT(\frac{n}{b}) + f(n)

With constraints over a, b and f(n). The recurrence relation form limits the usability of the Master’s theorem. Following are three recurrences that cannot be solved directly using master’s theorem:

  1. \large{T(n) = 3T(\frac{n}{5}) + 2T(\frac{n}{5}) + \theta(n)}
  2. \large{T(n) = \frac{1}{3}T(\frac{n}{3}) + \theta(\frac{1}{n})}
  3. \large{T(n) = 9T(\frac{n}{3}+\log n) + \theta(n)}

Akra-Bazzi Method: This article explores another method for solving such recurrences that were developed by Mohammad Akra and Louay Bazzi in 1998. The Akra-Bazzi method can be applied to the recurrences of the following form: 

T(n)=\sum_{i=1}^{k} a_{i} T(b_{i}n+h_{i}(n))+g(n)



 where,  a_{i}   and b_{i}   are constants such that:

  1. \ a_{i}>0
  2. \ 0<b_{i}<1 \textnormal{ i.e. in the range (0, 1)}
  3. \ g(n)\ \epsilon\ O(n^{c}) \textnormal{ i.e. }g(n) \textnormal{ must have a polynomial upperbound}
  4. \ h_{i}(x)\ \epsilon\ O(\large\frac{n}{(logn)^{2}}\normalsize)

Next, find p such that

\large{\sum_{i=1}^k a_ib_i^p=1}
 

Then

\large{T(n)\epsilon\ \theta(n^{p}(1+\int_{1}^{n}\frac{g(u)}{u^{p+1}}du))}

Examples
Let’s consider the three recurrences discussed above and solve them using the method:

Example 1. 

 \large{T(n) = 3T(\frac{n}{5}) + 2T(\frac{n}{5}) + \theta(n)}



Here 

  1. a1 = 3
  2. b1 \large \frac{1}{5}
  3. a2 = 2
  4. b2 \large \frac{1}{5}
  5. b1 and b2 are in the range (0, 1)
  6. g(n) = \theta(n) which is O(nc), here c can be 1.

In this problem h1(n) and h2(n) are not present.
Here p=1 satisfies

3*\large {\frac{1}{5}^p}\normalsize +2*\large {\frac{1}{5}^p}\normalsize=1.

 Finally,

=> n^{1}(1+\int_{1}^{n}\large\frac{u}{u^{1+1}}\normalsize du)

=> n(1+\log{n}-\log{1})

=>n+n\log{n}

=>\theta(n\log{n})

Example 2.  

\large{T(n) = \frac{1}{3}T(\frac{n}{3}) + \theta(\frac{1}{n^{2}})}



Here

  1. a = \large \frac{1}{3}
  2. b = \large \frac{1}{3}
  3. g(n) = \theta(n^2)
  4. b is in the range (0, 1)
  5. g(n) = \theta(n^2) which is in O(nc), here c can be 1.

In this problem h(n) is not present.
Here p= – 1 satisfies

\large\frac{1}{2}\normalsize*\large\frac{1}{2}^p\normalsize=1

Finally,  

=> n^{-1}(1+\int_{1}^{n}\large\frac{\frac{1}{u}}{u^{-1+1}}\normalsize du)

=> \large\frac{1}{n}\normalsize(1+\int_{1}^{n}\large\frac{1}{u}\normalsize du)

=> \large\frac{1}{n}\normalsize(1+[\log u]_{1}^{n})

=> \large\frac{1}{n}\normalsize(1+\log n)

=> \theta(\large\frac{\log n}{n}\normalsize)

Example 3.

 \large{T(n) = 9T(\frac{n}{3}+\log n) + \theta(n)}

Here 

  1. a = 9
  2. b = \large\frac{1}{3}
  3. g(n) = \theta(n)\theta(n)
  4. b is in the range(0, 1)
  5. g(n) =  \theta(n)   which is O(nc), here c can be 1.
  6. h(n) =  \log{n}   which is O(\large\frac{n}{(logn)^{2}}\normalsize)

Here p=2 satisfies 

9*\large\frac{1}{3}^p\normalsize=1

Finally,

=> n^{2}(1+\int_{1}^{n}\large\frac{u}{u^{2+1}}\normalsize du)

=> n^{2}(1+\int_{1}^{n}\large\frac{1}{u^{2}}\normalsize du)

=> n^{2}(1+[-\large\frac{1}{u}\normalsize]_{1}^{n})

=> n^{2}(2-\large\frac{1}{n}\normalsize)

=> 2n^{2}-n

=> \theta(n^{2})\theta(n^{2})

Advantages:

  • Works for many divides and conquer algorithms.
  • Has a lesser constraint over the format of the recurrence than Master’s Theorem.
  • p can be calculated using numerical methods for complex recurrence relations.

Disadvantages:

  • Does not work when the growth of g(n) is not bounded polynomial. For Example g(N) = 2N.
  • Does not deal with ceil and floor functions.

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