Massive Algorithms: Analysis of Algorithm | Set 4 (Solving Recurrences)

Many algorithms are recursive in nature. 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.

Time complexity of Merge Sort can be written as T(n) = 2T(n/2) + cn.

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

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

There are following three cases:
1. If f(n) = $\Theta \left(n^{{c}}\right)$ where c < $\log _{b}a$ then T(n) = $\Theta \left(n^{{\log _{b}a}}\right)$

2. If f(n) = $\Theta \left(n^{{c}}\right)$ where c = $\log _{b}a$ then T(n) = $\Theta \left(n^{{c}}\logn\right)$

3.If f(n) = $\Theta \left(n^{{c}}\right)$ where c > $\log _{b}a$ then T(n) = $\Theta \left(f(n)\right)$

How does this work?
Master method is mainly derived from recurrence tree method. If we draw recurrence tree of T(n) = aT(n/b) + f(n), we can see that the work done at root is f(n) and work done at all leaves is $\Theta \left(n^{{c}}\right)$ where c is $\log _{b}a$ . And the height of recurrence tree is $\log _{b}n$

1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the 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
    <= cn/2Log(n/2) + n
    =  cnLogn - cnLog2 + n
    =  cnLogn - cn + n
    <= cnLogn

2) Recurrence Tree Method: In this method, we draw a recurrence tree and calculate the time taken by every level of 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 a arithmetic or geometric series.

For example consider the recurrence relation 
T(n) = T(n/4) + T(n/2) + cn²

     cn²
            /            \      
       c(n²)/16          c(n²)/4
       /      \            /      \
c(n²)/256   c(n²)/64  c(n²)/64    c(n²)/16
 /    \      /    \    /    \       /    \  

To know the value of T(n), we need to calculate 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 geometrical progression with ratio 5/16.

To get an upper bound, we can sum the infinite series. 
We get the sum as (n²)/(1 - 5/16) which is O(n²)

Read full article from Analysis of Algorithm | Set 4 (Solving Recurrences) | GeeksforGeeks

Analysis of Algorithm | Set 4 (Solving Recurrences) | GeeksforGeeks

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