Oct 10, 2018

[HDGEM] Algorithms: There are mainly three ways for solving recurrences


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           /      \       T(n/4)     T(n/2)    If we further break down the expression T(n/4) and T(n/2),   we get following recursion tree.                    cn2             /           \               c(n2)/16      c(n2)/4        /      \          /     \    T(n/16)     T(n/8)  T(n/8)    T(n/4)   Breaking down further gives us following                   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)
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) = Θ(nc) where c < Logba then T(n) = Θ(nLogba)
2. If f(n) = Θ(nc) where c = Logba then T(n) = Θ(ncLog n)
3.If f(n) = Θ(nc) where c > Logba then T(n) = Θ(f(n))


--
Posted By Blogger to HDGEM at 3/07/2017 11:01:00 AM