Recurrences Solving

Click here to Download PDF for this Topic

Recurrences Solving

Recurrence Relations

• A recurrence relation is an equation which is defined in terms of itself.
• Why are recurrences good things?

–        Many natural functions are easily expressed as recurrences:

•         a_n = a _n-1 + 1, a₁ = 1--> a_n = n (polynomial)

•         a_n = 2a _n-1  ,a₁ = 1--> a_n = 2ⁿ (exponential)

•         a_n = na _n-1  ,a₁ = 1--> a_n = n! (weird fun)

• It is often easy to find a recurrence as the solution of a counting problem.
• Recursion is Mathematical Induction!
• In both, we have general and boundary conditions, with the general condition breaking the problem into smaller and smaller pieces.
• The initial or boundary condition terminates the recursion.

Recurrence Solving

There are 3 general methods for solving recurrences

Substitution: Guess & Verify: Guess a solution and verify: It is correct with an inductive proof.

Recurrence Tree Method: Draw a recurrence tree and calculate the time taken by every level of tree.

Master Theorem: If the recurrence has the form T(n) =aT (n/b) +f(n), then there is a formula that can (often) be applied.

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.

 

 

Click here to Download PDF for this Topic