The notes of Algorithms ---- Solving Recurrences

There are three solutions to solve the recurrences.

 

Substitution method:

          1. Guess the form of the solution

          2. Verify by induction

          3. Slove for consts

Recursion-tree method:

                 1. Count the height

          2.  Count the leaves.

          3. Total: level by level

The master method:

          The master method applies to recurrences of the form T(n) = aT(n/b) + f (n), where a ≥ 1, b > 1, and f is asymptotically positive.

          Method:Compare f (n) with n^log(b)a:

          case1:  f(n) = O(n^[log(b)a – ε]) for some constant ε > 0.
                      Explain: f(n) grows polynomially slower than n^log(b)a (by an n^ε factor).
                      Solution: T(n) = Θ(n^log(b)a) .

                      Ex: T(n)=4T(n/2)+n    ----------   a=4,b=2,f(n)=n.

                            n^log(b)a=n^2,  f(n)=n=O(n^2),   So the T(n)=Θ(n^2)

          case2:   f(n) = Θ(n^log(b)a*(lgn)^k) for some constant k ≥ 0.

                        Explain: f(n) and n^log(b)a grow at similar rates.

                        Solution: T(n) = Θ(n^log(b)a*(lgn)^(k+1)).

                        Ex: T(n)=4T(n/2)+n^2 ---------- a=4,b=2,f(n)=n^2.

                              n^log(b)a=n^2, f(n)=n^2=Θ(n^2*(lgn)^0), So the T(n)=Θ(n^2*lgn)

           case3:  f(n) = Ω(n^[log(b)a + ε]) for some constant ε > 0.

                        and f(n) satisfies the regularity condition that af(n/b) ≤ cf(n) for some constant c < 1.

                        (Under this condition, it can make sure that the  consts decrease geometrically, Dominate: f(n))

                        Explain: f(n) grows polynomially faster than nlogba (by an n^ε factor),

                                        Solution: T(n) = Θ(f (n)) .

                        Ex:  T(n)=4T(n/2)+n^3 ---------- a=4,b=2,f(n)=n^3.

                               n^log(b)a=n^2, f(n)=n^3=Ω(n^2), So the T(n)=Θ(n^3)