1. Master Method Some of the slides are from Prof. Plaisted’s resources at University of North Carolina at Chapel Hill

2. COT5407 The Master Method  Based on the Master theorem.  “Cookbook” approach for solving recurrences of the form T(n) = aT(n/b) + f(n) a  1, b > 1 are constants. f(n) is asymptotically positive. n/b may not be an integer, but we ignore floors and ceilings.  Requires memorization of three cases.

3. COT5407 The Master Theorem Theorem 4.1 Let a  1 and b > 1 be constants, let f(n) be a function, and Let T(n) be defined on nonnegative integers by the recurrence T(n) = aT(n/b) + f(n), where we can replace n/b by  n/b  or  n/b . T(n) can be bounded asymptotically in three cases: 1.If f(n) = O(n log b a–  ) for some constant  > 0, then T(n) =  ( n log b a ). 2.If f(n) =  (n log b a ), then T(n) =  ( n log b a lg n). 3.If f(n) =  (n log b a+  ) for some constant  > 0, and if, for some constant c < 1 and all sufficiently large n, we have a·f(n/b)  c f(n), then T(n) =  (f(n)). Theorem 4.1 Let a  1 and b > 1 be constants, let f(n) be a function, and Let T(n) be defined on nonnegative integers by the recurrence T(n) = aT(n/b) + f(n), where we can replace n/b by  n/b  or  n/b . T(n) can be bounded asymptotically in three cases: 1.If f(n) = O(n log b a–  ) for some constant  > 0, then T(n) =  ( n log b a ). 2.If f(n) =  (n log b a ), then T(n) =  ( n log b a lg n). 3.If f(n) =  (n log b a+  ) for some constant  > 0, and if, for some constant c < 1 and all sufficiently large n, we have a·f(n/b)  c f(n), then T(n) =  (f(n)).

4. COT5407 Recursion tree view f(n)f(n) af(n/b) a 2 f(n/b 2 )  (n log b a ) Total:

5. COT5407 The Master Theorem Theorem 4.1 Let a  1 and b > 1 be constants, let f(n) be a function, and Let T(n) be defined on nonnegative integers by the recurrence T(n) = aT(n/b) + f(n), where we can replace n/b by  n/b  or  n/b . T(n) can be bounded asymptotically in three cases: 1.If f(n) = O(n log b a–  ) for some constant  > 0, then T(n) =  ( n log b a ). 2.If f(n) =  (n log b a ), then T(n) =  ( n log b a lg n). 3.If f(n) =  (n log b a+  ) for some constant  > 0, and if, for some constant c < 1 and all sufficiently large n, we have a·f(n/b)  c f(n), then T(n) =  (f(n)). Theorem 4.1 Let a  1 and b > 1 be constants, let f(n) be a function, and Let T(n) be defined on nonnegative integers by the recurrence T(n) = aT(n/b) + f(n), where we can replace n/b by  n/b  or  n/b . T(n) can be bounded asymptotically in three cases: 1.If f(n) = O(n log b a–  ) for some constant  > 0, then T(n) =  ( n log b a ). 2.If f(n) =  (n log b a ), then T(n) =  ( n log b a lg n). 3.If f(n) =  (n log b a+  ) for some constant  > 0, and if, for some constant c < 1 and all sufficiently large n, we have a·f(n/b)  c f(n), then T(n) =  (f(n)).

6. COT5407 Master Method – Examples  T(n) = 16T(n/4)+n  a = 16, b = 4, n log b a = n log 4 16 = n 2.  f(n) = n = O(n log b a-  ) = O(n 2-  ), where  = 1  Case 1.  Hence, T(n) =  (n log b a ) =  (n 2 ).  T(n) = T(3n/7) + 1  a = 1, b=7/3, and n log b a = n log 7/3 1 = n 0 = 1  f(n) = 1 =  (n log b a )  Case 2.  Therefore, T(n) =  (n log b a lg n) =  (lg n)

7. COT5407 Master Method – Examples  T(n) = 3T(n/4) + n lg n  a = 3, b=4, thus n log b a = n log 4 3 = O(n 0.793 )  f(n) = n lg n =  (n log 4 3 +  ) where   0.2  Case 3.  Therefore, T(n) =  (f(n)) =  (n lg n).  T(n) = 2T(n/2) + n lg n  a = 2, b=2, f(n) = n lg n, and n log b a = n log 2 2 = n  f(n) is asymptotically larger than n log b a, but not polynomially larger. The ratio lg n is asymptotically less than n  for any positive . Thus, the Master Theorem doesn’t apply here.

8. COT5407 Master Theorem – What it means?  Case 1: If f(n) = O(n log b a–  ) for some constant  > 0, then T(n) =  ( n log b a ).  n log b a = a log b n : Number of leaves in the recursion tree.  f(n) = O(n log b a–  )  Sum of the cost of the nodes at each internal level asymptotically smaller than the cost of leaves by a polynomial factor.  Cost of the problem dominated by leaves, hence cost is  ( n log b a ).

9. COT5407 Master Theorem – What it means?  Case 2: If f(n) =  (n log b a ), then T(n) =  ( n log b a lg n).  n log b a = a log b n : Number of leaves in the recursion tree.  f(n) =  (n log b a )  Sum of the cost of the nodes at each level asymptotically the same as the cost of leaves.  There are  (lg n) levels.  Hence, total cost is  ( n log b a lg n).

10. COT5407 Master Theorem – What it means?  Case 3: If f(n) =  (n log b a+  ) for some constant  > 0, and if, for some constant c < 1 and all sufficiently large n, we have a·f(n/b)  c f(n), then T(n) =  (f(n)).  n log b a = a log b n : Number of leaves in the recursion tree.  f(n) =  (n log b a+  )  Cost is dominated by the root. Cost of the root is asymptotically larger than the sum of the cost of the leaves by a polynomial factor.  Hence, cost is  ( f(n)).

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12. COT5407 Master Theorem – Proof for exact powers  Proof when n is an exact power of b.  Three steps. 1.Reduce the problem of solving the recurrence to the problem of evaluating an expression that contains a summation. 2.Determine bounds on the summation. 3.Combine 1 and 2.

13. 13 Iterative “Proof” of the Master Theorem  Using iterative substitution, let us see if we can find a pattern:  We then distinguish the three cases as  The first term is dominant  Each part of the summation is equally dominant  The summation is a geometric series

14. COT5407 f(n)f(n) af(n/b) a 2 f(n/b 2 )  (n log b a ) Total:

15. Integer Multiplication  Algorithm: Multiply two n-bit integers I and J.  Divide step: Split I and J into high-order and low-order bits  We can then define I*J by multiplying the parts and adding:  So, T(n) = 4T(n/2) + n, which implies T(n) is O(n 2 ).  But that is no better than the algorithm we learned in grade school.

16. An Improved Integer Multiplication Algorithm  Algorithm: Multiply two n-bit integers I and J.  Divide step: Split I and J into high-order and low-order bits  Observe that there is a different way to multiply parts:  So, T(n) = 3T(n/2) + n, which implies T(n) is O(n log 2 3 ), by the Master Theorem.  Thus, T(n) is O(n 1.585 ).