Master Method

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

2. 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. COT5407

3. The Master Theorem • Theorem 4.1 • Let a  1 and b > 1be 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: • If f(n) = O(nlogba–) for some constant  > 0, then T(n) = (nlogba). • If f(n) = (nlogba), then T(n) = (nlogbalg n). • If f(n) = (nlogba+) 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)). COT5407

4. Recursion tree view f(n) a f(n/b) f(n/b) … f(n/b) a a a … … … f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) a a a a a a … … … … … … Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) f(n) af(n/b) a2f(n/b2) (nlogba) Total: COT5407

5. The Master Theorem • Theorem 4.1 • Let a  1 and b > 1be 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: • If f(n) = O(nlogba–) for some constant  > 0, then T(n) = (nlogba). • If f(n) = (nlogba), then T(n) = (nlogbalg n). • If f(n) = (nlogba+) 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)). COT5407

6. Master Method – Examples • T(n) = 16T(n/4)+n • a = 16, b = 4, nlogba = nlog416 = n2. • f(n) = n = O(nlogba-) = O(n2- ), where  = 1  Case 1. • Hence, T(n) = (nlogba ) = (n2). • T(n) = T(3n/7) + 1 • a = 1, b=7/3, and nlogba = nlog7/31 = n0 = 1 • f(n) = 1 = (nlogba) Case 2. • Therefore, T(n) = (nlogba lg n) = (lg n) COT5407

7. Master Method – Examples • T(n) = 3T(n/4) + n lg n • a = 3, b=4, thus nlogba = nlog43 = O(n0.793) • f(n) = n lg n = (nlog43 + )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 nlogba = nlog22 = n • f(n)is asymptotically larger thannlogba, but not polynomially larger. The ratio lg n is asymptotically less than n for any positive . Thus, the Master Theorem doesn’t apply here. COT5407

8. Master Theorem – What it means? • Case 1:If f(n) = O(nlogba–) for some constant  > 0, then T(n) = (nlogba). • nlogba = alogbn: Number of leaves in the recursion tree. • f(n) = O(nlogba–)  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 (nlogba). COT5407

9. Master Theorem – What it means? • Case 2:If f(n) = (nlogba), then T(n) = (nlogbalg n). • nlogba = alogbn : Number of leaves in the recursion tree. • f(n) = (nlogba)  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 (nlogba lg n). COT5407

10. Master Theorem – What it means? • Case 3:If f(n) = (nlogba+) 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)). • nlogba = alogbn: Number of leaves in the recursion tree. • f(n) = (nlogba+)  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)). COT5407

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

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. f(n) a f(n/b) f(n/b) … f(n/b) a a a … … … f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) f(n/b2) a a a a a a … … … … … … Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) Θ(1) f(n) af(n/b) a2f(n/b2) (nlogba) Total: COT5407

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(n2). • 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(nlog23), by the Master Theorem. • Thus, T(n) is O(n1.585).