Divide and Conquer

Divide and Conquer 2– Run-time determined by the size and number of subproblems to be solved in addition to the time required fordecomposition– General complexity computationT n =where a and b are known constants; assume that n = b k• Example, mergesort recurrenceSolution for the mergesort recurrence: Θ(n lg n){T1 n = 1aT n/b + f(n) n > 1{ Θ(1) if n = 1T n =2T n + Θ(n) if n > 12• You can ignore extreme details like floor, ceiling, and boundary in recurrence description.• Solving recurrence relations by substitution method– Guess the form of solution and use mathematical induction to find constants– Determine upper bound on the recurrenceGuess the solution as: T n = O(n lg n)Now, prove that T n ≤ cn lg n for some c > 0Assume that the bound holds for ⌊ ⌋n2Substituting into the recurrenceT n = 2T ⌊ n2 ⌋ + n⌊ n⌋T n ≤ 2(c2Boundary condition: Let the only bound be T 1 = 1≤cn lg⌊ n⌋lg(2( n2)+ n)) + n= cn lg n − cn lg 2 + n= cn lg n − cn + n≤ cn lg n ∀c ≥ 1̸ ∃c | T 1 ≤ c1 lg 1 = 0Problem overcome by the fact that asymptotic notation requires us to proveInclude T 2 and T 3 as boundary conditions for the proofChoose c such that T 2 ≤ c2 lg 2 and T 3 ≤ c3 lg 3True for any c ≥ 2– Making a good guessT n ≤ cn lg n for n ≥ n 0T 2 = 4 T 3 = 5∗ If a recurrence is similar to a known recurrence, it is reasonable to guess a similar solutionT n = 2T ⌊ n2 ⌋ + nIf n is large, difference between T ⌊ n2 ⌋ and T ⌊ n2 ⌋+17 is relatively small∗ Prove upper and lower bounds on a recurrence and reduce the range of uncertaintyStart with a lower bound of T n = Ω(n) and an initial upper bound of T n = O(n 2 ). Gradually lower the upperbound and raise the lower bound to get asymptotically tight solution of T n = Θ(n lg n)