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Induction: Basis: n = 1 ⇒ n lg n + n = 1 = T (n) Inductive step: Inductive hypothesis is that T (k) = k lg k + k for all k < n. We’ll use this inductive hypothesis for T (n/2). n +n T (n) = 2T 2 n n n lg + +n (by inductive hypothesis) = 2 2 2 2 n = n lg + n + n 2 = n(lg n − lg 2) + n + n = n lg n − n + n + n = n lg n + n . Lecture Notes for Chapter 4: Recurrences 4-3 Generally, we use asymptotic notation: • • • • We would write T (n) = 2T (n/2) + (n). We assume T (n) = O(1) for sufÞciently small n.

Interpret 2n2 + (n) = (n 2 ) as meaning for all functions f (n) ∈ (n), there exists a function g(n) ∈ (n2 ) such that 2n2 + f (n) = g(n). Can chain together: 2n 2 + 3n + 1 = 2n2 + (n) = (n 2 ) . Interpretation: • • First equation: There exists f (n) ∈ (n) such that 2n2 + 3n + 1 = 2n2 + f (n). Second equation: For all g(n) ∈ (n) (such as the f (n) used to make the Þrst equation hold), there exists h(n) ∈ (n2 ) such that 2n2 + g(n) = h(n). 3-4 Lecture Notes for Chapter 3: Growth of Functions o-notation o(g(n)) = { f (n) : for all constants c > 0, there exists a constant n 0 > 0 such that 0 ≤ f (n) < cg(n) for all n ≥ n0 } .

1 if n = 1 , T (n/3) + T (2n/3) + n if n > 1 . Solution: T (n) = (n lg n). ] Many technical issues: • • • Floors and ceilings [Floors and ceilings can easily be removed and don’t affect the solution to the recurrence. ] Exact vs. asymptotic functions Boundary conditions In algorithm analysis, we usually express both the recurrence and its solution using asymptotic notation. 4-2 Lecture Notes for Chapter 4: Recurrences • • • • Example: T (n) = 2T (n/2) + (n), with solution T (n) = (n lg n). ” When we desire an exact, rather than an asymptotic, solution, we need to deal with boundary conditions.

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Algorithms for Programmers: Ideas and Source Code by Jörg Arndt

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