Master theorem for divide-and-conquer recurrences
For T(n)=aT(n/b)+f(n), T(1)=Θ(1), let q=log_b a.
Key knowledge
For T(n)=aT(n/b)+f(n), T(1)=Θ(1), let q=log_b a. If f=O(n^(q−ε)), then T=Θ(n^q). If f=Θ(n^q(log n)^k), then T=Θ(n^q(log n)^(k+1)). If f=Ω(n^(q+ε)) and af(n/b)≤c f(n) eventually, then T=Θ(f(n)).
Conditions and scope
- Here a≥1,b≥2 are fixed integers, n ranges over powers of b, f is asymptotically nonnegative, ε>0, k≥0 and 0<c<1 are constants.
- The regularity inequality is required in the third case; unequal or subtractive recursive sizes require another recurrence analysis.
Sources · 1
References
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Evidence and review
Recitation3 Master Theorem table, printed5; restricted exact-power sizes avoid unstated rounding claims; recursion-tree framework read.
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