AppliedMultiple choice

What does the recurrence T(n)=4T(n/2)+Θ(n2)T(n) = 4T(n/2) + \Theta(n^2) solve to?

  1. AΘ(n2)\Theta(n^2), because the combine step already costs n2n^2
  2. BΘ(n2log⁡n)\Theta(n^2\log n)
  3. CΘ(n4)\Theta(n^4), from four subproblems of quadratic cost
  4. DΘ(nlog⁡n)\Theta(n\log n), as for merge sort

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