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# Lecture11 - 2 n for n>= 2 Induction Base Case Induction...

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CSE 421 Algorithms Richard Anderson Lecture 11 Recurrences

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Divide and Conquer Recurrences, Sections 5.1 and 5.2 Algorithms Counting Inversions (5.3) Closest Pair (5.4) Multiplication (5.5) FFT (5.6)
Divide and Conquer Array Mergesort(Array a){ n = a.Length; if (n <= 1) return a; b = Mergesort(a[0 .. n/2]); c = Mergesort(a[n/2+1 .. n-1]); return Merge(b, c); }

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Algorithm Analysis Cost of Merge Cost of Mergesort
T(n) <= 2T(n/2) + cn; T(2) <= c;

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Recurrence Analysis Solution methods Unrolling recurrence Guess and verify Plugging in to a “Master Theorem”
Unrolling the recurrence

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Substitution Prove T(n) <= cn log

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Unformatted text preview: 2 n for n >= 2 Induction: Base Case: Induction Hypothesis: T(n/2) <= c(n/2) log 2 (n/2) A better mergesort (?) • Divide into 3 subarrays and recursively sort • Apply 3-way merge What is the recurrence? Unroll recurrence for T(n) = 3T(n/3) + dn T(n) = aT(n/b) + f(n) T(n) = T(n/2) + cn Where does this recurrence arise? Solving the recurrence exactly T(n) = 4T(n/2) + cn T(n) = 2T(n/2) + n 2 T(n) = 2T(n/2) + n 1/2 Recurrences • Three basic behaviors – Dominated by initial case – Dominated by base case – All cases equal – we care about the depth...
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Lecture11 - 2 n for n>= 2 Induction Base Case Induction...

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