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Recurrence, Summations

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Recurrence, Summations

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    1. Recurrence, Summations Reading: Chap 4

    2. Recurrence Solve for n=2k: T(n)=lg n + 1=T(lg n)

    3. Boundary condition Solve: T(n)=(T(n/2))2 T(1)=2 ? T(n)=T(2n) T(1)=3 ? T(n)=T(3n) T(1)=1 ? T(n)=T(1) Different B.C. ? different solution.

    4. Solving recurrences Substitution method Iterating the recurrence Recursion tree Master method Generating function

    5. Changing variables Let m = lg n, i.e. 2m=n. ? T(2m)=2T(2m/2)+m Let S(m)=T(2m). ? S(m)=2S(m/2)+m By recursion tree method: S(m)=O(m lg m) T(n)=T(2m)=S(m)=O(m lg m)=O(lg n lglg n)

    6. Substitution method (4.1) Guess the solution Verify by induction and solve for const eg. T(n)=4T(n/2)+n Guess T(n)=O(n3) Assume T(k)=ck3 for k<n Prove T(n)=cn3 by induction. But O(n3) is not a tight bound! Can show O(n2)!

    7. Fallacious argument Assume T(k)=ck2 for k<n

    8. Correct Solution Substract a lower-order term. Assume T(k)=c1k2-c2k for k<n

    9. Iterating the recurrence (4.2)

    10. Recursion tree Visualizing iterating method eg. T(n)=T(n/4)+T(n/2)+n2

    11. Recursion tree (cont.)

    12. Master Method For T(n)=aT(n/b)+f(n), T(1)=T(1) Total :

    13. Case 1: nlogba/f(n)=O(ne), for some const e>0. weight of each level increases geometrically from root to leaves. Leaves contain constant fraction of total weight. T(n)=(nlogba)

    14. Case 2: f(n)/nlogba=T(logkn), for some k=0. I.e. f(n) and are within a polylogarithmic factor. Weight decreases T(n)=T(nlogbalogk+1n)

    15. Case 3: f(n)/nlogba=O(ne), for some const e>0. Weight is geometrically decreasing. Root contains constant fraction of total weight. T(n)=T(f(n)) Note: the above cases doesnt cover all possible cases!

    16. Master method examples Merge sort: T(n)=2T(n/2)+T(n) a=b=2, f(n)=n nlog22=n f(n)/n=T(1)=T(log0n), case 2. T(n)= T(nlog22log0+1n)= T(n log n) T(n)=4T(n/2)+n3 n3/nlog24=n, case 3. T(n)=T(n3)

    17. An Exception T(n)=4T(n/2)+n2/lg n No case applies! Sol: T(n)=(n2lglg n) ~ by substitution method!

    18. Generating function Fibonacci numbers: F0=0, F1=1, Fi=Fi-1+Fi-2 for i = 2. 0, 1, 1, 2, 3, 5, 8, (1-z-z2)G(z)=z, G(z)=z/(1-z-z2)

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