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Introduction to Algorithms 6.046J Lecture 2 Prof. Shafi Goldwasser

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L2.2 Solving recurrences The analysis of integer multiplication from Lecture 1 required us to solve a recurrence. Recurrences are a major tool for analysis of algorithms Today: Learn a few methods. Lecture 3: Divide and Conquer algorithms which are analyzable by recurrences.

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L2.3 Recall: Integer Multiplication Let X = A B and Y = C D where A,B,C and D are n/2 bit integers Simple Method: XY = (2 n/2 A+B)(2 n/2 C+D) Running Time Recurrence T(n) < 4T(n/2) + 100n How do we solve it?

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L2.4 Substitution method 1.Guess the form of the solution. 2.Verify by induction. 3.Solve for constants. The most general method: Example: T(n) = 4T(n/2) + 100n [Assume that T(1) = (1).] Guess O(n 3 ). (Prove O and separately.) Assume that T(k) ck 3 for k < n. Prove T(n) cn 3 by induction.

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L2.5 Example of substitution desired – residual whenever (c/2)n 3 – 100n 0, for example, if c 200 and n 1. desired residual 3 33 3 3 ) )2/(( )2/( )2/(4 )2/(4)( cn 100n nccn 100nnc nc nTnT

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L2.6 Example (continued) We must also handle the initial conditions, that is, ground the induction with base cases. Base: T(n) = (1) for all n < n 0, where n 0 is a suitable constant. For 1 n < n 0, we have “ (1)” cn 3, if we pick c big enough. This bound is not tight!

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L2.7 A tighter upper bound? We shall prove that T(n) = O(n 2 ). Assume that T(k) ck 2 for k < n: for no choice of c > 0. Lose! )2/(4)( 2 100ncn 100nnTnT 2 cn

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L2.8 A tighter upper bound! I DEA : Strengthen the inductive hypothesis. Subtract a low-order term. Inductive hypothesis: T(k) c 1 k 2 – c 2 k for k < n. if c 2 > 100. Pick c 1 big enough to handle the initial conditions. ) ( 2 ))2/()2/((4 )2/(4)( 2 2 1 22 2 1 2 2 1 2 2 1 ncnc 100n ncncnc ncnc ncnc nTnT

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L2.9 Recursion-tree method A recursion tree models the costs (time) of a recursive execution of an algorithm. The recursion tree method is good for generating guesses for the substitution method. The recursion-tree method can be unreliable, just like any method that uses ellipses (…). The recursion-tree method promotes intuition, however.

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L2.10 Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 :

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L2.11 Example of recursion tree T(n)T(n) Solve T(n) = T(n/4) + T(n/2) + n 2 :

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L2.12 Example of recursion tree T(n/4) T(n/2) n2n2 Solve T(n) = T(n/4) + T(n/2) + n 2 :

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L2.13 Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : n2n2 (n/4) 2 (n/2) 2 T(n/16)T(n/8) T(n/4)

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L2.14 Example of recursion tree (n/16) 2 (n/8) 2 (n/4) 2 (n/2) 2 (1) … Solve T(n) = T(n/4) + T(n/2) + n 2 : n2n2

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L2.15 Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2 (n/2) 2 (1) … n2n2

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L2.16 Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2 (n/2) 2 (1) … n2n2

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L2.17 Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2 (1) … n2n2 (n/2) 2 …

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L2.18 Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2 (1) … … Total = = (n 2 ) n2n2 (n/2) 2 geometric series

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L2.19 Appendix: geometric series for |x| < 1 for x 1 Return to last slide viewed.

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L2.20 The master method The master method applies to recurrences of the form T(n) = a T(n/b) + f (n), where a 1, b > 1, and f is asymptotically positive.

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L2.21 f (n/b) Idea of master theorem f (n/b) (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … #leaves = a h = a log b n = n log b a n log b a (1)

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L2.22 Three common cases Compare f (n) with n log b a : 1. f (n) = O(n log b a – ) for some constant > 0. f (n) grows polynomially slower than n log b a (by an n factor). Solution: T(n) = (n log b a ).

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L2.23 f (n/b) Idea of master theorem f (n/b) (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … n log b a (1) C ASE 1: The weight increases geometrically from the root to the leaves. The leaves hold a constant fraction of the total weight. (n log b a )

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L2.24 Three common cases Compare f (n) with n log b a : 2. f (n) = (n log b a lg k n) for some constant k 0. f (n) and n log b a grow at similar rates. Solution: T(n) = (n log b a lg k+1 n).

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L2.25 f (n/b) Idea of master theorem f (n/b) (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … n log b a (1) C ASE 2: (k = 0) The weight is approximately the same on each of the log b n levels. (n log b a lg n)

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L2.26 Three common cases (cont.) Compare f (n) with n log b a : 3. f (n) = (n log b a + ) for some constant > 0. f (n) grows polynomially faster than n log b a (by an n factor), and f (n) satisfies the regularity condition that a f (n/b) c f (n) for some constant c < 1. Solution: T(n) = ( f (n) ).

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L2.27 f (n/b) Idea of master theorem f (n/b) (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … n log b a (1) C ASE 3: The weight decreases geometrically from the root to the leaves. The root holds a constant fraction of the total weight. ( f (n))

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L2.28 Examples Ex. T(n) = 4T(n/2) + n a = 4, b = 2 n log b a = n 2 ; f (n) = n. C ASE 1: f (n) = O(n 2 – ) for = 1. T(n) = (n 2 ). Ex. T(n) = 4T(n/2) + n 2 a = 4, b = 2 n log b a = n 2 ; f (n) = n 2. C ASE 2: f (n) = (n 2 lg 0 n), that is, k = 0. T(n) = (n 2 lg n).

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L2.29 Examples Ex. T(n) = 4T(n/2) + n 3 a = 4, b = 2 n log b a = n 2 ; f (n) = n 3. C ASE 3: f (n) = (n 2 + ) for = 1 and 4(cn/2) 3 cn 3 (reg. cond.) for c = 1/2. T(n) = (n 3 ). Ex. T(n) = 4T(n/2) + n 2 /lg n a = 4, b = 2 n log b a = n 2 ; f (n) = n 2 /lg n. Master method does not apply. In particular, for every constant > 0, we have n (lg n).

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L2.30 Conclusion Next time: applying the master method. For proof of master theorem, goto section

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