Data Structures and Algorithms CMPSC 465
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1 Data Structures and Algorithms CMPSC 465 LECTURE 9 Solving recurrences Substitution method Adam Smith S. Raskhodnikova and A. Smith; based on slides by E. Demaine and C. Leiserson
2 Review question Draw the recursion tree for the following recurrence: T(n) = 4T(n/2) + c What is the depth? How many leaves? Guess T(n)?
3 Methods for solving recurrences Recursion tree Examples in last two lectures Substitution method This lecture Master theorem
4 Substitution method The most general method: 1.Guess the form of the solution. 2.Verify by induction. 3.Solve for constants.
5 Substitution method The most general method: 1.Guess the form of the solution. 2.Verify by induction. 3.Solve for constants. EXAMPLE: T(n) = 4T(n/2) + n [Assume that T(1) = Θ(1).] Guess O(n 3 ). (Prove O and Ω separately.) Assume that T(k) ck 3 for k < n. Prove T(n) cn 3 by induction.
6 Example of substitution T ( n) = 4T ( n / 2) + n 4c( n / 2) 3 + n = ( c / 2) n3 + n = cn3 (( c / 2) n3 n) desired residual cn3 desired whenever (c/2)n 3 n 0, for example, if c 2 and n 1. residual
7 Example (continued) We must also handle the initial conditions, that is, ground the induction with base cases. Base: T(n) = Θ(1) for all n < n 0, where n 0 is a suitable constant. For 1 n < n 0, we have Θ(1) cn 3, if we pick c big enough.
8 Example (continued) We must also handle the initial conditions, that is, ground the induction with base cases. Base: T(n) = Θ(1) for all n < n 0, where n 0 is a suitable constant. For 1 n < n 0, we have Θ(1) cn 3, if we pick c big enough. This bound is not tight!
9 A tighter upper bound? We shall prove that T(n) = O(n 2 ).
10 A tighter upper bound? We shall prove that T(n) = O(n 2 ). Assume that T(k) ck 2 for k < n: T ( n) = 4T ( n / 2) + n 2 4c( n / 2) + n 2 = cn + n 2 = O( n )
11 A tighter upper bound? We shall prove that T(n) = O(n 2 ). Assume that T(k) ck 2 for k < n: T ( n) = 4T ( n / 2) + n 2 4c( n / 2) + n 2 = cn + n 2 = O( n ) Wrong! We must prove the inductive hypothesis
12 A tighter upper bound? We shall prove that T(n) = O(n 2 ). Assume that T(k) ck 2 for k < n: T ( n) = 4T ( n / 2) + n 2 4c( n / 2) + n 2 = cn + n 2 = O( n ) Wrong! We must prove the I.H. = cn2 ( n) [ desired residual ] cn2 for no choice of c > 0. Lose!
13 A tighter upper bound! IDEA: Strengthen the inductive hypothesis. Subtract a low-order term. Inductive hypothesis: T(k) c 1 k 2 c 2 k for k < n.
14 A tighter upper bound! IDEA: Strengthen the inductive hypothesis. Subtract a low-order term. Inductive hypothesis: T(k) c 1 k 2 c 2 k for k < n. T(n) = 4T(n/2) + n 4(c 1 (n/2) 2 c 2 (n/2)) + n = c 1 n 2 2c 2 n + n = c 1 n 2 c 2 n (c 2 n n) c 1 n 2 c 2 n if c 2 1.
15 A tighter upper bound! IDEA: Strengthen the inductive hypothesis. Subtract a low-order term. Inductive hypothesis: T(k) c 1 k 2 c 2 k for k < n. T(n) = 4T(n/2) + n 4(c 1 (n/2) 2 c 2 (n/2)) + n = c 1 n 2 2c 2 n + n = c 1 n 2 c 2 n (c 2 n n) c 1 n 2 c 2 n if c 2 1. Pick c 1 big enough to handle the initial conditions.
16 A matching lower bound IDEA: Formulate inductive hypothesis with Inductive hypothesis: T(k) ck 2 for all k < n. T(n) = 4T(n/2) + n 4 cn 2 /4 + n = cn 2 + n cn 2 for any c 0. Pick c small enough to handle the initial conditions.
17 Appendix: geometric series x x x = for x < x x x x x n n = for x 1
18 Data Structures and Algorithms CMPSC 465 LECTURE 10? Solving recurrences Master theorem Adam Smith S. Raskhodnikova and A. Smith; based on slides by E. Demaine and C. Leiserson
19 Review questions Guess the solution to the recurrence: T(n)=2T(n/3)+n 3/2. (Answer: Θ(n 3/2 ).) Draw the recursion tree for this recurrence. a. What is its height? b. What is the number of leaves in the tree? (Answer: h=log 3 n.) (Answer: n (1/log 3).)
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, that is f (n) >0 for all n > n 0.
21 Three common cases Compare f (n) with n log ba : 1. f (n) = O(n log ba ε ) for some constant ε > 0. f (n) grows polynomially slower than n log ba (by an n ε factor). Solution: T(n) = Θ(n log ba ).
22 Three common cases Compare f (n) with n log ba : 1. f (n) = O(n log ba ε ) for some constant ε > 0. f (n) grows polynomially slower than n log ba (by an n ε factor). Solution: T(n) = Θ(n log ba ). 2. f (n) = Θ(n log ba lg k n) for some constant k 0. f (n) and n log ba grow at similar rates. Solution: T(n) = Θ(n log ba lg k+1 n).
23 Three common cases (cont.) Compare f (n) with n log ba : 3. f (n) = Ω(n log ba + ε ) for some constant ε > 0. f (n) grows polynomially faster than n log ba (by an n ε factor), and f (n) satisfies the regularity condition that a f (n/b) c f (n) for some constant c < 1. Solution: T(n) = Θ( f (n) ).
24 Idea of master theorem Recursion tree: f (n) a f (n/b) f (n/b) a f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) f (n/b) Τ (1)
25 Idea of master theorem Recursion tree: f (n) a f (n/b) f (n/b) a f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) f (n/b) f (n) a f (n/b) a 2 f (n/b 2 ) Τ (1)
26 Idea of master theorem h = log b n Recursion tree: f (n) a f (n/b) f (n/b) a f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) f (n/b) f (n) a f (n/b) a 2 f (n/b 2 ) Τ (1)
27 Idea of master theorem h = log b n Recursion tree: f (n) a f (n/b) f (n/b) a f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) f (n/b) f (n) a f (n/b) a 2 f (n/b 2 ) Τ (1) #leaves = a h = a log bn = n log ba n log ba Τ (1)
28 Idea of master theorem h = log b n Recursion tree: f (n) a f (n/b) f (n/b) a f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) f (n/b) f (n) a f (n/b) a 2 f (n/b 2 ) Τ (1) CASE 1: The weight increases geometrically from the root to the leaves. The leaves hold a constant fraction of the total weight. n log ba Τ (1) Θ(n log ba )
29 Idea of master theorem h = log b n Recursion tree: f (n) a f (n/b) f (n/b) a f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) f (n/b) f (n) a f (n/b) a 2 f (n/b 2 ) Τ (1) CASE 2: (k = 0) The weight is approximately the same on each of the log b n levels. n log ba Τ (1) Θ(n log ba lg n)
30 Idea of master theorem h = log b n Recursion tree: f (n) a f (n/b) f (n/b) a f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) f (n/b) f (n) a f (n/b) a 2 f (n/b 2 ) Τ (1) CASE 3: The weight decreases geometrically from the root to the leaves. The root holds a constant fraction of the total weight. n log ba Τ (1) Θ( f (n))
31 Examples EX. T(n) = 4T(n/2) + n a = 4, b = 2 n log ba = n 2 ; f (n) = n. CASE 1: f (n) = O(n 2 ε ) for ε = 1. T(n) = Θ(n 2 ).
32 Examples EX. T(n) = 4T(n/2) + n a = 4, b = 2 n log ba = n 2 ; f (n) = n. CASE 1: f (n) = O(n 2 ε ) for ε = 1. T(n) = Θ(n 2 ). EX. T(n) = 4T(n/2) + n 2 a = 4, b = 2 n log ba = n 2 ; f (n) = n 2. CASE 2: f (n) = Θ(n 2 lg 0 n), that is, k = 0. T(n) = Θ(n 2 lg n).
33 Examples EX. T(n) = 4T(n/2) + n 3 a = 4, b = 2 n log ba = n 2 ; f (n) = n 3. CASE 3: f (n) = Ω(n 2 + ε ) for ε = 1 and 4(n/2) 3 cn 3 (reg. cond.) for c = 1/2. T(n) = Θ(n 3 ).
34 Examples EX. T(n) = 4T(n/2) + n 3 a = 4, b = 2 n log ba = n 2 ; f (n) = n 3. CASE 3: f (n) = Ω(n 2 + ε ) for ε = 1 and 4(n/2) 3 cn 3 (reg. cond.) for c = 1/2. T(n) = Θ(n 3 ). EX. T(n) = 4T(n/2) + n 2 /lg n a = 4, b = 2 n log ba = n 2 ; f (n) = n 2 /lg n. Master method does not apply. In particular, for every constant ε > 0, we have n ε = ω(lg n).
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