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02 asymptotic-notation-and-recurrences
1.
Introduction to Algorithms
6.046J/18.401J LECTURE 2 Asymptotic Notation • O-, -, and -notation Recurrences • Substitution method • Iterating the recurrence • Recursion tree • Master method Prof. Erik Demaine September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.1
2.
Asymptotic notation O-notation
(upper bounds): We write f(n) = O(g(n)) if there exist constants c > 0, n0 > 0 such that 0 f(n) cg(n) for all n n0. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.2
3.
Asymptotic notation O-notation
(upper bounds): We write f(n) = O(g(n)) if there exist constants c > 0, n0 > 0 such that 0 f(n) cg(n) for all n n0. EXAMPLE: 2n2 = O(n3) (c = 1, n0 = 2) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.3
4.
Asymptotic notation O-notation
(upper bounds): We write f(n) = O(g(n)) if there exist constants c > 0, n0 > 0 such that 0 f(n) cg(n) for all n n0. EXAMPLE: 2n2 = O(n3) (c = 1, n0 = 2) functions, not values September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.4
5.
Asymptotic notation O-notation
(upper bounds): We write f(n) = O(g(n)) if there exist constants c > 0, n0 > 0 such that 0 f(n) cg(n) for all n n0. EXAMPLE: 2n2 = O(n3) (c = 1, n0 = 2) funny, “one-way” functions, equality not values September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.5
6.
Set definition of
O-notation O(g(n)) = { f(n) : there exist constants c > 0, n0 > 0 such that 0 f(n) cg(n) for all n n0 } September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.6
7.
Set definition of
O-notation O(g(n)) = { f(n) : there exist constants c > 0, n0 > 0 such that 0 f(n) cg(n) for all n n0 } EXAMPLE: 2n2 O(n3) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.7
8.
Set definition of
O-notation O(g(n)) = { f(n) : there exist constants c > 0, n0 > 0 such that 0 f(n) cg(n) for all n n0 } EXAMPLE: 2n2 O(n3) (Logicians: n.2n2 O(n.n3), but it’s convenient to be sloppy, as long as we understand what’s really going on.) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.8
9.
Macro substitution Convention: A
set in a formula represents an anonymous function in the set. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.9
10.
Macro substitution Convention: A
set in a formula represents an anonymous function in the set. EXAMPLE: f(n) = n3 + O(n2) means f(n) = n3 + h(n) for some h(n) O(n2) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.10
11.
Macro substitution Convention: A
set in a formula represents an anonymous function in the set. EXAMPLE: n2 + O(n) = O(n2) means for any f(n) O(n): n2 + f(n) = h(n) for some h(n) O(n2) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.11
12.
-notation (lower bounds) O-notation
is an upper-bound notation. It makes no sense to say f(n) is at least O(n2). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.12
13.
-notation (lower bounds) O-notation
is an upper-bound notation. It makes no sense to say f(n) is at least O(n2). g(n)) = { f(n) : there exist constants c > 0, n0 > 0 such that 0 cg(n) f(n) for all n n0 } September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.13
14.
-notation (lower bounds) O-notation
is an upper-bound notation. It makes no sense to say f(n) is at least O(n2). g(n)) = { f(n) : there exist constants c > 0, n0 > 0 such that 0 cg(n) f(n) for all n n0 } EXAMPLE: n (lg n) (c = 1, n0 = 16) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.14
15.
-notation (tight bounds)
g(n)) = O(g(n)) (g(n)) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.15
16.
-notation (tight bounds)
g(n)) = O(g(n)) (g(n)) 1 2 2 EXAMPLE: 2 n 2 n ( n ) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.16
17.
-notation and -notation O-notation
and -notation are like and . o-notation and -notation are like < and >. g(n)) = { f(n) : for any constant c > 0, there is a constant n0 > 0 such that 0 f(n) cg(n) for all n n0 } EXAMPLE: 2n2 = o(n3) (n0 = 2/c) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.17
18.
-notation and -notation O-notation
and -notation are like and . o-notation and -notation are like < and >. g(n)) = { f(n) : for any constant c > 0, there is a constant n0 > 0 such that 0 cg(n) f(n) for all n n0 } EXAMPLE: n (lg n) (n0 = 1+1/c) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.18
19.
Solving recurrences
• The analysis of merge sort from Lecture 1 required us to solve a recurrence. • Recurrences are like solving integrals, differential equations, etc. Learn a few tricks. • Lecture 3: Applications of recurrences to divide-and-conquer algorithms. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.19
20.
Substitution method
The most general method: 1. Guess the form of the solution. 2. Verify by induction. 3. Solve for constants. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.20
21.
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) = (1).] • Guess O(n3) . (Prove O and separately.) • Assume that T(k) ck3 for k < n . • Prove T(n) cn3 by induction. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.21
22.
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)n3 – n 0, for example, if c 2 and n 1. residual September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.22
23.
Example (continued) •
We must also handle the initial conditions, that is, ground the induction with base cases. • Base: T(n) = (1) for all n < n0, where n0 is a suitable constant. • For 1 n < n0, we have “ (1)” cn3, if we pick c big enough. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.23
24.
Example (continued) •
We must also handle the initial conditions, that is, ground the induction with base cases. • Base: T(n) = (1) for all n < n0, where n0 is a suitable constant. • For 1 n < n0, we have “ (1)” cn3, if we pick c big enough. This bound is not tight! September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.24
25.
A tighter upper
bound? We shall prove that T(n) = O(n2). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.25
26.
A tighter upper
bound? We shall prove that T(n) = O(n2). Assume that T(k) ck2 for k < n: T (n) 4T (n / 2) n 4c ( n / 2) 2 n cn 2 n 2 O(n ) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.26
27.
A tighter upper
bound? We shall prove that T(n) = O(n2). Assume that T(k) ck2 for k < n: T (n) 4T (n / 2) n 4c ( n / 2) 2 n cn 2 n 2 O(n ) Wrong! We must prove the I.H. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.27
28.
A tighter upper
bound? We shall prove that T(n) = O(n2). Assume that T(k) ck2 for k < n: T (n) 4T (n / 2) n 4c ( n / 2) 2 n cn 2 n 2 O(n ) Wrong! We must prove the I.H. cn 2 ( n) [ desired – residual ] cn 2 for no choice of c > 0. Lose! September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.28
29.
A tighter upper
bound! IDEA: Strengthen the inductive hypothesis. • Subtract a low-order term. Inductive hypothesis: T(k) c1k2 – c2k for k < n. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.29
30.
A tighter upper
bound! IDEA: Strengthen the inductive hypothesis. • Subtract a low-order term. Inductive hypothesis: T(k) c1k2 – c2k for k < n. T(n) = 4T(n/2) + n = 4(c1(n/2)2 – c2(n/2)) + n = c1n2 – 2c2n + n = c1n2 – c2n – (c2n – n) c1n2 – c2n if c2 1. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.30
31.
A tighter upper
bound! IDEA: Strengthen the inductive hypothesis. • Subtract a low-order term. Inductive hypothesis: T(k) c1k2 – c2k for k < n. T(n) = 4T(n/2) + n = 4(c1(n/2)2 – c2(n/2)) + n = c1n2 – 2c2n + n = c1n2 – c2n – (c2n – n) c1n2 – c2n if c2 1. Pick c1 big enough to handle the initial conditions. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.31
32.
Recursion-tree method • A
recursion tree models the costs (time) of a recursive execution of an algorithm. • The recursion-tree method can be unreliable, just like any method that uses ellipses (…). • The recursion-tree method promotes intuition, however. • The recursion tree method is good for generating guesses for the substitution method. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.32
33.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.33
34.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: T(n) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.34
35.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 T(n/4) T(n/2) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.35
36.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 T(n/16) T(n/8) T(n/8) T(n/4) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.36
37.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.37
38.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.38
39.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 5 n2 (n/4)2 (n/2)2 16 (n/16)2 (n/8)2 (n/8)2 (n/4)2 (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.39
40.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 5 n2 (n/4)2 (n/2)2 16 25 n 2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 256 … (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.40
41.
Example of recursion
tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 5 n2 (n/4)2 (n/2)2 16 25 n 2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 256 … (1) Total = n 2 5 5 2 1 16 16 5 3 16 = (n2) geometric series September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.41
42.
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. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.42
43.
Three common cases Compare
f (n) with nlogba: 1. f (n) = O(nlogba – ) for some constant > 0. • f (n) grows polynomially slower than nlogba (by an n factor). Solution: T(n) = (nlogba) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.43
44.
Three common cases Compare
f (n) with nlogba: 1. f (n) = O(nlogba – ) for some constant > 0. • f (n) grows polynomially slower than nlogba (by an n factor). Solution: T(n) = (nlogba) . 2. f (n) = (nlogba lgkn) for some constant k 0. • f (n) and nlogba grow at similar rates. Solution: T(n) = (nlogba lgk+1n) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.44
45.
Three common cases
(cont.) Compare f (n) with nlogba: 3. f (n) = (nlogba + ) for some constant > 0. • f (n) grows polynomially faster than nlogba (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)) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.45
46.
Examples EX. T(n) =
4T(n/2) + n a = 4, b = 2 nlogba = n2; f (n) = n. CASE 1: f (n) = O(n2 – ) for = 1. T(n) = (n2). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.46
47.
Examples EX. T(n) =
4T(n/2) + n a = 4, b = 2 nlogba = n2; f (n) = n. CASE 1: f (n) = O(n2 – ) for = 1. T(n) = (n2). EX. T(n) = 4T(n/2) + n2 a = 4, b = 2 nlogba = n2; f (n) = n2. CASE 2: f (n) = (n2lg0n), that is, k = 0. T(n) = (n2lg n). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.47
48.
Examples EX. T(n) =
4T(n/2) + n3 a = 4, b = 2 nlogba = n2; f (n) = n3. CASE 3: f (n) = (n2 + ) for = 1 and 4(n/2)3 cn3 (reg. cond.) for c = 1/2. T(n) = (n3). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.48
49.
Examples EX. T(n) =
4T(n/2) + n3 a = 4, b = 2 nlogba = n2; f (n) = n3. CASE 3: f (n) = (n2 + ) for = 1 and 4(n/2)3 cn3 (reg. cond.) for c = 1/2. T(n) = (n3). EX. T(n) = 4T(n/2) + n2/lg n a = 4, b = 2 nlogba = n2; f (n) = n2/lg n. Master method does not apply. In particular, for every constant > 0, we have n (lg n). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.49
50.
Idea of master
theorem Recursion tree: a f (n) f (n/b) f (n/b) … f (n/b) a f (n/b2) f (n/b2) … f (n/b2) (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.50
51.
Idea of master
theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.51
52.
Idea of master
theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.52
53.
Idea of master
theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) #leaves = ah … (1) = alogbn nlogba (1) = nlogba September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.53
54.
Idea of master
theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) CASE 1: The weight increases … geometrically from the root to the (1) leaves. The leaves hold a constant nlogba (1) fraction of the total weight. (nlogba) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.54
55.
Idea of master
theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … CASE 2: (k = 0) The weight (1) is approximately the same on nlogba (1) each of the logbn levels. (nlogbalg n) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.55
56.
Idea of master
theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) CASE 3: The weight decreases … geometrically from the root to the (1) leaves. The root holds a constant nlogba (1) fraction of the total weight. ( f (n)) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.56
57.
Appendix: geometric series
1 x n 1 1 x x2 xn for x 1 1 x 1 2 1 x x for |x| < 1 1 x Return to last slide viewed. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.57
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