Data Structures and Algorithms
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1 Data Structures and Algorithms CS S-03 Recursive Function Analysis David Galles Department of Computer Science University of San Francisco
2 03-0: Algorithm Analysis for (i=1; i<=n*n; i++) for (j=0; j<i; j++) sum++;
3 03-1: Algorithm Analysis for (i=1; i<=n*n; i++) Executed n*n times for (j=0; j<i; j++) Executed <= n*n times sum++; O(1) Running Time: O(n 4 )
4 03-2: Algorithm Analysis for (i=1; i<=n*n; i++) for (j=0; j<i; j++) sum++; Exact # of times sum++ is executed: n 2 i=1 i = n2 (n 2 +1) 2 = n4 +n 2 2 Θ(n 4 )
5 03-3: Recursive Functions long power(long x, long n) { if (n == 0) return 1; else return x * power(x, n-1); }
6 03-4: Recurrence Relations T(n) = Time required to solve a problem of size n Recurrence relations are used to determine the running time of recursive programs recurrence relations themselves are recursive T(0) = time to solve problem of size 0 Base Case T(n) = time to solve problem of size n Recursive Case
7 03-5: Recurrence Relations long power(long x, long n) { if (n == 0) return 1; else return x * power(x, n-1); } T(0) = c 1 for some constant c 1 T(n) = c 2 +T(n 1) for some constant c 2
8 03-6: Solving Recurrence Relations T(0) = c 1 T(n) = T(n 1)+c 2 If we knew T(n 1), we could solve T(n). T(n) = T(n 1)+c 2
9 03-7: Solving Recurrence Relations T(0) = c 1 T(n) = T(n 1)+c 2 If we knew T(n 1), we could solve T(n). T(n) = T(n 1)+c 2 T(n 1) = T(n 2)+c 2 = T(n 2)+c 2 +c 2 = T(n 2)+2c 2
10 03-8: Solving Recurrence Relations T(0) = c 1 T(n) = T(n 1)+c 2 If we knew T(n 1), we could solve T(n). T(n) = T(n 1)+c 2 T(n 1) = T(n 2)+c 2 = T(n 2)+c 2 +c 2 = T(n 2)+2c 2 T(n 2) = T(n 3)+c 2 = T(n 3)+c 2 +2c 2 = T(n 3)+3c 2
11 03-9: Solving Recurrence Relations T(0) = c 1 T(n) = T(n 1)+c 2 If we knew T(n 1), we could solve T(n). T(n) = T(n 1)+c 2 T(n 1) = T(n 2)+c 2 = T(n 2)+c 2 +c 2 = T(n 2)+2c 2 T(n 2) = T(n 3)+c 2 = T(n 3)+c 2 +2c 2 = T(n 3)+3c 2 T(n 3) = T(n 4)+c 2 = T(n 4)+4c 2
12 03-10: Solving Recurrence Relations T(0) = c 1 T(n) = T(n 1)+c 2 If we knew T(n 1), we could solve T(n). T(n) = T(n 1)+c 2 T(n 1) = T(n 2)+c 2 = T(n 2)+c 2 +c 2 = T(n 2)+2c 2 T(n 2) = T(n 3)+c 2 = T(n 3)+c 2 +2c 2 = T(n 3)+3c 2 T(n 3) = T(n 4)+c 2 = T(n 4)+4c 2 = = T(n k)+kc 2
13 03-11: Solving Recurrence Relations T(0) = c 1 T(n) = T(n k)+k c 2 for all k If we set k = n, we have: T(n) = T(n n)+nc 2 = T(0)+nc 2 = c 1 +nc 2 Θ(n)
14 03-12: Building a Better Power long power(long x, long n) { if (n==0) return 1; if (n==1) return x; if ((n % 2) == 0) return power(x*x, n/2); else return power(x*x, n/2) * x; }
15 03-13: Building a Better Power long power(long x, long n) { if (n==0) return 1; if (n==1) return x; if ((n % 2) == 0) return power(x*x, n/2); else return power(x*x, n/2) * x; } T(0) = c 1 T(1) = c 2 T(n) = T(n/2)+c 3 (Assume n is a power of 2)
16 03-14: Solving Recurrence Relations T(n) = T(n/2)+c 3
17 03-15: Solving Recurrence Relations T(n) = T(n/2)+c 3 T(n/2) = T(n/4)+c 3 = T(n/4)+c 3 +c 3 = T(n/4)+2c 3
18 03-16: Solving Recurrence Relations T(n) = T(n/2)+c 3 T(n/2) = T(n/4)+c 3 = T(n/4)+c 3 +c 3 = T(n/4)+2c 3 T(n/4) = T(n/8)+c 3 = T(n/8)+c 3 +2c 3 = T(n/8)+3c 3
19 03-17: Solving Recurrence Relations T(n) = T(n/2)+c 3 T(n/2) = T(n/4)+c 3 = T(n/4)+c 3 +c 3 = T(n/4)+2c 3 T(n/4) = T(n/8)+c 3 = T(n/8)+c 3 +2c 3 = T(n/8)+3c 3 T(n/8) = T(n/16)+c 3 = T(n/16)+c 3 +3c 3 = T(n/16)+4c 3
20 03-18: Solving Recurrence Relations T(n) = T(n/2)+c 3 T(n/2) = T(n/4)+c 3 = T(n/4)+c 3 +c 3 = T(n/4)+2c 3 T(n/4) = T(n/8)+c 3 = T(n/8)+c 3 +2c 3 = T(n/8)+3c 3 T(n/8) = T(n/16)+c 3 = T(n/16)+c 3 +3c 3 = T(n/16)+4c 3 T(n/16) = T(n/32)+c 3 = T(n/32)+c 3 +4c 3 = T(n/32)+5c 3
21 03-19: Solving Recurrence Relations T(n) = T(n/2)+c 3 T(n/2) = T(n/4)+c 3 = T(n/4)+c 3 +c 3 = T(n/4)+2c 3 T(n/4) = T(n/8)+c 3 = T(n/8)+c 3 +2c 3 = T(n/8)+3c 3 T(n/8) = T(n/16)+c 3 = T(n/16)+c 3 +3c 3 = T(n/16)+4c 3 T(n/16) = T(n/32)+c 3 = T(n/32)+c 3 +4c 3 = T(n/32)+5c 3 = = T(n/2 k )+kc 3
22 03-20: Solving Recurrence Relations T(0) = c 1 T(1) = c 2 T(n) = T(n/2)+c 3 T(n) = T(n/2 k )+kc 3 We want to get rid of T(n/2 k ). Since we know T(1) n/2 k = 1 n = 2 k lgn = k
23 03-21: Solving Recurrence Relations T(1) = c 2 T(n) = T(n/2 k )+kc 3 Set k = lgn: T(n) = T(n/2 lgn )+(lgn)c 3 = T(n/n)+c 3 lgn = T(1)+c 3 lgn = c 2 +c 3 lgn Θ(lgn)
24 03-22: Power Modifications long power(long x, long n) { if (n==0) return 1; if (n==1) return x; if ((n % 2) == 0) return power(x*x, n/2); else return power(x*x, n/2) * x; }
25 03-23: Power Modifications long power(long x, long n) { if (n==0) return 1; if (n==1) return x; if ((n % 2) == 0) return power(power(x,2), n/2); else return power(power(x,2), n/2) * x; } This version of power will not work. Why?
26 03-24: Power Modifications long power(long x, long n) { if (n==0) return 1; if (n==1) return x; if ((n % 2) == 0) return power(power(x,n/2), 2); else return power(power(x,n/2), 2) * x; } This version of power also will not work. Why?
27 03-25: Power Modifications long power(long x, long n) { if (n==0) return 1; if (n==1) return x; if ((n % 2) == 0) return power(x,n/2) * power(x,n/2); else return power(x,n/2) * power(x,n/2) * x; } This version of power does work. What is the recurrence relation that describes its running time?
28 03-26: Power Modifications long power(long x, long n) { if (n==0) return 1; if (n==1) return x; if ((n % 2) == 0) return power(x,n/2) * power(x,n/2); else return power(x,n/2) * power(x,n/2) * x; } T(0) = c 1 T(1) = c 2 T(n) = T(n/2)+T(n/2)+c 3 = 2T(n/2)+c 3 (Again, assume n is a power of 2)
29 03-27: Solving Recurrence Relations T(n) = 2T(n/2)+c 3 T(n/2) = 2T(n/4)+c 3 = 2[2T(n/4)+c 3 ]c 3 = 4T(n/4)+3c 3 T(n/4) = 2T(n/8)+c 3 = 4[2T(n/8)+c 3 ]+3c 3 = 8T(n/8)+7c 3 = 8[2T(n/16)+c 3 ]+7c 3 = 16T(n/16)+15c 3 = 32T(n/32)+31c 3 = 2 k T(n/2 k )+(2 k 1)c 3
30 03-28: Solving Recurrence Relations T(0) = c 1 T(1) = c 2 T(n) = 2 k T(n/2 k )+(2 k 1)c 3 Pick a value for k such that n/2 k = 1: n/2 k = 1 n = 2 k lgn = k T(n) = 2 lgn T(n/2 lgn )+(2 lgn 1)c 3 = nt(n/n)+(n 1)c 3 = nt(1)+(n 1)c 3 = nc 2 +(n 1)c 3 Θ(n)
31 03-29: Recursion Trees We can also do this substitution visually, leads to Recursion Trees Consider: T(n) = 2T(n/2)+Cn T(1) = C 2 T(0) = C 2
32 03-30: Recursion Trees Start with the recursive definition T(n) = Cn + 2T(n/2)
33 03-31: Recursion Trees Move the equation around a bit to get: T(n) = Cn + + T(n/2) T(n/2) Repalce each occurance of T(n/2) with T(n/4)+T(n/4)+C(n/2)
34 03-32: Recursion Trees T(n) = Cn + + C(n/2) C(n/2) T(n/4) T(n/4) T(n/4) T(n/4) Replace again, using T(n) = 2T(n/2) + Cn
35 03-33: Recursion Trees T(n) = Cn + + C(n/2) C(n/2) C(n/4) C(n/4) C(n/4) C(n/4) T(n/8) T(n/8) T(n/8) T(n/8) T(n/8) T(n/8) T(n/8) If we continue replacing T(n/8)
36 03-34: Recursion Trees T(n) = Cn Totals for each level in the tree: Cn C(n/2) C(n/2) Cn C(n/4) C(n/4) C(n/4) C(n/4) Cn C(n/8) C(n/8) C(n/8) C(n/8) C(n/8) C(n/8) C(n/8) C(n/8) Cn C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C 2 n 2 lg n = n leaves Cn( lg n-1) + C 2n Θ(n lg n )
37 03-35: Recursion Trees T(1) = C 1 T(n) = T(n 1)+C 2
38 03-36: Recursion Trees T(0) = C 1 T(1) = C 1 T(n) = T(n/2)+C 2
39 03-37: Substitution Method We can prove that a bound is correct using induction, this is the substituion method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) O(? )
40 03-38: Substitution Method We can prove that a bound is correct using induction, this is the substituion method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) O(n), that is: T(n) C n for all n > n 0, for some pair of constants C,n 0
41 03-39: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) O(n), that is, T(n) C n Base case: T(1) = C 1 C 1 for some constant C This is true as long as C C 1.
42 03-40: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) O(n), that is, T(n) C n Recursive case: T(n) = T(n 1)+C 2 Recurrence definition
43 03-41: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) O(n), that is, T(n) C n Recursive case: T(n) = T(n 1)+C 2 Recurrence definition C(n 1)+C 2 Inductive hypothesis
44 03-42: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) O(n), that is, T(n) C n Recursive case: T(n) = T(n 1)+C 2 Recurrence definition C(n 1)+C 2 Inductive hypothesis Cn+(C 2 C) Algebra Cn If C > C 2 This is true as long as C C 1.
45 03-43: Substitution Method We can prove that a bound is correct using induction, this is the substituion method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) Ω(n) T(n) C n for all n > n 0, for some pair of constants C,n 0
46 03-44: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) Ω(n), that is, T(n) C n Base case: T(1) = C 1 C 1 for some constant C This is true as long as C C 1.
47 03-45: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) Ω(n), that is, T(n) C n Recursive case: T(n) = T(n 1)+C 2 Recurrence definition
48 03-46: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) Ω(n), that is, T(n) C n Recursive case: T(n) = T(n 1)+C 2 Recurrence definition C(n 1)+C 2 Inductive hypothesis
49 03-47: Substitution Method T(1) = C 1 T(n) = T(n 1)+C 2 Show: T(n) Ω(n), that is, T(n) C n Recursive case: T(n) = T(n 1)+C 2 Recurrence definition C(n 1)+C 2 Inductive hypothesis Cn+(C 2 C) Algebra Cn If C C 2 This is true as long as C C 1.
50 03-48: Substitution Method T(0) = C 2 T(1) = C 2 T(n) = 2T(n/2)+C 1 n Show: T(n) O(nlgn), that is, T(n) C nlgn
51 03-49: Substitution Method T(0) = C 2 T(1) = C 2 T(n) = 2T(n/2)+C 1 n Show: T(n) O(nlgn), that is, T(n) C nlgn Base cases: T(0) = C 1 C 0lg0 for some constant C T(1) = C 1 C 1lg1 for some constant C Hmmm.
52 03-50: Substitution Method T(0) = C 2 T(1) = C 2 T(n) = 2T(n/2)+C 1 n Show: T(n) O(nlgn), that is, T(n) C nlgn Only care about n > n 0. We can pick 2, 3 as base cases (why?) T(2) = C 1 C 2lg2 for some constant C T(3) = C 1 C 3lg3 for some constant C
53 03-51: Substitution Method T(0) = C 2 T(1) = C 2 T(n) = 2T(n/2)+C 1 n T(n) = 2T(n/2)+C 1 n Recurrence Definition
54 03-52: Substitution Method T(0) = C 2 T(1) = C 2 T(n) = 2T(n/2)+C 1 n T(n) = 2T(n/2)+C 1 n Recurrence Definition 2C(n/2)lg(n/2)+C 1 n Inductive hypothesis
55 03-53: Substitution Method T(0) = C 2 T(1) = C 2 T(n) = 2T(n/2)+C 1 n T(n) = 2T(n/2)+C 1 n Recurrence Definition 2C(n/2)lg(n/2)+C 1 n Inductive hypothesis Cnlgn/2+C 1 n Algebra Cnlgn Cnlg2+C 1 n Algebra Cnlgn Cn+C 1 n Algebra
56 03-54: Substitution Method T(0) = C 2 T(1) = C 2 T(n) = 2T(n/2)+C 1 n T(n) = 2T(n/2)+C 1 n Recurrence Definition 2C(n/2)lg(n/2)+C 1 n Inductive hypothesis Cnlgn/2+C 1 n Algebra Cnlgn Cnlg2+C 1 n Algebra Cnlgn Cn+C 1 n Algebra Cnlgn If C > C 1
57 03-55: Substitution Method Sometimes, the math doesn t work out in the substitution method: T(1) = 1 T(n) = 2T ( n 2) +1 (Work on board)
58 03-56: Substitution Method Try T(n) cn: ( n T(n) = 2T 1 ( 2) n 2c +1 2) cn+1 We did not get back T(n) cn that extra +1 term means the proof is not valid. We need to get back exactly what we started with (see invalid proof of n i=1i O(n) for why this is true)
59 03-57: Substitution Method Try T(n) cn b: ( n T(n) = 2T +1 ( ( 2) n ) 2 c b 2) cn 2b+1 cn b +1 As long as b 1
60 03-58: Master Method Recursion Tree for: T(n) = 2T(n/4)+C 2
61 03-59: Master Method Totals for each level in the tree: T(n) = C 2 C 2 C 2 C 2 2 C 2 C 2 C 2 C 2 C 2 4 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 8 C 2 4 C C C C C C C C C C C C k 2 log 4 n C 2 C 2 Θ( n 1/2 ) 2 log 4 n = n log 4 2 = n 1/2 leaves in the tree
62 03-60: Master Method Recursion Tree for: T(n) = 2T(n/2)+n 2
63 03-61: Master Method Totals for each level in the tree: T(n) = n 2 n2 + + log 2 n levels in the tree (n/8) (n/8) (n/2) 2 2 (n/2) (n/4) 2 (n/4) 2 (n/4) 2 (n/4) (n/8) (n/8) (n/8) (n/8) (n/8) (n/8) C 1 C 1 C 1 C 1 C 1 C 1 C 1 C 1 2 log 4 n C1 + n 2 2 n 2 4 n 2 8 n 2 2 k 2 log 2 n = n log 2 2 = n leaves in the tree Θ(n 2 )
64 03-62: Master Method Recursion Tree for: T(n) = 4T(n/2)+n 2
65 03-63: Master Method Totals for each level in the tree: T(n) = n 2 n 2 (n/2) 2 (n/2) 2 (n/2) 2 (n/2) 2 n 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 n 2 n n 2 4 log 2 n = n log = n leaves in the tree 2 n lg n
66 03-64: Master Method Recursion Tree for: T(n) = at(n/b)+f(n)
67 03-65: Master Method Totals for each level in the tree: f(n) # of children = a f(n) f (n/b) f (n/b) f (n/b) # of chldren = a a f (n/b) f (n/b 2 ) f (n/b ) f (n/b ) f (n/b 2 ) f (n/b ) f (n/b ) f (n/b 2 ) f (n/b 2 ) f (n/b 2 ) # of chldren = a a 2 f (n/b 2 ) C C C C C C C C a k f (n/b k ) n log b C a a log b n = n log b a leaves in the tree
68 03-66: Master Method T(n) = at(n/b)+f(n) 1. if f(n) O(n log ba ǫ ) for some ǫ > 0, then T(n) Θ(n log b a ) 2. if f(n) Θ(n log ba ) then T(n) Θ(n log ba lgn) 3. if f(n) Ω(n log b a+ǫ ) for some ǫ > 0, and if af(n/b) cf(n) for some c < 1 and large n, then T(n) Θ(f(n))
69 03-67: Master Method T(n) = 9T(n/3)+n
70 03-68: Master Method T(n) = 9T(n/3)+n a = 9,b = 3,f(n) = n n log ba = n log 39 = n 2 n O(n 2 ǫ ) T(n) = Θ(n 2 )
71 03-69: Master Method T(n) = T(2n/3)+1
72 03-70: Master Method T(n) = T(2n/3)+1 a = 1,b = 3/2,f(n) = 1 n log b a = n log3/21 = n 0 = 1 1 O(1) T(n) = Θ(1 lgn) = Θ(lgn)
73 03-71: Master Method T(n) = 3T(n/4)+nlgn
74 03-72: Master Method T(n) = 3T(n/4)+nlgn a = 3,b = 4,f(n) = nlgn n log ba = n log 43 = n nlgn Ω(n ǫ ) 3(n/4)lg(n/4) c nlgn T(n) Θ(nlgn)
75 03-73: Master Method T(n) = 2T(n/2)+nlgn
76 03-74: Master Method T(n) = 2T(n/2)+nlgn a = 2,b = 2,f(n) = nlgn n log ba = n log 22 = n 1 Master method does not apply! n 1+ǫ grows faster than nlgn for any ǫ > 0 Logs grow incredibly slowly! lgn o(n ǫ ) for any ǫ > 0
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