recurrence

1. Introduction to Algorithms
Chapter 4: Recurrences

2. 2
Solving Recurrences
 A recurrence is an equation that describes a
function in terms of itself by using smaller inputs
 The expression:
 Describes the running time for a function contains
recursion.
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3. 3
Solving Recurrences
 Examples:
 T(n) = 2T(n/2) + (n)  T(n) = (n lg n)
 T(n) = 2T(n/2) + n  T(n) = (n lg n)
 T(n) = 2T(n/2) + 17 + n  T(n) = (n lg n)
 Three methods for solving recurrences
 Substitution method
 Iteration method
 Master method

4. 4
Recurrence Examples
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5. 5
Substitution Method
 The substitution method
 “making a good guess method”
 Guess the form of the answer, then
 use induction to find the constants and show
that solution works
 Our goal: show that
T(n) = 2T(n/2) + n = O(n lg n)

6. 6
Substitution Method
T(n) = 2T(n/2) + n = O(n lg n)
 Thus, we need to show that T(n)  c n lg n
with an appropriate choice of c
 Inductive hypothesis: assume
T(n/2)  c (n/2) lg (n/2)
 Substitute back into recurrence to show that
T(n)  c n lg n follows, when c  1
 T(n) = 2 T(n/2) + n
 2 (c (n/2) lg (n/2)) + n
= cn lg(n/2) + n
= cn lg n – cn lg 2 + n
= cn lg n – cn + n
 cn lg n for c  1
= O(n lg n) for c  1

7. 8
Iteration Method
 Iteration method:
 Expand the recurrence k times
 Work some algebra to express as a
summation
 Evaluate the summation

8. 9
 T(n) = c + T(n-1) = c + c + T(n-2)
= 2c + T(n-2) = 2c + c + T(n-3)
= 3c + T(n-3) … kc + T(n-k)
= ck + T(n-k)
 So far for n  k we have
 T(n) = ck + T(n-k)
 To stop the recursion, we should have
 n - k = 0  k = n
 T(n) = cn + T(0) = cn
 Thus in general T(n) = O(n)
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9. 10
 T(n) = n + T(n-1)
= n + n-1 + T(n-2)
= n + n-1 + n-2 + T(n-3)
= n + n-1 + n-2 + n-3 + T(n-4)
= …
= n + n-1 + n-2 + n-3 + … + (n-k+1) + T(n-k)
= for n  k
 To stop the recursion, we should have n - k = 0  k = n
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10. 11
 T(n) = 2 T(n/2) + c 1
= 2(2 T(n/2/2) + c) + c 2
= 22 T(n/22) + 2c + c
= 22(2 T(n/22/2) + c) + (22-1)c 3
= 23 T(n/23) + 4c + 3c
= 23 T(n/23) + (23-1)c
= 23(2 T(n/23/2) + c) + 7c 4
= 24 T(n/24) + (24-1)c
= …
= 2k T(n/2k) + (2k - 1)c k
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11. 12
 So far for n  k we have
 T(n) = 2k T(n/2k) + (2k - 1)c
 To stop the recursion, we should have
 n/2k = 1  n = 2k  k = lg n
 T(n) = n T(n/n) + (n - 1)c
= n T(1) + (n-1)c
= nc + (n-1)c
= nc + nc – c = 2cn – c
T(n) = 2cn – c = O(n)
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