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### Solving Recurrences Using Iteration and Master Method

• 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.                  1 2 2 1 ) ( n cn n T n c n T
• 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
• 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)         0 ) 1 ( 0 0 ) ( n n T c n n T
• 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         0 ) 1 ( 0 0 ) ( n n T n n n T ) ( 1 k n T i n k n i      2 1 0 ) 0 ( 1 1          n n i T i n i n i ) ( 2 1 ) ( 2 n O n n n T   
• 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                1 2 2 1 ) ( n c n T n c n T
• 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)                1 2 2 1 ) ( n c n T n c n T
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