8.-DAA-LECTURE-8-RECURRENCES-AND-ITERATION-METHOD.pdf
- 1. Analysis and Design of Algorithms UNIT-1 Recurrence Relations
- 2. Content Recurrence Relation Forming Recurrence Relation Solving Recurrence Relations – Iterative Method – Substitution Method – Recursion Tree Method – Master’s Method 2
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- 5. Examples of recurrence relations Example-1: – Initial condition a0 = 1 (BASE CASE) – Recursive formula: a n = 1 + 2a n-1 for n > 2 – First few terms are: 1, 3, 7, 15, 31, 63, … Example-2: – Initial conditions a0 = 1, a1 = 2 (BASE CASE) – Recursive formula: a n = 3(a n-1 + a n-2) for n > 2 – First few terms are: 1, 2, 9, 33, 126, 477, 1809, 6858, 26001,…
- 6. Example-3: Fibonacci sequence Initial conditions: (BASE CASE) – f1 = 1, f2 = 2 Recursive formula: – f n+1 = f n-1 + f n for n > 3 First few terms: n 1 2 3 4 5 6 7 8 9 10 11 fn 1 2 3 5 8 13 21 34 55 89 144
- 7. Example-4: Compound interest Given – P = initial amount (principal) – n = number of years – r = annual interest rate – A = amount of money at the end of n years At the end of: 1 year: A = P + rP = P(1+r) 2 years: A = P + rP(1+r) = P(1+r)2 3 years: A = P + rP(1+r)2 = P(1+r)3 … Obtain the formula A = P (1 + r) n
- 8. Recurrence Relations: Terms Recurrence relations have two parts: – recursive terms and – non-recursive terms T(n) = 2T(n-2) + n2 -10 Recursive terms come from when an algorithms calls itself Non-recursive terms correspond to the non-recursive cost of the algorithm: work the algorithm performs within a function First, we need to know how to solve recurrences.
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- 11. 11 Solving Recurrence Relations Iteration method Substitution method Recursion Tree Master method
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- 15. 15 1. Iteration Method Step-1: Expand the Recurrence. Step-2: Express the expansion as a summation, by plugging the Recurrence back into itself, until you see a Pattern. ( Use algebra to express as a summation) Step-3: Evaluate the summation. Also known as “Try back substituting until you know what is going on”.
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- 18. 18 Example-1 s(n) = c + s(n-1) c + c + s(n-2) 2c + s(n-2) 2c + c + s(n-3) 3c + s(n-3) … kc + s(n-k) = ck + s(n-k) 0 ) 1 ( 0 0 ) ( n n s c n n s
- 19. Example-1 So far for n >= k we have s(n) = ck + s(n-k) What if k = n? s(n) = cn + s(0) = cn 19
- 20. 20 So far for n >= k we have s(n) = ck + s(n-k) What if k = n? s(n) = cn + s(0) = cn So Thus in general s(n) = cn 0 ) 1 ( 0 0 ) ( n n s c n n s Example-1
- 21. 21 s(n) = n + s(n-1) = n + n-1 + s(n-2) = n + n-1 + n-2 + s(n-3) = n + n-1 + n-2 + n-3 + s(n-4) = … = n + n-1 + n-2 + n-3 + … + n-(k-1) + s(n-k) 0 ) 1 ( 0 0 ) ( n n s n n n s
- 22. 22 s(n) = n + s(n-1) = n + n-1 + s(n-2) = n + n-1 + n-2 + s(n-3) = n + n-1 + n-2 + n-3 + s(n-4) = … = n + n-1 + n-2 + n-3 + … + n-(k-1) + s(n-k) = 0 ) 1 ( 0 0 ) ( n n s n n n s ) ( 1 k n s i n k n i
- 23. 23 So far for n >= k we have What if k = n? Thus in general 0 ) 1 ( 0 0 ) ( n n s n n n s ) ( 1 k n s i n k n i 2 1 0 ) 0 ( 1 1 n n i s i n i n i 2 1 ) ( n n n s
- 24. Example-2 24 Solve T(n) = 2T(n/2) + n. Solution: Assume n = 2k (so k = log n). T(n) = 2T(n/2) + n = 2 ( 2T(n/22) + n/2 ) + n T(n/2) = 2T(n/22) + n/2 = 22 T(n/22) + 2n = 22 ( 2T(n/23) + n/22 ) + 2n T(n/22) = 2T(n/23) + n/22 = 23T(n/23) + 3n = … = 2kT(n/2k) + k n = n T(1) + n log n = (n log n)