Solving recurrences
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The document discusses methods for solving recurrences in algorithms, focusing on recursive and non-recursive algorithms. It presents various techniques such as the iteration method, substitution method, recursion tree method, and the master theorem, illustrated with examples. The content emphasizes finding explicit formulas for recurrences and understanding their running time in relation to algorithmic efficiency.
Solving recurrences
- 1. Data Structure and Algorithms Solving recurrences meghav@kannuruniv.ac.in Presented by Megha V Research scholar Dept. of IT Kannur University
- 2. Algorithm Algorithms are of two types: 1. Recursive 2. Non recursive meghav@kannuruniv.ac.in
- 3. Solving Recurrences Iteration method Substitution method Recursion Tree method Master’s Theorem meghav@kannuruniv.ac.in
- 4. Recurrences and Running Time • Recursive equation is an equation that defines a sequence recursively. It is normally in the form: T(n) = T(n-1) + n for all n≥0 T(0) = 0 initial condition Recurrences arise when an algorithm contains recursive calls to itself • • • What is the actual running time of the algorithm? Need to solve the recurrence – Find an explicit formula of the expression – Bound the recurrence by an expression that involves n meghav@kannuruniv.ac.in
- 5. Example Recurrences • T(n) = T(n-1) + n Θ(n2) – Recursive algorithm that loops through the input to eliminate one item • T(n) = T(n/2) + c Θ(lgn) – Recursive algorithm that halves the input in one step • T(n) = T(n/2) + n Θ(n) – Recursive algorithm that halves the input but must examine every item in the input • T(n) = 2T(n/2) + 1 Θ(n) – Recursive algorithm that splits the input into 2 halves and does a constant amount of other work meghav@kannuruniv.ac.in
- 6. Methods for Solving Recurrences • Iteration method • Substitution method • Recursion tree method • Master method meghav@kannuruniv.ac.in
- 7. The Iteration Method • Convert the recurrence into a summation and try to bound it using known series – Iterate the recurrence until the initial condition is reached. – Use back-substitution to express the recurrence in terms of n and the initial (boundary) condition. meghav@kannuruniv.ac.in
- 8. The Iteration Method T(n) = c + T(n/2) T(n) = c + T(n/2) = c + c + T(n/4) = c + c + c + T(n/8) Assume n = 2k T(n) = c + c + … + c + T(1) k times = clgn + T(1) = Θ(lgn) T(n/2) = c + T(n/4) T(n/4) = c + T(n/8) meghav@kannuruniv.ac.in
- 9. Iteration Method – Example T(n) = n + 2T(n/2) T(n) = n + 2T(n/2) = n + 2(n/2 + 2T(n/4)) = n + n + 4T(n/4) = n + n + 4(n/4 + 2T(n/8)) = n + n + n + 8T(n/8) … = in + 2iT(n/2i) = kn + 2kT(1) = nlgn + nT(1) = Θ(nlgn) Assume: n = 2k T(n/2) = n/2 + 2T(n/4) meghav@kannuruniv.ac.in
- 10. The substitution method 1. Guess a solution 2. Use induction to prove that the solution works meghav@kannuruniv.ac.in
- 11. Substitution method • Guess a solution – – T(n) = O(g(n)) Induction goal: apply the definition of the asymptotic notation • T(n) ≤ d g(n), for some d > 0 and n ≥ n0 – Induction hypothesis: T(k) ≤ d g(k) for all k < n • Prove the induction goal – Use the induction hypothesis to find some values of the constants d and n0 for which the induction goal holds (strong induction) meghav@kannuruniv.ac.in
- 12. Example: Binary Search T(n) = c + T(n/2) Guess: T(n) = O(lgn) • – – Induction goal: T(n) ≤ d lgn, for some d and n ≥ n0 Induction hypothesis: T(n/2) ≤ d lg(n/2) • Proof of induction goal: T(n) = T(n/2) + c ≤ d lg(n/2) + c = d lgn – d + c ≤ d lgn if: – d + c ≤ 0, d ≥ c • Base case? meghav@kannuruniv.ac.in
- 13. Example 2 T(n) = T(n-1) + n Guess: T(n) = O(n2) – Induction goal: T(n) ≤ c n2, for some c and n ≥ n0 • – Induction hypothesis: T(n-1) ≤ c(n-1)2 for all k < n • Proof of induction goal: T(n) = T(n-1) + n ≤ c (n-1)2 + n = cn2 – (2cn – c - n) ≤ cn2 if: 2cn – c – n ≥ 0 c ≥ n/(2n-1) c ≥ 1/(2 – 1/n) – For n ≥ 1 2 – 1/n ≥ 1 any c ≥ 1 will work meghav@kannuruniv.ac.in
- 14. Example 3 T(n) = 2T(n/2) + n Guess: T(n) = O(nlgn) • – Induction goal: T(n) ≤ cn lgn, for some c and n ≥ n0 – Induction hypothesis: T(n/2) ≤ cn/2 lg(n/2) • Proof of induction goal: T(n) = 2T(n/2) + n ≤ 2c (n/2)lg(n/2) + n = cn lgn – cn + n ≤ cn lgn if: - cn + n ≤ 0 c ≥ 1 meghav@kannuruniv.ac.in
- 15. Example 4 - Substitution T(n) = 3T(n/4) + cn2 • Guess: T(n) = O(n2) – Induction goal: T(n) ≤ dn2, for some d and n ≥ n0 – Induction hypothesis: T(n/4) ≤ d (n/4)2 • Proof of induction goal: T(n) = 3T(n/4) + cn2 ≤ 3d (n/4)2 + cn2 = (3/16) d n2+ cn2 ≤ d n2 if: d ≥ (16/13)c • Therefore: T(n) = O(n2) meghav@kannuruniv.ac.in
- 16. The recursion-tree method Convert the recurrence into a tree: – Each node represents the cost incurred at various levels of recursion – Sum up the costs of all levels Used to “guess” a solution for the recurrence meghav@kannuruniv.ac.in
- 17. Example 1 W(n) = 2W(n/2) + n2 • • • • Subproblem size at level i is: n/2i Subproblem size hits 1 when 1 = n/2i i = lgn Cost of the problem at level i = (n/2i)2 No. of nodes at level i = 2i Total cost: 2 1 1 1 1 i W (n) 2lgn W(1) n2 n n2 O(n) n2 O(n) 2n2 i0 2 i0 2 lgn1 n2 lgn1 1 i i0 2i W(n) = O(n2) meghav@kannuruniv.ac.in
- 18. Example 2 E.g.: T(n) = 3T(n/4) + cn2 • Subproblem size at level i is: n/4i • Subproblem size hits 1 when 1 = n/4i i = log4n • Cost of a node at level i = c(n/4i)2 • Number of nodes at level i = 3i last level has 3log 4 n = nlog 4 3 nodes • Total cost: T(n) = O(n2) 16 1 16 16 2 O(n ) 2 2 2 log4 3 log4 3 log4 3 1 cn n 3 cn n 3 i cn n T(n) i0 log4 n1 3 i i0
- 19. Master’s method • solving recurrences of the form: Idea: compare f(n) with 𝑛𝑙𝑜𝑔𝑏𝑎 • f(n) is asymptotically smaller or larger than 𝑛𝑙𝑜𝑔𝑏𝑎 by a polynomial factor n • f(n) is asymptotically equal with 𝑛𝑙𝑜𝑔𝑏𝑎 b where, a ≥ 1, b > 1, and f(n) > 0 T(n) aT n f (n) meghav@kannuruniv.ac.in
- 20. Master’s method Case 1: if f(n) = 𝑂(𝑛𝑙𝑜𝑔𝑏𝑎−)for some > 0, then: T(n) = (𝑛𝑙𝑜𝑔𝑏𝑎 ) Case 2: if f(n) = (𝑛𝑙𝑜𝑔𝑏𝑎), then: T(n) = (𝑛𝑙𝑜𝑔𝑏𝑎 lgn) Case 3: if f(n) = (𝑛𝑙𝑜𝑔𝑏𝑎+) for some > 0, and if af(n/b) ≤ cf(n) for some c < 1 and all sufficiently large n, then: T(n) = (f(n)) meghav@kannuruniv.ac.in
- 21. Example 1 T(n) = 2T(n/2) + n a = 2, b = 2, log22 = 1 Compare 𝑛𝑙𝑜𝑔22 with f(n) = n f(n) = (n) Case 2 T(n) = (nlgn) meghav@kannuruniv.ac.in
- 22. Example 2 T(n) = 2T(n/2) + n2 a = 2, b = 2, log22 = 1 Compare n with f(n) = n2 f(n) = (n1+) Case 3 verify regularity cond. a f(n/b) ≤ c f(n) 2 n2/4 ≤ c n2 c = ½ is a solution (c<1) T(n) = (n2) meghav@kannuruniv.ac.in
- 23. Examples 3 T(n) = 2T(n/2) + a = 2, b = 2, log22 = 1 Compare n with f(n) = n1/2 f(n) = O(n1-) Case 1 T(n) = (n) n meghav@kannuruniv.ac.in
- 24. Examples T(n) = 3T(n/4) + nlgn a = 3, b = 4, log43 = 0.793 Compare n0.793with f(n) = nlgn f(n) = (nlog 4 3+ ) Case 3 Check regularity condition: 3(n/4)lg(n/4) ≤ (3/4)nlgn = c f(n), c=3/4 T(n) = (nlgn) meghav@kannuruniv.ac.in
- 25. Assignment Solve the following recurrence using Master’s method 1. T(n) = 9T(n/3)+n 2. T(n) = 16T(n/4)+n3 3. T(n) = 8T(n/2)+n2 4. T(n) = 2T(n/2)+n meghav@kannuruniv.ac.in