Recurrence Relations

1. Recurrence Relations • Connection to recursive algorithms • Techniques for solving them

2. Recursion and Mathematical Induction • In both, we have general and boundary conditions: • The general conditions break the problem into smaller and smaller pieces. • The initial or boundary condition terminate the recursion. • Both take a Divide and Conquer approach to solving mathematical problems.

3. The Towers of Hanoi

4. What if we knew we could solve part of the problem? Assume we can move k (in this case, 4) different rings

5. Can we do one better?

6. Solved for one more!

7. Where do recurrence relations come from? • Analysis of a divide and conquer algorithm • Towers of Hanoi, Merge Sort, Binary Search • Analysis of a combinatorial object • up-down permutations • This is the key analysis step I want you to master • Use small cases to check correctness of your recurrence relation

8. Recurrence Relations Possible recurrence relations: an = an-1 + 1; a1 = 1 => an = n (linear) an = an-1+ 2n - 1; a1=1 => an = n2 (polynomial) an = 2an-1; a1 = 1 => an = 2n (exponential) an = n an-1; a1 = 1 => an = n! (others…)

9. Can recurrence relations be solved? • No general procedure for solving recurrence relations is known, which is why it is an art. • However, linear, finite history, constant coefficient recurrences always can be solved • Example: an = 2an-1 + 2an-2 +1 ; a1 = 1 ; a2 = 1 • degree = 1 • history = 2 • coefficients = 2, 2, and 1 • In the end, what is the best way to solve this? • Software like Mathematica or Maple, but I still want you to be able to solve some on your own without such aid

10. Solution Techniques • Guess a solution by looking at a few small examples and prove by induction. • Try back-substituting until you know what is going on. • Draw a recursion tree. • Master Theorem • General Technique

11. First step • Using the base case and the recursive case, calculate small values • Use these values to help guess a solution • Use these values to help verify correctness of your closed form solution

12. Guessing solution and proving by induction • We can use mathematical induction to prove that a general function solves for a recursive one. • Tn = 2Tn-1 + 1 ; T0 = 0 • n = 0 1 2 3 4 5 6 7 8 • Tn = • Guess what the solution is?

13. Guessing solution and proving by induction II • Prove: Tn = 2n - 1 by induction: • 1. Show the base case is true: T0 = 20 - 1 = 0 • 2. Now assume true for Tn-1 • 3. Substitute in Tn-1 in recurrence for Tn • Tn = 2Tn-1 + 1 • = 2 ( 2n-1 - 1 ) + 1 • = 2n -1

14. Back-substitution Example: T(n) = 3T(n/4) + n, T(1) = 1 = 3(3T(n/16)+n/4) + n = 9T(n/16) + 3n/4 + n = 9(3T(n/64) +n/16) + 3n/4 + n = 27T(n/64)+9n/16 + 3n/4 + n

15. Recursion Trees T(n) = 2 T(n/2) + n2 , T(1) = 1

16. Example Problem Use induction to prove that MergeSort is an O(n log n) algorithm. Mergesort(array) n = size(array) if ( n == 1) return array array1 = Mergesort(array[1 .. n/2]) array2 = Mergesort(array[n/2 .. n]) return Merge(array1, array2)

17. Induction Proof Example: Prove that T(n) = 2T( n/2 ) + n , T(1) = 1 is O(n log n). Base case: n=2, c>4. We need to prove that T(n) < c n log n , for all n greater than some value. Inductive step: Assume T(n/2) ≤ c (n/2) log (n/2) Show that T(n) ≤ c n log n .

18. Induction Proof Given : T(n/2) ≤ c (n/2) log (n/2) T(n) = 2T( n/2 ) + n ≤ 2( c(n/2) log (n/2) ) + n ≤ 2( c(n/2) log (n/2) ) + n (dropping floors makes it bigger!) = c n log(n/2) + n = c n ( log(n) - log(2) ) + n = c n log(n) - c n + n (log22 = 1) = c n log(n) - (c - 1) n < c n log(n) (c > 1)

19. Example Problem 2 T(n) = T(n/3) + T(2n/3) + n Show T(n) is (n log n) by appealing to the recursion tree.

20. Recursion Tree

21. Master Theorem • T(n) = a T(n/b) + f(n) • Ignore floors and ceilings for n/b • constants a >= 1 and b>1 • f(n) any function • If f(n) = O(nlog_ba-e) for constant e>0, T(n) = Q(nlog_b a) • If f(n) = Q(nlog_ba), T(n) = Q(nlog_b a lg n) • If f(n) = W(nlog_ba+e) for some constant e >0, and if af(n/b) <= c f(n) for some constant c < 1 and all sufficiently large n, T(n) = Q(f(n)). • Key idea: Compare nlog_b a with f(n)

22. Example • T(n) = 4 T(n/2) + n • a = • b = • nlog_ba = • f(n) =

23. Characteristic Equation Approach • tn = 3tn-1 + 4tn-2 for n > 1 • t0 = 0, t1 = 5 • Rewrite recurrence • tn - 3tn-1 - 4tn-2 = 0 • Properties • Homogeneous: no terms not involving tn • Linear: tn terms have no squares or worse • constant coefficients: 1, -3, -4

24. Characteristic Equation • tn - 3tn-1 - 4tn-2 = 0 • Rewrite assume solution of the form tn = xn • xn – 3xn-1 – 4xn-2 = 0 • xn-2 (x2 – 3x – 4) = 0 • Find roots of (x2 – 3x – 4) • (x+1)(x-4)  roots are -1 and 4 • Solution is of form c1(-1)n + c24n

25. Solving for constants • tn = c1(-1)n + c24n • Use base cases to solve for constants • t0 = 0 = c1(-1)0 + c240 = c1 + c2 • t1 = 5 = c1(-1)1 + c241 = -c1 + 4c2 • 5c2 = 5  c2 = 1  c1 = -1 • tn = (-1)n+1 + 4n • Always test solution on small values!

26. Repeated roots for char. eqn • tn - 5tn-1 + 8tn-2 – 4tn-3= 0 • boundary conditions: tn = n for n = 0, 1, 2 • x3 - 5x2 + 8x – 4 = 0 • (x-1)(x-2)2 roots are 1, 2, 2 • Solution is of form c1(1)n + c22n + c3n2n • If root is repeated third time, then n22n term, and so on

27. Solving for constants • tn = c1(1)n + c22n + c3n2n • Use base cases to solve for constants • t0 = 0 = c1(1)0 + c220 + c30 20 = c1 + c2 • t1 = 1 = c1(1)1 + c221 + c31 21 = c1 + 2c2 + 2c3 • t2 = 2 = c1(1)2 + c222 + c32 22 = c1 + 4c2 + 8c3 • c1 = -2, c2 = 2, c3 = -1/2 • tn = 2n+1 – n2n-1 – 2 • Test the solution on small values!

28. Inhomogeneous Equation • tn - 2tn-1 = 3n • base case value for t0only • (x – 2) (x-3) = 0 • (x-2) term comes from homogeneous solution • If rhs is of form bn poly(n) of degree d • In this case, b = 3, poly (n) = 1 is of degree 0 • Plug (x-b)d+1 into characteristic equation

29. Solving for constants • (x – 2) (x-3) = 0 • tn = c12n + c23n • Solve for c1 and c2 with only t0 base case • This is only 1 equation and 2 unknowns • Use recurrence to generate extra equations • tn - 2tn-1 = 3n t1 = 2t0 + 3 • Now we have two equations • t0 = c120 + c230 = c1 + c2 • t1 = 2t0 + 3 = c121 + c231 = 2c1 + 3c2 • c1 = t0 – 3 and c2 = 3 • tn = (t0-3)2n + 3n+1

30. Changing variable • tn – 3tn/2 = n if n is a power of 2 • t1 = 1 • Let n = 2i and si = tn • si – 3si-1 = 2i for i >= 1 • s0 = 1 • (x-3)(x-2) = 0 • x-3 from characteristic equation • x-2 bnpoly(n) rhs

31. Solving for constants • (x-3)(x-2) = 0 • si = c13i + c22i • Generating two equations • t1 = s0 = 1 = c130 + c220 = c1 + c2 • t2 = s1 = 3t1+2 = 5 = c131 + c221 = 3c1 + 2c2 • c1 = 3, c2 = -2 • si = 3i+1 – 2i+1 for i >= 0 • tn = 3nlg 3 – 2n for n a power of 2 >= 1

32. Example: Up-down Permutations • An up-down permutation is a permutation of the numbers from 1 to n such that the numbers strictly increase to n and then decrease thereafter • Examples • 1376542, 7654321, 3476521 • Let tn denote the number of up-down permutations • What is recurrence relation for tn? • What is solution?

33. Example: Towers of Hanoi • Suppose we have two disks of each size. • Let n be the number of sizes of disks • What is recurrence relation? • What is solution?

34. Example: Merge Sort • Merge sort breaking array into 3 pieces • What is recurrence relation? • What is solution? • How does this compare to breaking into 2 pieces?