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Recurrence Relations

- Connection to recursive algorithms
- Techniques for solving them

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.

What if we knew we could solve part of the problem?

Assume we can move k (in this case, 4) different rings

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

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…)

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

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

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

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?

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

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

Recursion Trees

T(n) = 2 T(n/2) + n2 , T(1) = 1

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)

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 .

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)

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.

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)

Example

- T(n) = 4 T(n/2) + n
- a =
- b =
- nlog_ba =
- f(n) =

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

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

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!

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

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!

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

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

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

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

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?

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?

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?

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