Recurrence Equations

1
Recurrence Equations

• Irena Pevac
• CS463
• CCSU

2
Recurrence Equations Use

• To measure cost of recursive subroutines
• To define mathematical functions
• Fibonacci
• F(1) 1
• F(2) 1
• F(n) F(n-1) F(n-2)
• Choose
• C(n,0) 1 C(k, k) 1 n gt m gt 0
• C(n,m) C(n-1,m) C(n-1,m-1)

3
Building Recurrence equations

• Specify some way to measure the size of the
  problem
• Call that size n.
• The left side of the recurrence equation will be
  T(n)
• To make up the right side
• Estimate how much the various blocks of the
  procedure will cost as a function of n.
• All constants can be 1 if we are satisfied with
  an answer within the constant factor.

4
Building Recurrence equations

• For linear sequence of blocks
• Add cost of each block
• For an alternation of blocks (if statement)
• Add maximum of the alternatives
• If a block contains method call that is not
  recursive with the one we analyze
• Express actual parameters as function of n
  ns(n)
• Cost of subroutine Ts( ns(n) )
• If a block has recursive calls
• Express actual parameters as function of n
  nR(n)
• Cost of subroutine T( nR(n) ) T is the
  same T on the left

5
Example

• Sequential search
• Page 131 Example 3.8

6
Recurrence Equations Types

• Divide and Conquer
• T(n)bT(n/c) f(n)
• Chip and Conquer
• T(n)T(n-c) f(n)
• Chip and Be Conquered
• T(n) b T(n-c) f(n)

7
Divide and Conquer T(n)bT(n/c) f(n)

• There are b subproblems
• Each subproblem has size that is fixed fraction
  of the size of the current problem such as n/c

8
Divide and Conquer ExamplesT(n)bT(n/c) f(n)

• b1 c2 T(n) T(n/2)1 binary search
• T(n) lg n
• b2 c2 T(n) 2 T(n/2)n merge sort
• T(n) n lg n
• b3 c2 T(n) 3 T(n/2)n Sierpinski
  triangles
• T(n) 3 nlg3-2n

9
Chip and Conquer ExamplesT(n)T(n-c)f(n)

• c1
• T(n) T(n-1)1 linear search , factorial
• T(n) n
• c1
• T(n) T(n-1) n Selection Sort
• T(n) n(n1)/2

10
Chip and Be Conquered T(n) b T(n-c) f(n)

• b2 c1 f(n) 1
• T(0) 0
• T(n) 2 T(n -1) 1 Tower of Hanoi
• T(n) 2n-1

11
Recursion Tree

• Tool for analyzing the cost of recursive methods
  for which we have formed recurrence equations.
• Each node has two fields
• The size field
• The nonrecursive cost field

12
Expansion of a node

• T(n) 2 T(n/2) n

13
Equation

• Size field of the root
• Sum of nonrecursive costs of expanded nodes
• Sum of size fields of incomplete nodes
• The same holds for root of any subtree.

14
Recursion TreeT(n) 2 T(n/2) n
15
Row sums

• Row at level 0
• Row at level 1
• Row at level 2
• Row at level depth
• n/2depth1
• where depthlg n

• Sum n
• Sum n
• Sum n
• Sum n
• ________
• Total sum of all row sums equals
• n lg n

16
Divide and Conquer

• T(n)bT(n/c) f(n)
• Size n decreases by c at each next level
• Size of the problem becomes n/c/c/c/c
• We are done when we reach the size of a base case
• n/cdepth 1 n cdepth ln n ln
  cdepth
• depth ln n/ ln c in ? (log n)
• The branching factor for recursion tree is b
• How many leaves has the tree ?
• At depth d there are bd nodes
• Number of leaves L bd

17

• lg(L) lg
• lg(L) (lgn/lg c) lg b (lg b / lg c) lg n
• Critical Exponent E lg b / lg c
• lg(L) (lg b / lg c) lg n lg n (lg b / lg c)

18
Lemma

• Lemma The number of leaves in the recursion
  tree for recurrence equation T(n)b(T(n/c)f(n)
  is nE, where E is critical exponent.
• If nonrecursive cost is 1 in the leaves the cost
  of the tree is at least nE.

19
Lemma

• The recursion tree has depth lg n/lg c
• The zero-th row-sum is f(n).
• The D-th row summ is nE assuming base case cost
  1 (or ?(nE) in any case).
• The T(n) solution is the sum of the nonrecursive
  costs of all nodes in the recusive tree, which is
  the sum of row-sums.

20
Special cases

• Row-sums form a geometric series
• Ratio r is not 1.
• The sum is in ? of its largest term

21
Little Master Theorem

• CASE 1 If the row sums form an increasing
  geometric series starting at the root of the
  tree then T(n) is in ? (nE). That is, the cost is
  proportional to the number of leaves in the
  recursion tree.
• CASE 2 If the row sums remain about constant,
  T(n) is in ? ( f(n) log n ).
• CASE 3 If row sums form a decreasing geometric
  series then T(n) is in ? (f(n) ), which means
  that cost is proportional to the cost of the root.

22
CASE 3 T(n) T(n/2) n/2

• Row sums form a decreasing geometric series.
• Row 0 n/2
• Row 1 n/4
• Row 2 n/8
• Row k n/2k1
• where n2k klgn
• Sum of all row sums n (1/21/41/81/2lgn)
  ?(n)

23
CASE 2 T(n)2T(n/2)n

• Row sums are constant and equal to n
• T(n) ?(f(n) log n) ?(n log n)

24
CASE 3 T(n)3T(n/2)n
25
Example CASE 3

• T(n)3T(n/2)n Sierpinski triangle
• Row sums form an increasing geometric series
• Level 0 sum n
• Level 1 sum 3n/2
• Level 2 sum 32n/22
• Level 3 sum 33n/23
• .
• Level k sum 3kn/2k
• Where n2k and klgn.

26
T(n) Sum all the row sums
Sum all the row sums 3 n lg3 2n ? (n
(lg3/lg2) )
27
Master Theorem

• Recurrence equation
• T(n)bT(n/c) f(n)
• has solutions as follows, where E is critical
  exponent
• E lg b / lg c

28
Master Theorem

• If f(n) in O(n E-e) for some positive e, then
  T(n) is in ? (nE), which is proportional to the
  number of leaves in the recursion tree.
• If f(n) in ? (nE), then T(n) in ?(f(n) log n),
  since all levels contribute equally.
• If f(n) in ?(n Ee) for some positive e, and
• if f(n) O(n Ed) for some ? gt e then
• T(n) is in ? (nE?),
• which is proportional to the nonrecursive
  cost at the root of the recursion tree.

29
Chip and ConquerChip and Be Conquered

• T(n) b T(n-c) f(n)
• For b2 f(n)1

30
T(n) 2 T(n-c) 1

• We are finished after k steps when we reach
  base case
• n k c 0 n kc kn/c
• 1248 2k2k1-1 22n/c-1 ?(2n/c)
• Even with the most favorable assumption when
  f(n)1 solution is exponential.

31
T(n) b T(n-c) f(n)

• In general
• Where the second sum uses h (n/c) - d.
• So, h is 0 at leaves and increases toward the
  root. When sum is ?(1), T(n) is in ?(bn/c).
  This is exponential growth.

32
T(n) T(n-c) f(n)

• Special case when branching factor b1
• If f(n) is a polynomial nk, then T(n) is in
  ?(nk1).
• If f(n) log n then T(n) is in ?(n log n).