1. Design and Analysis of Algorithms
Chp 4
Solving Recurrences

2. Recurrent Algorithms BINARY – SEARCH
For an ordered array A, finds if x is in the array A[lo…hi]
Alg.:BINARY-SEARCH (A, lo, hi, x)
if (lo > hi)
return FALSE
mid  (lo+hi)/2
if x = A[mid]
return TRUE
if ( x < A[mid] )
BINARY-SEARCH (A, lo, mid-1, x)
if ( x > A[mid] )
BINARY-SEARCH (A, mid+1, hi, x) 12 11 10 9 7 5 3 2 1 4 6 8
mid lo hi
Recurrences

3. Analysis of BINARY-SEARCH
Alg.:BINARY-SEARCH (A, lo, hi, x)
if (lo > hi)
returnFALSE
mid  (lo+hi)/2
ifx = A[mid]
return TRUE
if ( x < A[mid] )
BINARY-SEARCH (A, lo, mid-1, x)
if ( x > A[mid] )
BINARY-SEARCH (A, mid+1, hi, x)
T(n) = c +
T(n) – running time for an array of size n
constant time: c1
constant time: c2
constant time: c3
same problem of size n/2
same problem of size n/2
T(n/2)
Recurrences

4. Recurrences and Running Time
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 (the generic term of the sequence)
Recurrences

5. Example Recurrences T(n) = T(n-1) + n Θ(n2) T(n) = T(n/2) + c Θ(lgn)
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) Θ(n)
Recursive algorithm that splits the input into 2 halves and does a constant amount of other work
Recurrences

6. Methods for Solving Recurrences
Iteration method
Recursion tree method
Recurrences

7. Some Simple Summation Formulas
5/4/2018
Some Simple Summation Formulas
Arithmetic series:
Geometric series:
Special case: x < 1:
Harmonic series:
Other important formulas:

8. The Iteration Method T(n) = c + T(n/2) = c + c + T(n/4)
= c + c + c + T(n/8)
Assume n = 2k  k = lgn
T(n) = c + c + … + c + T(1)  = k . C + T(1) So
T(n) = clgn + T(1)
= Θ(lgn)
T(n/2) = c + T(n/4)
T(n/4) = c + T(n/8)
k times
Recurrences

9. Iteration Method – Example
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)
Recurrences

10. Iteration Method – Example
T(n) = n + T(n-1)
= n + (n-1) + T(n-2)
= n + (n-1) + (n-2) + T(n-3)
… = n + (n-1) + (n-2) + … T(1)
= n(n+1)/ T(1)
= n2+ T(1) = Θ(n2)
Recurrences

11. s(n) = c + s(n-1) c + c + s((n-1)-1) c + c + s(n-2) 2c + s(n-2)…
kc + s(n-k) = ck + s(n-k)
Recurrences

12. So far for n >= k we have What if k = n?
s(n) = ck + s(n-k)
What if k = n?
s(n) = cn + s(0) = cn
Recurrences

13. So far for n >= k we have What if k = n? So Thus in general
s(n) = ck + s(n-k)
What if k = n?
s(n) = cn + s(0) = cn So
Thus in general
s(n) = cn
Recurrences

14. = n + n-1 + n-2 + n-3 + … + n-(k-1) + s(n-k)
= 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)
Recurrences

15. = n + n-1 + n-2 + n-3 + … + n-(k-1) + s(n-k) =
= 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) =
Recurrences

16. So far for n >= k we have
Recurrences

17. So far for n >= k we have
What if k = n?
Recurrences

18. So far for n >= k we have
What if k = n?
Recurrences

19. So far for n >= k we have
What if k = n?
Thus in general
Recurrences

20. T(n) = 2T(n/2) + c 2(2T(n/2/2) + c) + c 22T(n/22) + 2c + c…
2kT(n/2k) + (2k - 1)c
Recurrences

21. So far for n > 2k we have What if k = lg n?
T(n) = 2kT(n/2k) + (2k - 1)c
What if k = lg n?
T(n) = 2lg n T(n/2lg n) + (2lg n - 1)c
= n T(n/n) + (n - 1)c
= n T(1) + (n-1)c
= nc + (n-1)c = (2n - 1)c
Recurrences

22. The recursion-tree method
Convert the recurrence into a tree:
Each node represents the cost incurred at that level of recursion
Sum up the costs of all levels
Used to “guess” a solution for the recurrence
Recurrences

23. 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:
 W(n) = O(n2)
Recurrences

24. 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:
 W(n) = O(n2)
Recurrences

25. Example 2
Solve T(n) = T(n/4) + T(n/2)+ n2
Recurrences

26. Example 2
Solve T(n) = T(n/4) + T(n/2)+ n2
Recurrences

27. Example 2
Solve T(n) = T(n/4) + T(n/2)+ n2
Recurrences

28. Example 2
Solve T(n) = T(n/4) + T(n/2)+ n2
Recurrences

29. Example 2
Solve T(n) = T(n/4) + T(n/2)+ n2
Recurrences

30. Example 2
Solve T(n) = T(n/4) + T(n/2)+ n2
Recurrences

31. Example 3
T(n) = 2T(n/2) + cn
Recurrences

32. Using the recursion tree to visualize a recurrence.
Recurrences