Practice in
Algorithm Analysis
1
Solving Recurrences
• Write recursive expression for time taken
• Keep substituting and see a pattern
• Find a value for k for which you know the solution
(base case)
– You may need to evaluate a series to get your
final answer
2
Recursion Example 1
long f(int n)
{
if( n <= 1 )
return 1;
else
return 1 + f(n-1);
}
In terms of big-Oh:
t(1) = 1
t(n) = 1 + t(n-1)
Note:
t(n-1) = 1 + t(n-2)
t(n-2) = 1 + t(n-3)
t(n-3) = 1 + t(n-4)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + t(n-1) = 1 + 1 + t(n-2)
= 1 + 1 + 1 + t(n-3) = ...
k + t(n-k)
If n-k = 1, we know the time
Choose k = n-1
t(n) = n-1 + t(1) = n-1 + 1 = O(n)
3
Recursion Example 2
f(1) = 5
f(n) = 5 + f(n-1)
In terms of big-Oh:
t(1) = ?
t(n) = ?
4
Recursion Example 2
f(1) = 5
f(n) = 5 + f(n-1)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + t(n-1)
5
Recursion Example 3
f(1) = 5
f(n) = 5 + 5*f(n-1)
In terms of big-Oh:
t(1) = ?
t(n) = ?
6
Recursion Example 3
f(1) = 5
f(n) = 5 + 5*f(n-1)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + t(n-1)
7
Recursion Example 4
f(1) = 5
f(n) = 5 + 5*f(n-1) + 2*f(n-1)
In terms of big-Oh:
t(1) = ?
t(n) = ?
8
Recursion Example 4
f(1) = 5
f(n) = 5 + 5*f(n-1) + 2*f(n-1)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + 2*t(n-1)
9
Recursion Example 5
f(1) = 5
f(n) = 5 + 5*f(n/3) + 2*f(n/3)
In terms of big-Oh:
t(1) = ?
t(n) = ?
10
Recursion Example 5
f(1) = 5
f(n) = 5 + 5*f(n/3) + 2*f(n/3)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + 2*t(n/3)
t(n/3) = ?
t(n/9) = ?
t(n/27) = ?
11
Recursion Example 5
f(1) = 5
f(n) = 5 + 5*f(n/3) + 2*f(n/3)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + 2*t(n/3)
t(n/3) = 1 + 2*t(n/9)
t(n/9) = 1 + 2*t(n/27)
t(n/27) = 1 + 2*t(n/81)
Find the general pattern for t(n) by repeated
substitution
12
Recursion Example 5
f(1) = 5
f(n) = 5 + 5*f(n/3) + 2*f(n/3)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + 2*t(n/3)
t(n) = 1 + 2*t(n/3) = 1 + 2[1+2*t(n/9)]
= 1 + 2 + 4*t(n/9)
= 1 + 2 + 4[1+ 2*t(n/27)]
= 1 + 2 + 4 + 8*t(n/27)
= 1 + 2 + 4 + ... 2k-1 + 2kt(n/3k)
= 2k-1 + 2kt(n/3k)
t(n/3) = 1 + 2*t(n/9)
t(n/9) = 1 + 2*t(n/27)
t(n/27) = 1 + 2*t(n/81)
What value of k should we choose?
13
Recursion Example 5
f(1) = 5
f(n) = 5 + 5*f(n/3) + 2*f(n/3)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + 2*t(n/3)
t(n/3) = 1 + 2*t(n/9)
t(n/9) = 1 + 2*t(n/27)
t(n/27) = 1 + 2*t(n/81)
t(n) = 1 + 2*t(n/3) = 1 + 2[1+2*t(n/9)]
= 1 + 2 + 4*t(n/9)
= 1 + 2 + 4[1+ 2*t(n/27)]
= 1 + 2 + 4 + 8*t(n/27)
= 1 + 2 + 4 + ... 2k-1 + 2kt(n/3k)
= 2k-1 + 2kt(n/3k)
n/3k = 1, so n = 3k so k = log3n
Substitute for k
14
Recursion Example 5
f(1) = 5
f(n) = 5 + 5*f(n/3) + 2*f(n/3)
In terms of big-Oh:
t(1) = 1
t(n) = 1 + 2*t(n/3)
t(n/3) = 1 + 2*t(n/9)
t(n/9) = 1 + 2*t(n/27)
t(n/27) = 1 + 2*t(n/81)
t(n) = 1 + 2*t(n/3) = 1 + 2[1+2*t(n/9)]
= 1 + 2 + 4*t(n/9)
= 1 + 2 + 4[1+ 2*t(n/27)]
= 1 + 2 + 4 + 8*t(n/27)
= 1 + 2 + 4 + ... 2k-1 + 2kt(n/3k)
= 2k-1 + 2kt(n/3k)
n/3k = 1, so n = 3k so k = log3n
t(n) = 2log3n - 1 + 2log3n t(1)
= O(2log3n) = O(2 log2n log32)
= O(nlog32)
15