Practice in Algorithm Analysis

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Practice inAlgorithm Analysis
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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

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Recursion Example 1

• long f(int n)
• if( n lt 1 )
• return 1
• else
• return 1 f(n-1)

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)
In terms of big-Oh t(1) 1 t(n) 1 t(n-1)
If n-k 1, we know the time Choose k n-1 t(n)
n-1 t(1) n-1 1 O(n)
Note t(n-1) 1 t(n-2) t(n-2) 1
t(n-3) t(n-3) 1 t(n-4)
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Recursion Example 2

• f(1) 5
• f(n) 5 f(n-1)

In terms of big-Oh t(1) ? t(n) ?
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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)
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Recursion Example 3

• f(1) 5
• f(n) 5 5f(n-1)

In terms of big-Oh t(1) ? t(n) ?
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Recursion Example 3

• f(1) 5
• f(n) 5 5f(n-1)

In terms of big-Oh t(1) 1 t(n) 1 t(n-1)
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Recursion Example 4

• f(1) 5
• f(n) 5 5f(n-1) 2f(n-1)

In terms of big-Oh t(1) ? t(n) ?
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Recursion Example 4

• f(1) 5
• f(n) 5 5f(n-1) 2f(n-1)

In terms of big-Oh t(1) 1 t(n) 1 2t(n-1)
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Recursion Example 5

• f(1) 5
• f(n) 5 5f(n/3) 2f(n/3)

In terms of big-Oh t(1) ? t(n) ?
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Recursion Example 5

• f(1) 5
• f(n) 5 5f(n/3) 2f(n/3)

In terms of big-Oh t(1) 1 t(n) 1 2t(n/3)
t(n/3) ? t(n/9) ? t(n/27) ?
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Recursion Example 5

• f(1) 5
• f(n) 5 5f(n/3) 2f(n/3)

In terms of big-Oh t(1) 1 t(n) 1 2t(n/3)
t(n/3) 1 2t(n/9) t(n/9) 1
2t(n/27) t(n/27) 1 2t(n/81)
Find the general pattern for t(n) by repeated
substitution
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Recursion Example 5

• f(1) 5
• f(n) 5 5f(n/3) 2f(n/3)

In terms of big-Oh t(1) 1 t(n) 1 2t(n/3)
t(n) 1 2t(n/3) 1 212t(n/9)
1 2 4t(n/9) 1 2 41
2t(n/27) 1 2 4 8t(n/27)
1 2 4 ... 2k-1 2kt(n/3k) 2k-1
2kt(n/3k)
t(n/3) 1 2t(n/9) t(n/9) 1
2t(n/27) t(n/27) 1 2t(n/81)
What value of k should we choose?
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Recursion Example 5

• f(1) 5
• f(n) 5 5f(n/3) 2f(n/3)

In terms of big-Oh t(1) 1 t(n) 1 2t(n/3)
t(n) 1 2t(n/3) 1 212t(n/9)
1 2 4t(n/9) 1 2 41
2t(n/27) 1 2 4 8t(n/27)
1 2 4 ... 2k-1 2kt(n/3k) 2k-1
2kt(n/3k)
t(n/3) 1 2t(n/9) t(n/9) 1
2t(n/27) t(n/27) 1 2t(n/81)
n/3k 1, so n 3k so k log3n
Substitute for k
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Recursion Example 5

• f(1) 5
• f(n) 5 5f(n/3) 2f(n/3)

In terms of big-Oh t(1) 1 t(n) 1 2t(n/3)
t(n) 1 2t(n/3) 1 212t(n/9)
1 2 4t(n/9) 1 2 41
2t(n/27) 1 2 4 8t(n/27)
1 2 4 ... 2k-1 2kt(n/3k) 2k-1
2kt(n/3k)
t(n/3) 1 2t(n/9) t(n/9) 1
2t(n/27) t(n/27) 1 2t(n/81)
n/3k 1, so n 3k so k log3n t(n) 2log3n -
1 2log3n t(1) O(2log3n) O(2 log2n
log32) O(nlog32)
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