1. Ch 4: Recurrences Ming-Te Chi
Algorithms
Ch 4: Recurrences
Ming-Te Chi
Ch4 Recurrences

2. Recurrences
When an algorithm contains a recursive call to itself, its running time can often described by a recurrence.
A recurrence is an equation or inequality that describes a function in terms of its value of smaller inputs.
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3. Recurrences─examples
Summation
Factorial
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4. Recurrences─examples
Summation
Factorial
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5. Termination conditions (boundary conditions)
Summation
Factorial
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6. Recurrences─CS oriented examples
Fibonacci number
Merge sort
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7. Influence of the boundary conditions
Fibonacci number
Merge sort
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8. Recurrences Substitution method Recursion-tree method Master method
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9. Technicalities
Neglect certain technical details when stating and solving recurrences.
A good example of a detail that is often glossed over is the assumption of integer arguments to functions.
Boundary conditions is ignored.
Omit floors, ceilings.
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10. The maximum-subarray problem
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13. Matrix multiplication
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14. A simple divide-and-conquer algorithm
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15. * Partition problem: Index calculation vs copy entries
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16. Strassen’s method
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17. Ch4 Recurrences

18. The substitution method: Mathematical induction
The substitution method for solving recurrence has two steps:
1. Guess the form of the solution.
2. Use mathematical induction to find the constants and show that the solution works.
Powerful but can be applied only in cases when it is easy to guess the form of solution.
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19. Can be used to establish either upper bound or lower bound on a recurrence.
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20. General principle of the Mathematical Induction
True for the trivial case,
problem size =1.
If it is true for the case of
problem at step k,
we can show that it is true for the case of
problem at step k+1.
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21. Example Determine the upper bound on the recurrence
(We may omit the initial condition later.)
Guess
Prove for some c > 0.
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22. Inductive hypothesis: assume this bound holds for
The recurrence implies
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23. Initial condition
However
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24. Making a good guess We guess
Making guess provides loose upper bound and lower bound. Then improve the gap.
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27. Subtleties
Guess
Assume
 However, assume
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28. Avoiding pitfalls Assume Hence (WRONG!) You cannot find such a c.
(WRONG!) You cannot find such a c.
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29. Changing variables
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30. 4.4 the Recursion-tree method
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31. Ch4 Recurrences

32. Subproblem size for a node at depth i Total level of tree
Number of nodes at depth i
Cost of each node at depth i
Total cost at depth i
Last level, depth , has nodes
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33. Prove of
We conclude,
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34. The cost of the entire tree
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35. Ch4 Recurrences

36. Substitution method
We want to Show that T(n) ≤ dn2 for some constant d > 0.
Using the same constant c > 0 as before, we have
Where the last step holds as long as d (16/13)c.
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38. Subproblem size for a node at depth i Total level of tree
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39. substitution method
As long as d  c/(lg3 – (2/3)).
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40. 4.5 Master method
Master Theorem
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41. Remarks 1: In the first case, f(n) must be polynomailly smaller than
That is, f(n) must be asymptotically smaller than by a factor of
Similarly, in the third case, f(n) must be polynomailly larger than
Also, the condition must be hold.
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42. Remarks 2:
The three cases in the master theorem do not cover all the possibilities.
There is a gap between cases 1 and 2 when f(n) is smaller than but not polynomailly smaller.
Similarly, there is a gap between cases 2 and 3 when f(n) is larger than but not polynomailly larger (or the additional condition does not hold.)
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43. Example 1:
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44. Example 2:
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45. Example 3:
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46. Example 4:
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47. Example 5:
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48. Remarks 3: The master theory does not apply to the recurrence
even though it has the proper form: a = 2, b=2, f(n)= n lgn, and
It might seem that case 3 should apply, since f(n)= n lgn is asymptotically larger than
But it is not polynomially larger.
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49. Remarks 3 (continue)
However, the master method augments the conditions
Case 2 can be applied.
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50. Example 6:
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51. Example 7:
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