David Luebke 1
10/21/16
CS 332: Algorithms
Solving Recurrences Continued
The Master Theorem
Introduction to heapsort
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Mastering Recurrences: Analyzing the Running Time of Heapify() and Heapsort Algorithms, Lecture notes of Algorithms and Programming
The master theorem and its application to the analysis of heapify() and heapsort algorithms. The basics of the master theorem, its use in analyzing recursive functions, and the advantages of merge sort and insertion sort. Additionally, it discusses the implementation of heaps as complete binary trees and arrays, basic heap operations, and heapify() example. Lastly, it analyzes the running time of heapify() and heapsort algorithms.
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2015/2016
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CS 332: Algorithms
Solving Recurrences Continued
The Master Theorem
Introduction to heapsort
Review: Merge Sort
MergeSort(A, left, right) {
if (left < right) {
mid = floor((left + right) / 2);
MergeSort(A, left, mid);
MergeSort(A, mid+1, right);
Merge(A, left, mid, right);
}
}
// Merge() takes two sorted subarrays of A and
// merges them into a single sorted subarray of A.
// Code for this is in the book. It requires O(n)
// time, and does require allocating O(n) space
Review: Solving Recurrences
●
Substitution method
●
Iteration method
●
Master method
Review: Solving Recurrences
●
The substitution method
■ A.k.a. the “making a good guess method”
■
Guess the form of the answer, then use induction to
find the constants and show that solution works
■ Run an example : merge sort
○ T(n) = 2T(n/2) + cn
○ We guess that the answer is O(n lg n)
○ Prove it by induction
■ Can similarly show T(n) = Ω(n lg n), thus Θ(n lg n)
● T(n) =
aT(n/b) + cn
a(aT(n/b/b) + cn/b) + cn
a
2
T(n/b
2
) + cna/b + cn
a
2
T(n/b
2
) + cn(a/b + 1)
a
2
(aT(n/b
2
/b) + cn/b
2
) + cn(a/b + 1)
a
3
T(n/b
3
) + cn(a
2
/b
2
) + cn(a/b + 1)
a
3
T(n/b
3
) + cn(a
2
/b
2
- a/b + 1)
…
a
k
T(n/b
k
) + cn(a
k-
/b
k-
- a
k-
/b
k-
- … + a
2
/b
2
- a/b + 1)
cn n
b
n
aT
T n
●
So we have
■ T(n) = a
k
T(n/b
k
) + cn(a
k-
/b
k-
- ... + a
2
/b
2
- a/b + 1)
● For k = log
b
n
■ n = b
k
■ T(n) = a
k
T(1) + cn(a
k-
/b
k-
- ... + a
2
/b
2
- a/b + 1)
= a
k
c + cn(a
k-
/b
k-
- ... + a
2
/b
2
- a/b + 1)
= ca
k
- cn(a
k-
/b
k-
- ... + a
2
/b
2
- a/b + 1)
= cna
k
/b
k
- cn(a
k-
/b
k-
- ... + a
2
/b
2
- a/b + 1)
= cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
cn n
b
n
aT
T n
●
So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a < b?
cn n
b
n
aT
T n
●
So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a < b?
■ Recall that (x
k
- x
k-
- … + x + 1) = (x
k+
-1)/(x-1)
cn n
b
n
aT
T n
●
So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a < b?
■ Recall that (x
k
- x
k-
- … + x + 1) = (x
k+
-1)/(x-1)
■ So:
■ T(n) = cn ·(1) = (n)
cn n
b
n
aT
T n
a b a b
a b
a b
a b
b
a
b
a
b
a
k k
k
k
k
k
1
1
1
1
1
1
1
1 1
1
1
●
So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a > b?
cn n
b
n
aT
T n
●
So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a > b?
■ T(n) = cn · (a
k
/ b
k
)
cn n
b
n
aT
T n
k
k
k
k
k
k
a b
a b
a b
b
a
b
a
b
a
1
1
1
1
1
1
●
So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a > b?
■ T(n) = cn · (a
k
/ b
k
)
= cn · (a
log n
/ b
log n
) = cn · (a
log n
/ n)
cn n
b
n
aT
T n
k
k
k
k
k
k
a b
a b
a b
b
a
b
a
b
a
1
1
1
1
1
1
● So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a > b?
■ T(n) = cn · (a
k
/ b
k
)
= cn · (a
log n
/ b
log n
) = cn · (a
log n
/ n)
recall logarithm fact: a
log n
= n
log a
= cn · (n
log a
/ n) = (cn · n
log a
/ n)
cn n
b
n
aT
T n
k
k
k
k
k
k
a b
a b
a b
b
a
b
a
b
a
1
1
1
1
1
1
● So with k = log
b
n
■ T(n) = cn(a
k
/b
k
- ... + a
2
/b
2
- a/b + 1)
●
What if a > b?
■ T(n) = cn · (a
k
/ b
k
)
= cn · (a
log n
/ b
log n
) = cn · (a
log n
/ n)
recall logarithm fact: a
log n
= n
log a
= cn · (n
log a
/ n) = (cn · n
log a
/ n)
= (n
log a
)
cn n
b
n
aT
T n
k
k
k
k
k
k
a b
a b
a b
b
a
b
a
b
a
1
1
1
1
1
1