CSE 2331 / 5331 Foundations II: Data Structures and Algorithms Topic 2: -- Analyzing algorithms --...

CSE 2331 / 5331

Foundations II: Data Structures and Algorithms

Topic 2:

-- Analyzing algorithms

-- Solving recurrences

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Example: for Loop (1)

Textbook A.1 for summation formulas

What if i (or j) starts from 1000 ?

What if i starts from n/2 ?

What if j ends with i2 ?

What if j ends with ?

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Example: for Loop (2)

What if we change the first line to n > 100000 ?

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Evaluating Summations

Evaluate No fixed technique Be familiar with some basics

Textbook Appendix A.1and A.2

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Two Useful Techniques

Splitting summations E.g, ,

Using integral: If f is monotonically increasing,

E.g,

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Example: while Loop

What if we change the to ?

What if we change the to ?

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Analysis of while Loop

Stops when , that is, after iterations.

IChanges to since for any

constant c.

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More Examples (0)

What if we change the to ?

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More Examples (1)

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More Examples (2)

What if we change the to ?

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More Examples (3)

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Recursive Algorithms

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Sorting Revisited (1)

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T(n): worst case time complexity for any input of size n

Recurrence: T(n) = T(n-1) + cn

Solving recurrence: T(n) = T(n-1) + cn

= T(n-2) + c(n-1) + cn

= T(n-k) + c(n-k+1) + .. + cn

= T(1) + c[2 + … + n-1 + n]= Θ (n2)

Expansion into a series.

Usual assumption: T(n) is constant for small n, say n = 1, 2, 3

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More examples (1)

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More Examples (2)

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Sorting Revisited (2)

Can we do better than ? Yes. Many ways.

We will talk about Merge-sort

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Merge Sort

Divide

Conquer

?

?

Conquer Step Merge(B, C)

Input: Given two sorted arrays B and C Output: Merge into a single sorted array

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16

10

8

3

7

5

4

2

22 3 4 5 7 8 10 16 2 32 3 4 2 3 4 5 2 3 4 5 7

If size of B and C are s and t:

Running time: O ( s + t )

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Pseudocode

MergeSort ( A, r, s ) if ( r ≥ s) return;

m = (r+s) / 2;

A1 = MergeSort ( A, r, m );

A2 = MergeSort ( A, m+1, s );

Merge (A1, A2);

Recurrence relation for MergeSort(A, 1, n) T(n) = 2T(n/2) + cn

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Solve Recurrence

Expansion into a series Recursion tree Substitution method

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Expansion

T(n) = 2 T(n/2) + n

= 2 (2 T (n/4) + n/2) + n = 4 T(n/4) + n + n

= 4 (2 T(n/8) + n/8)) + n + n

…….

= 2k T(n/2k) + n + … + n Let . Then:

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Recursion-tree Method

Solve T(n) = 2 T(n/2) + n

T(n)n

n/2 n/2

T(n/4) T(n/4) T(n/4) T(n/4)

n

n/2 n/2

n/4 n/4 n/4 n/4

(1) (1)

n

n/2+n/2

4 (n/4)

n (1)

T(n) ≤ n height(tree) = n lg n

n

T(n/2) T(n/2)T(n) =

n

T(n/2) T(n/2)

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Another MergeSort

MergeSort ( A, r, s )

if ( r ≥ s) return;

m1 = r+(s-r) / 3;

m2 = r+2(s-r) / 3;

A1 = MergeSort ( A, r, m1 );

A2 = MergeSort ( A, m1 +1, m2 );

A3 = MergeSort ( A, m2 +1, s );

Merge (A1, A2, A3); Recurrence relation for MergeSort(A,1,n)

T(n) = 3T(n/3) + cn

What if we change 3 to k ?

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Another MergeSort

MergeSort ( A, r, s ) if ( r ≥ s) return;

m = r+(s-r) / 3;

A1 = MergeSort ( A, r, m );

A2 = MergeSort ( A, m+1, s );

Merge (A1, A2);

Recurrence relation for MergeSort(A,1,n) T(n) = T(n/3) + T(2n/3) + cn

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Recursion Tree

full levels and total levels

Total cost of each level is at most cn

Upper bound Lower bound

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More Examples of Recurrences (1)

T(n) = T(n-10) + c ?

T(n) = T(n-10) + n2 + 3n ?

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More Examples (2)

?

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More Examples (3)

Sometimes, it is hard to get tight bound

Easy to prove that and

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Summary

No good formula for solving recurrences Practice!

Expansion and recursion tree are two most intuitive These two are quite similar

Substitution method more rigorous and powerful But require good intuition and use of induction

Other methods: Master Theorem (not covered)