Download presentation

Presentation is loading. Please wait.

Published byAnais Duford Modified over 4 years ago

1
Algorithms Algorithm: what is it ?

2
Algorithms Algorithm: what is it ? Some representative problems : - Interval Scheduling

3
Algorithms Algorithm: what is it ? Some representative problems : - Interval Scheduling - Bipartite Matching

4
Algorithms Algorithm: what is it ? Some representative problems : - Interval Scheduling - Bipartite Matching - Independent Set

5
Algorithms Algorithm: what is it ? Some representative problems : - Interval Scheduling - Bipartite Matching - Independent Set - Area of a Polygon

6
Algorithms How to decide which algorithm is better ? Search problem Input : a sequence of n numbers (in an array A) and a number x Output : YES, if A contains x, NO otherwise

7
Algorithms How to decide which algorithm is better ? Search problem Input : a sequence of n numbers (in an array A) and a number x Output : YES, if A contains x, NO otherwise What if A is already sorted ?

8
Running Time O(n) – running time of the linear search O(log n) – running time of the binary search Def : Big-Oh (asymptotic upper bound) f(n) = O(g(n)) if there exists a constant c > 0 and a constant n 0 such that for every n ¸ n 0 we have f(n) · c g(n) Examples : n, n 3, log n, 2 n, 7n 2 + n 3 /3, 1, 1 + log n, n log n, n + log n

9
Running Time Def : Big-Oh (asymptotic upper bound) f(n) = O(g(n)) if there exists a constant c > 0 and a constant n 0 such that for every n ¸ n 0 we have f(n) · c g(n) Example: Prove that n = O(n 3 )

10
Running Time Def : Big-Oh (asymptotic upper bound) f(n) = O(g(n)) if there exists a constant c > 0 and a constant n 0 such that for every n ¸ n 0 we have f(n) · c g(n) Example: Prove that n 3 = O(7n 2 +n 3 /3)

11
Running Time Def : Big-Oh (asymptotic upper bound) f(n) = O(g(n)) if there exists a constant c > 0 and a constant n 0 such that for every n ¸ n 0 we have f(n) · c g(n) Example: Prove that log 10 n = O(log n)

12
Running Time Def : Big-Oh (asymptotic upper bound) f(n) = O(g(n)) if there exists a constant c > 0 and a constant n 0 such that for every n ¸ n 0 we have f(n) · c g(n) Example: what about 3 n and 2 n

13
Running Time O(n) – running time of the linear search O(log n) – running time of the binary search Def : Big-Omega (asymptotic lower bound) f(n) = (g(n)) if there exists a constant c > 0 and a constant n 0 such that for every n ¸ n 0 we have f(n) ¸ c g(n) Examples : n, n 3, log n, 2 n, 7n 2 + n 3 /3, 1, 1 + log n, n log n, n + log n

14
Running Time O(n) – running time of the linear search O(log n) – running time of the binary search Def : Theta (asymptotically tight bound) f(n) = (g(n)) if there exists constants c 1, c 2 > 0 and a constant n 0 such that for every n ¸ n 0 we have c 1 g(n) · f(n) · c 2 g(n) Examples : n, n 3, log n, 2 n, 7n 2 + n 3 /3, 1, 1 + log n, n log n, n + log n

15
A survey of common running times 1. for i=1 to n do 2. something Linear Also linear : 1. for i=1 to n do 2. something 3. for i=1 to n do 4. something else

16
A survey of common running times Example (linear time): Given is a point A=(a x, a y ) and n points (x 1,y 1 ), (x 2,y 2 ), …, (x n,y n ) specifying a polygon. Decide if A lies inside or outside the polygon.

17
A survey of common running times Example (linear time): Given are n points (x 1,y 1 ), (x 2,y 2 ), …, (x n,y n ) specifying a polygon. Compute the area of the polygon.

18
A survey of common running times Example (linear time): Given are n points (x 1,y 1 ), (x 2,y 2 ), …, (x n,y n ) specifying a polygon. Compute the area of the polygon.

19
A survey of common running times O(n log n) Or: 1. for i=1 to n do 2. for j=1 to log(n) do 3. something 1. for i=1 to n do 2. j=n 3. while j>1 do 4. something 5. j = j/2

20
A survey of common running times Quadratic 1. for i=1 to n do 2. for j=1 to n do 3. something

21
A survey of common running times Cubic

22
A survey of common running times O(n k ) – polynomial (if k is a constant)

23
A survey of common running times Exponential, e.g., O(2 k )

Similar presentations

OK

Algorithm Analysis. Math Review – 1.2 Exponents –X A X B = X A+B –X A /X B =X A-B –(X A ) B = X AB –X N +X N = 2X N ≠ X 2N –2 N+ 2 N = 2 N+1 Logarithms.

Algorithm Analysis. Math Review – 1.2 Exponents –X A X B = X A+B –X A /X B =X A-B –(X A ) B = X AB –X N +X N = 2X N ≠ X 2N –2 N+ 2 N = 2 N+1 Logarithms.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on wind power in india Ppt on chapter 12 electricity suppliers Ppt on rich heritage of india Ppt on my school life Ppt on self awareness and leadership Ppt on biodegradable and non biodegradable definition Ppt on matter in our surroundings Ppt on natural resources download Ppt on carburetor systems Ppt on business letter writing