Big-Oh Notation

Agenda  What is Big-Oh Notation?  Example  Guidelines  Theorems

Big-Oh Notation (O)  f(x) is O(g(x))iff there exists constants ‘c’and ‘k’ such that f(x) k  Pronounced as f(x) is Big-Oh of g(x)  This gives the upper bound value of a function

Example  f(n) = 10n + 5 and g(n) = n  To show f(n) is O(g(n)) we must show constants c and k such that f(n) =k or 10n+5 = k  We are allowed to choose c and k to be integers we want as long as they are positive.

Contd..  They can be as big as we want, but they can't be functions of n.  Try c = 15. Then we need to show: 10n + 5 <= 15n. Solving for n we get: 5 <= 5n or 1 <= n. So f(n) = 10+5 = 1. (c = 15, k = 1). Therefore we have shown f(n) is O(g(n)).

How do we calculate big-O? 1Loops 2Nested loops 3Consecutive statements 4If-then-else statements 5Logarithmic complexity 6 Five guidelines for finding out the time complexity of a piece of code

Guideline 1: Loops 7 The running time of a loop is, at most, the running time of the statements inside the loop (including tests) multiplied by the number of iterations. for (i=1; i<=n; i++) { m = m + 2; } constant time executed n times Total time = a constant c * n = cn = O(N)

Guideline 2: Nested loops 8 Analyse inside out. Total running time is the product of the sizes of all the loops. for (i=1; i<=n; i++) { for (j=1; j<=n; j++) { k = k+1; } constant time outer loop executed n times inner loop executed n times Total time = c * n * n * = cn 2 = O(N 2 )

Guideline 3: Consecutive statements 9 Add the time complexities of each statement. Total time = c 0 + c 1 n + c 2 n 2 = O(N 2 ) x = x +1; for (i=1; i<=n; i++) { m = m + 2; } for (i=1; i<=n; i++) { for (j=1; j<=n; j++) { k = k+1; } inner loop executed n times outer loop executed n times constant time executed n times constant time

Guideline 4: If-then-else statements 10 Worst-case running time: the test, plus either the then part or the else part (whichever is the larger). if (depth( ) != otherStack.depth( ) ) { return false; } else { for (int n = 0; n < depth( ); n++) { if (!list[n].equals(otherStack.list[n])) return false; } then part: constant else part: (constant + constant) * n test: constant another if : constant + constant (no else part) Total time = c 0 + c 1 + (c 2 + c 3 ) * n = O(N)

Guideline 5: Logarithmic complexity 11 An algorithm is O(log N) if it takes a constant time to cut the problem size by a fraction (usually by ½) Example algorithm (binary search): finding a word in a dictionary of n pages Look at the centre point in the dictionary Is word to left or right of centre? Repeat process with left or right part of dictionary until the word is found

Theorems  Let d(n),e(n),f(n) and g(n) be functions mapping non negative integers real. Then  If d(n) is O(f(n)),then ad(n) is O(f(n)),for any constant a>0  If d(n) is O(f(n)) and e(n) is O(g(n)),then d(n)+e(n) is O(f(n)+g(n))  If d(n) is O(f(n)) and e(n) is O(g(n)),then d(n).e(n) is O(f(n) g(n))  If d(n) is O(f(n)) and f(n) is O(g(n)),then d(n) is O(g(n)).

Contd..  If f(n) is a polynomial of degree d i.e, f(n)=(a 0 +a 1 n+…+a d n d ) then f(n) is O(n d ).  n x is O(a n ) for any fixed x>0 and a>1  log n x is O(log n) for any fixed x>0  log x n is O(n y ) for any fixed constants x>0 and y>0

Example  2n 3 +4n 2 logn is O(n 3 )  Proof: log n is O(n) – Rule 8 4n 2 logn is O(4n 3 ) – Rule 3 2n 3 +4n 2 log n is O(2n 3 +4 n 3 ) – Rule 2 2n 3 +4n 2 log n is O(n 3 ) – Rule 1 2n 3 +4n 2 log n is O(n 3 ) – Rule 4 Hence,Proved.

Relatives of Big-Oh  big-Omega f(n) is Ω(g(n)) if there is a constant c > 0 and an integer constant n 0 ≥ 1 such that f(n) ≥ c*g(n) for n ≥ n 0  big-Theta f(n) is θ(g(n)) if there are constants c ’ > 0 and c’’ > 0 and an integer constant n 0 ≥ 1 such that c’g(n) > f(n) > c’’g(n) for n ≥ n 0

Contd…  little-oh f(n) is o(g(n)) if, for any constant c > 0, there is an integer constant n 0 ≥ 0 such that f(n) ≤ cg(n) for n ≥ n 0  little-omega f(n) is w(g(n)) if, for any constant c > 0, there is an integer constant n ≥ 0 such that f(n) ≥cg(n) for n ≥n 0

Intuition for Asymptotic Notation  Big-Oh : f(n) is O(g(n)) if f(n) is asymptotically less than or equal to g(n)  big-Omega : f(n) is W(g(n)) if f(n) is asymptotically greater than or equal to g(n)  big-Theta :f(n) is Q(g(n)) if f(n) is asymptotically equal to g(n)  little-oh : f(n) is o(g(n)) if f(n) is asymptotically strictly less than g(n)  little-omega: f(n) is w(g(n)) if is asymptotically strictly greater than g(n)

Bubble Sort – O(n) 2 main() { int a[6]={7,5,3,4,2,1}; int i=0,j=0; int n=6; for(j=n-1;j>0;j--) { for(i=0;i<j;i++) { if(a[i]>a[i+1]) { int temp=a[i]; //swap a[i]=a[i+1]; a[i+1]=temp; } }}

Assignment  Find the complexity of the following Algorithm in Big-Oh – Binary Search for (i=1;i<10 ;i++ ) { for (j=0;j<i ;j++ ) { if (arr[j]>arr[i]) { temp=arr[j]; arr[j]=arr[i]; for (k=i;k>j ;k-- ) arr[k]=arr[k-1]; arr[k+1]=temp; } }}

References  Fundamentals of Computer Algorithms Ellis Horowitz,Sartaj Sahni,Sanguthevar Rajasekaran  Algorithm Design Micheal T. GoodRich,Robert Tamassia  Analysis of Algorithms Jeffrey J. McConnell

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