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I am trying to learn the Big O notation through the example below, can you help?

Let S be a set of n vertexes of a graph G and R be a set that the set of edges of G. Specify an upper bound of the size m of R with the big O notation. The bound shall be as tight as possible.

Are upper bound of the size m and runtime two different concept?

How to proof that the m is O(n^2) if that is the case or else?

Thank you for your attention!

Harry
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    Perhaps this question will help: https://cs.stackexchange.com/questions/23068/how-do-o-and-%CE%A9-relate-to-worst-and-best-case. – Yuval Filmus Feb 25 '18 at 18:10

2 Answers2

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Are upper bound of the size m and runtime two different concept?

The concept is just upper bounding. It doesn't matter whether you're bounding the number of vertices in a graph, the number of steps a program takes to execute, or the number of elephants in India – it's just an upper bound.

Probably the first time most CS students see big-O notation is in the context of runtimes and it's a common mistake to assume that big-O somehow means runtime. But it doesn't. This is the same mistake as, "The first time I saw numbers, they referred to somebody's height. Therefore, numbers mean height."

David Richerby
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you can prove it by taking some set of vertices say four (v1,v2,v3,v4). Now you are asking of Big Oh(greatest upper bound ...tightest most precisely )

note:this prove is for simple graph(see wiki)

now place the vertices like:

v1 v2

v4 v3

{in shape like rectangle}

just think what will be maximum number of edges possible? hmmm answer is :

if we draw an edge between every pair of vertices.

for v1 : edges can be v(1->2) ,v(1->3),v(1->4)

for v2: edges can be v(2->1){already included therefore don't write it} ,v(2->3),v(2->4)

for v3:edges can be v(3->4)

for v4: all are already included

what u see? total no of edges are coming in form:3+2+1=6

for n vertices(v1,v2......,vn) same could be done:

for v1: (n-1) choices of vertices are available.{excluding v1 therefore n-1}

for v2:(n-2)

. . . .

for vn-1:1

for vn:zero

total edges:n+(n-1)+(n-2)+...+1+0={n(n+1)}/2=0.5{n^2+n}

since we are talking of Big Oh only term with max degree will make difference Hence, it is O(n^2) where n is no. of vertices

ravindra singh
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