Graph Theory: Independent sets and My blog feactured in feedSpost!!!

Continuing from the previous post ( which is here ). This is a bit short post. I think this short post is not at all helpful in olys but then let's do it cause I find them interesting :P. Here are my notes..

Book referred: Daniel A Marcus Graph theory

Chapter I:

Independence Sets:

• Independent vertex set: An independent vertex set in a graph is a set of vertices that does not include any adjacent vertices.
  
  
• Independent edge set: An independent edge set in a graph is a set of edges that does not include any edges that share an endpoint.
  
  
• Maximal Independent sets: Largest set possible.
  
  
• $\alpha =$ the number of vertices in a maximal independent set
• $\alpha ' =$ the number of edges in a maximal independent edge set

Theorem1: In a graph with $n$ vertices and $e$ edges,

a) $\alpha'$ is less than $n/2$

b) if $d_1,d_2,\dots$ be the degress in a maximal independent vertex set, then the sum $d_1+d_2+\dots$ is less than $e.$

Proof: Note that every edge will have distinct vertex endpoints and no two edge share the same. So $|\alpha'|\le n/2.$ 

Let $\{v_1,v_2,\dots, v_{\alpha}\}$ denote the maximal independent set. Note that there is no $a$ in the set such that $v_i$ and $a$ are connected.

Now, note that sum of degrees is always $2e$ where $2$ is there cause each edge give $2$ to the sum. But since  there is no edge, we get $d_1+d_2+\dots \le e$

Corollary: In any $d-$ regular graph with $n$ vertices and $d\ge 1,$ $\alpha$ is less than or equal to  $n/2.$

Proof: Note that the number of edges is $n\cdot d$ and we know that $$d_1+d_2+\dots +d_{\alpha} \le e=\frac{ dn}{2}\text{ but each have degree } d \implies \alpha \le \frac{n}{2}.$$  

Theorem2: In any graph with $n$ vertices, $\alpha +\alpha'$ is always less than or equal to $n.$

Proof: Let $S$ be a maximal independent vertex set and let $S'$ be the maximal independent edge set. 

Note that $S$ contains $\alpha$ vertices and the $S'$ will have $\alpha'$ vertices. 

So the vertices in $S'$ are $2\alpha.$ But for atleast $\alpha$ vertices are not in $S.$

Hence $\alpha+\alpha'\le n$

A nice question would be is "when is the equality"? Turns out our favourite Bipartite graphs satisfy!

Theorem3: For bipartite graph with $n$ vertices, $\alpha+\alpha'=n$

The proof is beyond my knowledge :pleading:

Corollary: In any $d$ regular bipartite graph with $n$ vertices and $d\ge 1,\alpha=\alpha'=n/2.$

Proof: We know that $\alpha' \le \frac{n}{2}$ and $$d_1+d_2+\dots +d_{\alpha}\le \frac{nd}{2}\implies \alpha\le \frac{n}{2}$$  

Hence $$\alpha=\alpha'=n/2.$$

Coversing Sets: 

• Covering vertex set: A covering vertex set in a graph is a set of edges such that any edge has atleast one endpoint.
  
  
• Covering edge set: A covering edge set in a graph is a set of edges such that any vertex is endpoint of atleast one edge.
  
  
• Minimal Covering sets: smallest set possible.
  
  
• $\beta  =$ the number of vertices in a minimal vertex set
• $\beta ' =$ the number of edges in a minimal vertex edge set

Now, a very curious guess would be if $\alpha+\beta =n?$ Which is true by the collorary!

Theorem1: In any graph $G$, let $S$ be a maximal independent vertex set. Then a minimal covering vertex set can be obtained by taking all vertices are not in $S.$

Proof: Let $S=\{v_1,v_2\dots, v_{\alpha}\}$ be the maximal independent vertex set. Consider the complement of $S=\{a_1,a_2,\dots,a_n\}$ ( say $S^c$) we get that for every $e\in V$ we get that $e$  has atleast one endpoint in $S^c$ ( because $S$ is independent ). Hence $S^c$ is a covering set.

Now, to show the minimality, consider the minimal covering set $X.$ We are assuming ( for contradiction) that $|X|\le |S^c|.$ Then, consider the complement of $X.$ The complement will also be an independent set. But the independent set's cardinality is $>|S|.$ Contradiction.

Hence a minimal covering vertex set can be obtained by taking all vertices are not in $S.$

Corollary: In any graph with $n$ vertices, $\alpha+\beta =n$

Next theorem was very hard to prove and the idea was very non trivial.

Theorem2: In any graph with minimum degree $\delta\ge 1,$ let $S'$ be a maximal independent edge set. Then a minimal covering edge set can be obtained in the following way: Start by including all the edges in $S'.$ Then, for each vertex that is not covered by $S',$ select any one edge at that vertex and include that edge.

Proof: We know that the set is a covering edge set ( say $C$) as every vertex is being include. We start with the claim.

Claim: $C$ has $n-\alpha'$ edges.

Proof: It's just $\alpha' +(n-2\alpha')$ where the second term follows are the number of vertices which arent included in $S'.$

Let $C'$ be any minimal covering edge set and $H$ the subgraph containg all vertices and the edges from $C'$ only. Then $C'$ is the minimal covering edge set of $H.$ And hence $$|C'|=n-\alpha'(H)\ge n-\alpha'$$ where  $\alpha'(H)$ is maximal independece number.

 

And we are done!

Hence we get $\alpha'+\beta=\alpha'+(n-\alpha') =n.$

Yeah that's it for today's post! Also, my blog is featured in FeedSpot's Top 10 Geometry Blogs and Website!! And it is ranked $4$!! I am so happy and I have literally no words T_T.. Really very grateful to FeedSpot! Also thankyou to everyone who reads the blog! 

Hi Sunaina,

My name is Anuj Agarwal, I'm the Founder of Feedspot.

I would like to personally congratulate you as your blog Sunaina thinks absurd has been selected by our panelist as one of the Top 10 Geometry Blogs on the web.

https://blog.feedspot.com/geometry_blogs/

I personally give you a high-five and want to thank you for your contribution to this world. This is the most comprehensive list of Top 10 Geometry Blogs on the internet and I'm honored to have you as part of this!

We'd be grateful if you can help us spread the word by briefly mentioning about the Top 10 Geometry Blogs list in any of your upcoming post.

Please let me know.

Best,
Anuj

Hopeyou enjoyed the blogpost! 

See you soon

Sunaina 💜