Skip to main content

Graph Theory: Euler, hamilton

Heyy! Welcome Back to my blog!

 Today we are doing Graph Theory. This post is a continuation of a previous post which if you are interested in here. I somehow always get scared when someone mentions Graph theory. I have to overcome the fear. So here we go! I am writing some important theorems, terms, problems, etc that I learnt from the GT books! In short, notes.

Book referred: Daniel A Marcus, Graph Theory, A Problem Oriented Approach

Chapter F:

  • Euler Path: An Euler path in a graph $G$ is a path that includes every edge of $G$ exactly once. ( Not all graphs have Euler Paths)
  • Closed Euler Path: An Euler path is closed if it starts and ends at the same vertex.
  • Open Euler Path: Not closed.
Problem 1: Show that if a graph has at least one edge, and if all vertices have an even degree, then the graph must contain at least one cycle.
Proof: Suppose, it doesn't. Then consider a component of the graph $G.$ Then the component is a tree ( since it has no cycles). But we know a tree has at least $2$ nodes, which have an odd degree.

Problem 2: If a graph contains an Euler path, then it has $0$ or $2$ odd vertices.
Proof: Suppose there are more than $2$ odd vertices, then there is a vertex $A$ which is not the endpoints of the Euler path ( if it's not closed). Now, each time one visits $A,$ we will have to travel in and then out, in order to go to the next vertex.  Implying $A$ must be an even degree.

Problem 3: If a graph contains an Euler path and has $2$ odd vertices implies that the Euler path is open.
Proof: By the above problem, we get that "Non-end points" must-have degree even. So the two odd vertices must be endpoints of the euler path $\implies$ Euler path is open.

The Euler Path Theorem: 

1. A graph or multigraph contains a closed Euler path iff 
a. Every vertex has an even degree
b. All edges are in the component.

2.  A graph or multigraph contains an open Euler path iff 
a. Every vertex has an odd degree
b. All edges are in the component.

Proof: 
1. We know that the graph $G$ will contain a cycle $C.$ Now delete the cycle $C.$ Note that the new graph $G'$ will still have even degree vertices. And we will have a new cycle. Delete it and so on. Repeat the process. Note that this will continue till our graph becomes a cycle.

Now, we merge all these cycles. 
Say we have a cycle, $V_1\rightarrow V_2\rightarrow \dots \rightarrow V_k\rightarrow$ and we have another cycle, $V_i\rightarrow D_1\rightarrow \dots \rightarrow D_k\rightarrow V_i.$

Then consider this path, 
$$V_1\rightarrow V_2\rightarrow \dots V_{i-1}\rightarrow V_i\rightarrow D_1\rightarrow \dots \rightarrow D_k\rightarrow V_i\rightarrow \dots V_k$$

And we can repeat this process.

2. Let the two odd vertices be $A$ and $B.$ And the number of edges to be greater than $1.$ So $A$ and $B$ are not adjacent. Then join $AB.$ By part 1, we know there is a closed Euler path. So removing $AB$ we get an open Euler path.

The Euler Path Theorem directed version: 

1. A graph or multigraph contains a closed Euler path iff 
a. At each vertex, the in degrees $=$ out-degree 
b. All edges are in the component.

2.  A graph or multigraph contains an open Euler path iff 
a. At one vertex, in degree $-$ out degree $=1,$ at one vertex, out degree $-$ indegree$= 1,$ rest vertex have indegree$=$outdegree.
b. All edges are in the component.

Chapter G:

  • Hamilton Path: A hamilton path in a graph $G$ is a simple path that contains every vertex of $G.$
  • Hamilton Cycle: A cycle containing every vertex of $G.$


Problem 1: Suppose $G$ is a bipartite graph with $m$ vertices of one colour and $n$ vertices of the other colour. If $m\ne n,$ $G$ contains no hamilton cycle.

Proof: Let $X=\{v_1,\dots,v_m\},Y=\{d_1,\dots,d_n\}$ be the vertices. If hamilton cycle is being formed, the cycle would be of the form $(v_{i_1},d_{j_1},v_{i_2},\dots d_{j_x},v_{i_1}). $ Note that vertices of different colour are alternating, since all vertices are included once only, we get $m=n.$

Problem 2:  Suppose $G$ is a bipartite graph with $m$ vertices of one colour and $n$ vertices of the other colour. If $m-n\ge 2$ $G$ contains no hamilton cycle.

Proof: Similar proof, but we get $\max(m-n)=1.$

Using the above two problems, we get the following.

 $K_{m,n}$ contains a hamilton cycle iff $m=n.$

Problem 3: If $k$ vertices are removed from a graph that contains a hamilton path, then the number of components in the resulting subgraph is at most $k+1.$

Proof: Before removing the $k$ vertices, note that the graph was connected. Let $(d_1,d_2,\dots d_n)$ be the hamilton path. We remove $v_1,\dots,v_k.$ Then the hamilton path divides into $k+1$ parts. Consider $(d_1,\dots, d_i)$ such that $d_i$ was just before $v_1.$ Then the subgraph $(d_1,\dots, d_i)$ is connected. Similarly for the other subgraphs. Hence the number of components in the resulting subgraph is at most $k+1.$

Problem 4: If $k$ vertices are removed from a graph that contains a hamilton cycle, then the number of components in the resulting subgraph is at most $k.$

Proof: The proof is similar to the above. But then since it's a cycle, after removing we get $k$ parts.



Following are some fancy results that help us prove that the Hamilton cycle will exist. 

The path cyle principle:
Let $G$ be a graph with $n$ vertices, where $n\ge3,$ suppose that $G$ contains a hamilton path whose endpoints ( $A$ and $B$) satisfy the condition $d(A)+d(b)\ge n.$

Proof: Let the hamilton path be $(A,v_1,v_2,\dots,v_{n-2},B).$ 
If $A$ and $B$ are adjacent then we are done, so let's assume $A$ and $B$ are not adjacent. 
We try to get something of this form.

We want to get edges $Av_{i+1}$ and $Bv_i$ to exist in graph $G.$ And then we can consider the hamilton cycle $(v_{i+1},A,v_1,v_2\dots, v_i, B, v_{n-2},\dots, v_{i+1}).$


Now consider the set $S$ be the set of $i$ such $Av_{i+1}$ are adjacent and $1\le i\le n-3.$
And consider the set $T$ be the set of $i$ such $Bv_{i}$ are adjacent and $1\le i\le n-3.$

Then note that $|S|=d(A)-1,|T|=d(B)-1.$ And $d(a)+d(b)-2\ge n-2.$ So $S\cap T\ge 1.$


The Bondy-Chvatal Theorem: 
Let $G$ be a graph with $n$ vertices, where $n\ge 3$ and suppose that $G$ can be expanded to a larger graph $G'$ that contains a hamilton cycle, by adding edges once at a time in such a way that the following condition is satisfied.

The bondy-chvatal condition: An edge can be added joining vertices $A$ and $B$ if $A$ and $B$ are not already adjacent and if their degrees satisfy $d(A)+d(B)\ge n.$

Proof: Suppose $G$ does not contain any hamilton cycle but $G'$ is. Then there will be a stage when $G_k$ does not contain a hamilton cycle but $G_{k+1}$ does. Say edge we add while converting $G_k$ to $G_{k+1}$ is $CD.$ Then we have $CD$ part of the hamilton cycle, hence we have a hamilton path in $G_k$ with $CD$ endpoints. But we also have $d(C)+d(D)\ge n\implies$ hamilton cycle exist in $G_k$ by path cycle principle.


Ore's theorem:
Let $G$ be a graph with $n$ vertices, where $n\ge 3$ and suppose that $G$ satisfies 
Ore's adjacency condition Whenever $d(A)+d(B)<n$ for two vertices $A$ and $B,$ $A$ and $B$ are adjacent.
Then $G$ contains a hamilton cycle.

Proof: Consider two vertices  ( say $C$ and $D$) which are not adjacent in $G.$ Then we must have $d(C )+d(D)\ge n.$ And we can continue this process and get $K_n.$ But $K_n$ contains  hamilton cycle, so by bondy-chvatal theorem, we are done.


Dirac's theorem can be skipped since Posa's theorem is a stronger version anyways.

Dirac's theorem: 
Let $G$ be a graph with $n$ vertices, where $n\ge 3,$ and suppose that each vertex greater than or equal to $n/2.$ Then $G$ contains a hamilton cycle.

Proof: Note that we can join any two vertices as any two vertice degree sum is greater than n-1. Then at the end, we get a $K_n.$  But $K_n$ contains  hamilton cycle, so by bondy-chvatal theorem, we are done.

Posa's theorem: 
Let $G$ be a graph with $n$ vertices, where $n\ge 3$ and $d_1\le d_2\le \dots \le d_n$ be the degrees satisfy the condition

Posa's degree condition: $d_1>1,d_2>2,\dots d_i>i$ for all $i<n/2.$

Proof: We will try to complete the graph by joining to non adjacent vertices $A$ and $B$ with $d(A)+d(B)\ge n$ and we will eventually get $K_n.$ And we will be done by bondy-chvatal theorem.

If $d_1+d_2\ge n$ we are done, so let's assume $d_1+d_2<n.$
Note that $d_r+d_{r+1}\ge n$ for $r=[n/2].$
So there must be $2\le i\le r$ such that $$d_{i-1}+d_i< n, ~~ d_i+d_{i+1}\ge n.$$ So $d_i<n-i.$
Now let $A$ be the vertex having degree $d_i.$ Now consider the $n-i$ vertices which have degree $d_{i+1},d_{i+2},\dots, d_{n}.$ 
Since $d_i<n-1\implies $ there is a vertex which have degree $\ge d_{i+1}$ and is not adjacent to $A.$ We let that vertex be $B$ and join them. We are done.

 

Phew! That was a lot for me. I will post the sequel tomorrow!

Sunaina 💜



Labels:

Comments

  1. AnandDecember 24, 2021 at 9:38 AM

    Woah a lot of new stuff up there; nice post!

    PS: iirc Ore's theorem made it's appearance in INMO 21

    ReplyDelete
    Replies
    1. Sunaina Pati December 24, 2021 at 11:05 AM

      Hi Anand, long time no talk :pleading:.. Thanks a lot for commenting! And yes, I heard people mentioning Ore's theorem in INMO 21 was equivalent to 10+ marks... :O

      Delete

Post a Comment

Popular posts from this blog

Geometry ( Finally!!!)

 This is just such an unfair blog.  Like if one goes through this blog, one can notice how dominated  Algebra is!! Like 6 out of 9 blog post is Algebra dominated -_- Where as I am not a fan of Algebra, compared to other genres of Olympiad Math(as of now). And this was just injustice for Synthetic Geo. So this time , go geo!!!!!!!!!!!  These problems are randomly from A Beautiful Journey through Olympiad Geometry.  Also perhaps I will post geo after March, because I am studying combi.  Problem:  Let $ABC$ be an acute triangle where $\angle BAC = 60^{\circ}$. Prove that if the Euler’s line of $\triangle ABC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively, then $\triangle ADE$ is equilateral. Solution:  Since $\angle A=60^{\circ}$ , we get $AH=2R\cos A=R=AO$. So $\angle EHA=\angle DOA.$ Also it's well known that $H$ and $O $ isogonal conjugates.$\angle OAD =\angle EAH.$ By $ASA$ congruence, we get $AE=AD.$ Hence $\triangle ADE$ is equilateral....
3 comments
Read more

My experiences at EGMO, IMOTC and PROMYS experience

Yes, I know. This post should have been posted like 2 months ago. Okay okay, sorry. But yeah, I was just waiting for everything to be over and I was lazy. ( sorry ) You know, the transitioning period from high school to college is very weird. I will join CMI( Chennai Mathematical  Institue) for bsc maths and cs degree. And I am very scared. Like very very scared. No, not about making new friends and all. I don't care about that part because I know a decent amount of CMI people already.  What I am scared of is whether I will be able to handle the coursework and get good grades T_T Anyways, here's my EGMO PDC, EGMO, IMOTC and PROMYS experience. Yes, a lot of stuff. My EGMO experience is a lot and I wrote a lot of details, IMOTC and PROMYS is just a few paras. Oh to those, who don't know me or are reading for the first time. I am Sunaina Pati. I was IND2 at EGMO 2023 which was held in Slovenia. I was also invited to the IMOTC or International Mathematical Olympiad Training Cam...
11 comments
Read more

Introduction

  Hey Everyone!! This is my first Blog post. So let me give a brief introduction about myself. I am Sunaina Pati. I love solving Olympiad math problems,  learning crazy astronomical facts , playing hanabi and anti-chess, listening to Kpop , love making diagrams in Geogebra and  teaching other people maths 😊 . I love geometry , number theory and Combinatorics . I am starting this blog to keep myself a bit motivated in doing studies 😎 . Right now, I am planning to write walkthroughs on some of the best problems I tried over the week which can refer for hints 'cause solutions contain some major spoilers and one learns a lot while solving the problem on his own rather than seeing solutions . Also, there will be some reviews about Kpop songs, study techniques, my day to day lifestyles,exam reviews and ofc some non-sense surprises 😂.  I am planning to  try  posting every week on Sundays or Saturdays ( most probably) ! Though there is no guarantee about when I ...
20 comments
Read more

Problems I did this week [Jan8-Jan14]

Yeyy!! I am being so consistent with my posts~~ Here are a few problems I did the past week and yeah INMO going to happen soon :) All the best to everyone who is writing!  I wont be trying any new problems and will simply revise stuffs :) Some problems here are hard. Try them yourself and yeah~~Solutions (with sources) are given at the end! Problems discussed in the blog post Problem1: Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic. Problem 2: Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ ...
Post a Comment
Read more

Orders and Primitive roots

 Theory  We know what Fermat's little theorem states. If $p$ is a prime number, then for any integer $a$, the number $a^p − a$ is an integer multiple of $p$. In the notation of modular arithmetic, this is expressed as \[a^{p}\equiv a{\pmod {p}}.\] So, essentially, for every $(a,m)=1$, ${a}^{\phi (m)}\equiv 1 \pmod {m}$. But $\phi (m)$ isn't necessarily the smallest exponent. For example, we know $4^{12}\equiv 1\mod 13$ but so is $4^6$. So, we care about the "smallest" exponent $d$ such that $a^d\equiv 1\mod m$ given $(a,m)=1$.  Orders Given a prime $p$, the order of an integer $a$ modulo $p$, $p\nmid a$, is the smallest positive integer $d$, such that $a^d \equiv 1 \pmod p$. This is denoted $\text{ord}_p(a) = d$. If $p$ is a primes and $p\nmid a$, let $d$ be order of $a$ mod $p$. Then $a^n\equiv 1\pmod p\implies d|n$. Let $n=pd+r, r\ll d$. Which implies $a^r\equiv 1\pmod p.$ But $d$ is the smallest natural number. So $r=0$. So $d|n$. Show that $n$ divid...
7 comments
Read more

Some problems in Olympiad Graph theory!

Hello there! It has been a long time since I uploaded a post here. I recently took a class at the European Girls' Mathematical Olympiad Training Camp 2024, held at CMI. Here are a few problems that I discussed! My main references were Po-Shen Loh's Graph theory Problem set (2008), Adrian tang's Graph theory problem set (2012) and Warut Suksompong's Graph Cycles and Olympiad Problems Handout and AoPS. I also referred to Evan Chen's Graph theory Otis Problem set for nice problems! Text Book Problems which are decent A connected graph $G$ is said to be $k$-vertex-connected (or $k$-connected) if it has more than $k$ vertices and remains connected whenever fewer than $k$ vertices are removed. Show that every $k$-connected graph of order atleast $2k$ contains a cycle of length at least $2k$. We begin with a lemma. Prove that a graph $G$ of order $n \geq 2k$ is $k$ connected then every 2 disjoint set $V_1$ and $V_2$ of $k$ distinct vertices each, there exist $k$...
2 comments
Read more

Solving Random ISLs And Sharygin Solutions! And INMO happened!!

Some of the ISLs I did before INMO :P  [2005 G3]:  Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$ Solution: Note that $$\Delta LDK \sim \Delta XBK$$ and $$\Delta ADY\sim \Delta XCY.$$ So we have $$\frac{BK}{DY}=\frac{XK}{LY}$$ and $$\frac{DY}{CY}=\frac{AD}{XC}=\frac{AY}{XY}.$$ Hence $$\frac{BK}{CY}=\frac{AD}{XC}\times \frac{XK}{LY}\implies \frac{BK}{BC}=\frac{CY}{XC}\times \frac{XK}{LY}=\frac{AB}{BC}\times \frac{XK}{LY} $$ $$\frac{AB}{LY}\times \frac{XK}{BK}=\frac{AB}{LY}\times \frac{LY}{DY}=\frac{AB}{DL}$$ $$\implies \Delta CBK\sim \Delta LDK$$ And we are done. We get that $$\angle KCL=360-(\angle ACB+\angle DKC+\angle BCK)=\angle DAB/2 +180-\angle DAB=180-\angle DAB/2$$ Motivation: I took a hint on this. I had other angles but I did...
4 comments
Read more

IMO 2023 P2

IMO 2023 P2 Well, IMO 2023 Day 1 problems are out and I thought of trying the geometry problem which was P2.  Problem: Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$. Well, here's my proof, but I would rather call this my rough work tbh. There are comments in the end! Proof Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$.  Claim1: $L-B-E$ collinear Proof: Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$ So $$\angle SPA=90-A/2-C\implies \ang...
1 comment
Read more

Let's complex bash Part 1

I have to learn complex bash. And almost everyone knows that I am notes taking girl so thought why not make a post on complex bash ( so that I don't get emotionally demotivated lol).😇 There wasn't any need for learning complex bash, but it was in my dream checklist i.e " To learn a bash." And since I am not loaded with exams, I think it's high time to learn Bash and new topics.  Also if anyone from the "anti-bash" community is reading, sorry in advance and R.I.P.  Notes:- 1. Complex numbers are of the form $z=a+ib,$ where $a$ and $b$ are real numbers and $i^2=-1.$ 2. In polar form, $z=r(\cos \theta+~~i\sin\theta)=~~re^{i\theta},$ where $r=~~|z|=~~\sqrt{a^2+b^2},$ which is called the magnitude. 3. Here we used euler's formula i.e $\cos \theta+~~i\sin\theta=~~e^{i\theta}.$ 4. The $\theta $ is called the argument of $z,$ denored $\arg z.$ ( $\theta$ can be considered in $\mod 360$ and it is  measured anti-clockwise). 5. The complex conjugate of $z$ is ...
4 comments
Read more

How to prepare for INMO

Since INMO is coming up, it would be nice to write a post about it! A lot of people have been asking me for tips. To people who are visiting this site for the first time, hi! I am Sunaina Pati, an undergrad student at Chennai Mathematical Institute. I was an INMO awardee in 2021,2022,2023. I am also very grateful to be part of the India EGMO 2023 delegation. Thanks to them I got a silver medal!  I think the title of the post might be clickbait for some. What I want to convey is how I would have prepared for INMO if I were to give it again. Anyway, so here are a few tips for people! Practice, practice, practice!! I can not emphasize how important this is. This is the only way you can realise which areas ( that is combinatorics, geometry, number theory, algebra) are your strength and where you need to work on. Try the problems as much as you want, and make sure you use all the ideas you can possibly think of before looking at a hint. So rather than fixing time as a measure to dec...
9 comments
Read more