INTRODUCTION to GRAPH THEORY - DISCRETE MATHEMATICS

We introduce a bunch of terms in graph theory like edge, vertex, trail, walk, and path.
#DiscreteMath #Mathematics #GraphTheory
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Introduction to Graph Theory. We cover a lot of definitions today, specifically walks, closed walks, paths, cycles, trails, circuits, adjacency, incidence, isolated vertices, and more.
Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.

[@anisivaram9617]

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[@hats5864]

You saved me last year with your tutorials, now it repeats, I was so happy when I saw that you have a course for Graph Theory too. Love you man

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[@bestyoueverhad.2408]

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In just few minutes and I understand this better than my maths teacher..thank you

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[@giridharkrcomputerscience2848]

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[@kaizen1496]

6:38 Walk 10:09 Trail & Circuit 13:23 Path & Cycle 24:27

[@jackmenirons4989]

Case 1: If there is a minimal walk from x to y with no repeated vertices, then it is a path by definition. We are given that w is an element of an x,y walk. Since there are no repeated vertices in this walk, x,y is also an element of the x-y path. Case 2: If there is a walk with repeated vertices, then that walk is not the shortest path (minimal walk) between that x,y pair. We can begin the walk at the repeated vertex to get a shorter walk since that walk will sill include the x,y pair as the start and end points. Then we end up with a walk with no repeated vertices and can apply Case 1.

[@Accelerate_CHAOS]

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[@xiaoraven2885]

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[@sperera5916]

8 isolated vertices disliked this video.

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[@evanknowles4780]

If you have a repeated vertex vi, you may modify the walk by connecting the first edge touching vi with the last one. If this changes the walk then it becomes a shorter walk. If this does not modify the walk, then the first edge is the same as the last one. We can thus make the walk shorter by removing that edge.

[@haukwantung9832]

This is quoted from my college's lecture notes: " A cycle (or circuit) is a path of nonzero length from v to v with no repeated edges." (It says cycle and circuit share the same meaning if I don't interpret wrongly.) But from your video, this is the definition of circuit ONLY. (For cycle, it should be with no repeated VERTICES). So, can I treat these two terms as the same or not? Thanks! :) By the way, your videos can really help me catch up all the things I can't understand in the lesson. Really great! :)

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[@Abohafsah]

Thank you so much trevor, You have just unload a big burden off my shoulder. As i saw the first ten minutes i subscribed it. I will have along way with you from now on... There still one question that really i cant get the idea on how to solve, The questions says: In every directed complete graph, prove that the sums of the squares of the in-degrees (overall vertices) is equal to the sum of the squares of the out-degrees (over all vertices).

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[@Trevtutor]

+Newton Cazzaro (Brazilian Channel) Accents are just one of the many challenges you must overcome to learn math. I wish I were joking :(

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@@darkdudironaji my math prof is native speaker. his issue: he hates his students. he literally sighs once every 2 sentences 🤦‍♀️, it s so depressing

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[@mitaanshuagarwal007]

In the example you solved in the end since we have to prove for a simple graph 2e

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[@moayyadarz2965]

thank you .. it is a bery good way of explaining such of complex concepts in very simple way

[@moayyadarz2965]

but I have a question about the terms what do you exactly mean by saying ********(v choose 2)=(v .(v-1 ) / 2! ) so my question how do you solve in this way what is the rule that controls the way thanks in advanced

[@souritraghosh1054]

Thank you sir.... Hopefully I will take a good exam tomorrow.

[@alperen32]

seven years and still saving lives.

[@DanielAlmeida499]

great video, thank you!

[@AbhishekTiwarics]

For the question asked for all walk there exists a path..... for i) point i.e. no repeated vertex then the walk is already a path for ii) when we get a vertex again we delete all vertex from the walk since the vertex was first discovered as from a vertex x->x the x itself is the minimal walk and addition of any edge to traverse same vertex will increase the walk length (contradiction as we are talking about minimal walk) ....thus we get a path for every walk(no vertex repeated) is above explanation correct...?? and your videos are really great...:) thank you

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[@ace4x3]

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[@ritikyadav3470]

I understand whole thing clearly . Niceeeee......... :)

[@killssingasuka7819]

For all vertices there is an edge. If there was a single vertex there would be no edge and therefore no graph. This continuous way of looking at graphs is more adaptive than viewing them as static things were two vertices are assigned to an edge. Yet two vertices are necessarily how an edge is defined. This superposition of equally true states is what creates the world.

[@farscamidediniz]

Thank you soo much

[@lyn8964]

thank you very much, from a german uni student :)

[@johndouros3759]

Im dreading this so much

[@anchanakrishnangp2086]

Thank you sir Thanks a lot

[@Legendofmudkip]

My professor taught us that (at 4:52) if there is an arrow going in one direction, then it is an undirected graph. So a to b would be undirected. And if there is a set of two arrows going in opposite directions from the same point, then that is a directed graph. So a to c would be a directed graph.

[@Trevtutor]

+xdarkness22x If there are arrow heads, it's directed. If there are no arrow heads, it's undirected. This is standard convention. For the sake of your course, it might be a little different.

[@Legendofmudkip]

TheTrevTutor Eh, my professor is probably teaching us wrong material. It wouldnt be the first time. Which is why I am every so grateful for your videos

[@vizziiiiob]

What should I know before getting into graph theory?

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Thank you

[@hisabin]

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In the problem where you have to prove they 2e

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you're a hero

[@Collegebook]

Great sir #graphDiscrete

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what a legend

[@mengzhuwang7233]

hi for the proof question , if the walk has repeated vertex , it can simply skip the walk around that traingle and get out directly; so clearly there is a shorter walk than the one with repeated vertex. In this case, it contradicts the assumption in saying that w is the shortest walk. Hence, The minimal walk can be path ( no repeated vertex) and thereby , 'there exist x-y path' can be thus proven. Is my thought correct? Thank you!

[@hchen31]

Does path have to be the shortest walk? 26:28 Can path be: b -> a -> c -> d (no repeated vertices, but not the shortest walk)? Thank you!

[@gabrielvillegas2033]

@18:00 that is quite the graphic graph

[@DOMINANTbeats]

Proof attempt at 24:00 mark: Suppose there does not exist a path in any x-y walk. Suppose we have w to be a minimal walk from x to y. Easy Case: If there is no repeated vertex, then by definition there is a path. Other Case: Assume any of those edges connects x to y must have repeated vertices. Each edge must then go back to a certain vertices more than once, where we note that a minimal walk does not repeat a path. Therefore a contradiction.

[@aaronpaulbarsana1702]

thank you :)

[@cybervigilante]

Can you increase the number of possible trails by starting at a different vertex, or does the number always remain the same? I think it's the latter, but I don't have a proof.

[@joshuaomotosho7463]

You are a great tutor How did you get e choose v2

[@LorddGray]

If vertex has a single edge that just goes to itself, is it still "isolated"?

[@Trevtutor]

by definition, no, since it doesn't have degree = 0.

[@EDROCKSWOO]

What does it mean for a vertex to have five neighbors. If i draw a vertex with five edges connecting with five other vertices, are those considered neighbors? Also, when I draw five neighborhoods does that satisfy condition two of this problem? Draw a graph that has 2) at least three vertices that are all adjacent to each other, and 3) a vertex with five neighbors.

[@XDGamingShow]

In the first example of the trail, is it acceptible to say that a -> a is just a? Thanks a lot

[@issatay8876]

Good video

[@sthevenrodriguez140]

Muchas gracias

[@markkennedy9767]

At 18:50 how do loops affect connectivity. If you go between say a and a, is this a path since a is used twice (so a is not connected to itself? Or do we accept cycles (closed paths) here as ways of connecting a to a). Or do we only take paths between vertices where x!=y to deal with this.

[@keithramanjooloo7464]

God bless you my g

[@antonheimdal8445]

For the dense people who are having a bit of a difficult time finding the solution to the last problem. Is there any answer somewhere. Step by step explanations on solving these sort of problems, both simple and complicated ones. I am feeling frustrated with this topic since I can't seem to get the application of it. I understand why and how but proofing mathematically is something I have struggled in Graph Theory. Help someone!

[@deehee1181]

is it possible for you to go back and forth in a directed graph?

[@luisdfernandez2601]

24:35 "A trail has repeated edges" but trails don't have repeated edges, correct?

[@oguz-kagan]

yes. i didn't understand that part. now both of them are not repeated?

[@oguz-kagan]

ah i get it. in trail you can repeat edge. in path you can't repeat vertex.

[@baldjaebeom4424]

i skipped my lectures bc online classes are meh,, thanks for mansplaining this to me xx

[@chexblu]

where can I find hamiltonian? awesome video btw, thanks a lot

[@darkferiousity]

I laughed quite a bit when you drew a vector

[@njabulomahlalela2912]

it's interesting

[@itsmeyaw_id]

Hello, I just want to tell that maybe there is a mistake in the video. As I read, simple path means that there isn't any multiple verticies instead of edges. Anyway, thanks for the lesson. I can understand graph theory better now.

[@Highest__]

Bless🙏

[@ritsu7337]

Um, for the last demonstration, I've resolved it as follows : We know that E (number of edges) has to be

[@appatel8731]

Nice explained 😃 keep it up bro

[@mihackerz4268]

Sir the video was really good but i am having two doubts: 1. In first question is it possible to have a path which is not a trail if not can you u give one example. 2. v choose 2 is simply combination vC2 but it signifies nCr means n!/(r!(n-r)! then how you can write v(v-1)/2 simply. I didn't understand Please can someone help me out in this problem

[@yahya9889]

are you indian?

[@mihackerz4268]

Yaa

[@TheAmbionC]

vC2 = v! / (2! (v - 2)!) = v (v - 1) * (v - 2)! / (2 (v - 2)!) = v (v - 1) / 2

[@RepeaterCreeper]

What software is that for drawing? Seems a little more convenient than SmoothDraw

[@PubicGore]

Here is my proof that for all x-y walks, there exists an x-y path. Let G be a graph with vertices v_1, v_2, ..., v_n. Pick two arbitrary vertices v_r and v_s with 1

[@jeantortuga570]

why choose 2? Is 2 meant as a number of vertices that are connected with an edge?

[@masumbillah3075]

cause in a simple graph,we need two vertices to make an edge so,the number of ways by which we can get a pair of vertices,is equal to the maximum number of edges we can form with those vertices! the way of choosing 2vertices [pair of vertices] from N number of vertices is , NC2 ,so,maximum number of edges we can get is = NC2 so,number of edges

[@jeandanielyannickchloe8149]

Is there any book (pdf) i can used to work some exercise on graph please?

[@mattbekhterev1249]

14:27 it’s not a path but it iiiis a trail

[@claxxicmedia6963]

Is it a must I repeat vertices to find Trail

[@emperor8716]

your other videos have been amazing but this one just confusing. my class uses path/trail and cycle/circuit interchangeably.

[@ragn594]

Can I please get an explination as to what the 2 is in the last example/exercise. I don't understand what this 2 does is it the amount of connections that every vertex has? What is the meaning of it?

[@gabrielcastillo6907]

Sir in the last exercise. Where did you get the formula v(v-1)/2!

[@anibalc.ripollr.9643]

Note that C(V,2)= V! / 2! (V-2)! = V(V-1)(V-2)! / 2 (V-2)!, cancel out the term (V-2)! on the numerator and the denominator and you get V(V-1) / 2