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.
This video saved my butt. I cannot say enough good things about it. I have been staring at my textbook for 3 hours and I didn’t understand most of it. This video made all the difference in the world.
I'm about to write an online test for a bootcamp, i hope this explains everything i need. But your tutorials so far has been comprehensive especially on Induction
watching this 6 years later in 2021 and you still saved me with this video. i didnt understand anything when my teacher explained. thank you so much...
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.
Thank you so much! I wish my professor was clear and direct like you are, but unfortunately he stutters and has a thick accent from Russia. I can't understand him and I am taking this class again (with him because he is the only one teaching it in Las Vegas) and I will need you to learn discrete math.....
I feel you. I'm actually at UNLV right now struggling with this class. Part of it has to do with the Korean accent of my prof. Hope these videos helped you enough to get you a passing grade, cause I really need it!
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! :)
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.
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
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.
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).
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!
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.
+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.
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
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.
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
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.
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
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?
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!
To prove that for all x to y walks, there exists an x to y path, we can use a proof by contradiction. First, assume that there exists a walk, denoted as w, from x to y that is not a path. This means that w contains repeated vertices or edges, which makes it longer than the shortest possible path from x to y. Now, let's consider a minimal walk, denoted as w', from x to y. By definition, w' is the shortest walk from x to y. Since w is not a path, it must be longer than w'. Next, we will show that there exists an x to y path that is shorter than w. Consider the subsequence of w' that consists of only the distinct vertices. Since w' is a minimal walk, this subsequence is also a walk from x to y. Moreover, it is a path because it does not contain any repeated vertices. Since this subsequence is a path, it is shorter than w because it does not have any repeated vertices or edges. This contradicts our assumption that w is the shortest walk from x to y. Therefore, our assumption that there exists a walk that is not a path is false. Hence, for all x to y walks, there exists an x to y path.
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.
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
![[Discrete Mathematics] Subgraphs, Complements, and Complete Graphs](http://i.ytimg.com/vi/GHOHV6gTOd4/mqdefault.jpg)
![[Discrete Mathematics] Subgraphs, Complements, and Complete Graphs](/img/tr.png)
I am seeing this video after 5 years (2021) . For my semister exams, you explained very clearly . Thank you man
same here
same bro
Me too
Same here bro😭
Same here
On my final stretch with discrete maths. Thank you sir. for all you have done, God bles!
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
your welcome
your welcome.
In just few minutes and I understand this better than my maths teacher..thank you
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.
Thanks man
Thanks man.
Thanks man..
Thanks man...
Thanks man....
This video saved my butt. I cannot say enough good things about it. I have been staring at my textbook for 3 hours and I didn’t understand most of it. This video made all the difference in the world.
6:38 Walk
10:09 Trail & Circuit
13:23 Path & Cycle
24:27
I'm about to write an online test for a bootcamp, i hope this explains everything i need. But your tutorials so far has been comprehensive especially on Induction
watching this 6 years later in 2021 and you still saved me with this video. i didnt understand anything when my teacher explained. thank you so much...
lol i totally forgot that i watched this video and now im retaking this course and watching your video again.
Cant tell you how much this is helping i have an exam in an hour and didnt study this topic last night
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.
I watched these video's after six years . You tought in a very understandable manner thank you so much.
beautifully explained!
one of the best tutorial I have ever watched on youtube
Im getting ready to start descrete mathematics next tuesday for university and your videos are very helpful.
Thanks man , read Lecture notes and bamboozled myself . But this helped
Thank you man. You are such a brilliant teacher!
Thank you so much! I wish my professor was clear and direct like you are, but unfortunately he stutters and has a thick accent from Russia. I can't understand him and I am taking this class again (with him because he is the only one teaching it in Las Vegas) and I will need you to learn discrete math.....
+Newton Cazzaro (Brazilian Channel) Accents are just one of the many challenges you must overcome to learn math. I wish I were joking :(
I feel you. I'm actually at UNLV right now struggling with this class. Part of it has to do with the Korean accent of my prof.
Hope these videos helped you enough to get you a passing grade, cause I really need it!
@@darkdudironaji my math prof is native speaker. his issue: he hates his students. he literally sighs once every 2 sentences 🤦♀️, it s so depressing
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! :)
8 isolated vertices disliked this video.
lol well applied
8 dumbasses disliked this video
Haha
S Perera 😂😂😂
Make that 30 now m8
For reviewing purpose, and got an A in direcrete math. Thank you so much!
seven years and still saving lives.
If you ever update these videos, 24:35 "A trail has repeated edges" needs to change to "A trail has no repeated edges"
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.
thank you very much, from a german uni student :)
Amazing!
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
In the example you solved in the end since we have to prove for a simple graph 2e
24:35
......a minor correction. A trail has no repeated edges insted of "a trail has repeated edges"
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!
24:35 "A trail has repeated edges" but trails don't have repeated edges, correct?
yes. i didn't understand that part. now both of them are not repeated?
ah i get it. in trail you can repeat edge. in path you can't repeat vertex.
Thanks from Algeria ❤️✌️
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.
Simply awesome, thanks!
In the first example of the trail, is it acceptible to say that a -> a is just a? Thanks a lot
In the problem where you have to prove they 2e
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).
Everything is perfectly defined.. Thank you so much! Subscribed :D
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!
Thank you sir.... Hopefully I will take a good exam tomorrow.
this literally better than 2 hours wasting my time in class and learn nothing
same here
I am seeing this video after 9 years (2024). For my mid semister exams. Thank you man
Thank you so much you explained perfectly... I understood it ❤ You helped me a lot
what a legend
Told my classmates about your channel, and these videos will help us immensely!!
Thank you so much for these!
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.
+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.
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
this video helped me so much... thank youu
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.
This is sooo great ! Thank you so much!!!
That's some great mouse control. Your mouse-writing looks better than my handwriting
i doubt its a mouse. probably a light pen
@29:22 why did he start with 3 vertices?
you're a hero
Thank you so much. This is very helpful.😊❤️
@18:00 that is quite the graphic graph
You sir are a life saver
What should I know before getting into graph theory?
29:20 why do you start with 3 vertices?
3:05 I think I'm d
you are absolutely amazing ! Thank you
I understand whole thing clearly .
Niceeeee......... :)
What's "V choose 2"? Is that permutations? Combinations? I'm sorry for being stupid...
Is it a must I repeat vertices to find Trail
Very well explained sir..
why choose 2? Is 2 meant as a number of vertices that are connected with an edge?
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
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.
thank you .. it is a bery good way of explaining such of complex concepts in very simple way
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
16:35 - will ada be a cycle?
great video, thank you!
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?
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!
Is an isolated vertex a simple path please ?
Um, for the last demonstration, I've resolved it as follows :
We know that E (number of edges) has to be
i actually love you
Can u say that a in 9:01 is adjacent to a?
Very helpful,thank you sir !
que pasión! crazy mathemagician! ❤️❤️❤️
You are a great tutor
How did you get e choose v2
You are really great! I understand things clearly. I have subscribed you.
can we go back and forth in walks for directed graphs
Holy shiiiit, maximum thanks to this person!
Path b -> d, u use bcd. Thats also a trail? Dont u have to deny trail by using an edge twice before u can properly call it a path?
To prove that for all x to y walks, there exists an x to y path, we can use a proof by contradiction.
First, assume that there exists a walk, denoted as w, from x to y that is not a path. This means that w contains repeated vertices or edges, which makes it longer than the shortest possible path from x to y.
Now, let's consider a minimal walk, denoted as w', from x to y. By definition, w' is the shortest walk from x to y. Since w is not a path, it must be longer than w'.
Next, we will show that there exists an x to y path that is shorter than w. Consider the subsequence of w' that consists of only the distinct vertices. Since w' is a minimal walk, this subsequence is also a walk from x to y. Moreover, it is a path because it does not contain any repeated vertices.
Since this subsequence is a path, it is shorter than w because it does not have any repeated vertices or edges. This contradicts our assumption that w is the shortest walk from x to y.
Therefore, our assumption that there exists a walk that is not a path is false. Hence, for all x to y walks, there exists an x to y path.
Im dreading this so much
If vertex has a single edge that just goes to itself, is it still "isolated"?
by definition, no, since it doesn't have degree = 0.
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.
AWESOME!!!!! well done man. Terse and neat
+kasp HAHAHHA
Very clear thank you
Is the trivial walk a circuit?
What software is that for drawing? Seems a little more convenient than SmoothDraw
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
are you indian?
Yaa
vC2 = v! / (2! (v - 2)!) = v (v - 1) * (v - 2)! / (2 (v - 2)!) = v (v - 1) / 2
Thank you sir
Thanks a lot
is it possible for you to go back and forth in a directed graph?
Is there any book (pdf) i can used to work some exercise on graph please?
i skipped my lectures bc online classes are meh,, thanks for mansplaining this to me xx
where can I find hamiltonian?
awesome video btw, thanks a lot
+TrevTutor what would incident mean in a directed graph..??
Abhishek Tiwari The edge will be incident from starting vertex to terminal
your other videos have been amazing but this one just confusing. my class uses path/trail and cycle/circuit interchangeably.
Nice explained 😃 keep it up bro
u r soo good