Eulerian Path - Intro to Algorithms

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Udacity

Udacity

Күн бұрын

This video is part of an online course, Intro to Algorithms. Check out the course here: www.udacity.co....

Пікірлер: 40
@JoonasNakiVaivaa
@JoonasNakiVaivaa 8 жыл бұрын
Damn what a cliffhanger
@Hasblock
@Hasblock Жыл бұрын
I'm still hanging 7 years later
@sonictitan433
@sonictitan433 5 ай бұрын
idk why he didnt answer lol. The answer is if all vertices of a connected graph have even degree, then it has a Euler circuit
@larsaugustat7029
@larsaugustat7029 5 ай бұрын
you are better now?@@Hasblock
@cranberry_sam
@cranberry_sam 4 жыл бұрын
Wasting 3 minutes on a mediocre video only to not get the entire lecture, REALLY makes me wanna sign up for udacity!
@mariomatos2130
@mariomatos2130 4 жыл бұрын
By the way, your videos are good. People mainly comment the negative stuff. Keep it up. Although it has been 8 years, but still.
@recursion.
@recursion. 2 жыл бұрын
you the type of guy who says its okay to life sentenced prisoners
@amirjutt0
@amirjutt0 Жыл бұрын
with all even degree nodes , graph will definately have eulerian path as well as eulerian circuit :)
@root6b61
@root6b61 2 ай бұрын
Not true if it's not connected though.
@kerel7682
@kerel7682 Ай бұрын
Yeah if graph have eulerian cycle you can be sure that it also have the path
@het314
@het314 4 жыл бұрын
WTF? What's the answer ?
@rossocorsa6577
@rossocorsa6577 4 жыл бұрын
the answer is yes
@comedyvidz1785
@comedyvidz1785 3 жыл бұрын
Good. A clear and concise explanation.
@aviroxi
@aviroxi 2 жыл бұрын
you came here bcz you saw this video only less than 5 min and you thought that this video will explain better guess what you are right!!
@Back-Air
@Back-Air 2 жыл бұрын
why not just say the answer to the last question ?
@user-hl1jt9qu8w
@user-hl1jt9qu8w 2 ай бұрын
thanks
@steveneumeyer681
@steveneumeyer681 2 жыл бұрын
on the last graph with the green edges, it appears that if you remove the 'e' node and the internal edges, you will have a graph where every node has degree 2, and can transit the graph by a,c,d,b,a.
@rj11wacat
@rj11wacat 7 жыл бұрын
Path = no repeated vertices. Trail = no repeated edges.
@anandkrishvvenkat5644
@anandkrishvvenkat5644 7 жыл бұрын
George Foley i think ..Repeated vertices is not a problem ...
@arnabdas7145
@arnabdas7145 6 жыл бұрын
Anand krishv venkat no in Euler's path visit vertex exactly once
@dennissantiago4524
@dennissantiago4524 2 жыл бұрын
Path and trail is the same thing😐
@dennissantiago4524
@dennissantiago4524 2 жыл бұрын
If you mean by trail is circuit you're right, but you're wrong jn path you can repeat vertices as shown in the video and you can also repeat vertices in both euler graph but you cant repeat edges
@davidbeauvais1183
@davidbeauvais1183 8 жыл бұрын
Not sure what he means by the degree.
@mark-6572
@mark-6572 8 жыл бұрын
+David Beauvais the number of edges connected to each vertex(or node)
@basementhermit3607
@basementhermit3607 9 жыл бұрын
Too many cuts/transitions of the hand - gets rather annoying/distracting.
@claudiamendoza9093
@claudiamendoza9093 2 жыл бұрын
what if you start with vertex B
@Devilfairy1
@Devilfairy1 7 жыл бұрын
Its Eulerian trail not path.
@naemfield
@naemfield 7 жыл бұрын
You're confusing it with Hamiltonian.
@naixinzong5449
@naixinzong5449 6 жыл бұрын
what if i say it is Eulerian tour
@govinggoriani7765
@govinggoriani7765 6 жыл бұрын
Vanilla Ice cream, anyone?
@bellesapum9427
@bellesapum9427 5 жыл бұрын
Devilfairy1 they’re the same thing
@kshtof
@kshtof Жыл бұрын
@@bellesapum9427 Yup, the terms are used interchangeably
@mariomatos2130
@mariomatos2130 4 жыл бұрын
Drumroll..... 🥁🥁 🥁🥁... 🥁🥁 ..... 🥁🥁..... 🥁🥁... 🥁🥁 ..... 🥁🥁..... 🥁🥁... 🥁🥁 ..... 🥁🥁.....
@fzbartttt666
@fzbartttt666 10 ай бұрын
İf path closed Yes
@user-bl2yq1hl8j
@user-bl2yq1hl8j 8 ай бұрын
❤❤❤
@rahulisgreat4911
@rahulisgreat4911 7 ай бұрын
Damn
@subschallenge-nh4xp
@subschallenge-nh4xp 7 жыл бұрын
pensil loll
@RubioDanA
@RubioDanA 8 жыл бұрын
What about connecting C & B?
@oriagruber
@oriagruber 8 жыл бұрын
+RubioDanA If you would connect C & B then no Eulerian path \ cycle would exist.
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