Isn't that W example near the end hamiltonian? If you take the bottom 2 vertices as your u and v then it 6 which is greater then 5, I've found quite a few hamiltonian circuits as well.
If Ore's theorem is "D(U)+D(V)>=n ==> Hamiltonian circuit exists" then the converse is "Hamiltonian circuit exists ==> D(U)+D(V)>=n"? To disprove that shouldn't we provide an example of a graph with a Hamiltonian circuit for which D(U)+D(V)>/=n ? Your example is a graph with no Hamiltonian circuit, which is not what we need, right?
@@kingshukdutta2064 Oops. It only took him a few seconds to discuss that graph and I must have missed it. When I wrote my comment, I thought he was using the *next* graph he drew to consider the converse of Ore's theorem. Thanks.
Again! What a wonderful lesson.
It's always this kind of lessons make me fall in love with math even when there's a math exam.
Isn't that W example near the end hamiltonian? If you take the bottom 2 vertices as your u and v then it 6 which is greater then 5, I've found quite a few hamiltonian circuits as well.
Black on black outfit is fire
If Ore's theorem is "D(U)+D(V)>=n ==> Hamiltonian circuit exists" then the converse is "Hamiltonian circuit exists ==> D(U)+D(V)>=n"? To disprove that shouldn't we provide an example of a graph with a Hamiltonian circuit for which D(U)+D(V)>/=n ? Your example is a graph with no Hamiltonian circuit, which is not what we need, right?
The Professor made a graph(pentagon-shaped) where D(U)+D(V)
@@kingshukdutta2064 Oops. It only took him a few seconds to discuss that graph and I must have missed it. When I wrote my comment, I thought he was using the *next* graph he drew to consider the converse of Ore's theorem. Thanks.
Dear God. Please let me pass my oral exam about discrete mathematics tomorrow. Let my prayers be heard. Thank you, God. - A student
Beutiful