anonymous
  • anonymous
CAN SOMEBODY HELP ME FIND A HAMILTON PATH BELOW.
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
CAN SOMEBODY HELP ME FIND A HAMILTON PATH BELOW.
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
|dw:1435337695042:dw|
Astrophysics
  • Astrophysics
Since circles have no vertices is there a rule for it to? I know we can only go through each vertice once right, mhm this seems pretty interesting.
anonymous
  • anonymous
yes , it has to go through one vertix once

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
The hint given is that the graph is symmetric around the vertix p
ganeshie8
  • ganeshie8
no solution ?
Astrophysics
  • Astrophysics
How'd you come up with that
ganeshie8
  • ganeshie8
i bruteforced : Notice that the hamiltonian path, if it exists, contains exactly 15 edges. We have 27 edges, so total number of choices = \( \binom{27}{15}\)
ganeshie8
  • ganeshie8
all of them have repeated vertices, so...
Astrophysics
  • Astrophysics
Niceee
ganeshie8
  • ganeshie8
im not 100% sure of my method though @SithsAndGiggles
ganeshie8
  • ganeshie8
|dw:1435355081518:dw|
ganeshie8
  • ganeshie8
http://goo.gl/T1jynP
anonymous
  • anonymous
Mathematica agrees, the graph isn't Hamiltonian.
ganeshie8
  • ganeshie8
thnks for checking @SithsAndGiggles do the commands work in wolfram ? if so can you please share them, just want to check how much time wolfram takes..
anonymous
  • anonymous
I don't think W|A supports all the graph theory functions that Mma does, but it's worth a check. Snapshot below (some of the cluttered code is cut off).
1 Attachment
anonymous
  • anonymous
`HamiltonianGraphQ` returns `True` if the graph is Hamiltonian, `False` otherwise. As for W|A, I think the character count for the actual graph is too high.
anonymous
  • anonymous
Apparently it's almost instantaneous? I don't know how complex the algorithm being used for the Hamiltonian test is.
ganeshie8
  • ganeshie8
Oh then im sure Mma is using some kindof hueristics; the bruteforce method took more than 10 minutes to finish on my laptop

Looking for something else?

Not the answer you are looking for? Search for more explanations.