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CAN SOMEBODY HELP ME FIND A HAMILTON PATH BELOW.
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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.
yes , it has to go through one vertix once
The hint given is that the graph is symmetric around the vertix p
no solution ?
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How'd you come up with that
i bruteforced : Notice that the hamiltonian path, if it exists, contains exactly 15 edges. We have 27 edges, so total number of choices = \( \binom{27}{15}\)
all of them have repeated vertices, so...
Niceee
im not 100% sure of my method though @SithsAndGiggles
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