Build a clique tour for a given number of node n, if n is odd then the tour is eulerian otherwise the tour contains n/2 duplicate edges.
The algorithm is presented in a post from Aryabhata that can be found on stackexchange.
Installation:
pip install clique_tourUsage:
import clique_tour
odd_n = 7
even_n = odd_n-1
# When n is odd:
eulerian_tour = clique_tour.eulerian(odd_n)
# When it's even:
not_eulerian_tour = clique_tour.almost_eulerian(even_n)
# This function is a shortcut that test the parity of n and returns the correct tour:
odd_tour = clique_tour.build(odd_n)
even_tour = clique_tour.build(even_n)Results:
print(eulerian_tour)
print(odd_tour)
print(not_eulerian_tour)
print(even_tour)[0, 5, 1, 4, 2, 3, 6, 1, 0, 2, 5, 3, 4, 6, 2, 1, 3, 0, 4, 5, 6, 0]
[0, 5, 1, 4, 2, 3, 6, 1, 0, 2, 5, 3, 4, 6, 2, 1, 3, 0, 4, 5, 6, 0]
[0, 5, 1, 4, 2, 3, 1, 0, 2, 5, 3, 4, 2, 1, 3, 0, 4, 5, 0]
[0, 5, 1, 4, 2, 3, 1, 0, 2, 5, 3, 4, 2, 1, 3, 0, 4, 5, 0]