Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
One line speedy solution solution in Speedy category for Break Rings by sshishov
from itertools import product
def break_rings(rings):
return min(len(set(value)) for value in product(*rings))
if __name__ == '__main__':
# These "asserts" using only for self-checking and not necessary for auto-testing
assert break_rings(({1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {4, 6})) == 3, "example"
assert break_rings(({1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4})) == 3, "All to all"
assert break_rings(({5, 6}, {4, 5}, {3, 4}, {3, 2}, {2, 1}, {1, 6})) == 3, "Chain"
assert break_rings(({8, 9}, {1, 9}, {1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {8, 7})) == 5, "Long chain"
Jan. 25, 2016
