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First solution in Clear category for Break Rings by taras-sereda
from collections import Counter
def find_least_connected(rings):
return Counter([r for i in rings for r in i if len(i) > 1]).most_common()[-1][0]
def find_victim(least_connected, rings):
return list([i for i in rings if least_connected in i and len(i) > 1][0].difference({least_connected}))[0]
def break_rings(rings):
counter = 0
while any([len(i)>1 for i in rings]):
least_connected = find_least_connected(rings)
ring_to_break = find_victim(least_connected,rings)
for ring in rings:
ring.discard(ring_to_break)
counter += 1
return counter
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, 5}, {3, 6})) == 2, "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"
Feb. 18, 2015