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First solution in Clear category for Break Rings by _Chico_
"""
Trying out all possibilities was too slow. This approach uses pruning
and avoids repeated removal of the same rings in different order.
"""
def break_ring(links, marked, depth, min_depth):
"""
Parameters:
links: remaining links between rings
marked: rings which have already been tried in previous iterations
depth: number of rings broken
min_depth: best solution so far
"""
if depth == min_depth: # no improvement possible
return min_depth
if len(links) == 0: # new optimum
return depth
# Rings available for breaking (avoiding repeats)
available_rings = set.union(*links) - marked
# Sorted list, with most likely candidates first
sorted_rings = sorted(available_rings, key=lambda x:sum(x in l for l in links), reverse = True)
# Try breaking different rings in order, removing any attached links
# After a ring has been tried, mark it to avoid repeating in a different iteration deeper down
for r in sorted_rings:
# Recursion with remaining links, updated set of marked rings over depth level
min_depth = break_ring([l for l in links if r not in l], marked.copy(), depth+1, min_depth)
marked |= {r}
return min_depth
def break_rings(links):
size = len(set.union(*links))
return break_ring(links, set(), 0, size)
def run():
# 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"
break_rings(({3,4},{5,6},{2,7},{1,5},{2,6},{8,4},{1,7},{4,5},{9,5},{2,3},{8,2},{2,4},{9,6},{5,7},{3,6},{1,3},))
break_rings(({1,2},{2,3},{3,4},{4,5},{5,2},{1,6},{6,7},{7,8},{8,9},{9,6},{1,10},{10,11},))
break_rings(({1,2},{2,3},{3,4},{4,5},{5,2},{1,6},{6,7},{7,8},{8,9},{9,6},{1,10},{10,11},{11,12},{12,13},{13,10},))
break_rings(({1,2},{2,3},{3,4},{4,5},{5,2},{1,6},{6,7},{7,8},{8,9},{9,6},{1,10},{10,11},{11,12},{12,13},{13,10},{1,14},{14,15},{15,16},{16,17},{17,14},))
May 11, 2021
