So I am implementing an algorithm to prune leaf nodes from a graph before doing the heavier calculation of finding cycles within that graph. I am representing the graphs as adjacency lists, the example I will use is the graph corresponding to the carbons in methyl benzene:
[[0, 1], [0, 2], [1, 3], [2, 4], [3, 5], [4, 5], [5, 6]]
I want to remove entries corresponding to edges that connect leaf nodes, so the edge [5,6] needs to be 'pruned' to produce this new adjacency list:
[[0, 1], [0, 2], [1, 3], [2, 4], [3, 5], [4, 5]]
The code below (Python 3.4) appears to perform this operation correctly for when I print the adjacency list within the function the output is correct. However, when I call the function and print the returned 'value' it is always None
.
def prune_adj_list(adj_list):
print(adj_list)
all_indices = []
for bond in adj_list:
all_indices = all_indices + bond
leaf_nodes = []
for value in range(min(all_indices),max(all_indices)+1):
if all_indices.count(value) == 1: #count how many bonds to node
leaf_nodes.append(value)
if len(leaf_nodes) == 0: # no leaf nodes so done
print("done pruning")
print(adj_list) #<- the value here is correct
return adj_list
else:
new_adj_list = []
for bond in adj_list:
if not (leaf_nodes.count(bond[0]) > 0 or leaf_nodes.count(bond[1]) > 0):
new_adj_list.append(bond)
if len(new_adj_list) > 0:
prune_adj_list(new_adj_list)
else:
print("pruned away entire graph")
return []
Here is an example of a call that prints "None" instead of the correct value.
def find_cycles(compound):
adj_list = make_adj_list(compound) # this produces the list: [[0, 1], [0, 2], [1, 3], [2, 4], [3, 5], [4, 5], [5, 6]]
print(adj_list)
print(prune_adj_list(adj_list))
Can someone explain to me why the return value is incorrect but when I print it right before the return it is correct? Also any suggestions on how to fix this? I've played with returning copies of adj_list and have tested that the code seems to perform the operation I want correctly.
Thanks in advance, if additional information is needed please let me know.