CODEWITHSUNDEEP
pythonalgorithmgraph
AndrewSokolowski
AndrewSokolowski

Reputation: 703

My graph contraction algorithm in Python behaves strangely

I've written some code for this problem. (python27)

Graph is represented as dictionary with frozenset keys and sets of frozensets:

sample_graph = {frozenset([7]): set([frozenset([4]), frozenset([5]), frozenset([3])]), frozenset([5]): set([frozenset([7]), frozenset([2]), frozenset([1])]), frozenset([3]): set([frozenset([7]), frozenset([4]), frozenset([2]), frozenset([1])]), frozenset([6]): set([frozenset([4]), frozenset([2]), frozenset([1])]), frozenset([4]): set([frozenset([6]), frozenset([7]), frozenset([3]), frozenset([1])]), frozenset([1]): set([frozenset([6]), frozenset([4]), frozenset([5]), frozenset([2]), frozenset([3])]), frozenset([2]): set([frozenset([6]), frozenset([5]), frozenset([3]), frozenset([1])])}

Output should be a graph with only two nodes which are the frozensets of all the nodes in the graph. At this point it runs into KeyError.

def kargerMinCut(graph):
if len(graph) == 2:
    return graph
u = random.choice(graph.keys())   # u and v are frozensets, idea is that they form
v = random.choice(list(graph[u])) # a clique in a single frozenset
for node in graph:
    if node != u and node != v:
        links = graph[node]       
        if u in links or v in links:
            links.add(frozenset(tuple(u | v))) # combine u and v to form one link
            links.discard(u)                   # delete old links to u and v
            links.discard(v)            
            graph[node] = links
graph[u | v] = graph[u] | graph[v]             # new key for u and v 
del graph[u], graph[v]                         # u and v are no longer needed
return kargerMinCut(graph)

Upvotes: 0

Views: 688

Answers (1)

user
user

Reputation: 7333

I think the problem may be the use of the is keyword. Note that in python, is only returns true when the two arguments refer to the exact same object (equivalent to char* == char* in C++. The == operator returns true if the contents are the same (equivalent to string == string in C++).

So rather than is not try !=.

I once had this identical problem when traversing elements in a graph in python. : )

PS-- Also, I'd write the following line as a full if:

links.add(frozenset(tuple(u | v))) if u in links or v in links else None

Upvotes: 1

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