DFS algorithm implementation in Python with classes

Question

I am trying to implement DFS that will return a graph with all the nodes containing their predecessor, while having the color attribute to identify a cycle in the graph (there is a cycle iff (u,v) is a back edge iff v is gray and v.discovery < u.discovery).

The code:

# A class to represent a vertex object
class Vertex:
    def __init__(self, val, color="white", d_time=-1, f_time=-1, pred=None):
        self.val = val
        self.color = color
        self.d_time = d_time
        self.f_time = f_time
        self.pred = pred

# A class to represent a graph object
class Graph:
    def __init__(self, adjDict):
        self.vertexes = dict()
        for vertex in adjDict:
            self.vertexes[Vertex(vertex)] = [Vertex(neighbor) for neighbor in adjDict[vertex]]#problematic


def dfs(g):
    for vertex in g.vertexes:
        if vertex.color == "white":
            dfs_vist(g, vertex)
    return g


def dfs_vist(g, v, time=0):
    time += 1
    v.d_time = time
    v.color = "gray"
    for neighbor in g.vertexes[v]:
        if neighbor.color == "white":
            neighbor.pred = v
            dfs_vist(g, neighbor, time)
    v.color = "black"
    time += 1
    v.f_time = time

if __name__ == '__main__':
    graph = {
        0: [2, 4],
        1: [],
        2: [1, 5],
        3: [8],
        4: [7],
        5: [4],
        6: [3],
        7: [1],
        8: []
    }
    g = dfs(Graph(graph))
    print(g)

I am trying to initialize the graph when the graph is represented as dictionary with list of neighbors. The error is in the initializing phase: the same vertex gets created twice, once as a neighbor and once as a key, and then the object is different and the g.vertexes[v] result in a key error.
I want to keep my dfs and dfs_visit functions the same, is it possible to initialize it so that they remain the same?

Asked By: Kahalon

||

Source

Answer 1

Try doing this in your graph constructor:

class Graph:
    def __init__(self, adjDict):
        self.vertexes = dict()
        temp = {v: Vertex(v) for v in adjDict.keys()}
        for vertex, neighbors in adjDict.items():
            self.vertexes[temp[vertex]] = [temp[neighbor] for neighbor in neighbors]

If you create all of the vertexes prior to constructing the graph and then plugging them into their respective locations you avoid recreating the same vertex repeatedly.

This should solve your issue with creating duplicate vertexes, but I think your algorithm might still need a bit of tuning as well.

Answered By: Alexander

DFS algorithm implementation in Python with classes

Question:

Answers: