$**$Code is still not working, please help$**$

I attahced my class code:

class DFSTimeCounter:

def $\_\_$init$\_\_($self$)$:

self.count $=0$

def reset$($self$)$:

self.count $=0$

def increment$($self$)$:

self.count $=$ self.count $+1$

def get$($self$)$:

return self.count

class UndirectedGraph:

def $\_\_$init$\_\_($self$,$ n$)$:

self.n $=$ n

self.adj$\_$list $=[$ set$()$ for i in range$($self$.$n$)]$

def add$\_$edge$($self$,$ i$,$ j$)$:

assert i $$ self.n

assert j $$ self.n

assert i $!=$ j

# Make sure to add edge from i to j

self.adj$\_$list$[$i$].$add$($j$)$

# Also add edge from j to i

self.adj$\_$list$[$j$].$add$($i$)$

# get a set of all vertices that

# are neighbors of the

# vertex i

def get$\_$neighboring$\_$vertices$($self$,$ i$)$:

assert i $$ self.n

return self.adj$\_$list$[$i$]$

def dfs$\_$visit$($self$,$ i$,$ dfs$\_$timer, discovery$\_$times, finish$\_$times,

dfs$\_$tree$\_$parent, dfs$\_$back$\_$edges$)$:

assert i $$ self.n

assert discovery$\_$times$[$i$]==$ None

assert finish$\_$times$[$i$]==$ None

discovery$\_$times$[$i$]=$ dfs$\_$timer.get$()$

dfs$\_$timer.increment$()$

for v in self.get$\_$neighboring$\_$vertices$($i$)$:

if discovery$\_$times$[$v$]$ is not None and finish$\_$times$[$v$]$ is None:

dfs$\_$back$\_$edges.append$(($i$,$v$))$

if discovery$\_$times$[$v$]$ is None:

dfs$\_$tree$\_$parent$[$v$]=$ i

self.dfs$\_$visit$($v$,$ dfs$\_$timer, discovery$\_$times, finish$\_$times, dfs$\_$tree$\_$parent, dfs$\_$back$\_$edges$)$

finish$\_$times$[$i$]=$ dfs$\_$timer.get$()$

dfs$\_$timer.increment$()$

# Function: dfs$\_$traverse$\_$graph

# Traverse the entire graph.

def dfs$\_$traverse$\_$graph$($self$)$:

dfs$\_$timer $=$ DFSTimeCounter$()$

discovery$\_$times $=[$None$]*$self$.$n

finish$\_$times $=[$None$]*$self$.$n

dfs$\_$tree$\_$parents $=[$None$]*$self$.$n

dfs$\_$back$\_$edges $=[]$

for i in range$($self$.$n$)$:

if discovery$\_$times$[$i$]==$ None:

self.dfs$\_$visit$($i$,$dfs$\_$timer, discovery$\_$times, finish$\_$times,

dfs$\_$tree$\_$parents, dfs$\_$back$\_$edges$)$

# Clean up the back edges so that if $($i$,$j$)$ is a back edge then j cannot

# be i$\text{'}$s parent.

non$\_$trivial$\_$back$\_$edges $=[($i$,$j$)$ for $($i$,$j$)$ in dfs$\_$back$\_$edges if dfs$\_$tree$\_$parents$[$i$]!=$ j$]$

return $($dfs$\_$tree$\_$parents, non$\_$trivial$\_$back$\_$edges, discovery$\_$times, finish$\_$times$)$