I'm having certain Issues with my K Shortest Path algorithm. The code is given as:

```
def K_shortest_Paths(graph,S,T,K=4):
'''Initialize Variables Accordingly'''
B = {}
P = set()
count = {}
for U in graph.keys():
count[U] = 0
B[S] = 0
'''Algorithm Starts'''
while(len(B)>=1 and count[T]<K):
PU = min(B,key=lambda x:B[x])
cost = B[PU]
U = PU[len(PU)-1]
del B[PU]
count[U] += 1
if U==T:
P.add(PU)
if count[U]<=K:
V = graph[U].keys()
for v in V:
if v not in PU:
PV = PU+v
B[PV] = cost+1
return P
```

This is equivalent to https://en.wikipedia.org/wiki/K_shortest_path_routing which provides the Pseudo Code for Implementation. The Graph is given as: Now, It works well if I have Starting Node S<10, and Terminating Node T<10, but with S and T>10, it returns an empty set, Whereas it should return the Path. Please Note that I can't Use Networkx libraries. I only have to Use basic libraries in Python

Furthermore, the Code to Generate Graph is this:

```
def create_dictionary(graph):
D = {}
for item in graph.items():
temp = {}
connected = list(item[1])
key = item[0]
for V in connected:
temp[str(V)] = 1
D[str(key)] = temp
return D
def gen_p_graph(nodes,prob):
if prob>1:
er='error'
return er
graph_matrix=np.zeros([nodes,nodes])
num_of_connections=int(((nodes * (nodes-1)) * prob )/2)
num_list_row=list(range(nodes-1))
while(np.sum(np.triu(graph_matrix))!=num_of_connections):
row_num=random.choice(num_list_row)
num_list_col=(list(range(row_num+1,nodes)))
col_num=random.choice(num_list_col)
if graph_matrix[row_num,col_num]==0:
graph_matrix[row_num,col_num]=1
graph_matrix[col_num,row_num]=1
#create dictionary
df=pd.DataFrame(np.argwhere(graph_matrix==1))
arr=np.unique(df.iloc[:,0])
dct={}
for i in range(graph_matrix.shape[0]):
dct[str(i)]=set()
for val in arr:
dct[str(val)].update(df.loc[df.iloc[:,0]==val].iloc[:,1].values)
return pd.DataFrame(graph_matrix),dct
```

And I run it like this:

```
graph= create_dictionary(gen_p_graph(100,0.8)[1])
K_shortest_Paths(graph,'11','10')
```

return an empty set whereas it should return paths.

### Answer1:

What I presume you are trying to achieve. Try this.

```
def k_shortest_paths(graph, src_node, dest_node, k=4):
result = []
pathes = [[src_node]]
while len(pathes) > 0 and len(result) < k:
path = pathes.pop()
last_node = path[-1]
if last_node == dest_node:
result.append(path)
else:
for child_node in graph[last_node].keys():
if child_node not in path:
pathes.append(path + [child_node])
return result
```

### Answer2:

If you call `K_shortest_Pathes(graph, "11", "10")`

you will never add an element to the set `P`

. Read my inline comments.

```
def K_shortest_Paths(graph,S,T,K=4):
'''Initialize Variables Accordingly'''
B = {}
P = set()
count = {}
for U in graph.keys():
count[U] = 0
# currently the B has only one item, i.e. { S: 0 } => { "11": 0 }
B[S] = 0
'''Algorithm Starts'''
while(len(B)>=1 and count[T]<K):
# results in the only key in B, i.e. PU = S => PU = "11"
PU = min(B,key=lambda x:B[x])
cost = B[PU]
# U = PU[len(PU) - 1], where PU = "11" =>
# U = "11"[len("11")-1] =>
# *** U = "1"
U = PU[len(PU)-1]
del B[PU]
count[U] += 1
# *** U == T => "1" == T => "1" == "10" which is False
# Thus nothing is ever added to set P
if U==T:
P.add(PU)
if count[U]<=K:
V = graph[U].keys()
for v in V:
if v not in PU:
PV = PU+v
B[PV] = cost+1
return P
```