Revision history [back]
 | 1 | initial version |
A code from chatgpt maybe OK.
def is_exactly_one_path(G, start, end, length):
"""
Check if there exists exactly one path from start to end with length in graph G.
Stop searching when two paths are found.
Return True if exactly one path is found.
Return False if no path is found.
"""
paths = [] # Used to store found paths
stack = [(start, [start])] # Initialize stack with tuples of (current node, path)
while stack:
current, path = stack.pop()
if len(paths) > 1: # If more than one path is found, terminate early
return False
if len(path) - 1 == length: # Check path length
if current == end:
paths.append(path) # If valid path, add to paths
continue # Move to the next iteration
# Explore neighbors of the current node
for neighbor in G.neighbors(current):
if neighbor not in path: # Avoid cyclic paths
stack.append((neighbor, path + [neighbor])) # Add new state to stack
return len(paths) == 1 # Return True if exactly one path is found
# Assume G is a predefined graph object, for example:
# G = graphs.PetersenGraph()
# Call the function to check if exactly one path from 0 to 4 with length 3 exists
# result = is_exactly_one_path(G, 0, 4, 3)
# print(result)
| | 2 | No.2 Revision |
A code from chatgpt maybe seems OK.
def is_exactly_one_path(G, start, end, length):
"""
Check if there exists exactly one path from start to end with length in graph G.
Stop searching when two paths are found.
Return True if exactly one path is found.
Return False if no path is found.
"""
paths = [] # Used to store found paths
stack = [(start, [start])] # Initialize stack with tuples of (current node, path)
while stack:
current, path = stack.pop()
if len(paths) > 1: # If more than one path is found, terminate early
return False
if len(path) - 1 == length: # Check path length
if current == end:
paths.append(path) # If valid path, add to paths
continue # Move to the next iteration
# Explore neighbors of the current node
for neighbor in G.neighbors(current):
if neighbor not in path: # Avoid cyclic paths
stack.append((neighbor, path + [neighbor])) # Add new state to stack
return len(paths) == 1 # Return True if exactly one path is found
# Assume G is a predefined graph object, for example:
# G = graphs.PetersenGraph()
# Call the function to check if exactly one path from 0 to 4 with length 3 exists
# result = is_exactly_one_path(G, 0, 4, 3)
# print(result)
| | 3 | No.3 Revision |
A code from chatgpt seems OK.
def is_exactly_one_path(G, start, end, length):
"""
Check if there exists exactly one path from start to end with length in graph G.
Stop searching when two paths are found.
Return True if exactly one path is found.
Return False if no path is found.
"""
paths = [] # Used to store found paths
stack = [(start, [start])] # Initialize stack with tuples of (current node, path)
while stack:
current, path = stack.pop()
if len(paths) > 1: # If more than one path is found, terminate early
return False
if len(path) - 1 == length: # Check path length
if current == end:
paths.append(path) # If valid path, add to paths
continue # Move to the next iteration
# Explore neighbors of the current node
for neighbor in G.neighbors(current):
if neighbor not in path: # Avoid cyclic paths
stack.append((neighbor, path + [neighbor])) # Add new state to stack
return len(paths) == 1 # Return True if exactly one path is found
# Assume G is a predefined graph object, for example:
# G = graphs.PetersenGraph()
# Call the function to check if exactly one path from 0 to 4 with length 3 exists
# result = is_exactly_one_path(G, 0, 4, 3)
# print(result)
construct all paths:
def em_k_path(G, start, end, length):
"""
Check if there exists exactly one path from start to end with length in graph G.
Stop searching when two paths are found.
Return True if exactly one path is found.
Return False if no path is found.
"""
paths = [] # Used to store found paths
stack = [(start, [start])] # Initialize stack with tuples of (current node, path)
while stack:
current, path = stack.pop()
if len(path) - 1 == length: # Check path length
if current == end:
paths.append(path) # If valid path, add to paths
continue # Move to the next iteration
# Explore neighbors of the current node
for neighbor in G.neighbors(current):
if neighbor not in path: # Avoid cyclic paths
stack.append((neighbor, path + [neighbor])) # Add new state to stack
return paths
| | 4 | No.4 Revision |
A code from chatgpt seems OK.
def is_exactly_one_path(G, start, end, length):
"""
Check if there exists exactly one path from start to end with length in graph G.
Stop searching when two paths are found.
Return True if exactly one path is found.
Return False if no path is found.
"""
paths = [] # Used to store found paths
stack = [(start, [start])] # Initialize stack with tuples of (current node, path)
while stack:
current, path = stack.pop()
if len(paths) > 1: # If more than one path is found, terminate early
return False
if len(path) - 1 == length: # Check path length
if current == end:
paths.append(path) # If valid path, add to paths
continue # Move to the next iteration
# Explore neighbors of the current node
for neighbor in G.neighbors(current):
if neighbor not in path: # Avoid cyclic paths
stack.append((neighbor, path + [neighbor])) # Add new state to stack
return len(paths) == 1 # Return True if exactly one path is found
# Assume G is a predefined graph object, for example:
# G = graphs.PetersenGraph()
# Call the function to check if exactly one path from 0 to 4 with length 3 exists
# result = is_exactly_one_path(G, 0, 4, 3)
# print(result)
construct all paths:
def em_k_path(G, start, end, length):
"""
Check if there exists exactly one path from start to end with length in graph G.
Stop searching when two paths are found.
Return True if exactly one path is found.
Return False if no path is found.
"""
paths = [] # Used to store found paths
stack = [(start, [start])] # Initialize stack with tuples of (current node, path)
while stack:
current, path = stack.pop()
if len(path) - 1 == length: # Check path length
if current == end:
paths.append(path) # If valid path, add to paths
continue # Move to the next iteration
# Explore neighbors of the current node
for neighbor in G.neighbors(current):
if neighbor not in path: # Avoid cyclic paths
stack.append((neighbor, path + [neighbor])) # Add new state to stack
return paths
