S

A

I

```
"""
A Hamiltonian cycle (Hamiltonian circuit) is a graph cycle
through a graph that visits each node exactly once.
Determining whether such paths and cycles exist in graphs
is the 'Hamiltonian path problem', which is NP-complete.
Wikipedia: https://en.wikipedia.org/wiki/Hamiltonian_path
"""
def valid_connection(
graph: list[list[int]], next_ver: int, curr_ind: int, path: list[int]
) -> bool:
"""
Checks whether it is possible to add next into path by validating 2 statements
1. There should be path between current and next vertex
2. Next vertex should not be in path
If both validations succeed we return True, saying that it is possible to connect
this vertices, otherwise we return False
Case 1:Use exact graph as in main function, with initialized values
>>> graph = [[0, 1, 0, 1, 0],
... [1, 0, 1, 1, 1],
... [0, 1, 0, 0, 1],
... [1, 1, 0, 0, 1],
... [0, 1, 1, 1, 0]]
>>> path = [0, -1, -1, -1, -1, 0]
>>> curr_ind = 1
>>> next_ver = 1
>>> valid_connection(graph, next_ver, curr_ind, path)
True
Case 2: Same graph, but trying to connect to node that is already in path
>>> path = [0, 1, 2, 4, -1, 0]
>>> curr_ind = 4
>>> next_ver = 1
>>> valid_connection(graph, next_ver, curr_ind, path)
False
"""
# 1. Validate that path exists between current and next vertices
if graph[path[curr_ind - 1]][next_ver] == 0:
return False
# 2. Validate that next vertex is not already in path
return not any(vertex == next_ver for vertex in path)
def util_hamilton_cycle(graph: list[list[int]], path: list[int], curr_ind: int) -> bool:
"""
Pseudo-Code
Base Case:
1. Check if we visited all of vertices
1.1 If last visited vertex has path to starting vertex return True either
return False
Recursive Step:
2. Iterate over each vertex
Check if next vertex is valid for transiting from current vertex
2.1 Remember next vertex as next transition
2.2 Do recursive call and check if going to this vertex solves problem
2.3 If next vertex leads to solution return True
2.4 Else backtrack, delete remembered vertex
Case 1: Use exact graph as in main function, with initialized values
>>> graph = [[0, 1, 0, 1, 0],
... [1, 0, 1, 1, 1],
... [0, 1, 0, 0, 1],
... [1, 1, 0, 0, 1],
... [0, 1, 1, 1, 0]]
>>> path = [0, -1, -1, -1, -1, 0]
>>> curr_ind = 1
>>> util_hamilton_cycle(graph, path, curr_ind)
True
>>> path
[0, 1, 2, 4, 3, 0]
Case 2: Use exact graph as in previous case, but in the properties taken from
middle of calculation
>>> graph = [[0, 1, 0, 1, 0],
... [1, 0, 1, 1, 1],
... [0, 1, 0, 0, 1],
... [1, 1, 0, 0, 1],
... [0, 1, 1, 1, 0]]
>>> path = [0, 1, 2, -1, -1, 0]
>>> curr_ind = 3
>>> util_hamilton_cycle(graph, path, curr_ind)
True
>>> path
[0, 1, 2, 4, 3, 0]
"""
# Base Case
if curr_ind == len(graph):
# return whether path exists between current and starting vertices
return graph[path[curr_ind - 1]][path[0]] == 1
# Recursive Step
for next_ver in range(len(graph)):
if valid_connection(graph, next_ver, curr_ind, path):
# Insert current vertex into path as next transition
path[curr_ind] = next_ver
# Validate created path
if util_hamilton_cycle(graph, path, curr_ind + 1):
return True
# Backtrack
path[curr_ind] = -1
return False
def hamilton_cycle(graph: list[list[int]], start_index: int = 0) -> list[int]:
r"""
Wrapper function to call subroutine called util_hamilton_cycle,
which will either return array of vertices indicating hamiltonian cycle
or an empty list indicating that hamiltonian cycle was not found.
Case 1:
Following graph consists of 5 edges.
If we look closely, we can see that there are multiple Hamiltonian cycles.
For example one result is when we iterate like:
(0)->(1)->(2)->(4)->(3)->(0)
(0)---(1)---(2)
| / \ |
| / \ |
| / \ |
|/ \|
(3)---------(4)
>>> graph = [[0, 1, 0, 1, 0],
... [1, 0, 1, 1, 1],
... [0, 1, 0, 0, 1],
... [1, 1, 0, 0, 1],
... [0, 1, 1, 1, 0]]
>>> hamilton_cycle(graph)
[0, 1, 2, 4, 3, 0]
Case 2:
Same Graph as it was in Case 1, changed starting index from default to 3
(0)---(1)---(2)
| / \ |
| / \ |
| / \ |
|/ \|
(3)---------(4)
>>> graph = [[0, 1, 0, 1, 0],
... [1, 0, 1, 1, 1],
... [0, 1, 0, 0, 1],
... [1, 1, 0, 0, 1],
... [0, 1, 1, 1, 0]]
>>> hamilton_cycle(graph, 3)
[3, 0, 1, 2, 4, 3]
Case 3:
Following Graph is exactly what it was before, but edge 3-4 is removed.
Result is that there is no Hamiltonian Cycle anymore.
(0)---(1)---(2)
| / \ |
| / \ |
| / \ |
|/ \|
(3) (4)
>>> graph = [[0, 1, 0, 1, 0],
... [1, 0, 1, 1, 1],
... [0, 1, 0, 0, 1],
... [1, 1, 0, 0, 0],
... [0, 1, 1, 0, 0]]
>>> hamilton_cycle(graph,4)
[]
"""
# Initialize path with -1, indicating that we have not visited them yet
path = [-1] * (len(graph) + 1)
# initialize start and end of path with starting index
path[0] = path[-1] = start_index
# evaluate and if we find answer return path either return empty array
return path if util_hamilton_cycle(graph, path, 1) else []
```