|
| 1 | +""" |
| 2 | + A Hamiltonian cycle (Hamiltonian circuit) is a graph cycle |
| 3 | + through a graph that visits each node exactly once. |
| 4 | + Determining whether such paths and cycles exist in graphs |
| 5 | + is the 'Hamiltonian path problem', which is NP-complete. |
| 6 | + |
| 7 | + Wikipedia: https://en.wikipedia.org/wiki/Hamiltonian_path |
| 8 | +""" |
| 9 | +from typing import List |
| 10 | + |
| 11 | + |
| 12 | +def valid_connection( |
| 13 | + graph: List[List[int]], next_ver: int, curr_ind: int, path: List[int] |
| 14 | +) -> bool: |
| 15 | + """ |
| 16 | + Checks whether it is possible to add next into path by validating 2 statements |
| 17 | + 1. There should be path between current and next vertex |
| 18 | + 2. Next vertex should not be in path |
| 19 | + If both validations succeeds we return true saying that it is possible to connect this vertices |
| 20 | + either we return false |
| 21 | + |
| 22 | + Case 1:Use exact graph as in main function, with initialized values |
| 23 | + >>> graph = [[0, 1, 0, 1, 0], |
| 24 | + ... [1, 0, 1, 1, 1], |
| 25 | + ... [0, 1, 0, 0, 1], |
| 26 | + ... [1, 1, 0, 0, 1], |
| 27 | + ... [0, 1, 1, 1, 0]] |
| 28 | + >>> path = [0, -1, -1, -1, -1, 0] |
| 29 | + >>> curr_ind = 1 |
| 30 | + >>> next_ver = 1 |
| 31 | + >>> valid_connection(graph, next_ver, curr_ind, path) |
| 32 | + True |
| 33 | +
|
| 34 | + Case 2: Same graph, but trying to connect to node that is already in path |
| 35 | + >>> path = [0, 1, 2, 4, -1, 0] |
| 36 | + >>> curr_ind = 4 |
| 37 | + >>> next_ver = 1 |
| 38 | + >>> valid_connection(graph, next_ver, curr_ind, path) |
| 39 | + False |
| 40 | + """ |
| 41 | + |
| 42 | + # 1. Validate that path exists between current and next vertices |
| 43 | + if graph[path[curr_ind - 1]][next_ver] == 0: |
| 44 | + return False |
| 45 | + |
| 46 | + # 2. Validate that next vertex is not already in path |
| 47 | + return not any(vertex == next_ver for vertex in path) |
| 48 | + |
| 49 | + |
| 50 | +def util_hamilton_cycle(graph: List[List[int]], path: List[int], curr_ind: int) -> bool: |
| 51 | + """ |
| 52 | + Pseudo-Code |
| 53 | + Base Case: |
| 54 | + 1. Chceck if we visited all of vertices |
| 55 | + 1.1 If last visited vertex has path to starting vertex return True either return False |
| 56 | + Recursive Step: |
| 57 | + 2. Iterate over each vertex |
| 58 | + Check if next vertex is valid for transiting from current vertex |
| 59 | + 2.1 Remember next vertex as next transition |
| 60 | + 2.2 Do recursive call and check if going to this vertex solves problem |
| 61 | + 2.3 if next vertex leads to solution return True |
| 62 | + 2.4 else backtrack, delete remembered vertex |
| 63 | + |
| 64 | + Case 1: Use exact graph as in main function, with initialized values |
| 65 | + >>> graph = [[0, 1, 0, 1, 0], |
| 66 | + ... [1, 0, 1, 1, 1], |
| 67 | + ... [0, 1, 0, 0, 1], |
| 68 | + ... [1, 1, 0, 0, 1], |
| 69 | + ... [0, 1, 1, 1, 0]] |
| 70 | + >>> path = [0, -1, -1, -1, -1, 0] |
| 71 | + >>> curr_ind = 1 |
| 72 | + >>> util_hamilton_cycle(graph, path, curr_ind) |
| 73 | + True |
| 74 | + >>> print(path) |
| 75 | + [0, 1, 2, 4, 3, 0] |
| 76 | +
|
| 77 | + Case 2: Use exact graph as in previous case, but in the properties taken from middle of calculation |
| 78 | + >>> graph = [[0, 1, 0, 1, 0], |
| 79 | + ... [1, 0, 1, 1, 1], |
| 80 | + ... [0, 1, 0, 0, 1], |
| 81 | + ... [1, 1, 0, 0, 1], |
| 82 | + ... [0, 1, 1, 1, 0]] |
| 83 | + >>> path = [0, 1, 2, -1, -1, 0] |
| 84 | + >>> curr_ind = 3 |
| 85 | + >>> util_hamilton_cycle(graph, path, curr_ind) |
| 86 | + True |
| 87 | + >>> print(path) |
| 88 | + [0, 1, 2, 4, 3, 0] |
| 89 | + """ |
| 90 | + |
| 91 | + # Base Case |
| 92 | + if curr_ind == len(graph): |
| 93 | + # return whether path exists between current and starting vertices |
| 94 | + return graph[path[curr_ind - 1]][path[0]] == 1 |
| 95 | + |
| 96 | + # Recursive Step |
| 97 | + for next in range(0, len(graph)): |
| 98 | + if valid_connection(graph, next, curr_ind, path): |
| 99 | + # Insert current vertex into path as next transition |
| 100 | + path[curr_ind] = next |
| 101 | + # Validate created path |
| 102 | + if util_hamilton_cycle(graph, path, curr_ind + 1): |
| 103 | + return True |
| 104 | + # Backtrack |
| 105 | + path[curr_ind] = -1 |
| 106 | + return False |
| 107 | + |
| 108 | + |
| 109 | +def hamilton_cycle(graph: List[List[int]], start_index: int = 0) -> List[int]: |
| 110 | + r""" |
| 111 | + Wrapper function to call subroutine called util_hamilton_cycle, |
| 112 | + which will either return array of vertices indicating hamiltonian cycle |
| 113 | + or an empty list indicating that hamiltonian cycle was not found. |
| 114 | + Case 1: |
| 115 | + Following graph consists of 5 edges. |
| 116 | + If we look closely, we can see that there are multiple Hamiltonian cycles. |
| 117 | + For example one result is when we iterate like: |
| 118 | + (0)->(1)->(2)->(4)->(3)->(0) |
| 119 | + |
| 120 | + (0)---(1)---(2) |
| 121 | + | / \ | |
| 122 | + | / \ | |
| 123 | + | / \ | |
| 124 | + |/ \| |
| 125 | + (3)---------(4) |
| 126 | + >>> graph = [[0, 1, 0, 1, 0], |
| 127 | + ... [1, 0, 1, 1, 1], |
| 128 | + ... [0, 1, 0, 0, 1], |
| 129 | + ... [1, 1, 0, 0, 1], |
| 130 | + ... [0, 1, 1, 1, 0]] |
| 131 | + >>> hamilton_cycle(graph) |
| 132 | + [0, 1, 2, 4, 3, 0] |
| 133 | + |
| 134 | + Case 2: |
| 135 | + Same Graph as it was in Case 1, changed starting index from default to 3 |
| 136 | + |
| 137 | + (0)---(1)---(2) |
| 138 | + | / \ | |
| 139 | + | / \ | |
| 140 | + | / \ | |
| 141 | + |/ \| |
| 142 | + (3)---------(4) |
| 143 | + >>> graph = [[0, 1, 0, 1, 0], |
| 144 | + ... [1, 0, 1, 1, 1], |
| 145 | + ... [0, 1, 0, 0, 1], |
| 146 | + ... [1, 1, 0, 0, 1], |
| 147 | + ... [0, 1, 1, 1, 0]] |
| 148 | + >>> hamilton_cycle(graph, 3) |
| 149 | + [3, 0, 1, 2, 4, 3] |
| 150 | + |
| 151 | + Case 3: |
| 152 | + Following Graph is exactly what it was before, but edge 3-4 is removed. |
| 153 | + Result is that there is no Hamiltonian Cycle anymore. |
| 154 | + |
| 155 | + (0)---(1)---(2) |
| 156 | + | / \ | |
| 157 | + | / \ | |
| 158 | + | / \ | |
| 159 | + |/ \| |
| 160 | + (3) (4) |
| 161 | + >>> graph = [[0, 1, 0, 1, 0], |
| 162 | + ... [1, 0, 1, 1, 1], |
| 163 | + ... [0, 1, 0, 0, 1], |
| 164 | + ... [1, 1, 0, 0, 0], |
| 165 | + ... [0, 1, 1, 0, 0]] |
| 166 | + >>> hamilton_cycle(graph,4) |
| 167 | + [] |
| 168 | + """ |
| 169 | + |
| 170 | + # Initialize path with -1, indicating that we have not visited them yet |
| 171 | + path = [-1] * (len(graph) + 1) |
| 172 | + # initialize start and end of path with starting index |
| 173 | + path[0] = path[-1] = start_index |
| 174 | + # evaluate and if we find answer return path either return empty array |
| 175 | + return path if util_hamilton_cycle(graph, path, 1) else [] |
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