**Concept:**
A **Hamiltonian graph** is a graph that contains a Hamiltonian cycle, which is a closed loop that visits every vertex exactly once. In other words, the graph has a path that starts at one node, traverses through all other nodes without repetition, and returns to the starting node. Some edges may not be used, but no edge can be traversed more than once.
**Difference with Euler Graphs:**
An **Euler graph**, on the other hand, is concerned with the traversal of edges, ensuring that each edge is visited exactly once. A Hamiltonian graph focuses on visiting each vertex exactly once, regardless of how many edges are used or skipped.
**Determination:**
1. **Dirac's Theorem (Sufficient Condition):**
If an undirected graph has N vertices, and each vertex has a degree of at least N/2, then the graph must contain a Hamiltonian cycle.
2. **Basic Necessary Conditions:**
For a graph G = (V, E) to have a Hamiltonian cycle, for any non-empty subset S of V, if W(GS) represents the number of connected components in the subgraph obtained by removing the vertices in S, then W(GS) ≤ |S| must hold true.
3. **Tournament Graph (Hamiltonian Path):**
A tournament graph of order N (a directed graph where there is an edge between every pair of vertices) always contains a Hamiltonian path.
**Algorithm:**
**One: Constructing a Hamiltonian Cycle under Dirac’s Theorem**
**Process:**
1. Start by selecting two adjacent nodes S and T. Extend the path as long as possible without repeating nodes.
2. If S is adjacent to T, the path forms a cycle.
3. If S is not adjacent to T, find a node vi such that vi is adjacent to T and vi+1 is adjacent to S. Then restructure the path to form a cycle.
4. If the current path does not include all nodes, continue expanding it by adding new nodes until the full cycle is formed.
**Proof:**
The algorithm relies on the pigeonhole principle, ensuring that each expansion step adds at least one new node to the path.
**Pseudocode:**
Let s be the starting point of the Hamiltonian cycle, and t be the node before s. ans[] stores the final cycle. Reverse means reversing the elements in a given range.
1. Initialize s = 1, t as any neighbor of s.
2. While the length of ans[] is less than n, expand from t. If an unvisited node v is found, add it to the path.
3. If expansion fails, reverse the path, swap s and t, and continue expanding.
4. If s and t are not adjacent, find a suitable node to adjust the path and form a cycle.
5. Repeat until the entire cycle is constructed.
**Time Complexity:**
Each round adds at least one node to the path, resulting in O(n²) time complexity. Adjacency matrices are preferred for efficiency.
**Code Example (C++):**
```cpp
const int maxN = 100;
inline void reverse(int arr[maxN + 7], int s, int t) {
int temp;
while (s < t) {
temp = arr[s];
arr[s] = arr[t];
arr[t] = temp;
s++;
t--;
}
}
void Hamilton(int ans[maxN + 7], bool map[maxN + 7][maxN + 7], int n) {
int s = 1, t;
int ansi = 2;
int i, j;
int w;
int temp;
bool visit[maxN + 7] = {false};
for (i = 1; i <= n; i++) if (map[s][i]) break;
t = i;
visit[s] = visit[t] = true;
ans[0] = s;
ans[1] = t;
while (true) {
while (true) {
for (i = 1; i <= n; i++) {
if (map[t][i] && !visit[i]) {
ans[ansi++] = i;
visit[i] = true;
t = i;
break;
}
}
if (i > n) break;
}
w = ansi - 1;
reverse(ans, 0, w);
temp = s;
s = t;
t = temp;
while (true) {
for (i = 1; i <= n; i++) {
if (map[t][i] && !visit[i]) {
ans[ansi++] = i;
visit[i] = true;
t = i;
break;
}
}
if (i > n) break;
}
if (!map[s][t]) {
for (i = 1; i < ansi - 1; i++)
if (map[ans[i]][t] && map[s][ans[i + 1]]) break;
w = ansi - 1;
i++;
t = ans[i];
reverse(ans, i, w);
}
if (ansi == n) return;
for (j = 1; j <= n; j++) {
if (visit[j]) continue;
for (i = 1; i < ansi - 1; i++) if (map[ans[i]][j]) break;
if (map[ans[i]][j]) break;
}
s = ans[i - 1];
t = j;
reverse(ans, 0, i - 1);
reverse(ans, i, ansi - 1);
ans[ansi++] = j;
visit[j] = true;
}
}
```
**Two: Constructing a Hamiltonian Path in a Tournament Graph (N ≥ 2)**
A **tournament graph** is a directed graph where each pair of vertices has exactly one directed edge between them. It is guaranteed that a Hamiltonian path exists in any tournament graph.
**Proof via Mathematical Induction:**
- Base case: When N = 2, the path is trivial.
- Inductive step: Assume a Hamiltonian path exists for N = k. For N = k + 1, insert the new node into the existing path based on its connections.
**Example:**
For a tournament graph with 4 nodes, the Hamiltonian path can be constructed by inserting the fifth node into the correct position based on its directed edges.
**Code Example (C++):**
```cpp
#include
#include
using namespace std;
const int maxN = 200;
void Hamilton(int ans[maxN + 7], bool map[maxN + 7][maxN + 7], int n) {
int ansi = 1;
ans[ansi++] = 1;
for (int i = 2; i <= n; i++) {
if (map[i][ans[ansi - 1]] == 1)
ans[ansi++] = i;
else {
int flag = 0;
for (int j = ansi - 2; j >= 0; --j) {
if (map[i][ans[j]] == 1) {
flag = 1;
for (int k = ansi; k > j + 1; k--)
ans[k] = ans[k - 1];
ans[j + 1] = i;
ansi++;
break;
}
}
if (!flag) {
for (int k = ansi; k > 0; k--)
ans[k] = ans[k - 1];
ans[0] = i;
ansi++;
}
}
}
}
```
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