<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

468. A bit of classic
Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



Everybody loves classical problems! Real professional can solve them quickly and gaily come to more difficult problems, and amateurs are just able to solve them, which is also important for popularization of computer programming contests. Here you are required to solve classical problem not only in chess but in computer science as well. Given integer N, find a way to bypass every cell of a chessboard N x N exactly once with a chess knight.

Input
Input file contains integer number N — size of the board (1 ≤ N ≤ 250).

Output
If there is no solution for the given size of the board, write to the output file the only message No solution. (without quotes). Otherwise, write to the first line of the output file message There is solution:, and then to every of the N following lines write N numbers separated by spaces — order of traversal of the board. Each of the numbers from 1 to N2 should occur in this sequence exactly once. Knight may start and end its trip at any cell of the chessboard.

Example(s)
sample input
sample output
3
No solution.

sample input
sample output
5
There is solution:
 1 14 9 20 3
 24 19 2 15 10
 13 8 25 4 21
 18 23 6 11 16
 7 12 17 22 5

<|response|>
1. Abridged problem statement  
Given an integer N (1 ≤ N ≤ 250), find an open knight’s tour on an N×N chessboard: a sequence of knight moves that visits every cell exactly once. If no such tour exists, print “No solution.” Otherwise print “There is solution:” on the first line, then an N×N grid of integers 1..N² giving the visitation order.

2. Key observations  
- A knight’s tour does not exist for N = 2, 3 or 4 (except the trivial 1×1 case).  
- For N = 1, the tour is just the single cell.  
- For all N ≥ 5, an open knight’s tour exists.  
- Warnsdorff’s heuristic is a simple greedy rule that usually finds a full tour: always move to the unvisited neighbor that has the fewest onward moves (minimum “degree”).  
- To avoid early dead-ends, randomize ties, randomize the starting cell, and if the tour gets stuck before covering all N² squares, restart the procedure.

3. Full solution approach  
a. Handle trivial / no-tour cases:
   - If N is 2, 3 or 4, output “No solution.” and stop.  
   - If N = 1, output “There is solution:” then “ 1”.  
b. Prepare the eight knight moves as a list of (dx,dy) pairs.  
c. Seed a random number generator (fixed seed for reproducibility, or system random).  
d. Repeat until a full tour is found:  
   1. Initialize an N×N board of zeros (0 = unvisited).  
   2. Shuffle the global list of knight moves.  
   3. Pick a random starting cell (cx,cy), mark it as move 1.  
   4. For move = 2 to N²:  
      - Collect all valid knight moves from (cx,cy) to an unvisited cell (nx,ny).  
      - For each candidate (nx,ny), compute its Warnsdorff degree = number of onward moves available from (nx,ny).  
      - If there are no candidates, we’re stuck: abort this attempt and restart.  
      - Otherwise choose the candidate with minimum degree (break ties randomly by prior shuffling).  
      - Move to that cell, mark it with the current move number, update (cx,cy).  
   5. If we completed all moves up to N², we have a solution; break out of the retry loop.  
e. Print “There is solution:” and then the board, each row on its own line, cells separated by spaces.

This runs in O(N²·8) per attempt, and empirically the number of attempts before success is small even for N up to 250.

4. C++ implementation with detailed comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Board size
int N;

// The eight knight moves (dx, dy)
vector<pair<int,int>> knight_moves = {
    {1, 2}, {1, -2}, {2, 1}, {2, -1},
    {-1, 2}, {-1, -2}, {-2, 1}, {-2, -1}
};

// Random number generator (fixed seed for reproducibility)
mt19937 rng(42);

// Check whether (x,y) is inside the board and unvisited (board[x][y]==0)
bool is_valid(int x, int y, const vector<vector<int>>& board) {
    return x >= 0 && x < N
        && y >= 0 && y < N
        && board[x][y] == 0;
}

// Compute Warnsdorff’s degree from (x,y):
// number of onward moves to unvisited squares
int get_degree(int x, int y, const vector<vector<int>>& board) {
    int d = 0;
    for (auto &mv : knight_moves) {
        int nx = x + mv.first;
        int ny = y + mv.second;
        if (is_valid(nx, ny, board))
            ++d;
    }
    return d;
}

// Try one pass of Warnsdorff’s heuristic.
// Return true if a full tour is found (fills board with 1..N*N), false if stuck.
bool find_tour(vector<vector<int>>& board) {
    // Random starting cell
    uniform_int_distribution<int> dist(0, N-1);
    int cx = dist(rng);
    int cy = dist(rng);
    board[cx][cy] = 1;  // first move

    // Place moves 2..N*N
    for (int step = 2; step <= N*N; ++step) {
        vector<pair<int,int>> candidates;
        // For each knight move, if valid, compute its degree
        for (int i = 0; i < 8; ++i) {
            int nx = cx + knight_moves[i].first;
            int ny = cy + knight_moves[i].second;
            if (is_valid(nx, ny, board)) {
                int deg = get_degree(nx, ny, board);
                candidates.emplace_back(deg, i);
            }
        }
        // No valid onward move → stuck
        if (candidates.empty())
            return false;

        // Pick the candidate with minimum degree
        // move index is in second of the pair
        auto best = *min_element(
            candidates.begin(),
            candidates.end(),
            [](auto &a, auto &b){ return a.first < b.first; }
        );
        int best_i = best.second;

        // Apply the chosen move
        cx += knight_moves[best_i].first;
        cy += knight_moves[best_i].second;
        board[cx][cy] = step;
    }
    return true;  // full tour completed
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    // Read input
    if (!(cin >> N)) return 0;

    // No solutions for N=2,3,4
    if (N == 2 || N == 3 || N == 4) {
        cout << "No solution.\n";
        return 0;
    }

    // Special case N=1
    if (N == 1) {
        cout << "There is solution:\n 1\n";
        return 0;
    }

    // Prepare an empty board
    vector<vector<int>> board(N, vector<int>(N, 0));

    // Retry until we find a full tour
    while (true) {
        // Reset board
        for (int i = 0; i < N; ++i)
            fill(board[i].begin(), board[i].end(), 0);

        // Randomize the global move order for tie-breaking
        shuffle(knight_moves.begin(), knight_moves.end(), rng);

        // Attempt one pass
        if (find_tour(board))
            break;  // success
    }

    // Output the solution
    cout << "There is solution:\n";
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            // Leading space for each number, as per sample
            cout << ' ' << board[i][j];
        }
        cout << '\n';
    }
    return 0;
}
```

5. Python implementation with detailed comments  
```python
import sys
import random

# Read N
data = sys.stdin.read().strip().split()
if not data:
    sys.exit(0)
N = int(data[0])

# No solutions for N=2,3,4
if N in (2, 3, 4):
    print("No solution.")
    sys.exit(0)

# Special case N=1
if N == 1:
    print("There is solution:")
    print(" 1")
    sys.exit(0)

# The eight knight moves
knight_moves = [
    (1,  2), (1, -2), (2,  1), (2, -1),
    (-1, 2), (-1,-2), (-2, 1), (-2,-1)
]

# Fix the random seed for reproducibility
random.seed(42)

def is_valid(x, y, board):
    """Check that (x,y) is on board and unvisited."""
    return 0 <= x < N and 0 <= y < N and board[x][y] == 0

def degree(x, y, board):
    """Count onward moves from (x,y)."""
    cnt = 0
    for dx, dy in knight_moves:
        if is_valid(x+dx, y+dy, board):
            cnt += 1
    return cnt

def find_tour():
    """Attempt one Warnsdorff pass. Return board or None if stuck."""
    # Initialize board to zeros
    board = [[0]*N for _ in range(N)]
    # Random starting cell
    cx = random.randrange(N)
    cy = random.randrange(N)
    board[cx][cy] = 1

    # Place moves 2..N*N
    for step in range(2, N*N + 1):
        candidates = []
        # Gather valid next moves with their onward degree
        for i, (dx, dy) in enumerate(knight_moves):
            nx, ny = cx + dx, cy + dy
            if is_valid(nx, ny, board):
                candidates.append((degree(nx, ny, board), i))

        # If no candidates, stuck → fail
        if not candidates:
            return None

        # Randomize the candidates to break ties
        random.shuffle(candidates)
        # Pick the one with minimal degree
        _, move_i = min(candidates, key=lambda x: x[0])
        dx, dy = knight_moves[move_i]
        cx, cy = cx + dx, cy + dy
        board[cx][cy] = step

    return board

# Main retry loop
while True:
    # Shuffle global move order to vary tie-breaking
    random.shuffle(knight_moves)
    res = find_tour()
    if res is not None:
        board = res
        break

# Output the tour
print("There is solution:")
for row in board:
    # Leading space on each line, as in sample
    print(' ' + ' '.join(str(x) for x in row))
```

Explanation of the Python key steps:  
- We read N and handle trivial cases.  
- We define knight moves and helper functions `is_valid` and `degree`.  
- In `find_tour()`, we pick a random start, then for each step pick the unvisited neighbor with minimum onward degree (Warnsdorff’s rule), breaking ties by shuffling.  
- If we get stuck before filling the board, we return None and retry.  
- Once we complete all N² visits, we print the result.