1. Abridged Problem Statement
Given an integer N (1 ≤ N ≤ 250), find an open knight's tour on an N×N chessboard (a sequence visiting every cell exactly once). If no tour exists, print "No solution." Otherwise print "There is solution:" followed by an N×N grid of visitation order from 1 to N².

2. Detailed Editorial
The knight's tour is a classic problem where a knight must visit every square of an N×N board exactly once. It is known that no tour exists for N=2,3,4 (except the trivial 1×1 case). For all other N≥1, an open tour exists. We implement a heuristic called Warnsdorff's rule:

- Warnsdorff's heuristic: Always move the knight to the square from which it would have the fewest onward moves (minimum "degree"). This is a greedy rule that often succeeds in producing a full tour in one pass for moderate N.
- Tie-breaking: When multiple next moves share the same minimum degree, pick one at random. We also randomize the initial starting square and shuffle the move order to avoid pathological sequences.
- Repetition: Though Warnsdorff's rule commonly succeeds, it can get stuck early. We wrap the procedure in a loop: if we get stuck, we reset the board, reshuffle the global move list, pick a new random start, and retry. Empirically, this converges extremely quickly even for N up to 250.

Complexities: Each attempt takes O(N²·8) to build the tour (constant 8 knight moves per square). With random restarts, the expected attempts are constant for N≥5.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int N;
vector<pair<int, int>> knight_moves = {{1, 2},  {1, -2},  {2, 1},  {2, -1},
                                       {-1, 2}, {-1, -2}, {-2, 1}, {-2, -1}};
mt19937 rng(42);

void read() { cin >> N; }

bool is_valid_move(int x, int y, const vector<vector<int>>& board) {
    return x >= 0 && x < N && y >= 0 && y < N && board[x][y] == 0;
}

int get_degree(int x, int y, const vector<vector<int>>& board) {
    int degree = 0;
    for(const auto& [dx, dy]: knight_moves) {
        int next_x = x + dx, next_y = y + dy;
        if(is_valid_move(next_x, next_y, board)) {
            degree++;
        }
    }
    return degree;
}

bool find_tour(vector<vector<int>>& board) {
    int current_x = uniform_int_distribution<int>(0, N - 1)(rng);
    int current_y = uniform_int_distribution<int>(0, N - 1)(rng);
    board[current_x][current_y] = 1;

    for(int position = 2; position <= N * N; position++) {
        vector<pair<int, int>> next_moves;
        for(int i = 0; i < knight_moves.size(); i++) {
            int next_x = current_x + knight_moves[i].first;
            int next_y = current_y + knight_moves[i].second;
            if(is_valid_move(next_x, next_y, board)) {
                int degree = get_degree(next_x, next_y, board);
                next_moves.emplace_back(degree, i);
            }
        }

        if(next_moves.empty()) {
            return false;
        }

        auto [_, move_index] =
            *min_element(next_moves.begin(), next_moves.end());

        current_x += knight_moves[move_index].first;
        current_y += knight_moves[move_index].second;
        board[current_x][current_y] = position;
    }
    return true;
}

void solve() {
    // Uses Warnsdorff's heuristic to greedily select the next move with the
    // fewest onward moves Randomly breaks ties among moves with the same
    // minimum degree. This is not guaranteed to be polynomial time but highly
    // effective for N ≤ 250.

    if(N == 2 || N == 3 || N == 4) {
        cout << "No solution." << endl;
        return;
    }

    vector<vector<int>> board(N, vector<int>(N, 0));
    while(true) {
        board.assign(N, vector<int>(N, 0));
        shuffle(knight_moves.begin(), knight_moves.end(), rng);
        if(find_tour(board)) {
            break;
        }
    }

    cout << "There is solution:" << endl;
    for(const auto& row: board) {
        for(int i = 0; i < N; i++) {
            cout << " " << row[i];
        }
        cout << endl;
    }
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution
```python
import sys
import random

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

# Quick no-solution cases
if N in (2, 3, 4):
    print("No solution.")
    sys.exit(0)

# All eight knight moves
knight_moves = [
    (1,2), (1,-2), (2,1), (2,-1),
    (-1,2), (-1,-2), (-2,1), (-2,-1)
]
random.seed(42)

def is_valid(x, y, board):
    "Check if (x,y) is in bounds 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():
    "Try one Warnsdorff pass; return board or None if stuck."
    # Initialize empty board
    board = [[0]*N for _ in range(N)]
    # Pick random start
    cx = random.randrange(N)
    cy = random.randrange(N)
    board[cx][cy] = 1

    for step in range(2, N*N + 1):
        candidates = []
        # Collect all 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 next move, fail
        if not candidates:
            return None
        # Select the move with minimum degree
        # random.shuffle ensures random tie-breaking
        random.shuffle(candidates)
        # min by 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

# Keep retrying until success
while True:
    random.shuffle(knight_moves)
    result = find_tour()
    if result is not None:
        board = result
        break

# Output
print("There is solution:")
for row in board:
    # Print each cell with a leading space for formatting
    print(' ' + ' '.join(str(x) for x in row))
```

5. Compressed Editorial
Use Warnsdorff's heuristic: always move to the unvisited square with the fewest onward moves, randomizing ties, the start, and move order. If it gets stuck, restart. This method quickly finds an open knight's tour for N≥5 (no tours for N=2,3,4).
