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. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

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

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

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

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

int N;  // Board size N×N
// All 8 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}
};
mt19937 rng(42);  // Fixed seed for reproducibility

// Read input
void read() {
    cin >> N;
}

// Check if (x,y) is on board and not yet visited (board[x][y] == 0)
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;
}

// Compute Warnsdorff degree: # of valid onward moves from (x,y)
int get_degree(int x, int y, const vector<vector<int>>& board) {
    int degree = 0;
    for(const auto& [dx, dy] : knight_moves) {
        int nx = x + dx, ny = y + dy;
        if(is_valid_move(nx, ny, board)) {
            degree++;
        }
    }
    return degree;
}

// Attempt to build one knight’s tour using Warnsdorff’s heuristic
// Returns true if full tour (1..N*N) is found, false if stuck early.
bool find_tour(vector<vector<int>>& board) {
    // Pick a random starting square
    int cx = uniform_int_distribution<int>(0, N - 1)(rng);
    int cy = uniform_int_distribution<int>(0, N - 1)(rng);
    board[cx][cy] = 1;  // Mark first move

    // Place moves 2..N*N
    for(int pos = 2; pos <= N * N; pos++) {
        vector<pair<int,int>> candidates;
        // Explore all 8 knight moves
        for(int i = 0; i < knight_moves.size(); i++) {
            int nx = cx + knight_moves[i].first;
            int ny = cy + knight_moves[i].second;
            if(is_valid_move(nx, ny, board)) {
                // Compute onward degree for this candidate
                int d = get_degree(nx, ny, board);
                // Store (degree, moveIndex) for sorting
                candidates.emplace_back(d, i);
            }
        }
        // No valid moves ⇒ fail to complete tour
        if(candidates.empty()) {
            return false;
        }
        // Pick the move with minimum degree (Warnsdorff’s rule)
        auto [_, best_i] = *min_element(candidates.begin(), candidates.end());
        // Apply the chosen move
        cx += knight_moves[best_i].first;
        cy += knight_moves[best_i].second;
        board[cx][cy] = pos;  // Mark visitation order
    }
    return true;  // Completed all N*N moves
}

void solve() {
    // Handle small N with no solutions
    if(N == 2 || N == 3 || N == 4) {
        cout << "No solution." << "\n";
        return;
    }
    // Board will hold visitation order; 0=unvisited
    vector<vector<int>> board(N, vector<int>(N, 0));

    // Repeat until find a full tour
    while(true) {
        // Reset board
        for(int i = 0; i < N; i++)
            fill(board[i].begin(), board[i].end(), 0);
        // Shuffle global move ordering for random tie-breaking
        shuffle(knight_moves.begin(), knight_moves.end(), rng);
        if(find_tour(board)) break;
    }

    // Output the result
    cout << "There is solution:" << "\n";
    for(const auto& row : board) {
        for(int x : row) {
            cout << " " << x;
        }
        cout << "\n";
    }
}

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

    int T = 1;
    // If there were multiple test cases:
    // cin >> T;
    while(T--) {
        read();
        solve();
    }
    return 0;
}
```

4. Python Solution 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])

# 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).