Podano celo število n naloga je najti vse posebne rešitve do problem n-kraljic kjer n kraljice so postavljene na an n * n šahovnico tako, da nobeni kraljici ne moreta napadati ena druge.
Opomba: Vsaka rešitev je edinstvena konfiguracija n kraljice, predstavljene kot permutacija [123....n] . Številka na i th položaj označuje vrsto kraljice v i th stolpec. Na primer [3142] prikazuje eno takšno postavitev.
primer:
Vnos: n = 4
Izhod: [2 4 1 3] [3 1 4 2]greibach normalna oblika
Pojasnilo: To sta 2 možni rešitvi.Vnos: n = 2
Izhod: []
Pojasnilo: Ni rešitve, saj lahko kraljice napadajo druga drugo v vseh možnih konfiguracijah.kaj je datotečni sistem linux
Kazalo vsebine
- [Naivni pristop] Z generiranjem vseh permutacij z uporabo rekurzije
- [Pričakovan pristop] Uporaba povratnega sledenja z obrezovanjem
- [Nadomestni pristop] Sledenje nazaj z bitnim maskiranjem
[Naivni pristop] - Uporaba rekurzije - O(n! * n) časa in O(n) prostora
Preprosta ideja za rešitev Problem N-Queens je ustvariti vse možne permutacije [1 2 3 ... n] in nato preverite, ali predstavlja veljavno konfiguracijo N-Queens. Ker mora biti vsaka kraljica v drugi vrstici in stolpcu samodejno uporablja permutacije skrbi za ta pravila. Še vedno pa moramo preveriti, da ni dveh kraljic na enako diagonalo.
Spodaj je podano izvedba:
C++//C++ program to find all solution of N queen problem //using recursion #include
//Java program to find all solution of N queen problem //using recursion import java.util.ArrayList; class GfG { // Check if placement is safe static boolean isSafe(ArrayList<Integer> board int currRow int currCol) { for(int i = 0; i < board.size(); i++) { int placedRow = board.get(i); int placedCol = i + 1; // Check diagonals if(Math.abs(placedRow - currRow) == Math.abs(placedCol - currCol)) { return false; // Not safe } } return true; // Safe to place } // Recursive utility to solve static void nQueenUtil(int col int n ArrayList<Integer> board ArrayList<ArrayList<Integer>> res boolean[] visited) { // If all queens placed add to res if(col > n) { res.add(new ArrayList<>(board)); return; } // Try each row in column for(int row = 1; row <= n; row++) { // If row not used if(!visited[row]) { // Check safety if(isSafe(board row col)) { // Mark row visited[row] = true; // Place queen board.add(row); // Recur for next column nQueenUtil(col + 1 n board res visited); // Backtrack board.remove(board.size()-1); visited[row] = false; } } } } // Function to solve N-Queen static ArrayList<ArrayList<Integer>> nQueen(int n) { ArrayList<ArrayList<Integer>> res = new ArrayList<>(); ArrayList<Integer> board = new ArrayList<>(); boolean[] visited = new boolean[n +1]; nQueenUtil(1 n board res visited); return res; } public static void main(String[] args) { int n = 4; ArrayList<ArrayList<Integer>> res = nQueen(n); for(ArrayList<Integer> row : res) { System.out.print('['); for(int i = 0; i < row.size(); i++) { System.out.print(row.get(i)); if(i != row.size()-1) System.out.print(' '); } System.out.println(']'); } } }
#Python program to find all solution of N queen problem #using recursion # Function to check if placement is safe def isSafe(board currRow currCol): for i in range(len(board)): placedRow = board[i] placedCol = i + 1 # Check diagonals if abs(placedRow - currRow) == abs(placedCol - currCol): return False # Not safe return True # Safe to place # Recursive utility to solve N-Queens def nQueenUtil(col n board res visited): # If all queens placed add to res if col > n: res.append(board.copy()) return # Try each row in column for row in range(1 n+1): # If row not used if not visited[row]: # Check safety if isSafe(board row col): # Mark row visited[row] = True # Place queen board.append(row) # Recur for next column nQueenUtil(col+1 n board res visited) # Backtrack board.pop() visited[row] = False # Main N-Queen solver def nQueen(n): res = [] board = [] visited = [False] * (n + 1) nQueenUtil(1 n board res visited) return res if __name__ == '__main__': n = 4 res = nQueen(n) for row in res: print(row)
//C# program to find all solution of N queen problem //using recursion using System; using System.Collections.Generic; class GfG { // Check if placement is safe static bool isSafe(List<int> board int currRow int currCol){ for (int i = 0; i < board.Count; i++) { int placedRow = board[i]; int placedCol = i + 1; // Check diagonals if (Math.Abs(placedRow - currRow) == Math.Abs(placedCol - currCol)) { return false; // Not safe } } return true; // Safe to place } // Recursive utility to solve static void nQueenUtil(int col int n List<int> board List<List<int> > res bool[] visited){ // If all queens placed add to res if (col > n) { res.Add(new List<int>(board)); return; } // Try each row in column for (int row = 1; row <= n; row++) { // If row not used if (!visited[row]) { // Check safety if (isSafe(board row col)) { // Mark row visited[row] = true; // Place queen board.Add(row); // Recur for next column nQueenUtil(col + 1 n board res visited); // Backtrack board.RemoveAt(board.Count - 1); visited[row] = false; } } } } // Main N-Queen solver static List<List<int>> nQueen(int n){ List<List<int> > res = new List<List<int> >(); List<int> board = new List<int>(); bool[] visited = new bool[n + 1]; nQueenUtil(1 n board res visited); return res; } static void Main(string[] args) { int n = 4; List<List<int>> res = nQueen(n); foreach (var temp in res) { Console.WriteLine('[' + String.Join(' ' temp) + ']'); } } }
//JavaScript program to find all solution of N queen problem //using recursion // Function to check if placement is safe function isSafe(board currRow currCol){ for (let i = 0; i < board.length; i++) { let placedRow = board[i]; let placedCol = i + 1; // Check diagonals if (Math.abs(placedRow - currRow) === Math.abs(placedCol - currCol)) { return false; // Not safe } } return true; // Safe to place } // Recursive utility to solve N-Queens function nQueenUtil(col n board res visited){ // If all queens placed add to res if (col > n) { res.push([...board ]); return; } // Try each row in column for (let row = 1; row <= n; row++) { // If row not used if (!visited[row]) { // Check safety if (isSafe(board row col)) { // Mark row visited[row] = true; // Place queen board.push(row); // Recur for next column nQueenUtil(col + 1 n board res visited); // Backtrack board.pop(); visited[row] = false; } } } } // Main N-Queen solver function nQueen(n){ let res = []; let board = []; let visited = Array(n + 1).fill(false); nQueenUtil(1 n board res visited); return res; } // Driver code let n = 4; let res = nQueen(n); res.forEach(row => console.log(row));
Izhod
[2 4 1 3] [3 1 4 2]
Časovna zapletenost: O(n!*n) n! za ustvarjanje vseh permutacije in O(n) za validacijo vsake permutacije.
Pomožni prostor: O(n)
[Pričakovan pristop] - Uporaba sledenja nazaj z obrezovanjem - O(n!) časa in O(n) prostora
Za optimizacijo zgornjega pristopa lahko uporabimo vračanje z obrezovanjem . Namesto generiranja vseh možnih permutacij, gradimo rešitev postopoma, pri tem pa se lahko prepričamo na vsakem koraku, da delna rešitev ostane veljavna. Če pride do konflikta, se bomo takoj vrnili nazaj, kar pomaga izogibanje nepotrebno izračuni .
Izvedba po korakih :
javascript onclick
- Začnite s prvim stolpcem in poskusite postaviti kraljico v vsako vrstico.
- Hranite nize, da sledite katerim vrstice so že zasedeni. Podobno za sledenje glavni in manjše diagonale so že zasedeni.
- Če je kraljica umestitev konflikti z obstoječimi kraljicami preskoči to vrstica in nazaj kraljica, da poskusi naslednjo mogoče vrstica (Obrezovanje in vračanje med konfliktom).
// C++ program to find all solution of N queen problem by // using backtracking and pruning #include
// Java program to find all solutions of the N-Queens problem // using backtracking and pruning import java.util.ArrayList; import java.util.List; class GfG { // Utility function for solving the N-Queens // problem using backtracking. static void nQueenUtil(int j int n ArrayList<Integer> board boolean[] rows boolean[] diag1 boolean[] diag2 ArrayList<ArrayList<Integer>> res) { if (j > n) { // A solution is found res.add(new ArrayList<>(board)); return; } for (int i = 1; i <= n; ++i) { if (!rows[i] && !diag1[i + j] && !diag2[i - j + n]) { // Place queen rows[i] = diag1[i + j] = diag2[i - j + n] = true; board.add(i); // Recurse to the next column nQueenUtil(j + 1 n board rows diag1 diag2 res); // Remove queen (backtrack) board.remove(board.size() - 1); rows[i] = diag1[i + j] = diag2[i - j + n] = false; } } } // Solves the N-Queens problem and returns // all valid configurations. static ArrayList<ArrayList<Integer>> nQueen(int n) { ArrayList<ArrayList<Integer>> res = new ArrayList<>(); ArrayList<Integer> board = new ArrayList<>(); // Rows occupied boolean[] rows = new boolean[n + 1]; // Major diagonals (row + j) and Minor diagonals (row - col + n) boolean[] diag1 = new boolean[2 * n + 1]; boolean[] diag2 = new boolean[2 * n + 1]; // Start solving from the first column nQueenUtil(1 n board rows diag1 diag2 res); return res; } public static void main(String[] args) { int n = 4; ArrayList<ArrayList<Integer>> res = nQueen(n); for (ArrayList<Integer> solution : res) { System.out.print('['); for (int i = 0; i < solution.size(); i++) { System.out.print(solution.get(i)); if (i != solution.size() - 1) { System.out.print(' '); } } System.out.println(']'); } } }
# Python program to find all solutions of the N-Queens problem # using backtracking and pruning def nQueenUtil(j n board rows diag1 diag2 res): if j > n: # A solution is found res.append(board[:]) return for i in range(1 n + 1): if not rows[i] and not diag1[i + j] and not diag2[i - j + n]: # Place queen rows[i] = diag1[i + j] = diag2[i - j + n] = True board.append(i) # Recurse to the next column nQueenUtil(j + 1 n board rows diag1 diag2 res) # Remove queen (backtrack) board.pop() rows[i] = diag1[i + j] = diag2[i - j + n] = False def nQueen(n): res = [] board = [] # Rows occupied rows = [False] * (n + 1) # Major diagonals (row + j) and Minor diagonals (row - col + n) diag1 = [False] * (2 * n + 1) diag2 = [False] * (2 * n + 1) # Start solving from the first column nQueenUtil(1 n board rows diag1 diag2 res) return res if __name__ == '__main__': n = 4 res = nQueen(n) for temp in res: print(temp)
// C# program to find all solutions of the N-Queens problem // using backtracking and pruning using System; using System.Collections.Generic; class GfG { // Utility function for solving the N-Queens // problem using backtracking. static void nQueenUtil(int j int n List<int> board bool[] rows bool[] diag1 bool[] diag2 List<List<int>> res) { if (j > n) { // A solution is found res.Add(new List<int>(board)); return; } for (int i = 1; i <= n; ++i) { if (!rows[i] && !diag1[i + j] && !diag2[i - j + n]) { // Place queen rows[i] = diag1[i + j] = diag2[i - j + n] = true; board.Add(i); // Recurse to the next column nQueenUtil(j + 1 n board rows diag1 diag2 res); // Remove queen (backtrack) board.RemoveAt(board.Count - 1); rows[i] = diag1[i + j] = diag2[i - j + n] = false; } } } // Solves the N-Queens problem and returns // all valid configurations. static List<List<int>> nQueen(int n) { List<List<int>> res = new List<List<int>>(); List<int> board = new List<int>(); // Rows occupied bool[] rows = new bool[n + 1]; // Major diagonals (row + j) and Minor diagonals (row - col + n) bool[] diag1 = new bool[2 * n + 1]; bool[] diag2 = new bool[2 * n + 1]; // Start solving from the first column nQueenUtil(1 n board rows diag1 diag2 res); return res; } static void Main(string[] args) { int n = 4; List<List<int>> res = nQueen(n); foreach (var temp in res) { Console.WriteLine('[' + String.Join(' ' temp) + ']'); } } }
// JavaScript program to find all solutions of the N-Queens problem // using backtracking and pruning // Utility function for solving the N-Queens // problem using backtracking. function nQueenUtil(j n board rows diag1 diag2 res) { if (j > n) { // A solution is found res.push([...board]); return; } for (let i = 1; i <= n; ++i) { if (!rows[i] && !diag1[i + j] && !diag2[i - j + n]) { // Place queen rows[i] = diag1[i + j] = diag2[i - j + n] = true; board.push(i); // Recurse to the next column nQueenUtil(j + 1 n board rows diag1 diag2 res); // Remove queen (backtrack) board.pop(); rows[i] = diag1[i + j] = diag2[i - j + n] = false; } } } // Solves the N-Queens problem and returns // all valid configurations. function nQueen(n) { const res = []; const board = []; // Rows occupied const rows = Array(n + 1).fill(false); // Major diagonals (row + j) and Minor diagonals (row - col + n) const diag1 = Array(2 * n + 1).fill(false); const diag2 = Array(2 * n + 1).fill(false); // Start solving from the first column nQueenUtil(1 n board rows diag1 diag2 res); return res; } // Driver Code const n = 4; const res = nQueen(n); res.forEach(temp => console.log(temp));
Izhod
[2 4 1 3] [3 1 4 2]
Časovna zahtevnost: O(n!) Za generiranje vseh permutacije .
Pomožni prostor: O(n)
kislinske lastnosti
[Nadomestni pristop] - Sledenje nazaj z bitnim maskiranjem
Za nadaljnjo optimizacijo vračanje nazaj pristop zlasti za večje vrednosti n lahko uporabimo bitno maskiranje za učinkovito sledenje zaseden vrstice in diagonale. Bitno maskiranje nam omogoča uporabo celih števil ( vrstice ld rd ), da sledite, katere vrstice in diagonale so zasedene, z uporabo hitrega bitne operacije za hitrejše izračuni. Pristop ostaja enak kot zgoraj.
Spodaj je podano izvedba:
C++//C++ program to find all solution of N queen problem //using recursion #include
// Java program to find all solution of N queen problem // using recursion import java.util.*; class GfG { // Function to check if the current placement is safe static boolean isSafe(int row int col int rows int ld int rd int n) { return !(((rows >> row) & 1) == 1) && !(((ld >> (row + col)) & 1) == 1) && !(((rd >> (row - col + n)) & 1) == 1); } // Recursive function to generate all possible permutations static void nQueenUtil(int col int n ArrayList<Integer> board ArrayList<ArrayList<Integer>> res int rows int ld int rd) { // If all queens are placed add into res if (col > n) { res.add(new ArrayList<>(board)); return; } // Try placing a queen in each row // of the current column for (int row = 1; row <= n; ++row) { // Check if it's safe to place the queen if (isSafe(row col rows ld rd n)) { // Place the queen board.add(row); // Recur to place the next queen nQueenUtil(col + 1 n board res rows | (1 << row) (ld | (1 << (row + col))) (rd | (1 << (row - col + n)))); // Backtrack: remove the queen board.remove(board.size() - 1); } } } // Main function to find all distinct // res to the n-queens puzzle static ArrayList<ArrayList<Integer>> nQueen(int n) { ArrayList<ArrayList<Integer>> res = new ArrayList<>(); // Current board configuration ArrayList<Integer> board = new ArrayList<>(); // Start solving from the first column nQueenUtil(1 n board res 0 0 0); return res; } public static void main(String[] args) { int n = 4; ArrayList<ArrayList<Integer>> res = nQueen(n); for (ArrayList<Integer> solution : res) { System.out.print('['); for (int j = 0; j < n; ++j) { System.out.print(solution.get(j)); if (j != n - 1) System.out.print(' '); } System.out.println(']'); } } }
# Python program to find all solution of N queen problem # using recursion # Function to check if the current placement is safe def isSafe(row col rows ld rd n): return not ((rows >> row) & 1) and not ((ld >> (row + col)) & 1) and not ((rd >> (row - col + n)) & 1) # Recursive function to generate all possible permutations def nQueenUtil(col n board res rows ld rd): # If all queens are placed add into res if col > n: res.append(board[:]) return # Try placing a queen in each row # of the current column for row in range(1 n + 1): # Check if it's safe to place the queen if isSafe(row col rows ld rd n): # Place the queen board.append(row) # Recur to place the next queen nQueenUtil(col + 1 n board res rows | (1 << row) (ld | (1 << (row + col))) (rd | (1 << (row - col + n)))) # Backtrack: remove the queen board.pop() # Main function to find all distinct # res to the n-queens puzzle def nQueen(n): res = [] # Current board configuration board = [] # Start solving from the first column nQueenUtil(1 n board res 0 0 0) return res if __name__ == '__main__': n = 4 res = nQueen(n) for solution in res: print('[' end='') for j in range(n): print(solution[j] end='') if j != n - 1: print(' ' end='') print(']')
// C# program to find all solution of N queen problem // using recursion using System; using System.Collections.Generic; class GfG { // Function to check if the current placement is safe static bool isSafe(int row int col int rows int ld int rd int n) { return !(((rows >> row) & 1) == 1) && !(((ld >> (row + col)) & 1) == 1) && !(((rd >> (row - col + n)) & 1) == 1); } // Recursive function to generate all possible permutations static void nQueenUtil(int col int n List<int> board List<List<int>> res int rows int ld int rd) { // If all queens are placed add into res if (col > n) { res.Add(new List<int>(board)); return; } // Try placing a queen in each row // of the current column for (int row = 1; row <= n; ++row) { // Check if it's safe to place the queen if (isSafe(row col rows ld rd n)) { // Place the queen board.Add(row); // Recur to place the next queen nQueenUtil(col + 1 n board res rows | (1 << row) (ld | (1 << (row + col))) (rd | (1 << (row - col + n)))); // Backtrack: remove the queen board.RemoveAt(board.Count - 1); } } } // Main function to find all distinct // res to the n-queens puzzle static List<List<int>> nQueen(int n) { List<List<int>> res = new List<List<int>>(); // Current board configuration List<int> board = new List<int>(); // Start solving from the first column nQueenUtil(1 n board res 0 0 0); return res; } static void Main() { int n = 4; List<List<int>> res = nQueen(n); foreach (var solution in res) { Console.Write('['); for (int j = 0; j < n; ++j) { Console.Write(solution[j]); if (j != n - 1) Console.Write(' '); } Console.WriteLine(']'); } } }
// JavaScript program to find all solution of N queen problem // using recursion // Function to check if the current placement is safe function isSafe(row col rows ld rd n) { return !((rows >> row) & 1) && !((ld >> (row + col)) & 1) && !((rd >> (row - col + n)) & 1); } // Recursive function to generate all possible permutations function nQueenUtil(col n board res rows ld rd) { // If all queens are placed add into res if (col > n) { res.push([...board]); return; } // Try placing a queen in each row // of the current column for (let row = 1; row <= n; ++row) { // Check if it's safe to place the queen if (isSafe(row col rows ld rd n)) { // Place the queen board.push(row); // Recur to place the next queen nQueenUtil(col + 1 n board res rows | (1 << row) (ld | (1 << (row + col))) (rd | (1 << (row - col + n)))); // Backtrack: remove the queen board.pop(); } } } // Main function to find all distinct // res to the n-queens puzzle function nQueen(n) { let res = []; // Current board configuration let board = []; // Start solving from the first column nQueenUtil(1 n board res 0 0 0); return res; } // Driver Code let n = 4; let res = nQueen(n); for (let i = 0; i < res.length; i++) { process.stdout.write('['); for (let j = 0; j < n; ++j) { process.stdout.write(res[i][j].toString()); if (j !== n - 1) process.stdout.write(' '); } console.log(']'); }
Izhod
[2 4 1 3] [3 1 4 2]
Časovna zapletenost: O(n!) za generiranje vseh permutacij.
Kompleksnost prostora: O(n)