This post demonstrates solving Sudoku using Backtracking. Algorithm and a working code sample is given. The code is meant to be self-explanatory, try to understand the code, it 's actually simple.
References:
Backtracking: http://en.wikipedia.org/wiki/Backtracking
Sudoku: http://en.wikipedia.org/wiki/Sudoku
Pre-condition:
# Read input grid (this is in matrix form)
# Grid has 9x9 cells. Cell is represented by (x,y)
# Each cell contains a value from 1 to 9.
# Empty cells are contain the value '0'
Algorithm:
1# solve(cell)
2# If cell not empty, call solve(nextCell)
3# If cell is empty (grid[cell.x][cell.y] == 0)
4# Add value v (v belongs to 1-9)
5# Check if v is valid for cell
6# if not valid, continue to #4
7# If valid, call solve(nextCell)
8# If sudoku solved, return true
9# Else return false
Explanation:
References:
Backtracking: http://en.wikipedia.org/wiki/Backtracking
Sudoku: http://en.wikipedia.org/wiki/Sudoku
Pre-condition:
# Read input grid (this is in matrix form)
# Grid has 9x9 cells. Cell is represented by (x,y)
# Each cell contains a value from 1 to 9.
# Empty cells are contain the value '0'
1# solve(cell)
2# If cell not empty, call solve(nextCell)
3# If cell is empty (grid[cell.x][cell.y] == 0)
4# Add value v (v belongs to 1-9)
5# Check if v is valid for cell
6# if not valid, continue to #4
7# If valid, call solve(nextCell)
8# If sudoku solved, return true
9# Else return false
Explanation:
- Create a function solve(currentCell) which takes a cell and returns true/false based on whether the Sudoku is solved or not.
- Remember if the cell already has a value, there is nothing to do.
- In each call our job is to find a suitable value from 1-9 for the cell, which will solve the sudoku.
- If we cannot find such value, we backtrack.
- First focus on simple conditions:
- If currentCell is null, we have reached the end. Return true. (We have solved the problem)
- If currentCell already has a value, no-op (Nothing to do here, call next)
- Now we know currentCell is empty, try each value from 1-9 (checkout the for loop in the code)
- For each value, check if its valid, then put the value else just keep putting next value.
- See if sudoku is solved; if solved, just return true, else return false
- If for loop ends, this means we did not find any valid value for currentCell, something is wrong in the previous value,
- Reset the value to 0, and just return false.
Working Code:
package com.example;
public class Sudoku {
// dimension of input
static int N = 9;
// sample input
static int grid[][] = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 }, //
{ 5, 2, 0, 0, 0, 0, 0, 0, 0 }, //
{ 0, 8, 7, 0, 0, 0, 0, 3, 1 }, //
{ 0, 0, 3, 0, 1, 0, 0, 8, 0 }, //
{ 9, 0, 0, 8, 6, 3, 0, 0, 5 }, //
{ 0, 5, 0, 0, 9, 0, 6, 0, 0 }, //
{ 1, 3, 0, 0, 0, 0, 2, 5, 0 }, //
{ 0, 0, 0, 0, 0, 0, 0, 7, 4 }, //
{ 0, 0, 5, 2, 0, 6, 3, 0, 0 } };
/**
* Class to abstract the representation of a cell. Cell => (x, y)
*/
static class Cell {
int row, col;
public Cell(int row, int col) {
super();
this.row = row;
this.col = col;
}
@Override
public String toString() {
return "Cell [row=" + row + ", col=" + col + "]";
}
};
/**
* Utility function to check whether @param value is valid for @param cell
*/
static boolean isValid(Cell cell, int value) {
if (grid[cell.row][cell.col] != 0) {
throw new RuntimeException(
"Cannot call for cell which already has a value");
}
// if v present row, return false
for (int c = 0; c < 9; c++) {
if (grid[cell.row][c] == value)
return false;
}
// if v present in col, return false
for (int r = 0; r < 9; r++) {
if (grid[r][cell.col] == value)
return false;
}
// if v present in grid, return false
// to get the grid we should calculate (x1,y1) (x2,y2)
int x1 = 3 * (cell.row / 3);
int y1 = 3 * (cell.col / 3);
int x2 = x1 + 2;
int y2 = y1 + 2;
for (int x = x1; x <= x2; x++)
for (int y = y1; y <= y2; y++)
if (grid[x][y] == value)
return false;
// if value not present in row, col and bounding box, return true
return true;
}
// simple function to get the next cell
// read for yourself, very simple and straight forward
static Cell getNextCell(Cell cur) {
int row = cur.row;
int col = cur.col;
// next cell => col++
col++;
// if col > 8, then col = 0, row++
// reached end of row, got to next row
if (col > 8) {
// goto next line
col = 0;
row++;
}
// reached end of matrix, return null
if (row > 8)
return null; // reached end
Cell next = new Cell(row, col);
return next;
}
// everything is put together here
// very simple solution
// must return true, if the soduku is solved, return false otherwise
static boolean solve(Cell cur) {
// if the cell is null, we have reached the end
if (cur == null)
return true;
// if grid[cur] already has a value, there is nothing to solve here,
// continue on to next cell
if (grid[cur.row][cur.col] != 0) {
// return whatever is being returned by solve(next)
// i.e the state of soduku's solution is not being determined by
// this cell, but by other cells
return solve(getNextCell(cur));
}
// this is where each possible value is being assigned to the cell, and
// checked if a solutions could be arrived at.
// if grid[cur] doesn't have a value
// try each possible value
for (int i = 1; i <= 9; i++) {
// check if valid, if valid, then update
boolean valid = isValid(cur, i);
if (!valid) // i not valid for this cell, try other values
continue;
// assign here
grid[cur.row][cur.col] = i;
// continue with next cell
boolean solved = solve(getNextCell(cur));
// if solved, return, else try other values
if (solved)
return true;
else
grid[cur.row][cur.col] = 0; // reset
// continue with other possible values
}
// if you reach here, then no value from 1 - 9 for this cell can solve
// return false
return false;
}
public static void main(String[] args) {
boolean solved = solve(new Cell(0, 0));
if (!solved) {
System.out.println("SUDOKU cannot be solved.");
return;
}
System.out.println("SOLUTION\n");
printGrid(grid);
}
// utility to print the grid
static void printGrid(int grid[][]) {
for (int row = 0; row < N; row++) {
for (int col = 0; col < N; col++)
System.out.print(grid[row][col]);
System.out.println();
}
}
}