![]() ![]() it happens that no solution can be generated. Check that no number 1.9 is present twice in a 3x3 blockīut notice. Check that no number 1.9 is present twice in a column Here is a way to generate a random suduko. So it looks like some kind of Undo stack is required. The easy way to do that would be recursion. However it still may be that a trial makes the puzzle insoluble and needs to go back a cell (or more) to come forward again. Here is a blank Sudoku grid you can print out. When solving only pick from the set of 'remaining possible values' rather than rand. If that leaves any cell with the empty set, the puzzle is insoluble. If you set a cell to (say) 5 remove 5 from all cells in that column, row and sub-square. Start with a 9x9 grid all cells initialised to the full range. When a value is set (at start) or tried (during solve) you check the constraints and remove the value being tried from its column, row and square. The 'common' approach is to hold a 9x9 matrix of sets holding a subset of 1-9 which are the untried values. You need some method of backtracking if you get 'stuck' or detecting that it will get stuck. You can have easily filled in numbers that while valid individually leave the puzzle insoluble. It's quite possible to get to a dead end here. USA Today also offers other games, like sudoku and mahjong. There's a loop while(1) where you pick a random number and determine if it is valid in the current position. There are two main strategies for getting crossword puzzle help: enter in the clue and have. Int squareCheck(int num, int board, int row, int col)įor (int i = row i < row + sqrt(MATRIX_SIZE) i++)įor (int j = col j < col + sqrt(MATRIX_SIZE) j++) Int columnCheck(int num, int board, int col)įor (int row = 0 row < MATRIX_SIZE row++) Int rowCheck(int num, int board, int row)įor (int col = 0 col < MATRIX_SIZE col++) SqrCh = squareCheck(num, board, row, col-square) Int squareCheck(int num, int board, int row, int col) ĬolCh = columnCheck(num, board, row, col) Int columnCheck(int num, int board, int row) Int rowCheck(int num, int board, int row) ![]() I thought maybe somehow to use a histogram of numbers from 1-9 that maps which numbers already used to somehow change the use of fetching random numbers, but I'm not really sure how to use it and if it's even right to do so. I didn't see a solution of my code yet, even after leaving it to run more than 5 minutes. My problem is that, that way it takes the code a very long time to solve, and I don't know if I'm doing it right. So far I thought to use two dimension array for the board and go over every row in a nested "for" loop.Įvery time I fetch a number with a random function and check a row, a column and a square (3X3), and if all of them pass then I fill the number. This book contains 500 Blank 9x9 Sudoku Grids, 2 Blank Sudoku Grid per page, Large Print White Paper, 8,5 x 11 Paper Size. No recursion, no search and sort algorithms to improve the time complexity. The tools that we have are only functions, arrays and pointers. I got an assignment to write a program that fills an empty sudoku board and prints it out. ![]()
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