#Pseudocode for solving sudoku how to#
#Pseudocode for solving sudoku movie#

We know that each digit can occur only once for each row, column and sub-grid. See that we’re dividing the original grid into groups of 3×3 sub-grids from left to right, top to bottom. This will return the following sub-grids for our original puzzle: > subgrids = get_subgrids(grid) We can group the values of each 3×3 sub-grid as follows: def get_subgrids(grid): Part of the analysis has to do with checking the values of each sub-grid. Which will return the grid as expected: > grid = create_grid(puzzle) We can populate the 9×9 Sudoku grid with the following function: def create_grid(puzzle_str): As an example, we will work with: puzzle = '''043080250 That means we will have 9 lines of 9 digits each. Empty cases will be denoted by 0 and each line of the string will represent a row of the grid. Using Numpy for the gridįor the 9×9 grid, we will use Numpy arrays (the experience I got using Numpy to implement a tic-tac-toe board came in handy here). The goal is to fill the grid’s empty cases, having each digit between 1-9 appear only once in each row, column and 3×3 sub-grid. The grid is also divided into nine 3×3 sub-grids. If you’re not familiar with it, Sudoku is a mathematical puzzle where you have a 9×9 grid partially filled with digits from 1 to 9. Setting up the Sudoku grid Sudoku: What is it? For a more efficient solution based on this one, see here. It will take too long to solve difficult Sudoku puzzles. Note: This solver is a simplified implementation.

#Pseudocode for solving sudoku how to#

In this post we will see how to implement a simple version of the solver. After much work, I was finally able to program a solver that recursively tried filling the empty cases and tracked back when necessary. The meat of the problem was implementing a Sudoku solver. I would absolutely not recommend BFS for Sudoku.During this week I was working on one of Project Euler’s problems. While running these algorithms on some test puzzles, I found that DFS does decently well onĮasy and medium difficulty puzzles, while struggles on the harder ones with very large numbers of steps.īFS works fine on the 4x4 puzzles, but is incredibly slow otherwise because it checks every intermediate possibilityĪll the way to the solution. I will still attempt to make functional solvers, however I think that I will favour the use of stack safe solvers This function is essentially the same as the earlier recursive function, just without the recursion. These are simple algorithms, they can really just start looking on the very first empty cell in the grid, starting from theĬonst solve2 = isBfs => grid => When deciding what value to put in each cell, we first have to decide which cell we want to try filling first. Also, because the Sudoku grid is anĪcyclic graph where every move we take will lead to a uniqueīoard state, we can omit keeping a record of discovered cells that are separate from We willĪctually use this to our advantage and make a single function able to use either algorithm viaĬurrying.

#Pseudocode for solving sudoku movie#

Whereas a queue being the opposite of it is a 'First In, First Out' order (like a queue to buy movie tickets). The gist of it is items stored in a stack are retrieved in a 'Last In, First Out' order (like a stack of plates), The nodes it will expand, whereas BFS uses a queue. When looking at the pseudocode on Wikipedia for both algorithms, they are extremely similar with only one key difference:ĭFS uses a stack data structure to store the order of Search path sufficiently to go back and expand other paths. Both algorithms make use ofīacktracking once they have explored a branch in their The two most basic methods of search are Depth First(DFS) andīreadth First Search(BFS). The simplest way to solve a Sudoku puzzle would be to simply search for the answer one cell at a time.