![]() Here is the pseudo-code: function Solve(table)Ĭ = cell with smallest number of candidates It boils down to a function which solves a single cell and then calls itself recursively, trying another candidate each time when recursive call returns failure condition. When properly applied, this heuristic can cause such tremendous drop in number of search steps that once unfeasible search scheme becomes fast enough to be used in practice.Īlgorithm which solves the Sudoku table is now quite simple. This search strategy is called "most constrained first" heuristic, deriving its name from the occasion that cell with most constraints upon itself is chosen first in each step of the algorithm. By looking for such heavily constrained cells we hope to provoke collisions early in the search, while tree is still shallow and number of nodes reasonably small. digits that do not generate collision immediately. At every step we should choose the cell in the table which offers smallest number of candidates, i.e. There is one simple solution which is applicable to many search problems. So what should we do to reduce number of tables generated by the algorithm? Going down fifty or so steps into depth sounds quite unfeasible. How many tables should we consider before the solution is found? With every step deeper down the search tree, number of tables considered becomes larger and larger - actually it grows exponentially with depth of the search tree. Sudoku table contains 81 cells, most of which are initially blank (57 in the example shown in the picture above). On the bitter end, we might finish the search without encountering the fully populated table, in which case we simply conclude that initial position was not valid.īut now there appears to be a problem. In perspective, we hope to search all possibilities until final solution to the Sudoku is found. ![]() Should this process come to a dead-end, in which there is such a cell in which no digit can be placed lest to cause a violation, we unroll the process back one step and try to replace the last digit with the next viable candidate, if such digit exists otherwise, we unroll one step more to try next candidate one level further up the line. If successful, we simply repeat the operation recursively, on a table which has one empty cell less than before. In each step we pick up one empty cell in the table and try to put any digit that wouldn't cause a constraint violation. ![]() One possible solution is to search through partial solutions in depth-first manner. The problem here is how to solve a given Sudoku so that final state of the table satisfies the validation criteria. Picture shows starting position and corresponding Sudoku table when solved (source: In order to be solved correctly, Sudoku table must satisfy seemingly simple criteria: on every row, every column and every 3x3 block, each digit must appear exactly once. Sub-tables consisting of 3x3 cells are specifically grouped, as shown on the picture below. I am just very lost right now as I literally a Fine Arts major and I have no solid knowledge of CS, so some help will be greatly appreciated.Sudoku table consists of 9x9 cells, each cell receiving a single digit between 1 and 9, inclusive. I also need help with solving the puzzle after printing it, using depth first search. I just need a little boost, as I am very lost right now. But I am not sure how to implement the code. Return true //returns true when all cells are filled with correct valuesĪs you can see, I have to create a for loop that goes through each element in the sol array, then place it in the corresponding grid. Return false //this is returned if no number works so we backtrack ![]() Set grid cell back to 0 //try next number If(solve() = true) //make a recursive call to solve If //number satisfies Sudoku requirements check if number satisfies Sudoku requirements Here is index.html: Īnd here is the sudoku.js file: //global variable
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