How to Solve Sudoku with R – Open Source Automation

In this post we discuss how to write an R script to solve any Sudoku puzzle. There are some R packages to handle this, but in our case, we’ll write our own solution. For our purposes, we’ll assume the input Sudoku is a 9×9 grid. At the end result, each row, column, and 3×3 box needs to contain exactly one of each integer 1 through 9.

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Step 0) Define a sample board

Let’s define a sample Sudoku board for testing. Empty cells will be represented as zeroes.

board <- matrix(
     c(0,0,0,0,0,6,0,0,0,
     0,9,5,7,0,0,3,0,0,
     4,0,0,0,9,2,0,0,5,
     7,6,4,0,0,0,0,0,3,
     0,0,0,0,0,0,0,0,0,
     2,0,0,0,0,0,9,7,1,
     5,0,0,2,1,0,0,0,9,
     0,0,7,0,0,5,4,8,0,
     0,0,0,8,0,0,0,0,0),

     byrow = T,
     ncol = 9
)

Step 1) Find the empty cells

In the first step, let’s write a function that will find all of the empty cells on the board.

     find_empty_cells <- function(board) {

      which(board == 0, arr.ind = TRUE)

}

Step 2) Make sure cell placement is valid

Next, we need a function that will check if a cell placement is valid. In other words, if we try putting a number into a particular cell, we need to ensure that the number appears only once in that row, column, and box. Otherwise, the placement would not be valid.

is_valid <- function(board, num, row, col) {

    # Check if any cell in the same row has value = num
    if(any(board[row, ] == num)) {

        return(FALSE)

    }

    # Check if any cell in the same column has value = num
    if(any(board[, col] == num)) {

        return(FALSE)

    }

    # Get cells in num’s box
    box_x <- floor((row - 1) / 3) + 1
    box_y <- floor((col - 1) / 3) + 1

    # Get subset of matrix containing num’s box
    box <- board[(3 * box_x - 2):(3 * box_x), (3 * box_y - 2):(3 * box_y)]

    # Check if the number appears elsewhere in its box
    if(any(box == num)) {

        return(FALSE)

    }

        return(TRUE)

    }

Step 3) Recursively solve the Sudoku

In the third step, we write our function to solve the Sudoku. This function will return TRUE if the input Sudoku is solvable. Otherwise, it will return FALSE. The final result will be stored in a separate variable.

solve_sudoku <- function(board, needed_cells = NULL, index = 1) {

    # Find all empty cells
    if(is.null(needed_cells))
        needed_cells <- find_empty_cells(board)

    if(index > nrow(needed_cells)) {

        # Set result equal to current value of board
        # and return TRUE
        result <<- board
        return(TRUE)

    } else {

        row <- needed_cells[index, 1]
        col <- needed_cells[index, 2]
    }

    # Solve the Sudoku
    for(num in 1:9) {
        # Test for valid answers
        if(!is_valid(board, num, row, col)) {next} else{
            board2 = board
            board2[row, col] <- num

        # Retest with input
        if(solve_sudoku(board2, needed_cells, index + 1)) {
        return(TRUE)

            }

        }

    }

    # If not solvable, return FALSE
    return(FALSE)

}

Calling the Sudoku solver

Lastly, we call our Sudoku solver. The result is stored in the variable “result”, as can be seen below.

solve_sudoku(board)

Conclusion

That’s it for this post! If you enjoyed reading this and want to learn more about R or Python, check out the great data science program at 365 Data Science.

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