Sudoku board viewer and solver in IPython/Jupyter, using NumPy arrays
by Kunal Marwaha
bd = Board() will instantiate a new Sudoku board. Can add optional parameter
n (such as
bd = Board(n)) to create a n2 by n2 board. (By default, n=3, like in normal Sudoku.)
bd.view() will return a copy of the current Sudoku board (as a NumPy array). The value "0" is unknown.
bd.set_val(val,pos) will put a value at position (either tuple or int). Position is indexed from 1. If value is not nonzero, then sets to "0" (unknown). A full list of commands is here:
## Assuming bd.size = n**2 = 9 bd.set_board(vals) #expects len(vals) == 81 bd.set_row(vals,row) #expects len(vals) == 9; row is int from 1 to 9 bd.set_column(vals,col) #expects len(vals) == 9; col is int from 1 to 9 bd.set_val(val,pos) #expects len(pos) == 1 (int from 1 to 81) or 2 (tuple with row and column indices)
bd.check_solved() will check if the Sudoku board is complete.
bd.check_legal() will check if the entries on the board are legal. A full list of checks are here:
bd.check_solved() #board solved? bd.check_filled() #board has all nonzero entries? bd.check_legal() #board has all legal entries? bd.check_rows() #for each row, is each nonzero entry distinct? bd.check_columns() #for each col, is each nonzero entry distinct? bd.check_boxes() #for each box, is each nonzero entry distinct?
bd.solve() will attempt to solve the board with its given entries. On completion, the procedure will print its status (success or failure). If failure, the procedure has no effect. (Although the solver is mostly brute-force, it is reasonably fast for many boards)
Let me know if there are any bugs or comments. Enjoy :-)