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sudoku

Binder

Sudoku board viewer and solver in IPython/Jupyter, using NumPy arrays

by Kunal Marwaha

View the notebook on Github or nbviewer. Forks and comments are welcome!

Controls

Reset

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.)

View

bd.view() will return a copy of the current Sudoku board (as a NumPy array). The value "0" is unknown.

Adding/Removing Values

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)

Checking Board

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?

Solving

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)

Have fun!

Let me know if there are any bugs or comments. Enjoy :-)

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sudoku board viewer and solver in IPython/Jupyter

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