solution.py

from utils import *
import copy
import itertools
rows = 'ABCDEFGHI'
cols = '123456789'
history = []

def cross(A, B):
    "Cross product of elements in A and elements in B."
    return [s+t for s in A for t in B]

boxes = [r + c for r in rows for c in cols]
row_units = [cross(r, cols) for r in rows]
column_units = [cross(rows, c) for c in cols]
square_units = [cross(rs, cs) for rs in ('ABC','DEF','GHI') for cs in ('123','456','789')]
unitlist = row_units + column_units + square_units

diagonal_units=[[s+t for s,t in zip(rows,cols)],[s+t for s,t in zip(rows,cols[::-1])]]
# TODO: Update the unit list to add the new diagonal units
unitlist = unitlist + diagonal_units

# Must be called after all units (including diagonals) are added to the unitlist
units = dict((s, [u for u in unitlist if s in u]) for s in boxes)
peers = dict((s, set(sum(units[s],[]))-set([s]))for s in boxes)


def assign_value(values, box, value):
    """You must use this function to update your values dictionary if you want to
    try using the provided visualization tool. This function records each assignment
    (in order) for later reconstruction.

    Parameters
    ----------
    values(dict)
        a dictionary of the form {'box_name': '123456789', ...}

    Returns
    -------
    dict
        The values dictionary with the naked twins eliminated from peers
    """
    # Don't waste memory appending actions that don't actually change any values
    if values[box] == value:
        return values

    values[box] = value
    if len(value) == 1:
        history.append(values.copy())
    return values

    
def naked_twins(values):
    """Eliminate values using the naked twins strategy.

    The naked twins strategy says that if you have two or more unallocated boxes
    in a unit and there are only two digits that can go in those two boxes, then
    those two digits can be eliminated from the possible assignments of all other
    boxes in the same unit.

    Parameters
    ----------
    values(dict)
        a dictionary of the form {'box_name': '123456789', ...}

    Returns
    -------
    dict
        The values dictionary with the naked twins eliminated from peers

    Notes
    -----
    Your solution can either process all pairs of naked twins from the input once,
    or it can continue processing pairs of naked twins until there are no such
    pairs remaining -- the project assistant test suite will accept either
    convention. However, it will not accept code that does not process all pairs
    of naked twins from the original input. (For example, if you start processing
    pairs of twins and eliminate another pair of twins before the second pair
    is processed then your code will fail the PA test suite.)

    The first convention is preferred for consistency with the other strategies,
    and because it is simpler (since the reduce_puzzle function already calls this
    strategy repeatedly).

    See Also
    --------
    Pseudocode for this algorithm on github:
    https://github.com/udacity/artificial-intelligence/blob/master/Projects/1_Sudoku/pseudocode.md
    """
    # TODO: Implement this function!
    # First we find those boxes that have only 2 options that could possibly be used to fill it
    for unit in unitlist:
        twooptionboxes = [box for box in unit if len(values[box])==2]
    # then we pair up the boxes to make multiple possible combinations
        possibletwins = [list(pair) for pair in itertools.combinations(twooptionboxes,2)] 
    # Now finding the naked twins
        for pair in possibletwins:
            if values[pair[0]] == values[pair[1]]:
                # Eliminating the naked twins from values
                    for box in unit:
                        if box!= pair[0] and box!= pair[1]:
                            for digit in values[pair[0]]:
                                values[box] = values[box].replace(digit,'')
    return values                                                        
    raise NotImplementedError

def grid_values(grid):
    """
    Convert grid into a dict of {square: char} with '123456789' for empties.
    Args:
        grid(string) - A grid in string form.
    Returns:
        A grid in dictionary form
            Keys: The boxes, e.g., 'A1'
            Values: The value in each box, e.g., '8'. If the box has no value, then the value will be '123456789'.
    """
    chars = []
    digits = '123456789'
    for c in grid:
        if c in digits:
            chars.append(c)
        if c == '.':
            chars.append(digits)
    assert len(chars) == 81
    return dict(zip(boxes, chars))


def display(values):
    """
    Display the values as a 2-D grid.
    Args:
        values(dict): The sudoku in dictionary form
    """
    width = 1+max(len(values[s]) for s in boxes)
    line = '+'.join(['-'*(width*3)]*3)
    for r in rows:
        print(''.join(values[r+c].center(width)+('|' if c in '36' else '')
                      for c in cols))
        if r in 'CF': print(line)
    return

def eliminate(values):
    """Apply the eliminate strategy to a Sudoku puzzle

    The eliminate strategy says that if a box has a value assigned, then none
    of the peers of that box can have the same value.

    Parameters
    ----------
    values(dict)
        a dictionary of the form {'box_name': '123456789', ...}

    Returns
    -------
    dict
        The values dictionary with the assigned values eliminated from peers
    """
    # TODO: Copy your code from the classroom to complete this function
    solved_values = [box for box in values.keys() if len(values[box]) == 1]
    for box in solved_values:
        digit = values[box]
        if len(digit) == 1:
            for peer in peers[box]:
                values = assign_value(values,peer,values[peer].replace(digit, ''))
    return values


def only_choice(values):
    """Apply the only choice strategy to a Sudoku puzzle

    The only choice strategy says that if only one box in a unit allows a certain
    digit, then that box must be assigned that digit.

    Parameters
    ----------
    values(dict)
        a dictionary of the form {'box_name': '123456789', ...}

    Returns
    -------
    dict
        The values dictionary with all single-valued boxes assigned

    Notes
    -----
    You should be able to complete this function by copying your code from the classroom
    """
    # TODO: Copy your code from the classroom to complete this function
    for unit in unitlist:
        for digit in '123456789':
            dplaces = [box for box in unit if digit in values[box]]
            if len(dplaces) == 1:
                values[dplaces[0]] = digit
    return values


def reduce_puzzle(values):
    """Reduce a Sudoku puzzle by repeatedly applying all constraint strategies

    Parameters
    ----------
    values(dict)
        a dictionary of the form {'box_name': '123456789', ...}

    Returns
    -------
    dict or False
        The values dictionary after continued application of the constraint strategies
        no longer produces any changes, or False if the puzzle is unsolvable 
    """
    # TODO: Copy your code from the classroom and modify it to complete this function
    solved_values = [box for box in values.keys() if len(values[box]) == 1]
    stalled = False
    while not stalled:
        solved_values_before = len([box for box in values.keys() if len(values[box]) == 1])
        # Using Eliminate strategy
        values = eliminate(values)
        # Using only choice strategy
        values = only_choice(values)
        values = naked_twins(values)
        solved_values_after = len([box for box in values.keys() if len(values[box]) == 1])
        stalled = solved_values_before == solved_values_after
        if len([box for box in values.keys() if len(values[box]) == 0]):
            return False
    return values


def search(values):
    values = reduce_puzzle(values)
    if values is False:
        return False ## Failed earlier
    if all(len(values[s]) == 1 for s in boxes): 
        return values ## Solved!
    # Choose one of the unfilled squares with the fewest possibilities
    n,s = min((len(values[s]), s) for s in boxes if len(values[s]) > 1)
    # Now use recurrence to solve each one of the resulting sudokus, and 
    for value in values[s]:
        new_sudoku = values.copy()
        new_sudoku[s] = value
        attempt = search(new_sudoku)
        if attempt:
            return attempt

def solve(grid):
    """Find the solution to a Sudoku puzzle using search and constraint propagation

    Parameters
    ----------
    grid(string)
        a string representing a sudoku grid.
        
        Ex. '2.............62....1....7...6..8...3...9...7...6..4...4....8....52.............3'

    Returns
    -------
    dict or False
        The dictionary representation of the final sudoku grid or False if no solution exists.
    """
    values = grid_values(grid)
    values = search(values)
    return values

# below code from http://www.analyzingdata.org/portfolio/300-sudoku/
if __name__ == '__main__':
    diag_sudoku_grid = '2.............62....1....7...6..8...3...9...7...6..4...4....8....52.............3'
    display(solve(diag_sudoku_grid))

    try:
        from visualize import visualize_assignments
        visualize_assignments(assignments)

    except SystemExit:
        pass
    except:
        print('We could not visualize your board due to a pygame issue. Not a problem! It is not a requirement.')