from utils import *

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')]

def grid_values(grid):
    # In this function, you will take a sudoku as a string
    # and return a dictionary where the keys are the boxes,
    # for example 'A1', and the values are the digit at each
    # box (as a string) or '.' if the box has no value
    # assigned yet.
    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 eliminate(values):
    """
    Go through all the boxes, and whenever there is a box with a value, eliminate this value from the values of all its peers.
    Input: A sudoku in dictionary form.
    Output: The resulting sudoku in dictionary form.
    """
    solved_values = [box for box in values.keys() if len(values[box]) == 1]
    for box in solved_values:
        digit = values[box]
        for peer in peers[box]:
            values[peer] = values[peer].replace(digit,'')
    return values

def only_choice(values):
    """
    Go through all the units, and whenever there is a unit with a value that only fits in one box, assign the value to this box.
    Input: A sudoku in dictionary form.
    Output: The resulting sudoku in dictionary form.
    """
    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):
    """
    Iterate eliminate() and only_choice(). If at some point, there is a box with no available values, return False.
    If the sudoku is solved, return the sudoku.
    If after an iteration of both functions, the sudoku remains the same, return the sudoku.
    Input: A sudoku in dictionary form.
    Output: The resulting sudoku in dictionary form.
    """
    solved_values = [box for box in values.keys() if len(values[box]) == 1]
    stalled = False
    while not stalled:
        # Check how many boxes have a determined value
        solved_values_before = len([box for box in values.keys() if len(values[box]) == 1])
        # Use the Eliminate Strategy
        values = eliminate(values)
        # Use the Only Choice Strategy
        values = only_choice(values)
        # Check how many boxes have a determined value, to compare
        solved_values_after = len([box for box in values.keys() if len(values[box]) == 1])
        # If no new values were added, stop the loop.
        stalled = solved_values_before == solved_values_after
        # Sanity check, return False if there is a box with zero available values:
        if len([box for box in values.keys() if len(values[box]) == 0]):
            return False
    return values

def solve(values):
    "Using depth-first search and propagation, try all possible values."
    # First, reduce the puzzle using the previous function
    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!
    # Chose one of the unfilled square s 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 = solve(new_sudoku)
        if attempt:
            return attempt
# grid1 = '..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..';
# values = grid_values('..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..');
# display(reduce_puzzle(values));
grid2 = '4.....8.5.3..........7......2.....6.....8.4......1.......6.3.7.5..2.....1.4......'
values = grid_values(grid2)
display(solve(values));
