Tag: time

How can I decrease the time complexity and increase efficiency, without writing a new algorithm. My solution solves the majority of sudoku puzzles in a fast time, but for some difficult ones it can take over a minute! I have added some of the difficult puzzles below the code.

import numpy as np
from skimage.util import view_as_blocks  # pip install scikit-image

#input: 9x9 numpy array, empty cell = 0 
#output: 9x9 numpy array: if not solution all array entries should be -1
#backtracking depth-first search with constraint propagation




#searches for 0 value in board
def zero_search(sudoku):
    for i in range(len(sudoku)):
        for j in range(len(sudoku(0))):
            if sudoku(i)(j) == 0:
                return (i, j)  # row, col

    return False



#parameters: board, num = inserted value, pos = board position(vector/tuple)
def valid(sudoku, num, pos):
    # Check row
    for i in range(len(sudoku(0))):
        if sudoku(pos(0))(i) == num and pos(1) != i:
            return False

    # Check column
    for i in range(len(sudoku)):
        if sudoku(i)(pos(1)) == num and pos(0) != i:
            return False

    # Check box
    box_x = pos(1) // 3
    box_y = pos(0) // 3

    #y-axis/columns 
    for i in range(box_y*3, box_y*3 + 3):
        #x-axis/rows
        for j in range(box_x * 3, box_x*3 + 3):
            if sudoku(i)(j) == num and (i,j) != pos:
                return False

    return True


def solver(sudoku):
    
    find = zero_search(sudoku)
    if not find:
        return True
    
    else:
        row, col = find

    #check insertion values 1-9
    for i in range(1,10):
        if valid(sudoku, i, (row, col)):
            sudoku(row)(col) = i
            
            if solver(sudoku):
                return True

            sudoku(row)(col) = 0
            
    return False 
    
    
    
def initial_invalid(sudoku):

    # Check row
    for i in range(9):
        dup_lst=()
        for j in range(9):
            if sudoku(j)(i)!=0:
                if sudoku(j)(i) in dup_lst:
                    return True

                else:
                    dup_lst.append(sudoku(j)(i))
     
    #check column
    for i in range(9):
        dup_lst=()
        for j in range(9):
            if sudoku(i)(j)!=0:
                if sudoku(i)(j) in dup_lst:
                    return True

                else:
                    dup_lst.append(sudoku(i)(j))
                    
                    
    

#not needed
def final_valid (sudoku):

    subgrids = view_as_blocks(sudoku, (3, 3))

    sums = (np.sum(subgrids(i)(j)) for j in range(3) for i in range(3))
    if sum(sums) !=405:
        return False
    
    else:
        return True
    
    #row_sum = sudoku.sum(axis=1)
    #if sum(row_sum) != 405:
    #   return False

    #col_sum = sudoku.sum(axis=0)
    #if sum(col_sum) != 405:
    #    return False
        

    
def sudoku_solver(sudoku):

    if initial_invalid(sudoku):
        return np.full((9,9),-1)
        
    if solver(sudoku):
            return sudoku
    else:
        return np.full((9,9),-1)

((0 8 0 4 3 0 0 0 0)
 (0 0 5 0 0 9 0 0 0)
 (6 0 0 0 8 0 0 7 0)
 (0 0 0 0 9 0 0 0 3)
 (0 0 0 8 0 7 0 0 0)
 (9 0 0 0 0 0 0 5 4)
 (0 6 0 0 0 0 0 0 5)
 (0 0 8 0 0 0 4 0 0)
 (0 4 0 0 0 6 0 1 0))


((0 0 2 0 0 0 0 0 4)
 (0 5 0 0 1 3 7 0 0)
 (7 9 0 0 0 0 0 5 0)
 (0 0 9 0 0 0 0 6 0)
 (0 0 0 0 3 0 5 0 8)
 (5 0 7 0 0 0 4 0 0)
 (0 0 0 0 6 0 8 0 0)
 (0 6 0 0 2 7 0 4 0)
 (8 0 0 0 0 0 0 2 0))


((0 0 0 6 0 0 2 0 0)
 (0 0 0 0 0 9 0 6 0)
 (0 8 0 0 0 5 0 0 3)
 (1 0 0 4 0 0 9 0 0)
 (8 3 0 0 0 0 0 0 0)
 (0 2 0 0 0 6 0 0 0)
 (0 0 0 0 6 0 0 0 0)
 (2 5 0 3 0 7 0 9 0)
 (0 0 1 0 0 0 0 8 4))