Python-opencv+KNN求解数独

最近一直在玩数独，突发奇想实现图像识别求解数独，输入到输出平均需要0.5s。

整体思路大概就是识别出图中数字生成list，然后求解。

输入输出demo

数独采用的是微软自带的Microsoft sudoku软件随便截取的图像，如下图所示：

经过程序求解后，得到的结果如下图所示：

程序具体流程

程序整体流程如下图所示：

读入图像后，根据求解轮廓信息找到数字所在位置，以及不包含数字的空白位置，提取数字信息通过KNN识别，识别出数字；无数字信息的在list中置0；生成未求解数独list，之后求解数独，将信息在原图中显示出来。

# -*-coding:utf-8-*-

import os

import cv2 as cv

import numpy as np

import time

####################################################

#寻找数字生成list

def find_dig_(img, train_set):

if img is None:

print("无效的图片！")

os._exit(0)

return

_, thre = cv.threshold(img, 230, 250, cv.THRESH_BINARY_INV)

_, contours, hierarchy = cv.findContours(thre, cv.RETR_TREE, cv.CHAIN_APPROX_SIMPLE)

sudoku_list = []

boxes = []

for i in range(len(hierarchy[0])):

if hierarchy[0][i][3] == 0: # 表示父轮廓为 0

boxes.append(hierarchy[0][i])

# 提取数字

nm = []

for j in range(len(boxes)): # 此处len(boxes)=81

if boxes[j][2] != -1:

x, y, w, h = cv.boundingRect(contours[boxes[j][2]])

nm.append([x, y, w, h])

# 在原图中框选各个数字

cropped = img[y:y + h, x:x + w]

im = img_pre(cropped)#预处理

AF = incise(im)#切割数字图像

result = identification(train_set, AF, 7)#knn识别

sudoku_list.insert(0, int(result))#生成list

else:

sudoku_list.insert(0, 0)

if len(sudoku_list) == 81:

sudoku_list= np.array(sudoku_list)

sudoku_list= sudoku_list.reshape((9, 9))

print("old_sudoku -> \n", sudoku_list)

return sudoku_list, contours, hierarchy

else:

print("无效的图片！")

os._exit(0)

######################################################

#KNN算法识别数字

def img_pre(cropped):

# 预处理数字图像

im = np.array(cropped) # 转化为二维数组

for i in range(im.shape[0]): # 转化为二值矩阵

for j in range(im.shape[1]):

# print(im[i, j])

if im[i, j] != 255:

im[i, j] = 1

else:

im[i, j] = 0

return im

# 提取图片特征

def feature(A):

midx = int(A.shape[1] / 2) + 1

midy = int(A.shape[0] / 2) + 1

A1 = A[0:midy, 0:midx].mean()

A2 = A[midy:A.shape[0], 0:midx].mean()

A3 = A[0:midy, midx:A.shape[1]].mean()

A4 = A[midy:A.shape[0], midx:A.shape[1]].mean()

A5 = A.mean()

AF = [A1, A2, A3, A4, A5]

return AF

# 切割图片并返回每个子图片特征

def incise(im):

# 竖直切割并返回切割的坐标

a = [];

b = []

if any(im[:, 0] == 1):

a.append(0)

for i in range(im.shape[1] - 1):

if all(im[:, i] == 0) and any(im[:, i + 1] == 1):

a.append(i + 1)

elif any(im[:, i] == 1) and all(im[:, i + 1] == 0):

b.append(i + 1)

if any(im[:, im.shape[1] - 1] == 1):

b.append(im.shape[1])

# 水平切割并返回分割图片特征

names = locals();

AF = []

for i in range(len(a)):

names['na%s' % i] = im[:, range(a[i], b[i])]

if any(names['na%s' % i][0, :] == 1):

c = 0

else:

for j in range(names['na%s' % i].shape[0]):

if j < names['na%s' % i].shape[0] - 1:

if all(names['na%s' % i][j, :] == 0) and any(names['na%s' % i][j + 1, :] == 1):

c = j

break

else:

c = j

if any(names['na%s' % i][names['na%s' % i].shape[0] - 1, :] == 1):

d = names['na%s' % i].shape[0] - 1

else:

for j in range(names['na%s' % i].shape[0]):

if j < names['na%s' % i].shape[0] - 1:

if any(names['na%s' % i][j, :] == 1) and all(names['na%s' % i][j + 1, :] == 0):

d = j + 1

break

else:

d = j

names['na%s' % i] = names['na%s' % i][range(c, d), :]

AF.append(feature(names['na%s' % i])) # 提取特征

for j in names['na%s' % i]:

pass

return AF

# 训练已知图片的特征

def training():

train_set = {}

for i in range(9):

value = []

for j in range(15):

ima = cv.imread('E:/test_image/knn_test/{}/{}.png'.format(i + 1, j + 1), 0)

im = img_pre(ima)

AF = incise(im)

value.append(AF[0])

train_set[i + 1] = value

return train_set

# 计算两向量的距离

def distance(v1, v2):

vector1 = np.array(v1)

vector2 = np.array(v2)

Vector = (vector1 - vector2) ** 2

distance = Vector.sum() ** 0.5

return distance

# 用最近邻算法识别单个数字

def knn(train_set, V, k):

key_sort = [11] * k

value_sort = [11] * k

for key in range(1, 10):

for value in train_set[key]:

d = distance(V, value)

for i in range(k):

if d < value_sort[i]:

for j in range(k - 2, i - 1, -1):

key_sort[j + 1] = key_sort[j]

value_sort[j + 1] = value_sort[j]

key_sort[i] = key

value_sort[i] = d

break

max_key_count = -1

key_set = set(key_sort)

for key in key_set:

if max_key_count < key_sort.count(key):

max_key_count = key_sort.count(key)

max_key = key

return max_key

# 生成数字

def identification(train_set, AF, k):

result = ''

for i in AF:

key = knn(train_set, i, k)

result = result + str(key)

return result

######################################################

######################################################

#求解数独

def get_next(m, x, y):

# 获得下一个空白格在数独中的坐标。

:param m 数独矩阵

:param x 空白格行数

:param y 空白格列数

"""

for next_y in range(y + 1, 9): # 下一个空白格和当前格在一行的情况

if m[x][next_y] == 0:

return x, next_y

for next_x in range(x + 1, 9): # 下一个空白格和当前格不在一行的情况

for next_y in range(0, 9):

if m[next_x][next_y] == 0:

return next_x, next_y

return -1, -1 # 若不存在下一个空白格，则返回 -1，-1

def value(m, x, y):

# 返回符合"每个横排和竖排以及九宫格内无相同数字"这个条件的有效值。

i, j = x // 3, y // 3

grid = [m[i * 3 + r][j * 3 + c] for r in range(3) for c in range(3)]

v = set([x for x in range(1, 10)]) - set(grid) - set(m[x]) - \

set(list(zip(*m))[y])

return list(v)

def start_pos(m):

# 返回第一个空白格的位置坐标

for x in range(9):

for y in range(9):

if m[x][y] == 0:

return x, y

return False, False # 若数独已完成，则返回 False, False

def try_sudoku(m, x, y):

# 试着填写数独

for v in value(m, x, y):

m[x][y] = v

next_x, next_y = get_next(m, x, y)

if next_y == -1: # 如果无下一个空白格

return True

else:

end = try_sudoku(m, next_x, next_y) # 递归

if end:

return True

m[x][y] = 0 # 在递归的过程中，如果数独没有解开，

# 则回溯到上一个空白格

def sudoku_so(m):

x, y = start_pos(m)

try_sudoku(m, x, y)

print("new_sudoku -> \n", m)

return m

###################################################

# 将结果绘制到原图

def draw_answer(img, contours, hierarchy, new_sudoku_list ):

new_sudoku_list = new_sudoku_list .flatten().tolist()

for i in range(len(contours)):

cnt = contours[i]

if hierarchy[0, i, -1] == 0:

num = new_soduku_list.pop(-1)

if hierarchy[0, i, 2] == -1:

x, y, w, h = cv.boundingRect(cnt)

cv.putText(img, "%d" % num, (x + 19, y + 56), cv.FONT_HERSHEY_SIMPLEX, 1.8, (0, 0, 255), 2) # 填写数字

cv.imwrite("E:/answer.png", img)

if __name__ == '__main__':

t1 = time.time()

train_set = training()

img = cv.imread('E:/test_image/python_test_img/Sudoku.png')

img_gray = cv.cvtColor(img, cv.COLOR_BGR2GRAY)

sudoku_list, contours, hierarchy = find_dig_(img_gray, train_set)

new_sudoku_list = sudoku_so(sudoku_list)

draw_answer(img, contours, hierarchy, new_sudoku_list )

print("time :",time.time()-t1)

PS：

使用KNN算法需要创建训练集，数独中共涉及9个数字，“1,2,3,4,5,6,7,8,9”各15幅图放入文件夹中，如下图所示。

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