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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