How to get the cells of a sudoku grid with OpenCV?

Question

I've been trying for the last few days to get a sudoku grid from a picture, and I have been struggling on getting the smaller squares of the grid. I am working on the picture below. I thought processing the image with a canny filter would work fine, but it didn't and I couldn't get every contour of each square. I then put adaptive threshold, otsu, and a classic thresholding to the test, but every time, it just could not seem to capture every small square.

The final goal is to get the cells containing a number, and recognize the numbers with pytorch, so I would really like to have some clean images of the numbers, so the recognition doesn't screw up :)

Would anyone have an idea on how to achieve this? Thanks a lot in advance! :D

Octennial

Did you try searching using a popular search engine for opencv sudoku ? – Carvalho

I did, but I did not find an example using a very distorted grid. Hence, the pieces of code I searched online did not work for this picture. – Octennial

Can’t you take better pictures? – Carvalho

Or even just fix the contrast better so it’s a binary image With black numbers then you don’t need to bother with the grid, just use tesseract to pick out the numbers - did you try that? If you did, please also summarise in your question the other things you’ve tried and rejected, so people reading your question aren’t wasting their time suggesting things – Carvalho

There have been many post on this forum about finding grid cells, especially for checkerboards. Try searching and reviewing that code. – Manson

You can check out this answer too. https://mcmap.net/q/536574/-extract-boxes-from-sudoku-in-opencv-duplicate – Countdown

Answer 1

Here's a potential solution:

1. Obtain binary image. Convert image to grayscale and adaptive threshold

2. Filter out all numbers and noise to isolate only boxes. We filter using contour area to remove the numbers since we only want each individual cell

3. Fix grid lines. Perform morphological closing with a horizontal and vertical kernel to repair grid lines.

4. Sort each cell in top-to-bottom and left-to-right order. We organize each cell into a sequential order using imutils.contours.sort_contours() with the top-to-bottom and left-to-right parameter

---

Here's the initial binary image (left) and filtered out numbers + repaired grid lines + inverted image (right)

Here's a visualization of the iteration of each cell

The detected numbers in each cell

Code

import cv2
from imutils import contours
import numpy as np

# Load image, grayscale, and adaptive threshold
image = cv2.imread('1.png')
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
thresh = cv2.adaptiveThreshold(gray,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C, cv2.THRESH_BINARY_INV,57,5)

# Filter out all numbers and noise to isolate only boxes
cnts = cv2.findContours(thresh, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
for c in cnts:
    area = cv2.contourArea(c)
    if area < 1000:
        cv2.drawContours(thresh, [c], -1, (0,0,0), -1)

# Fix horizontal and vertical lines
vertical_kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (1,5))
thresh = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, vertical_kernel, iterations=9)
horizontal_kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (5,1))
thresh = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, horizontal_kernel, iterations=4)

# Sort by top to bottom and each row by left to right
invert = 255 - thresh
cnts = cv2.findContours(invert, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
(cnts, _) = contours.sort_contours(cnts, method="top-to-bottom")

sudoku_rows = []
row = []
for (i, c) in enumerate(cnts, 1):
    area = cv2.contourArea(c)
    if area < 50000:
        row.append(c)
        if i % 9 == 0:  
            (cnts, _) = contours.sort_contours(row, method="left-to-right")
            sudoku_rows.append(cnts)
            row = []

# Iterate through each box
for row in sudoku_rows:
    for c in row:
        mask = np.zeros(image.shape, dtype=np.uint8)
        cv2.drawContours(mask, [c], -1, (255,255,255), -1)
        result = cv2.bitwise_and(image, mask)
        result[mask==0] = 255
        cv2.imshow('result', result)
        cv2.waitKey(175)

cv2.imshow('thresh', thresh)
cv2.imshow('invert', invert)
cv2.waitKey()

Note: The sorting idea was adapted from an old previous answer in Rubrik cube solver color extraction.

Sudra

Answer 2

Steps:

1. Image PreProcessing ( closing operation )
2. Finding Sudoku Square and Creating Mask Image
3. Finding Vertical lines
4. Finding Horizontal Lines
5. Finding Grid Points
6. Correcting the defects
7. Extracting the digits from each cell

Code:

# ==========import the necessary packages============
import imutils
import numpy as np
import cv2
from transform import four_point_transform
from PIL import Image
import pytesseract
import math
from skimage.filters import threshold_local

# =============== For Transformation ==============
def order_points(pts):
    """initialzie a list of coordinates that will be ordered
    such that the first entry in the list is the top-left,
    the second entry is the top-right, the third is the
    bottom-right, and the fourth is the bottom-left"""

    rect = np.zeros((4, 2), dtype = "float32")
 
    # the top-left point will have the smallest sum, whereas
    # the bottom-right point will have the largest sum
    s = pts.sum(axis = 1)
    rect[0] = pts[np.argmin(s)]
    rect[2] = pts[np.argmax(s)]
 
    # now, compute the difference between the points, the
    # top-right point will have the smallest difference,
    # whereas the bottom-left will have the largest difference
    diff = np.diff(pts, axis = 1)
    rect[1] = pts[np.argmin(diff)]
    rect[3] = pts[np.argmax(diff)]
 
    # return the ordered coordinates
    return rect


def four_point_transform(image, pts):
    # obtain a consistent order of the points and unpack them
    # individually
    rect = order_points(pts)
    (tl, tr, br, bl) = rect
 
    # compute the width of the new image, which will be the
    # maximum distance between bottom-right and bottom-left
    # x-coordiates or the top-right and top-left x-coordinates
    widthA = np.sqrt(((br[0] - bl[0]) ** 2) + ((br[1] - bl[1]) ** 2))
    widthB = np.sqrt(((tr[0] - tl[0]) ** 2) + ((tr[1] - tl[1]) ** 2))
    maxWidth = max(int(widthA), int(widthB))
 
    # compute the height of the new image, which will be the
    # maximum distance between the top-right and bottom-right
    # y-coordinates or the top-left and bottom-left y-coordinates
    heightA = np.sqrt(((tr[0] - br[0]) ** 2) + ((tr[1] - br[1]) ** 2))
    heightB = np.sqrt(((tl[0] - bl[0]) ** 2) + ((tl[1] - bl[1]) ** 2))
    maxHeight = max(int(heightA), int(heightB))
 
    # now that we have the dimensions of the new image, construct
    # the set of destination points to obtain a "birds eye view",
    # (i.e. top-down view) of the image, again specifying points
    # in the top-left, top-right, bottom-right, and bottom-left
    # order
    dst = np.array([
        [0, 0],
        [maxWidth - 1, 0],
        [maxWidth - 1, maxHeight - 1],
        [0, maxHeight - 1]], dtype = "float32")
 
    # compute the perspective transform matrix and then apply it
    M = cv2.getPerspectiveTransform(rect, dst)
    warped = cv2.warpPerspective(image, M, (maxWidth, maxHeight))
 
    # return the warped image
    return warped

############## To show image ##############
def show_image(img,title):
    cv2.imshow(title, img) 
    cv2.waitKey(0) 
    cv2.destroyAllWindows()  


def find_largest_feature(inp_img, scan_tl=None, scan_br=None):
    """
    Uses the fact the `floodFill` function returns a bounding box of the area it filled to find the biggest
    connected pixel structure in the image. Fills this structure in white, reducing the rest to black.
    """
    img = inp_img.copy()  # Copy the image, leaving the original untouched
    height, width = img.shape[:2]

    max_area = 0
    seed_point = (None, None)

    if scan_tl is None:
        scan_tl = [0, 0]

    if scan_br is None:
        scan_br = [width, height]

    # Loop through the image
    for x in range(scan_tl[0], scan_br[0]):
        for y in range(scan_tl[1], scan_br[1]):
            # Only operate on light or white squares
            if img.item(y, x) == 255 and x < width and y < height:  # Note that .item() appears to take input as y, x
                area = cv2.floodFill(img, None, (x, y), 64)
                if area[0] > max_area:  # Gets the maximum bound area which should be the grid
                    max_area = area[0]
                    seed_point = (x, y)

    # Colour everything grey (compensates for features outside of our middle scanning range
    for x in range(width):
        for y in range(height):
            if img.item(y, x) == 255 and x < width and y < height:
                cv2.floodFill(img, None, (x, y), 64)

    mask = np.zeros((height + 2, width + 2), np.uint8)  # Mask that is 2 pixels bigger than the image

    # Highlight the main feature
    if all([p is not None for p in seed_point]):
        cv2.floodFill(img, mask, seed_point, 255)

    

    for x in range(width):
        for y in range(height):
            if img.item(y, x) == 64:  # Hide anything that isn't the main feature
                cv2.floodFill(img, mask, (x, y), 0)
    
    return img


################# Preprocessing of sudoku image ###############
def preprocess(image,case):
    ratio = image.shape[0] / 500.0
    orig = image.copy()
    image = imutils.resize(image, height = 500)

    if case == True:
    
        gray = cv2.GaussianBlur(image,(5,5),0)
        gray = cv2.cvtColor(gray,cv2.COLOR_BGR2GRAY)
        mask = np.zeros((gray.shape),np.uint8)
        kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))

        close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
        div = np.float32(gray)/(close)
        res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
        res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)
        edged = cv2.Canny(res, 75, 200)
    
        cnts = cv2.findContours(edged.copy(), cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
        cnts = cnts[0] if imutils.is_cv2() else cnts[1]
        cnts = sorted(cnts, key = cv2.contourArea, reverse = True)[:5]
 
        # loop over the contours
        for c in cnts:
            # approximate the contour
            rect = cv2.boundingRect(c)
            area = cv2.contourArea(c)

            cv2.rectangle(edged.copy(), (rect[0],rect[1]), (rect[2]+rect[0],rect[3]+rect[1]), (0,0,0), 2)
            peri = cv2.arcLength(c, True)
            approx = cv2.approxPolyDP(c, 0.02 * peri, True)
 
            # if our approximated contour has four points, then we
            # can assume that we have found our screen
            if len(approx) == 4:
                screenCnt = approx
                #print(screenCnt)
                break
         
        # show the contour (outline) of the piece of paper
        #print(screenCnt)
        cv2.drawContours(image, [screenCnt], -1, (0, 255, 0), 2)
    
        # apply the four point transform to obtain a top-down
        # view of the original image    
        warped = four_point_transform(orig, screenCnt.reshape(4, 2) * ratio)
        warped1 = cv2.resize(warped,(610,610))
        warp = cv2.cvtColor(warped, cv2.COLOR_BGR2GRAY) 
        T = threshold_local(warp, 11, offset = 10, method = "gaussian")
        warp = (warp > T).astype("uint8") * 255
        th3 = cv2.adaptiveThreshold(warp,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C,\
            cv2.THRESH_BINARY_INV,11,2) 
        kernel = np.ones((5,5),np.uint8)
        dilation =cv2.GaussianBlur(th3,(5,5),0)

    else :
        
        warped = image
        warped1 = cv2.resize(warped,(610,610))
        warp = cv2.cvtColor(warped, cv2.COLOR_BGR2GRAY) 
        T = threshold_local(warp, 11, offset = 10, method = "gaussian")
        warp = (warp > T).astype("uint8") * 255
        th3 = cv2.adaptiveThreshold(warp,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C,\
            cv2.THRESH_BINARY_INV,11,2)

    #show_image(warped1,"preprocessed")

    return th3,warped1,warped

def grids(img,warped2):
    #print("im:",img.shape)
    img2 = img.copy()
    img = np.zeros((500,500,3), np.uint8)

    ratio2 = 3
    kernel_size = 3
    lowThreshold = 30

    frame = img

    img = cv2.resize(frame,(610,610))

    for i in range(10):
        cv2.line(img, (0,(img.shape[0]//9)*i),(img.shape[1],(img.shape[0]//9)*i), (255, 255, 255), 3, 1)
        cv2.line(warped2, (0,(img.shape[0]//9)*i),(img.shape[1],(img.shape[0]//9)*i), (125, 0, 55), 3, 1)
    
    for j in range(10):
        cv2.line(img, ((img.shape[1]//9)*j, 0), ((img.shape[1]//9)*j, img.shape[0]), (255, 255, 255), 3, 1)
        cv2.line(warped2, ((img.shape[1]//9)*j, 0), ((img.shape[1]//9)*j, img.shape[0]), (125, 0, 55), 3, 1)
  
    #show_image(warped2,"grids")
    return img

############### Finding out the intersection pts to get the grids #########
def grid_points(img,warped2):
    img1 = img.copy()
    kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))

    dx = cv2.Sobel(img,cv2.CV_16S,1,0)
    dx = cv2.convertScaleAbs(dx)
    c=cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
    c = cv2.morphologyEx(c,cv2.MORPH_DILATE,kernelx,iterations = 1)
    cy = cv2.cvtColor(c,cv2.COLOR_BGR2GRAY)
    closex = cv2.morphologyEx(cy,cv2.MORPH_DILATE,kernelx,iterations = 1)

    kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
    dy = cv2.Sobel(img,cv2.CV_16S,0,2)
    dy = cv2.convertScaleAbs(dy)
    c = cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
    c = cv2.morphologyEx(c,cv2.MORPH_DILATE,kernely,iterations = 1)
    cy = cv2.cvtColor(c,cv2.COLOR_BGR2GRAY)
    closey = cv2.morphologyEx(cy,cv2.MORPH_DILATE,kernelx,iterations = 1)

    res = cv2.bitwise_and(closex,closey)
    #gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
    ret, thresh = cv2.threshold(res,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)

    kernel = np.ones((6,6),np.uint8)


    # Perform morphology
    se = np.ones((8,8), dtype='uint8')
    image_close = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, se)
    image_close = cv2.morphologyEx(image_close, cv2.MORPH_OPEN, kernel)

    contour, hier = cv2.findContours        (image_close,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
    cnts = sorted(contour, key=cv2.contourArea, reverse=True)[:100]
    centroids = []
    for cnt in cnts:
    
        mom = cv2.moments(cnt)
        (x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00'])
        cv2.circle(img1,(x,y),4,(0,255,0),-1)
        cv2.circle(warped2,(x,y),4,(0,255,0),-1)
        centroids.append((x,y))

    #show_image(warped2,"grid_points")


    Points = np.array(centroids,dtype = np.float32)
    c = Points.reshape((100,2))
    c2 = c[np.argsort(c[:,1])]

    b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in range(10)])
    bm = b.reshape((10,10,2))

    return c2,bm,cnts

############ Recognize digit images to number #############
def image_to_num(c2):     
    img = 255-c2
    text = pytesseract.image_to_string(img, lang="eng",config='--psm 6 --oem 3') #builder=builder)
    return list(text)[0]

###### To get the digit at the particular cell #############
def get_digit(c2,bm,warped1,cnts):
    num = []
    centroidx = np.empty((9, 9))
    centroidy = np.empty((9, 9))
    global list_images
    list_images = []
    for i in range(0,9):
        for j in range(0,9):

            x1,y1 = bm[i][j] # bm[0] row1 
            x2,y2 = bm[i+1][j+1]
            
            coordx = ((x1+x2)//2)
            coordy = ((y1+y2)//2)
            centroidx[i][j] = coordx
            centroidy[i][j] = coordy
            crop = warped1[int(x1):int(x2),int(y1):int(y2)]
            crop = imutils.resize(crop, height=69,width=67)
            c2 = cv2.cvtColor(crop, cv2.COLOR_BGR2GRAY)
            c2 = cv2.adaptiveThreshold(c2,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C,\
                cv2.THRESH_BINARY_INV,11,2)
            kernel = np.ones((2,2),np.uint8)
            #c2 = cv2.morphologyEx(c2, cv2.MORPH_OPEN, kernel)
            c2= cv2.copyMakeBorder(c2,5,5,5,5,cv2.BORDER_CONSTANT,value=(0,0,0))
            no = 0
            shape=c2.shape
            w=shape[1]
            h=shape[0]
            mom = cv2.moments(c2)
            (x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00']) 
            c2 = c2[14:70,15:62]
            contour, hier = cv2.findContours (c2,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
            if cnts is not None:
                cnts = sorted(contour, key=cv2.contourArea,reverse=True)[:1]

            for cnt in cnts:
                x,y,w,h = cv2.boundingRect(cnt)
                aspect_ratio = w/h
#               print(aspect_ratio)
                area = cv2.contourArea(cnt)
                #print(area)
                if area>120 and cnt.shape[0]>15 and aspect_ratio>0.2 and aspect_ratio<=0.9 : 
                    #print("area:",area)
                    c2 = find_largest_feature(c2)
                    #show_image(c2,"box2")
                    contour, hier = cv2.findContours (c2,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
                    cnts = sorted(contour, key=cv2.contourArea,reverse=True)[:1]
                    for cnt in cnts:
                        rect = cv2.boundingRect(cnt)
                        #cv2.rectangle(c2, (rect[0],rect[1]), (rect[2]+rect[0],rect[3]+rect[1]), (255,255,255), 2)
                        c2 = c2[rect[1]:rect[3]+rect[1],rect[0]:rect[2]+rect[0]]
                        c2= cv2.copyMakeBorder(c2,5,5,5,5,cv2.BORDER_CONSTANT,value=(0,0,0))
                        list_images.append(c2)
                    #show_image(c2,"box")
                    no = image_to_num(c2)
            num.append(no)
    centroidx = np.transpose(centroidx)
    centroidy = np.transpose(centroidy)
    return c2, num, centroidx, centroidy

######## creating matrix and filling numbers exist in the orig image #######
def sudoku_matrix(num):
    c = 0
    grid = np.empty((9, 9))
    for i in range(9):
        for j in range(9):
            grid[i][j] = int(num[c])
            
            c += 1
    grid = np.transpose(grid)
    return grid

######## Creating board to show the puzzle result in terminal##############
def board(arr):
    for i in range(9):
    
        if i%3==0 :
                print("+",end="")
                print("-------+"*3)
                
        for j in range(9):
            if j%3 ==0 :
                print("",end="| ")
            print(int(arr[i][j]),end=" ")
      
        print("",end="|")       
        print()
      
    print("+",end="")
    print("-------+"*3)
    return arr      
   
def check_col(arr,num,col):
    if  all([num != arr[i][col] for i in range(9)]):
        return True
    return False
    

def check_row(arr,num,row):
    if  all([num != arr[row][i] for i in range(9)]):
        return True
    return False


def check_cell(arr,num,row,col):
    sectopx = 3 * (row//3)
    sectopy = 3 * (col//3)
          
    for i in range(sectopx, sectopx+3):
        for j in range(sectopy, sectopy+3):
            if arr[i][j] == num:
                return True
    return False


def empty_loc(arr,l):
    for i in range(9):
        for j in range(9):
            if arr[i][j] == 0:
                l[0]=i
                l[1]=j
                return True              
    return False

#### Solving sudoku by back tracking############
def sudoku(arr):
    l=[0,0]

    if not empty_loc(arr,l):
        return True
    
    row = l[0]
    col = l[1]
                
    for num in range(1,10):
        if check_row(arr,num,row) and check_col(arr,num,col) and not check_cell(arr,num,row,col):
            arr[row][col] = int(num) 
            
            if(sudoku(arr)):
                return True
 
            # failure, unmake & try again
            arr[row][col] = 0
                    
    return False

def overlay(arr,num,img,cx,cy):
    no = -1
    for i in range(9):
        for j in range(9):
            no += 1 
            #cv2.putText(img,str(no), (int(cx[i][j]),int(cy[i][j])),cv2.FONT_HERSHEY_SIMPLEX, 0.5, (0, 0, 0), 2)
            if num[no] == 0:
                
                cv2.putText(img,str(int(arr[j][i])), (int(cx[i][j]-4),int(cy[i][j])+8),cv2.FONT_HERSHEY_SIMPLEX, 1, (0, 255, 0), 4)
                
    cv2.imshow("Sudoku",img)
    cv2.waitKey(0)

---

case = "False" # If transformation is required set True 
image = cv2.imread("QupKb.png")

th3,warped1,warped = preprocess(image,case)
warped2 = warped1.copy()
img = grids(warped,warped2)
c2,bm,cnts = grid_points(img,warped2)
c2,num,cx,cy = get_digit(c2,bm,warped1,cnts)
grid = sudoku_matrix(num)
if(sudoku(grid)):
    arr = board(grid)
    overlay(arr,num,warped1,cx,cy)

else:
    print("There is no solution")

---

warped:

th3:

warped2:

sudoku result:

---

All the extracted digits:

########## To view all the extracted digits ###############
_, axs = plt.subplots(1, len(list_images), figsize=(24, 24))
axs = axs.flatten()
for img, ax in zip(list_images, axs):
    ax.imshow(cv2.resize(img,(64,64)))
plt.show()

References:

• grid points

• get features (of digits)

• Images samples

Goldberg

Answer 3

If image contains just the tightly fitted sudoku grid, one crude way to achieve it would be to divide image into equal 9X9 grid and then try to extract number in each of that grid.

Lakin

That's actually the first thing I tried. The problem is, most of the time, I can't make the grid fit perfectly as a square. Hence a cell would look like half a number with a line on top. That's typically what happens with the 4 or the 6 at the top of the grid. But if you have a technique to undistort the image to make it a perfect square, I would gladly take it! – Octennial