This is a simple solver for a very specific and easy-to-solve CAPTCHAs like the one proposed here. Use the force to find something more complex.
In this example we are going to use the following images.
First of all, a fixed size (monospace) font has been used. This makes extracting the letters and using them as masks to check each digit, one by one, very easy. Also, the alphabet is simple lowercase hexadecimal letters. Thus, we had to extract only 16 letters.
The first part was to to extract all the letters. To achieve that, first of all we sampled several images so as to be sure that the images we have contains all the 16 letters. Then, using a simple image editor we cropped all the letters, one by one. You have to be very careful so all the letters be aligned properly. Here is the final mask.
Now we can focus on the CAPTCHA. As you can notice there is some noise which we have to remove (lines and stuff). After playing with several techniques we finally ended to the following. We turned the image to greyscale, then used a threshold to remove some of the noise. Here is the example after the filtering (cropping also applied).
So, now we have the image almost cleared and some letters to play with.
Move each letter across the image and take the difference of the pixels for each position and sum them. Thus for each position we have a score of how much the letter (mask) fits the letter behind it. Then, store for each letter the position where the maximum score found. Then sort by score, take the top five results (our captcha is five letters) and finally sort by position. The result is the CAPTCHA text.
More formally
Let
![d(I,l,o)=\sum_{0\leq i \leq W \ 0 \leq j \leq H}{[I(o+i, j)-l(i,j)]}](http://s0.wp.com/latex.php?latex=d%28I%2Cl%2Co%29%3D%5Csum_%7B0%5Cleq+i+%5Cleq+W+%5C%5C+0+%5Cleq+j+%5Cleq+H%7D%7B%5BI%28o%2Bi%2C+j%29-l%28i%2Cj%29%5D%7D&bg=T&fg=333333&s=0 "d(I,l,o)=\sum_{0\leq i \leq W \ 0 \leq j \leq H}{[I(o+i, j)-l(i,j)]}")
be the distance of the image 
Then
Thus, we need 5 letters 
def p(img, letter):
A = img.load()
B = letter.load()
mx = 1000000
max_x = 0
x = 0
for x in range(img.size[0] - letter.size[0]):
_sum = 0
for i in range(letter.size[0]):
for j in range(letter.size[1]):
_sum = _sum + abs(A[x+i, j][0] - B[i, j][0])
if _sum < mx :
mx = _sum
max_x = x
return mx, max_xHere is the code which implements this method. You can browse and download everything from https://github.com/ptigas/simple-CAPTCHA-solver
from PIL import Image
def ocr(im, threshold=200, alphabet="0123456789abcdef"):
img = Image.open(im)
img = img.convert("RGB")
box = (8, 8, 58, 18)
img = img.crop(box)
pixdata = img.load()
letters = Image.open('letters.bmp')
ledata = letters.load()
# Clean the background noise, if color != white, then set to black.
for y in range(img.size[1]):
for x in range(img.size[0]):
if (pixdata[x, y][0] > threshold) \
and (pixdata[x, y][1] > threshold) \
and (pixdata[x, y][2] > threshold):
pixdata[x, y] = (255, 255, 255, 255)
else:
pixdata[x, y] = (0, 0, 0, 255)
counter = 0;
old_x = -1;
letterlist = []
for x in range(letters.size[0]):
black = True
for y in range(letters.size[1]):
if ledata[x, y][0] <> 0 :
black = False
break
if black :
if True :
box = (old_x + 1, 0, x, 10)
letter = letters.crop(box)
t = p(img, letter);
print counter, x, t
letterlist.append((t[0], alphabet[counter], t[1]))
old_x = x
counter += 1
box = (old_x + 1, 0, 140, 10)
letter = letters.crop(box)
t = p(img, letter)
letterlist.append((t[0], alphabet[counter], t[1]))
t = sorted(letterlist)
t = t[0:5] # 5-letter captcha
final = sorted(t, key=lambda e: e[2])
answer = ""
for l in final:
answer = answer + l[1]
return answer
print(ocr(sys.argv[1]))