import os, sys

#The program begins here
#provisorial name count_noise1_0.py
# It calculates the noise as number of pixels that are different between two images recorded from the same subject
#  Provided under GPL licence
# Author Alessio Papini, Department of Plant Biology University of Florence Italy, Via La Pira, 4 Firenze, mail alpapiniATunifi.it


print('It calculates the number of different pixels between 2 images taken consecutively from the same subject')
print('supported formats bmp, jpg, tif')
print('The images must be of the same dimensions (widhxheight) and in frame one with the other')
# It explains what the program does


i1=raw_input('Write the name of the first of the images (with its path to directory if different from the current working directory): ')
i2=raw_input('Write the name of the second of the images (with its path to directory if different from the current working directory): ')
# It gets the names of the two images 


import Image
im1 = Image.open(i1)

im2 = Image.open(i2)

# it loads the two images 

x, y = im1.size
# width and height for the two nested cycle needed to go through the image: the pixel position will be in a(x,y)

#ora inizio i due cicli annidati lo esterno con la width e lo interno con la heighth della immagine

countnoise_im1_im2=0
#initializes counter

for i in range (x):
 for j in range (y):
  a=im1.getpixel((i,j))
  b=im2.getpixel((i,j))
  if a!=b:
   countnoise_im1_im2=countnoise_im1_im2 + 1 
#if the value of the pixel is different in position x,y in the two images the counter increases by 1

print('number of pixels that are different between the first and the second image (noise) is ')
print countnoise_im1_im2
print('of a total of image pixels of ')
print x*y
print ('fraction of differing pixels between the two images')
fract = float(countnoise_im1_im2)/(x*y)
print fract
# in the starting images, after the formula
#probab di due pixel divers (calcolata da countnoise) su due immagini dato rumore x nelle immagini di partenza n!/(k!(n-k)!)*(x**k)*((1-x)**(n-k))=2/2*(x**2)*((1-x)**0)=x**2 plus the probability di avere 1 solo pixel rumoroso: n!/(k!(n-k)!)*(x**k)*((1-x)**(n-k))=2/1*x*((1-x)**1)=2x(1-x)=2x-2x**2
#Total numero pixel da countnoise=y=2x-2x**2+x**2=2x-x**2 (since y will be a constant: x**2-2x+y=0) For instance per y=75% i will have x=0.5 (50% pixels rumorosi nella immagine di partenza)
# i will have x1=(2-(squareroot(4*countnoise_im1_im2)))/2   and x2=(2+(squareroot(4*countnoise_im1_im2)))/2 

delta = (4-(4*(float(countnoise_im1_im2)/(x*y))))

import math
sqdelta = math.sqrt(float(delta))

x_1 = float(2-sqdelta)/2
x_2 = float(2+sqdelta)/2
print('Evaluation of the fraction of noisy pixels in the starting images')
if x_1 < 1:
 print x_1
elif x_2 < 1:
 print x_2