像素是图像的组成部分。
每个像素包含三个值:(红色、绿色、蓝色)也称为 RGB 值。
每个 RGB 值的范围从 0 到 255。
对于数据中的每个字符,将其 ASCII 值转换为 8 位二进制 [1]。
一次读取三个像素,其总 RGB 值为 3*3=9 个。前八个 RGB 值用于存储一个转换为 8 位二进制的字符。
比较相应的RGB值和二进制数据。如果二进制数字为 1,则 RGB 值将转换为奇数,否则为偶数。
第 9 个值确定是否应该读取更多像素。如果有更多数据要读取,即编码或解码,则第 9 个像素变为偶数;否则,如果我们想停止进一步读取像素,那就让它变得奇数。
假设要隐藏的消息是‘Hii’。
[(27, 64, 164), (248, 244, 194), (174, 246, 250), (149, 95, 232),(188, 156, 169), (71, 167, 127), (132, 173, 97), (113, 69, 206),(255, 29, 213), (53, 153, 220), (246, 225, 229), (142, 82, 175)]
(27, 64, 164), (248, 244, 194), (174, 246, 250)
(26, 63, 164), (248, 243, 194), (174, 246, 250)
[(26, 63, 164), (248, 243, 194), (174, 246, 250), (148, 95, 231),(188, 155, 168), (70, 167, 126), (132, 173, 97), (112, 69, 206),(254, 29, 213), (53, 153, 220), (246, 225, 229), (142, 82, 175)]
同样,一次读取三个像素。前 8 个 RGB 值为我们提供了有关机密数据的信息,第 9 个值告诉我们是否继续前进。
对于前八个值,如果值为奇数,则二进制位为 1 ,否则为 0 。
这些位连接成一个字符串,每三个像素,我们得到一个字节的秘密数据,这意味着一个字符。
现在,如果第 9 个值是偶数,那么我们继续一次读取三个像素,否则,我们停止。
[(26, 63, 164), (248, 243, 194), (174, 246, 250), (148, 95, 231),(188, 155, 168), (70, 167, 126), (132, 173, 97), (112, 69, 206),(254, 29, 213), (53, 153, 220), (246, 225, 229), (142, 82, 175)]
[(26, 63, 164), (248, 243, 194), (174, 246, 250)
# Python program implementing Image Steganography# PIL module is used to extract# pixels of image and modify itfrom PIL import Image# Convert encoding data into 8-bit binary# form using ASCII value of charactersdef genData(data):# list of binary codes# of given datanewd = []for i in data:newd.append(format(ord(i), '08b'))return newd# Pixels are modified according to the# 8-bit binary data and finally returneddef modPix(pix, data):datalist = genData(data)lendata = len(datalist)imdata = iter(pix)for i in range(lendata):# Extracting 3 pixels at a timepix = [value for value in imdata.__next__()[:3] +imdata.__next__()[:3] +imdata.__next__()[:3]]# Pixel value should be made# odd for 1 and even for 0for j in range(0, 8):if (datalist[i][j] == '0' and pix[j]% 2 != 0):pix[j] -= 1elif (datalist[i][j] == '1' and pix[j] % 2 == 0):if(pix[j] != 0):pix[j] -= 1else:pix[j] += 1# pix[j] -= 1# Eighth pixel of every set tells# whether to stop ot read further.# 0 means keep reading; 1 means thec# message is over.if (i == lendata - 1):if (pix[-1] % 2 == 0):if(pix[-1] != 0):pix[-1] -= 1else:pix[-1] += 1else:if (pix[-1] % 2 != 0):pix[-1] -= 1pix = tuple(pix)yield pix[0:3]yield pix[3:6]yield pix[6:9]def encode_enc(newimg, data):w = newimg.size[0](x, y) = (0, 0)for pixel in modPix(newimg.getdata(), data):# Putting modified pixels in the new imagenewimg.putpixel((x, y), pixel)if (x == w - 1):x = 0y += 1else:x += 1# Encode data into imagedef encode():img = input("Enter image name(with extension) : ")image = Image.open(img, 'r')data = input("Enter data to be encoded : ")if (len(data) == 0):raise ValueError('Data is empty')newimg = image.copy()encode_enc(newimg, data)new_img_name = input("Enter the name of new image(with extension) : ")newimg.save(new_img_name, str(new_img_name.split(".")[1].upper()))# Decode the data in the imagedef decode():img = input("Enter image name(with extension) : ")image = Image.open(img, 'r')data = ''imgdata = iter(image.getdata())while (True):pixels = [value for value in imgdata.__next__()[:3] +imgdata.__next__()[:3] +imgdata.__next__()[:3]]# string of binary databinstr = ''for i in pixels[:8]:if (i % 2 == 0):binstr += '0'else:binstr += '1'data += chr(int(binstr, 2))if (pixels[-1] % 2 != 0):return data# Main Functiondef main():a = int(input(":: Welcome to Steganography ::\n""1. Encode\n2. Decode\n"))if (a == 1):encode()elif (a == 2):print("Decoded Word : " + decode())else:raise Exception("Enter correct input")# Driver Codeif __name__ == '__main__' :# Calling main functionmain()
https://www.geeksforgeeks.org/program-decimal-binary-conversion/
https://www.geeksforgeeks.org/working-images-python/
https://dev.to/erikwhiting88/let-s-hide-a-secret-message-in-an-image-with-python-and-opencv-1jf5
A code along with the dependencies can be found here:https://github.com/goelashwin36/image-steganography
分享
点收藏
点点赞
点在看
文章转发自AI科技大本营微信公众号,版权归其所有。文章内容不代表本站立场和任何投资暗示。
Copyright © 2021.Company 元宇宙YITB.COM All rights reserved.元宇宙YITB.COM