隐写术 是在任何图像/音频/视频中隐藏秘密数据的方法。简而言之,隐写术的主要目的是在任何图像/音频/视频中隐藏想要的信息,而这些图像/音频/视频看起来并不是秘密的。 基于图像的隐写术背后的想法非常简单。图像由数字数据(像素)组成,这些数据描述了图片中的内容,通常是所有像素的颜色。因为我们知道每个图像都是由像素组成的,每个像素包含3个值(红、绿、蓝)。
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对数据进行编码:
数据的每个字节都使用ASCII值转换为8位二进制代码。现在,从左到右读取像素,每组3个,共包含9个值。前8个值用于存储二进制数据。如果出现1,则该值为奇数,如果出现0,则该值为奇数。 例如: 假设要隐藏的消息是’ 嗨 ‘. 由于消息为3字节,因此,对数据进行编码所需的像素为3 x 3=9。考虑一个4×3的图像,总共有12个像素,这足以编码给定的数据。
[(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)]
‘的ASCII值 H ‘是72,其二进制等价物是01000。 将前3个像素(27、64、164)、(248、244、194)、(174、246、250)进行编码。现在将像素更改为奇数表示1,偶数表示0。因此,修改的像素是(26,63,164),(248,243,194),(174,246,250)。因为我们必须对更多数据进行编码,所以最后一个值应该是偶数。同样地,’ 我 “可以在这个图像中编码。 新图像将如下所示:
[(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)]
解码数据:
要解码,一次读取三个像素,直到最后一个值为奇数,这意味着消息结束。每3个像素包含一个二进制数据,可以通过相同的编码逻辑提取。如果值为奇数,则二进制位为1,否则为0。
以下是上述理念的实施情况:
python
# Python program implementing Image Steganography # PIL module is used to extract # pixels of image and modify it from PIL import Image # Convert encoding data into 8-bit binary # form using ASCII value of characters def genData(data): # list of binary codes # of given data newd = [] for i in data: newd.append( format ( ord (i), '08b' )) return newd # Pixels are modified according to the # 8-bit binary data and finally returned def modPix(pix, data): datalist = genData(data) lendata = len (datalist) imdata = iter (pix) for i in range (lendata): # Extracting 3 pixels at a time pix = [value for value in imdata.__next__()[: 3 ] + imdata.__next__()[: 3 ] + imdata.__next__()[: 3 ]] # Pixel value should be made # odd for 1 and even for 0 for j in range ( 0 , 8 ): if (datalist[i][j] = = '0' and pix[j] % 2 ! = 0 ): pix[j] - = 1 elif (datalist[i][j] = = '1' and pix[j] % 2 = = 0 ): if (pix[j] ! = 0 ): pix[j] - = 1 else : 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 ] - = 1 else : pix[ - 1 ] + = 1 else : if (pix[ - 1 ] % 2 ! = 0 ): pix[ - 1 ] - = 1 pix = 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 image newimg.putpixel((x, y), pixel) if (x = = w - 1 ): x = 0 y + = 1 else : x + = 1 # Encode data into image def 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 image def 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 data binstr = '' 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 Function def main(): a = int ( input ( ":: Welcome to Steganography ::" "1. Encode2. Decode" )) if (a = = 1 ): encode() elif (a = = 2 ): print ( "Decoded Word : " + decode()) else : raise Exception( "Enter correct input" ) # Driver Code if __name__ = = '__main__' : # Calling main function main() |
输出:
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