我正在为一个大学项目写这篇论文的代码.其思想是在灰度图像中插入一个不可见的水印，稍后可以提取该水印以验证图像的所有权.
import pywt
import numpy as np
import cv2
from PIL import Image
from math import sqrt, log10
from scipy.fftpack import dct, idct
def Get_MSB_LSB_Watermark () : #Function that separates the watermark into MSB and LSB images
MSBs = []
LSBs = []
for i in range (len(Watermark)) :
binary = '{:0>8}'.format(str(bin(Watermark[i]))[2:])
MSB = (binary[0:4])
LSB = (binary[4:])
MSB = int(MSB, 2)
LSB = int(LSB,2)
MSBs.append(MSB)
LSBs.append(LSB)
MSBs = np.array(MSBs)
LSBs = np.array(LSBs)
return MSBs.reshape(64,64), LSBs.reshape(64,64)
def split(array, nrows, ncols): #Split array into blocks of size nrows* ncols
r, h = array.shape
return (array.reshape(h//nrows, nrows, -1, ncols)
.swapaxes(1, 2)
.reshape(-1, nrows, ncols))
def unblockshaped(arr, h, w): #the inverse of the split function
n, nrows, ncols = arr.shape
return (arr.reshape(h//nrows, -1, nrows, ncols)
.swapaxes(1,2)
.reshape(h, w))
def ISVD (U,S,V): #the inverse of singular value decomposition
s = np.zeros(np.shape(U))
for i in range(4):
s[i, i] = S[i]
recon_image = U @ s @ V
return recon_image
def Watermark_Embedding (blocks, watermark) :
Watermarked_blocks = []
k1 = []
k2 = []
#convert the watermark to a list
w = list(np.ndarray.flatten(watermark))
for i in range (len(blocks)) :
B = blocks[i]
#Aplly singular value decoposition to the block
U, s, V = np.linalg.svd(B)
#Modify the singular values of the block
P = s[1] - s[2]
delta = abs(w[i]) - P
s[1] = s[1] + delta
if s[0] >= s[1] :
k1.append(1)
else :
k1.append(-1)
#the inverse of SVD after watermark embedding
recunstructed_B = ISVD(U, s, V)
Watermarked_blocks.append(recunstructed_B)
for j in range(len(w)):
if w[j] >= 0:
k2.append(1)
else:
k2.append(-1)
return k1,k2, np.array(Watermarked_blocks)
def apply_dct(image_array):
size = image_array[0].__len__()
all_subdct = np.empty((size, size))
for i in range (0, size, 4):
for j in range (0, size, 4):
subpixels = image_array[i:i+4, j:j+4]
subdct = dct(dct(subpixels.T, norm="ortho").T, norm="ortho")
all_subdct[i:i+4, j:j+4] = subdct
return all_subdct
def inverse_dct(all_subdct):
size = all_subdct[0].__len__()
all_subidct = np.empty((size, size))
for i in range (0, size, 4):
for j in range (0, size, 4):
subidct = idct(idct(all_subdct[i:i+4, j:j+4].T, norm="ortho").T, norm="ortho")
all_subidct[i:i+4, j:j+4] = subidct
return all_subidct
#read watermark
Watermark = Image.open('Copyright.png').convert('L')
Watermark = list(Watermark.getdata())
#Separate the watermark into LSB and MSB images
Watermark1, Watermark2 = Get_MSB_LSB_Watermark()
#Apply descrete cosine Transform on the two generated images
DCT_Watermark1 = apply_dct(Watermark1)
DCT_Watermark2 = apply_dct(Watermark2)
#read cover Image
Cover_Image = Image.open('10.png').convert('L')
#Apply 1 level descrete wavelet transform
LL1, (LH1, HL1, HH1) = pywt.dwt2(Cover_Image, 'haar')
#Split the LH1 and HL1 subbands into blocks of size 4*4
blocks_LH1 = split(LH1,4,4)
blocks_HL1 = split(HL1,4,4)
#Watermark Embedding in LH1 and HL1 and Keys generation
Key1, Key3, WatermarkedblocksLH1 = Watermark_Embedding(blocks_LH1,DCT_Watermark1)
Key2 ,Key4, WatermarkedblocksHL1 = Watermark_Embedding(blocks_HL1,DCT_Watermark2)
#Merge the watermzrked Blocks
reconstructed_LH1 = unblockshaped(WatermarkedblocksLH1, 256,256)
reconstructed_HL1 = unblockshaped(WatermarkedblocksHL1, 256,256)
#Apply the inverse of descrete wavelet transform to get the watermarked image
IDWT = pywt.idwt2((LL1, (reconstructed_LH1, reconstructed_HL1, HH1)), 'haar')
cv2.imwrite('Watermarked_img.png', IDWT)
这是我为提取过程编写的代码:
import pywt
from scipy import fftpack
import numpy as np
import cv2
from PIL import Image
import scipy
from math import sqrt, log10
from Watermark_Embedding import *
def Watermark_Extraction(blocks,key1, key2) :
Extracted_Watermark = []
for i in range(len(blocks)):
B = blocks[i]
#apply SVD on the Block
U, s, V = np.linalg.svd(B)
if key1[i] == 1 :
P = (s[1] - s[2])
Extracted_Watermark.append(P)
else :
P = (s[0] - s[2])
Extracted_Watermark.append(P)
for j in range(len(Extracted_Watermark)) :
if key2[j] == 1 :
Extracted_Watermark[j] = Extracted_Watermark[j]
else :
Extracted_Watermark[j] = - (Extracted_Watermark[j])
return np.array(Extracted_Watermark)
def Merge_W1_W2 ():
Merged_watermark = []
w1 = list(np.ndarray.flatten(IDCTW1))
w2 = list(np.ndarray.flatten(IDCTW2))
for i in range (len(w2)):
bw1 = '{:0>4}'.format((bin(int(abs(w1[i]))))[2:])
bw2 = '{:0>4}'.format((bin(int(abs(w2[i]))))[2:])
P = bw1+bw2
pixel = (int(P,2))
Merged_watermark.append(pixel)
return Merged_watermark
Watermarked_Image = Image.open('Watermarked_img.png')
LL1, (LH1, HL1, HH1) = pywt.dwt2(Watermarked_Image, 'haar')
blocks_LH1 = split(LH1,4,4)
blocks_HL1 = split(HL1,4,4)
W1 = Watermark_Extraction(blocks_LH1, Key1,Key3)
W2 = Watermark_Extraction(blocks_HL1, Key2, Key4)
W1 = W1.reshape(64,64)
W2 = W2.reshape(64,64)
IDCTW1 = inverse_dct(W1)
IDCTW2 = inverse_dct(W2)
Merged = np.array(Merge_W1_W2())
Merged = Merged.reshape(64,64)
cv2.imwrite('Extracted_Watermark.png', Merged)
大小为512*512的封面图片:
我使用的64*64水印
水印图像:
我获得的提取水印:
我使用SSIM计算了两个水印之间的相似性:
from skimage.metrics import structural_similarity
original_Watermark = cv2.imread('Copyright.png')
extracted_watermark = cv2.imread('Extracted_Watermark.png')
# Convert images to grayscale
original_watermark = cv2.cvtColor(original_Watermark, cv2.COLOR_BGR2GRAY)
extracted_Watermark = cv2.cvtColor(extracted_watermark, cv2.COLOR_BGR2GRAY)
# Compute SSIM between two images
(score, diff) = structural_similarity(original_Watermark, extracted_Watermark, full=True)
print("SSIM = ", score)
我没有对水印图像进行任何修改，得到的SSIM是0.8445354561524052.然而，根据论文，提取的水印的SSIM应为0.99.