Why need shuffle before SVD operation? Fast mode does not works for white images #78
Comments
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I met the problem too |
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我觉得可以加掩码来解决这个问题 |
def extract_raw(self, img):
# 每个分块提取 1 bit 信息
self.read_img_arr(img=img)
self.init_block_index()
wm_block_bit = np.zeros(shape=(3, self.block_num)) # 3个channel，length 个分块提取的水印，全都记录下来
self.idx_shuffle = random_strategy1(seed=self.password_img,
size=self.block_num,
block_shape=self.block_shape[0] * self.block_shape[1], # 16
)
for channel in range(3):
wm_block_bit[channel, :] = self.pool.map(self.block_get_wm,
[(self.ca_block[channel][self.block_index[i]], self.idx_shuffle[i])
for i in range(self.block_num)])
return wm_block_bit
def extract_avg(self, wm_block_bit):
# 对循环嵌入+3个 channel 求平均
wm_avg = np.zeros(shape=self.wm_size)
for i in range(self.wm_size):
wm_avg[i] = wm_block_bit[:, i::self.wm_size].mean()
return wm_avg
def extract(self, img, wm_shape):
self.wm_size = np.array(wm_shape).prod()
# 提取每个分块埋入的 bit：
wm_block_bit = self.extract_raw(img=img)
# 做平均：
wm_avg = self.extract_avg(wm_block_bit)
return wm_avg
def extract_with_kmeans(self, img, wm_shape):
wm_avg = self.extract(img=img, wm_shape=wm_shape)
return one_dim_kmeans(wm_avg)
def one_dim_kmeans(inputs):
threshold = 0
e_tol = 10 ** (-6)
center = [inputs.min(), inputs.max()] # 1. 初始化中心点
for i in range(300):
threshold = (center[0] + center[1]) / 2
is_class01 = inputs > threshold # 2. 检查所有点与这k个点之间的距离，每个点归类到最近的中心
center = [inputs[~is_class01].mean(), inputs[is_class01].mean()] # 3. 重新找中心点
if np.abs((center[0] + center[1]) / 2 - threshold) < e_tol: # 4. 停止条件
threshold = (center[0] + center[1]) / 2
break
is_class01 = inputs > threshold
return is_class01 |
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my code with mask: import numpy as np
from blind_watermark import WaterMark # '0.4.1'
import hashlib
mask = bytearray()
m = hashlib.sha256()
m.update(b"1145141919810")
for i in range(1024 // 32):
mask.extend(m.digest())
m.update(m.digest())
def bitstring_to_bytes(s):
v = int(s, 2)
b = bytearray()
while v:
b.append(v & 0xff)
v >>= 8
return b[::-1]
# 1057226 - 1024 * 1032 = 458
with open("a.docx", "rb") as f:
data = f.read()
print(len(data))
i = 0
while True:
bwm1 = WaterMark(password_img=1, password_wm=1)
bwm1.bwm_core.fast_mode = True
bwm1.bwm_core.d1 = 72
bwm1.read_img('6b7858c3798b5d0d1afb9f67bf43e2d5.jpg')
kk = data[i * 1024:i * 1024 + 1024]
if i == 1032:
kk = data[i * 1024:i * 1024 + 458]
kk = bytearray(kk)
for k in range(len(kk)):
kk[k] ^= mask[k]
print(kk)
byte = bin(int(kk.hex(), base=16))
wm = np.array(list(byte)) == '1'
bwm1.read_wm(wm, mode='bit')
bwm1.embed('embedded-{}.jpg'.format(i))
len_wm = len(bwm1.wm_bit)
print({i: len_wm})
wm_extract = bwm1.extract('embedded-{}.jpg'.format(i), mode='bit', wm_shape=len_wm)
wm_extract = bitstring_to_bytes("".join(["1" if x else "0" for x in wm_extract]))
print(wm_extract)
for j in range(len(kk)):
if kk[j] != wm_extract[j]:
print(j)
print()
i += 1
if i == 1033:
breakoutput: |
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Hi,
I am curious why u shuffled DCT block before the SVD operation, what's the meaning of that ? For safty? And I found that the fast mode doest works for pure white image . Thank you so much for your time.
Best,
Weizhou
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