[QUICK NOTE] Jsteg系列算法速查手册

前言

从Jsteg到F5,这一系列算法是因应前面算法的缺点逐步改进的.所以在这里也使用增量的办法进行说明.

基础:DCT系数的特征

1. 基本关于0对称分布
2. 0系数最多,向两边逐渐减小.
   示例:
   

Jsteg算法

Jsteg是基于DCT的最低位进行隐写的.优点是正常情况下视觉差异几乎难以识别.

规则

Description:
Steg:

• 0/1: Skip
• Adjust to fit the lowest digit (0-even,1-odd) by keeping or $x\pm1$x±1
• Remember:2-1 can’t reach to 1, so it has to change to 3.
  Extract:
• 0/1: Skip
• Just read the lowest digit (0-even,1-odd)
  Note: It can be remembered as “|X|” structure.

F3算法

因应

Jsteg出现的值对现象,易被 $\chi^2$χ2攻击**.
Like This:

规则

Note:
Shrinkage means changing the coefficient but not regarding it as an “effective” embed.
Description:
Steg:

1. 0:Skip
2. Positive and negative parts are SYMMETRICAL.
3. 1/-1 and 0: Shrinkage
4. Adjust to fit the lowest digit (0-even,1-odd) by keeping or $|x|-=1$∣x∣−=1

Remember: $\pm 1$±1- $0$0 CAN reach to 0, so it lead to $|x|-1$∣x∣−1.
Extract:

• 0: Skip
• Just read the lowest digit (0-even,1-odd)
  Note: It can be remembered as symmetrical “N” structure.

F4

因应

1. F3制造了大量的新增0,这种异常可以用作确定隐写类型.
2. 重传导致DCT系数中偶数明显增多(除0外).原始图像一般来说DCT系数中奇数总数多于偶数总数(不包括零)，这样很容易区分开原始图像和载密图像.

规则

Steg:

1. 0:Skip
2. Positive and negative parts are CORRESPONDING.
3. 1-0 and -1-1: Shrinkage
4. For the positive: odd-1,even-0
5. For the negative: odd-0, even-1
6. Remember: -1-1->0, so -1-0->-1. And 1-0->0, so 1-1->1

Extract:

1. 0:Skip
2. For the positive: Read the lowest digit
3. For the negative: Read the lowest digit and REVERSE

Note: It can be remembered as corresponding “N” structure.

F5

因应

F4隐写是顺序嵌入的，这就使得LSB的修改集中在图像的某一部分，可能导致图像质量的不均匀；另外，当嵌入信息时，对系数的LSB更改的概率为1/2，所以在嵌入较多信息时，系数的更改会比较多，检测者据此很可能发现秘密信息的存在。

规则

矩阵编码

以k=2为例.
对象:2隐写信息 $\{x_i\}${xi​}+3非0DCT系数 $\{a_j\}${aj​}
过程:

1. bool p= (if $x_1$x1​==lowbit( $a_1\oplus a_3$a1​⊕a3​))
2. bool q= (if $x_2$x2​==lowbit( $a_2\oplus a_3$a2​⊕a3​))
3. p && q?keep:pass;
4. !p&&q?Change a1:pass;
5. p&&!q?Change a2:pass;
6. !p&&!q?Change a3:pass;

Change means abs(a)-1

In other words, if a non-match occurs in either side, change the exclusive “a” of this side; if both sides are mismatched, change the common one.

To extract:
For each group with 3 DCT coefficient, try to recover:
$x_1=a_1\oplus a_3\tag{1}$x1​=a1​⊕a3​(1)
$x_2=a_2\oplus a_3\tag{2}$x2​=a2​⊕a3​(2)
[Attention] lowbit(Negative XOR Positive)=Reverse(lowbit(Abs(Negative) XOR Positive))
e.g. -1 XOR 1=-1,not 0, and -1 XOR -1=0.

过程

1. Select non-0 DCT coefficients and shuffle
2. For each 3 DCT coefficients, try to use matrix coding to embed
3. [ATTENTION] Shrikage in this process: if any DCT coefficient bacame 0 after coding, cancel the embedding this time and load 3 DCT coefficients and restart.
   e.g.
   30 20 1->30 20 0(x)
   Then
   30 20 13(the next non-0 DCT coefficient)->30 20 12
   or 30 1 20->30 0 20
   Then
   30 20 13->30 19 13
4. Then Try to write in.

F5隐写后，图像的DCT系数直方图特性依然得到了保持.时间效率较高因为改动位数少.但是空间效率降低.

重要参考: https://fennudehaogua.top/2019/03/21/Steganography-basic/