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Adaptive Data Hiding in Edge
Areas of Images With Spatial
LSB Domain Systems
Cheng-Hsing Yang, Chi-Yao Weng,
Shiuh-Jeng Wang, Member, IEEE, and
Hung-Min Sun
學號:M99G0222
報告人:林福城
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Outline
 I. Introduction
 II. Literature review
 III. Our method
 IV. Fundamentals and discussions
 V. Experimental results
 VI. Conclusions and future work
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Introduction(1)
 The Internet is becoming increasingly popular as a
communication channel. However, message transmissions via
the Internet have some problems, such as information
security , copyright protection.
 A well-known steganographic method is the least-significantbit (LSB) substitution . This method embeds secret data by
replacing k LSBs of a pixel with k secret bits directly.
 It is an important topic to find an adaptive steganographic
method, which embeds data adaptively by considering the
concept of human vision, with the features of high capacity
and low distortion.
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Introduction(2)
 In 2005, Wu et al. proposed the pixel-value differencing (PVD) and
LSB replacement method. In their approach, two pixels are
embedded by the LSB replacement method if their difference value
falls into a lower level. Similarly , the PVD method is used if the
difference value falls into a higher level.
 In 2005, Park et al. proposed a steganographic scheme based on
the information of neighboring pixels. In this Approach ,the number
of bits embedded into a target pixel Px is determined by the
difference value of its neighboring pixels Pu and Pl, where Pu is
the upper pixel of and Pl is the left pixel of . Their approach is
similar to Chang and Tseng’s side-match method .
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Literature review-LSB(1)
Wu et al.’s PVD and LSB replacement method.
Di=|pi-pi+1|
If Di <=15 , Lower level , R1
3bit LSB substitution method .
If Di >15 , High level ,R2,R3,R4,R5
pixel–value different (PVD).
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Literature review-LSB(2)
For example
pi=56,pi+1=47,di=|56-47|=9
(Lower level)
Embed data is 111000,
After embed pi’=63 ,pi’+1=40,
di’=23>Div=15
After being readjusted
pi’=55,pi’+1=48,di’=55-48=7
(Lower level)
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Literature review-PVD
(1) 舉列說明：
pi=100,pi+1=125, diff=25,
在對照表一後發現差值25是位於標記15的地
方，而隱藏的資料是4個位元。而目前要隱
藏的位元流為6=0110(2)。
(2) 隱藏演算法：
首先將6加上標記15，6+15=21。現在目標是
將原本的差值25變成21，於是我們將差異量
4分配到兩個像素上，使得100變成102、125
變成123，造成兩相鄰像素的差值為21。
(3) 粹取演算法：
當我們收到兩相鄰像素102、123後，首先對
照表一，發現差值21是位於標記15的地方，
隱藏的資料是4位元。然後將21-15=6，
6=0110(2)，即可以得到隱藏位元流為0110。
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This paper’s method(1)
R1=16
R2=16
R3=224
l  log 2 R1
m  log 2 R 2
h  log 2 R 3
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This paper’s methodEmbedding Procedure
 Step 1. Calculate the difference value di=|pi-(pi+1)|
 Step 2. Find out the level to which di belongs to. Let k= l,m,h
 Step3. embed k secret bits into pi and k secret bits into pi+1 .
It represented by pi’, (pi+1)’
 Setp4. Apply the modified LSB substitution method to pi’ and
(pi+1)’
 Step 5. Calculate the new difference value di’=|pi’-(pi+1)’|
 Step 6. if di and di’ belong to different levels, then readjust.
case 1.di is L,di’is not L , case 2 di is M ,di’ is L
case 3. di is M, di’ is H, case 4. di is H, di’ is not H.
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This paper’s methodEmbedding Procedure
case 1.di is L , di’ is not L
if pi’>=(pi+1)’, readjust between (pi’,(pi+1)’+2^k) and (pi’-2^k,(pi+1)’),
else
(pi’,(pi+1)’-2^k) and (pi’+2^k,(pi+1)’)
case 2. di is M , di’ is L
if pi’>=(pi+1)’, readjust between (pi’ +2^k,(pi+1)’) and (pi’,(pi+1)’ -2^k),
else
(pi’,(pi+1)’-2^k) and (pi’+2^k,(pi+1)’
case 3. di is M, di’ is H
if pi’>=(pi+1)’, readjust between (pi’ +2^k,(pi+1)’) and (pi’,(pi+1)’ -2^k),
else
(pi’,(pi+1)’-2^k) and (pi’+2^k,(pi+1)’)
case 4. di is H, di’ is not H
if pi’>=(pi+1)’, readjust between (pi’,(pi+1)’-2^k) and (pi’+2^k,(pi+1)’)
else
(pi’ +2^k,(pi+1)’) and (pi’,(pi+1)’ -2^k)
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This paper’s methodEmbedding Procedure
 1.Pi=64=1000000(2),Pi+1=47=101111(2),





secret data bit=1010 0000(2)
2.di=|64-47|=17,it belong to middle level .So K=4, then change
4 bit.
3.Pi’=74=1001010(2), pi+1’=32=100000(2)
4.After the modified LSB substitution is applied (調整最高位元)
5.Pi’=58=111010(2), pi+1’=48=110000(2), di’=|58-48|=10,Low
level
6.Case 2,After Readjust
(58+2^4,48)=(74,48) range is 26 =(1001010(2),110000(2))
(58,48-2^4,)=(58,32)不使用
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This paper’s method-Extraction
Procedure(1)
 Step 1 Calculate the difference value di’ for each block with
consecutive pixels
 Step 2 From the l-m-h division find out the level to which di’
belongs to. Let k=l,m,and h if di’ belongs to the level.
 Step3 From the k-bit LSB of a pixel , extract k secret bits from
pi’ and (pi+1)’
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This paper’s method-Extraction
Procedure(2)
 (pi’,pi+1’)=(74,48)=(1001010(2),110000(2))
 Di’=|74-48|=26 (Middle level)
 Therefore k=4
 So we can get secret bits is 1010(2),0000(2).
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Fundamentals and discussions
 Six restrictions are
 1.l<log2|R1|
 2.m<=log2|R2|
 3.h<=log2|R3|
 4.|R1|<=|R2|<=|R3|
 5.|R1|+|R2|<=|R3|
 6.1<= k<=6
Definition 1:
For the k-bit modified LSB substitution.
The rage [0,2^k-1] is bottom .
[255-2^k-1,255] is top.
If 5 bit modified LSB
[0,16] and [239,255]
Definition 2:
For the k-bit modified LSB substitution,
Pixel p is modifiable, if p is not belong to
the extreme ranges.
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Lemma 1
Lemma 1 : p - 2
k -1
p  p2
'
i
k -1
p '  [p - (2 k  1), p  (2 k  1)]
k 1
k 1
1.[p - (2  1), p  2 ]  p  [p  1, p  2 ]
k
'
同加2^k
2.[p - 2 k-1 , p  2 k 1 ]
k 1
3.[p  2 , p  2 - 1]  p  [p - 2 , p  1]
同減2^k
p - 2 k-1  p '  p  2 k 1
得到
k -1
k
'
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Lemma 2
 From Lemma 1:
p i - 2 k -1  p i'  pi  2 k 1
p i 1 - 2 k -1  p i' 1  p i 1 2 k 1
d i  pi 1  pi
d i'  pi' 1  pi'
(p i 1 - 2 k -1 )  ( pi  2 k 1 )  p i' 1  p i'  ( p i 1 2 k 1 )  (p i - 2 k -1 )
 di  2 k  d i'  di  2 k
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Theorem 1
 Theorem 1: Suppose that pi and pi+1are modifiable
and are embedded by the k-bit modified LSB
substitution, where their difference value belongs to
range R and k<=log|R| . Then , the readjusting
phase works.
d i  2 k  d i'  di  2 k
| d i'  d i | 2 k
R  2k
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Different range
 Cases that di and di’ fall into difference ranges,
 (a) di ∈ lower level.
 (b) One case of
di ∈ middle level.
 (c) The other case of
di ∈ middle level.
 (d) di ∈ higher level.
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Corresponding moving operations
 Corresponding moving operations of p’i and p’i+1
 (a) di ∈ lower level
 (b) one case of
di ∈ middle level
 (c) the other case
of di ∈ middle level
 (d) di ∈ higher level.
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Lemma 3 and Lemma 4
From define 1
Lemma 3 :
1.p - 2 k -1  p'  p  (2 k - 1)( Botton)
2.p  (2 k  1)  p'  p  2 k -1 (Top)
Lemma 4 :
1 .0  p '  2 k  1
2.255  ( 2 k  1)  p '  255
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Lemma 5
From Lemma 1 and 3 :
1.pi - 2 k-1  pi '  pi  ( 2 k - 1)
2.pi 1 - 2 k-1  pi 1 '  pi 1  ( 2 k-1 )
pi 1 - 2 k-1  ( pi  ( 2 k - 1)  pi 1 ' pi '  pi 1  ( 2 k-1 ) - pi - 2 k-1
 di  2k  ( 2k 1  1)  di '  di  2k
From Lemma 1 and 3 :
1.pi - 2 k-1  pi '  pi  ( 2 k-1 )
2.pi 1 - (2 k  1)  pi 1 '  pi 1  ( 2 k-1 )
pi 1 - (2 k  1)  pi  ( 2 k-1 )  pi 1 ' pi '  pi 1  ( 2 k-1 ) - pi - 2 k-1
 di  2k  ( 2k 1  1)  di '  di  2k
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Lemma 5’s show
 Relative positions of di and d’i
di  2k  (2k  1)  di '  di  2k
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Experimental results
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Conclusions and future works
 This paper propose a new adaptive LSB substitution
method to embed secret data into gray images without
making perceptible distortion.
 The relative attacks to either destroy or detect the
embedding information are not given in this paper.
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