higher-order statistics and pairs · pdf filehigher-order statistics and pairs analysis csm25...

Higher-Order Statistics and Pairs AnalysisCSM25 Secure Information Hiding

Dr Hans Georg Schaathun

University of Surrey

Spring 2008

Dr Hans Georg Schaathun Higher-Order Statistics and Pairs Analysis Spring 2008 1 / 41

Outcomes

Understand the principle of higher-order statisticsBe able to implement at least one steganalysis technique usinghigher-order statistics.

Reading

Core Reading

«Quantitative steganalysis of digital images:estimating the secret message length»by Jessica Fridrich, Miroslav Goljan, Dorin Hogea, David Soukal,in Multimedia Systems 2003

Suggested Reading

Jessica Fridrich, Miroslav Goljan, and Rui Du (State University of NewYork, Binghamton) ‘Detecting LSB Steganography in Color andGray-Scale Images’ in Multimedia and Security 2001

Background

Outline

1 Background

2 Pairs analysisThe characteristic sequenceHomogenous pairs

3 Where Pairs Analysis failsDithered backgrounds

4 RS steganalysisThe ideaThe resultCounter-measures

Background

Where χ2 falls short

The χ2 test we have seenAnalyses histogram only.Detects embedding in consecutive pixels

What if the message is randomly spread across the image?Generalised χ2 analysis.

Yes/No answer; cannot estimate message lengthCan be fooled if the message is biased (more 0-s than 1-s or v.v.)

Background

Higher-order statisticsPixels in neighbourhoods

Pairs of Values counts single pixels→ first-order statistic

Higher-order statisticsCount pairs of (neighbour) pixels (2nd order)Pixel triplets (3rd order)

Study relations between pixels in a neighbourhood

Pairs analysis

Pairs Analysis

Pairs Analysis is quantitativei.e. estimates the message length

Originally designed for GIF.We present it for spatial, grayscale images.

Pairs analysis The characteristic sequence

Outline

1 Background

The characteristic sequence

Let c, c′ be two colours (grayscales).Read image row by row (left to right and top down).Assign 0 to c and 1 to c′.Ignore all other colours.Resulting sequence is denoted Z (c, c′).

Definition

Z = Z (0, 1)|Z (2, 3)|Z (4, 5)| . . . |Z (254, 255), (1)Z ′ = Z (1, 2)|Z (3, 4)|Z (5, 6)| . . . |Z (255, 0). (2)

The colour cut

Z ( , ) extracted from an imageExtracted column-wise (Matlab-style)Row-wise extraction is equally valid.

��

001111000000111111001101110010111001111110

Pairs analysis Homogenous pairs

Outline

1 Background

Second-order structure

Second-order structure (of Z and of Z ′)count pairs of consecutive bitsfour possible pairs 00,01,10,11

Homogenous pairs: 00, 11Let F be frequency of Homogenous pairs in Z .Let R = F/n be the relative frequency.

where n = N ·M − 1 is the number of pairs.

Example

0 0 0 1 1 0 1 1 1 1 0 1 1Homog. 1 1 0 1 0 0 1 1 1 0 0 1

F = 7N = 11R = 7/11 = 0.6364

Expected structure of Z

Let R(p) = E(R) beexpected, relative frequency of homogenous pairs in Zwhen a fraction p of pixel LSB-s have been flipped.(e.g. if a random unbiased bit string of length 2p has beenembedded)

TheoremR(p) is a parabola with minimum at R(1/2) = 1/2.

R(p) = ap2 + bp + c

for some constants a, b, and c.

Why parabola?

k1 k2 k3 k4 · · · kr

Z =︷︸︸︷0000

︷︸︸︷111

︷︸︸︷00 . . . 0

︷︸︸︷11 . . . 1 · · ·

︷︸︸︷11 . . . 1

nR(0) =∑r

i=0(ki − 1)

Homogenous pair remains homogenous: Pr = q2 + (1− q)2

Both change + Neither changes

Heterougenous pair becomes homogenous: Pr = 2q(1− q)

nR(q) =r∑

[q2 + (1− q)2](ki − 1) + 2q(1− q)(r − 1)

Why parabola?

k1 k2 k3 k4 · · · kr

Z =︷︸︸︷0000

︷︸︸︷111

︷︸︸︷00 . . . 0

︷︸︸︷11 . . . 1 · · ·

︷︸︸︷11 . . . 1

nR(0) =∑r

i=0(ki − 1)

nR(q) =r∑

[q2 + (1− q)2](ki − 1) + 2q(1− q)(r − 1)

Why parabola?

k1 k2 k3 k4 · · · kr

Z =︷︸︸︷0000

︷︸︸︷111

︷︸︸︷00 . . . 0

︷︸︸︷11 . . . 1 · · ·

︷︸︸︷11 . . . 1

nR(0) =∑r

i=0(ki − 1)

nR(q) =r∑

[q2 + (1− q)2](ki − 1) + 2q(1− q)(r − 1)

The R-function

Structure of the shifted pairs

Compare the pairs Z with shifted pairs Z ′

Two parabolas

Assumption

R′(p) are parabolic and symmetric around p = 1/2, i.e.

R′(p) = a′p2 + b′p + c′.

We will study D(p) = R(p)− R′(p).Difference between two parabolæ is a parabola

D(p) = ap2 + bp + c for unknown a, b, p.

Two parabolas

Assumption

R′(p) = a′p2 + b′p + c′.

Two parabolas

Assumption

R′(p) = a′p2 + b′p + c′.

Two parabolas

Assumption

R′(p) = a′p2 + b′p + c′.

Two parabolas

Assumption

R′(p) = a′p2 + b′p + c′.

Two parabolas

Assumption

R′(p) = a′p2 + b′p + c′.

The R′-function

Structure of the shifted pairs (II)

WriteZ ′ = b1, b2, b3, . . . , bn

Theorem

nR′(1/2) =n−1∑k=1

2−khk ,

where hk is number of homogenous pairs among

(b1, bk+1), (b2, bk+2), . . . , (bn−k , bn).

The estimate of R′(1/2)

Zero message assumption

D(p) = ap2 + bp + c

Assumption

Z and Z ′ have the same structure when no message is embedded.

R(0) = R′(0)

D(0) = 0c = 0

Zero message assumption

D(p) = ap2 + bp + c

Assumption

Z and Z ′ have the same structure when no message is embedded.

R(0) = R′(0)

D(0) = 0c = 0

Zero point

Symmetry

Swapping all bits does not change the statisticSwapping 1− q random bits means

Swapping all bits, and then (re)swap q random bits

Embedding q or 1− q bits is the same thing.We conclude, D(q) = D(1− q)

Symmetry

Swapping all bits does not change the statisticSwapping 1− q random bits means

Swapping all bits, and then (re)swap q random bits

Embedding q or 1− q bits is the same thing.We conclude, D(q) = D(1− q)

Symmetry

Solving a second-order equation

We can estimate R and R′ at q = 12

R( 12 )− R′( 1

2 ) = D( 12 ) = a/4 + b/2 (left side known)

We exploit symmetry0 = D(q)− D(1− q) = (aq2 + bq)− (a(1− q)2 + b(1− q)).

We solve for a and b, to get

4D(12)q − 4D(

12)q2 = D(q)

where D(12) and D(q) are known

Solving the quadratic estimate gives pEstimated message length is 2p

Where Pairs Analysis fails Dithered backgrounds

Outline

1 Background

Dithered backgrounds

Dithering is used to simulate additional coloursTwo colours c1 and c2 alternate over an area.The appearance would be a uniform colour somewhere inbetween.Colour cut (c1, c2) has many heterogenous pairs.The result is that R(0) 6= R′(0) and our assumption fails.This can be fixed by clever choice of colour cut.(I did not find any good examples.)

RS steganalysis The idea

Outline

1 Background

RS steganalysis

Proposed for true colour imagesUse information from all 8 bits of a pixelLinked to so-called lossless capacity

Pixel groups and smoothness

Divide image into pixel groups G1, G2, . . .

Disjoint groupsConsecutive pixels

Define the smoothness of a group G = (x1, x2, . . . , xn)

f (G) =n−1∑i=1

|xi − xi+1|.

High f (G) means sharp changes from pixel to pixel.Unusual for neighbour pixels in natural images

Compare f (F (G)) and f (G)

where F is flips pixels.

f (G) =n−1∑i=1

|xi − xi+1|.

f (G) =n−1∑i=1

|xi − xi+1|.

f (G) =n−1∑i=1

|xi − xi+1|.

Bit flipping

Maps on a single pixelF+ : 2i ↔ 2i + 1F− : 2i ↔ 2i − 1F0 is the identity.

Maps on a group, say of four, G = (x1, x2, x3, x4)

F = [F0F+F+F0] = [0110]F (G) = (F0(x1), F1(x2), F1(x3), F0(x4))

Shifted bit flip−F = [F0F−F−F0] = [0− 1− 10]

We will use one map F and the shift −F .

Bit flipping

Characteristic Groups

The groupsRegular group: f (G) < f (F (G))Singular group: f (G) > f (F (G))Useless group: f (G) = f (F (G))

The statisticsRF : number of regular groups under FSF : number of singular groups under FR−F : number of regular groups under −FS−F : number of singular groups under −F

RF , R−F , SF , S−F as functions of pp is number of pixels flipped by the embedding.

Statistics plot

RS steganalysis The result

Outline

1 Background

Approximations

Approximations based on experimental investigationSF and RF are parabolicS−F and R−F are linear

RF (p) = a1p2 + b1p + c1, (3)

SF (p) = a2p2 + b2p + c2, (4)R−F (p) = a3p + b3, (5)S−F (p) = a4p + b4. (6)

11 unknowns (10 coefficients and p)

Approximations

RF (p) = a1p2 + b1p + c1, (3)

Approximations

RF (p) = a1p2 + b1p + c1, (3)

Equations

ObservationsRF (p), SF (p), R−F (p), S−F (p) from stegogrammeRF (1− p), SF (1− p), R−F (1− p), S−F (1− p) by flipping all LSB-s.

AssumptionsRF (1/2) = SF (1/2)RF (0) = R−F (0)SF (0) = S−F (0).

11 equationsWith 11 unknowns, this can be solved

Equations

The message length

x − 1/2,

where x is the smaller root of

s(d3 + d1)x2 + (d2 − d3 − d4 − 3d1)x + d1 − d2 = 0,

d1 = RF (p/2)− SF (p/2),

d2 = R−F (p/2)− S−F (p/2),

d3 = RF (1− p/2)− SF (1− p/2),

d4 = R−F (1− p/2)− S−F (1− p/2).

Initial bias

Some experiments show estimates within ±1% of true lengthSome images have an initial bias

i.e. the cover image appear to have a short message.This must be taken into accountShort messages cannot be detected with certainty

Gaussian distribution: µ = 0, σ = 0.5%

Is it possible to estimate the initial bias?

Plot from Fridrich et al.

Example from Fridrich et al.

RS steganalysis Counter-measures

Outline

1 Background

RS steganalysis Counter-measures

Good stego-systems?How do we foil higher-order statistics?

Stegogramme should resemble cover-imagenot necesserally visually... but statistically

Statistics-aware steganographyDesigned for specific higher-order statisticsStegogramme resembles cover with respect to statisticStill ad hoc approach

higher-order statistics and pairs · pdf filehigher-order statistics and pairs analysis csm25...

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