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LevelBSc CSIT (TU)
StreamScience
SubjectImage Processing (BSc CSIT, CSC413)
Year2082 BS
Exam sessionRegular (annual)
Full marks60
Time allowed180 minutes
Questions12, all with step-by-step solutions
A

Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1Long answer10 marks

Distinguish between high frequency emphasis filter and Laplacian filter. Find the output of the following image using histogram equalization, where number of possible gray levels is 8.

3 3 3 3 3

2 3 4 3 2

2 4 4 4 2

2 3 4 3 2

3 3 3 3 3

High-Frequency Emphasis Filter vs Laplacian Filter

AspectHigh-Frequency Emphasis FilterLaplacian Filter
DomainMainly frequency domainMainly spatial domain (a mask)
DefinitionHhfe(u,v)=a+bHhp(u,v)H_{hfe}(u,v)=a+b\,H_{hp}(u,v) where HhpH_{hp} is a high-pass filter, a0a\ge 0, b>ab>a2f=2fx2+2fy2\nabla^2 f=\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2}, a second-order derivative
IdeaAdds back a fraction of the original so low frequencies (background) are not lost while high frequencies (edges) are boostedProduces only the edge/detail map; must be added to the original for sharpening: g=f2fg=f-\nabla^2 f
EffectSharpens and preserves overall tonality/brightnessStrongly highlights edges but darkens flat regions if used alone
OrderBuilt on a high-pass responsePure second-derivative operator
OutputEnhanced image with detail emphasizedEdge image (zero in flat areas)

In short, the high-frequency emphasis filter is essentially a high-pass filter with an added offset so the result is not a bare edge map, whereas the Laplacian is an isotropic second-derivative operator whose raw output is an edge map that is normally combined with the original for sharpening.

Histogram Equalization (L=8L=8)

Input image (5×55\times5):

3 3 3 3 3
2 3 4 3 2
2 4 4 4 2
2 3 4 3 2
3 3 3 3 3

Total pixels MN=25MN = 25, levels L=8L=8 so the scaling factor is L1=7L-1=7.

Step 1 – Histogram (count of each gray level):

rkr_knkn_k
26
314
45

(All other levels 0, 1, 5, 6, 7 have count 0.)

Step 2 – PDF, CDF and transformation sk=round[(L1)CDF]s_k=\text{round}\big[(L-1)\,\text{CDF}\big]:

rkr_knkn_kpr=nk/25p_r=n_k/25CDF(L1)(L-1)\cdotCDFsks_k
260.240.241.682
3140.560.805.606
450.201.007.007

Step 3 – Mapping: 22,  36,  472\to2,\; 3\to6,\; 4\to7.

Output (equalized) image:

6 6 6 6 6
2 6 7 6 2
2 7 7 7 2
2 6 7 6 2
6 6 6 6 6

The original values were clustered in {2,3,4}\{2,3,4\}; after equalization they are spread across {2,6,7}\{2,6,7\}, increasing the global contrast.

2Long answer10 marks

Define noise. Explain the models for image degradation and restoration process.

Definition of Noise

Noise in an image is any unwanted random variation in pixel intensity that is not part of the true scene. It is introduced during acquisition (sensor thermal noise, low light), digitization (quantization), or transmission (channel errors). Noise degrades quality and is usually modelled as a random process described by its probability density function (PDF).

Common noise models include Gaussian, Rayleigh, Erlang (gamma), exponential, uniform, and salt-and-pepper (impulse) noise.

Image Degradation / Restoration Model

The goal of restoration is to recover an estimate f^(x,y)\hat{f}(x,y) of the original image from a degraded observation g(x,y)g(x,y), using knowledge of the degradation process. (Restoration is objective, based on a model, unlike enhancement which is subjective.)

Degradation Model

Degradation is modelled as a degradation function HH acting on the input image f(x,y)f(x,y), followed by the addition of noise η(x,y)\eta(x,y):

 f(x,y) ---->[ H ]----> (+) ----> g(x,y) ----->[ Restoration ]----> f^(x,y)
 degradation ^ filter(s)
 function |
 noise η(x,y)

Spatial domain (H is linear and position-invariant ⇒ convolution):

g(x,y)=h(x,y)f(x,y)+η(x,y)g(x,y) = h(x,y) \star f(x,y) + \eta(x,y)

Frequency domain (convolution becomes multiplication):

G(u,v)=H(u,v)F(u,v)+N(u,v)G(u,v) = H(u,v)\,F(u,v) + N(u,v)

Restoration Process

Given gg, an estimate of HH, and statistics of η\eta, restoration applies an inverse/restoration filter RR to obtain f^\hat{f}. Typical filters:

  • Inverse filtering: F^(u,v)=G(u,v)/H(u,v)\hat{F}(u,v)=G(u,v)/H(u,v) — simple but amplifies noise where HH is small.
  • Wiener (minimum mean-square error) filter: balances inverse filtering against noise using the signal/noise power ratio:
F^=[1HH2H2+Sη/Sf]G\hat{F}=\left[\frac{1}{H}\,\frac{|H|^2}{|H|^2+S_\eta/S_f}\right]G
  • Mean / order-statistic (median) filters when only noise (no blur, H=1H=1) is present.

Summary

The more that is known about HH and η\eta, the better the restoration. When HH is identity, the problem reduces to noise removal; when noise is negligible, it reduces to deblurring.

3Long answer10 marks

What are the differences between spatial domain and frequency domain? Compute the 2D DFT of the following image.

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

Spatial Domain vs Frequency Domain

AspectSpatial DomainFrequency Domain
Works onImage pixels f(x,y)f(x,y) directlyFourier coefficients F(u,v)F(u,v)
OperationConvolution with a mask/kernelPoint-wise multiplication with H(u,v)H(u,v)
BasisSpatial coordinatesSinusoidal frequency components
InterpretationDirect manipulation of intensities/neighbourhoodsLow freq = smooth areas, high freq = edges/detail
CostCheap for small kernelsEfficient (via FFT) for large kernels
Example operationsSmoothing/sharpening masks, histogram opsIdeal/Butterworth/Gaussian low- and high-pass filtering
Relationg(x,y)=hfg(x,y)=h\star fG(u,v)=H(u,v)F(u,v)G(u,v)=H(u,v)F(u,v) — linked by the Fourier transform

2-D DFT of the Image

Input image (4×44\times4, all ones):

1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1

The 2-D DFT is

F(u,v)=x=03y=03f(x,y)ej2π(ux4+vy4)F(u,v)=\sum_{x=0}^{3}\sum_{y=0}^{3} f(x,y)\,e^{-j2\pi\left(\frac{ux}{4}+\frac{vy}{4}\right)}

DC term (u=v=0u=v=0): the exponential is 1, so

F(0,0)=xyf(x,y)=16.F(0,0)=\sum_{x}\sum_{y} f(x,y) = 16.

All other terms (u0u\neq0 or v0v\neq0): since ff is constant, each is the sum of a full period of a complex exponential, which is zero:

x=03ej2πux/4=0   for u=1,2,3,\sum_{x=0}^{3} e^{-j2\pi ux/4}=0 \;\text{ for } u=1,2,3,

so F(u,v)=0F(u,v)=0 for every (u,v)(0,0)(u,v)\neq(0,0).

Result:

F = 16 0 0 0
 0 0 0 0
 0 0 0 0
 0 0 0 0

Interpretation: a constant (uniform) image has no variation, so all its energy lies in the DC (zero-frequency) component F(0,0)=16F(0,0)=16 (equal to total intensity = average ×\times MN=1×16MN = 1\times16), with no higher-frequency content.

B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4Short answer5 marks

Discuss about chain code. Write the shape number of the object having chain code {0, 7, 5, 4, 3, 1}.

Chain Code

A chain code is a boundary (contour) representation that encodes a connected sequence of straight-line segments of specified length and direction. Starting from a chosen boundary pixel and tracing the boundary, each move to the next pixel is recorded as a direction number.

  • 4-directional code uses directions 0–3 (E, N, W, S).
  • 8-directional code uses directions 0–7 (counter-clockwise, with diagonals):
 3 2 1
 \ | /
 4 -- P -- 0
 / | \
 5 6 7

Advantages: compact boundary description; disadvantages: depends on the starting point and is not rotation-invariant, which is fixed by the difference code and shape number.

Shape Number of {0, 7, 5, 4, 3, 1}

Step 1 – First difference (number of counter-clockwise direction changes between consecutive codes, treating the code as circular, mod 8):

Differences for {0,7,5,4,3,1}\{0,7,5,4,3,1\} (also wrapping last→first):

transition707-0 wait, compute (nextprev)mod8(\text{next}-\text{prev}) \bmod 8
0→7(70)mod8=7(7-0)\bmod8 = 7
7→5(57)mod8=6(5-7)\bmod8 = 6
5→4(45)mod8=7(4-5)\bmod8 = 7
4→3(34)mod8=7(3-4)\bmod8 = 7
3→1(13)mod8=6(1-3)\bmod8 = 6
1→0 (wrap)(01)mod8=7(0-1)\bmod8 = 7

First difference (rotation-invariant): 7 6 7 7 6 7

Step 2 – Shape number = the first difference of smallest magnitude when treated as a circular sequence (rotate to the smallest possible integer):

Rotations of 7 6 7 7 6 7: the smallest starts with the smaller digit 6: 6 7 6 7 7 7 and 6 7 7 7 6 7. The numerically smallest is

Shape number=676777.\textbf{Shape number} = 6\,7\,6\,7\,7\,7.

Its order (length) is 6.

5Short answer5 marks

How logic operation on binary images can be used for masking and feature detection? Explain.

Logic Operations on Binary Images for Masking & Feature Detection

On binary images (pixels are 0 or 1), the basic logic operations AND, OR, NOT (complement), XOR are applied pixel-by-pixel between an image and a second image or mask. Because they are functionally complete, they let us select, combine, and isolate regions.

Masking (region selection)

A mask is a binary image whose foreground (1) marks the region of interest (ROI).

  • AND with a mask keeps only the pixels where the mask is 1 and forces everything else to 0 — i.e. it extracts/crops the ROI:
g=f  AND  maskg = f \;\text{AND}\; \text{mask}
  • OR is used to insert/merge a region or combine two binary shapes.
  • NOT inverts foreground and background (useful to build the complementary mask).

Example: to keep a circular object, AND the image with a circular mask of 1's; pixels outside the circle (mask = 0) become 0.

Feature Detection

  • XOR (difference): f1f2f_1 \oplus f_2 is 1 only where the two images differ, so XOR between two frames detects change / moving features, and XOR between an object and its eroded version yields the boundary.
  • AND with a template highlights where a pattern matches; combined with morphology (hit-or-miss) it detects specific shapes/corners.
  • AND/OR of bit-planes or thresholded feature maps isolates pixels satisfying multiple conditions simultaneously.

Summary

OperationUse
ANDMask/extract ROI
ORMerge/insert regions
NOTInvert mask/background
XORDetect differences, boundaries, change

Thus logic operations provide a simple, fast mechanism for isolating regions (masking) and for detecting features such as boundaries, changes, and template matches.

6Short answer5 marks

Define pattern and pattern class. How decision theoretic methods minimize the probability of misclassification?

Pattern and Pattern Class

  • A pattern is an arrangement of descriptors (features) that characterise an object. It is usually written as a feature vector x=(x1,x2,,xn)T\mathbf{x}=(x_1,x_2,\dots,x_n)^T in an nn-dimensional feature space (it may also be a string or a tree for structural recognition).
  • A pattern class is a family/category of patterns that share common properties, denoted ω1,ω2,,ωW\omega_1,\omega_2,\dots,\omega_W for WW classes. Pattern recognition assigns each pattern to its correct class.

Decision-Theoretic Methods and Minimizing Misclassification

Decision-theoretic recognition uses decision (discriminant) functions di(x)d_i(\mathbf{x}), one per class. A pattern x\mathbf{x} is assigned to class ωi\omega_i if

di(x)>dj(x)for all ji.d_i(\mathbf{x}) > d_j(\mathbf{x}) \quad \text{for all } j\neq i.

The boundary between classes ii and jj is di(x)dj(x)=0d_i(\mathbf{x})-d_j(\mathbf{x})=0.

Bayes Classifier — minimum-error decision

The probability of misclassification is minimized by the Bayes classifier, which assigns x\mathbf{x} to the class with the largest posterior probability. Using Bayes' rule the decision function becomes

di(x)=p(xωi)P(ωi),d_i(\mathbf{x}) = p(\mathbf{x}\mid\omega_i)\,P(\omega_i),

and we choose ωi\omega_i if p(xωi)P(ωi)>p(xωj)P(ωj)p(\mathbf{x}\mid\omega_i)P(\omega_i) > p(\mathbf{x}\mid\omega_j)P(\omega_j) for all jij\neq i.

Why it is optimal: at every point in feature space the classifier picks the most probable class, so the conditional error at that point is the smallest possible. Integrating over all x\mathbf{x}, the total average error (Bayes risk) is the minimum achievable by any classifier. Equivalently it places the decision boundary exactly where the posteriors of two classes are equal, allocating each region to the class that is more likely there.

(For Gaussian classes this reduces to comparing Mahalanobis distances; equal covariance gives linear, otherwise quadratic, boundaries.)

7Short answer5 marks

Given the following image, detect the edge using magnitude and direction of gradient, using Prewitt operator.

0 30 60

5 32 62

10 38 64

Edge Detection with the Prewitt Operator

Prewitt masks:

Gx=[101101101],Gy=[111000111]G_x=\begin{bmatrix}-1&0&1\\-1&0&1\\-1&0&1\end{bmatrix}, \qquad G_y=\begin{bmatrix}-1&-1&-1\\0&0&0\\1&1&1\end{bmatrix}

(GxG_x detects vertical edges / horizontal change; GyG_y detects horizontal edges / vertical change.)

Input image (3×33\times3):

 0 30 60
 5 32 62
 10 38 64

We compute the gradient at the centre pixel (value 32), where the full mask fits.

Horizontal gradient GxG_x (right column − left column)

Gx=(60+62+64)(0+5+10)=18615=171G_x = (60+62+64) - (0+5+10) = 186 - 15 = \mathbf{171}

Vertical gradient GyG_y (bottom row − top row)

Gy=(10+38+64)(0+30+60)=11290=22G_y = (10+38+64) - (0+30+60) = 112 - 90 = \mathbf{22}

Gradient Magnitude

M=Gx2+Gy2=1712+222=29241+484=29725172.4M=\sqrt{G_x^2+G_y^2}=\sqrt{171^2+22^2}=\sqrt{29241+484}=\sqrt{29725}\approx \mathbf{172.4}

Gradient Direction

θ=tan1 ⁣(GyGx)=tan1 ⁣(22171)7.3\theta=\tan^{-1}\!\left(\frac{G_y}{G_x}\right)=\tan^{-1}\!\left(\frac{22}{171}\right)\approx \mathbf{7.3^{\circ}}

Interpretation

The large magnitude (172\approx172) indicates a strong edge at the centre pixel. Since GxGyG_x \gg G_y, the intensity changes mainly in the horizontal direction (values increase left→right), so the edge is essentially vertical, with the gradient pointing almost horizontally (7\approx7^{\circ} above horizontal).

8Short answer5 marks

What is image segmentation? How do you detect horizontal and vertical line?

Image Segmentation

Image segmentation is the process of partitioning an image into multiple regions (sets of pixels) that are meaningful for analysis, so that pixels within a region are similar and adjacent regions differ. Formally, segmentation divides image RR into regions R1,,RnR_1,\dots,R_n such that:

  1. iRi=R\bigcup_i R_i = R (complete),
  2. each RiR_i is connected,
  3. RiRj=R_i \cap R_j = \varnothing for iji\neq j (disjoint),
  4. a predicate P(Ri)=TRUEP(R_i)=\text{TRUE} (each region is homogeneous), and
  5. P(RiRj)=FALSEP(R_i\cup R_j)=\text{FALSE} for adjacent regions.

Approaches are based on discontinuity (points, lines, edges) or similarity (thresholding, region growing).

Detecting Horizontal and Vertical Lines

Lines are detected by convolving the image with directional line masks that give a strong response when the local pattern matches that orientation.

Horizontal line mask (responds to a row of high values):

 -1 -1 -1
 2 2 2
 -1 -1 -1

Vertical line mask (responds to a column of high values):

 -1 2 -1
 -1 2 -1
 -1 2 -1

Procedure

  1. Convolve the image with the mask to obtain response R(x,y)R(x,y).
  2. The mask coefficients sum to zero, so flat regions give R0R\approx0 and a matching line gives a large R|R|.
  3. Threshold the response: if R(x,y)T|R(x,y)| \ge T, mark the pixel as a line point of that orientation.

Applying the horizontal mask isolates horizontal lines and the vertical mask isolates vertical lines; ±45° masks similarly detect diagonal lines.

9Short answer5 marks

Describe the steps in digital image processing.

Steps (Fundamental Stages) in Digital Image Processing

The standard pipeline (Gonzalez & Woods) proceeds as follows, with the knowledge base guiding every stage:

  1. Image Acquisition — capture the image with a sensor and digitize it (sampling + quantization). May include simple pre-processing like scaling.
  2. Image Enhancement — improve appearance subjectively (e.g. contrast stretch, histogram equalization, sharpening) to highlight features of interest.
  3. Image Restoration — objectively recover a degraded image using a mathematical model of the degradation and noise (e.g. inverse, Wiener filtering).
  4. Colour Image Processing — handle colour models (RGB, HSI, CMYK) and colour-based operations; important for web and multimedia.
  5. Wavelets and Multiresolution Processing — represent the image at multiple resolutions; basis for compression and pyramidal analysis.
  6. Compression — reduce storage/bandwidth by removing redundancy (e.g. JPEG, run-length, Huffman coding).
  7. Morphological Processing — extract structure/shape using set operations (dilation, erosion, opening, closing); bridges processing and analysis.
  8. Segmentation — partition the image into objects/regions (thresholding, edge/region methods). A critical and difficult step.
  9. Representation and Description — represent segmented regions by boundary (chain code) or region (texture) and compute descriptors/features.
  10. Object Recognition — assign a label to an object based on its descriptors (decision-theoretic or structural recognition).

Throughout, the Knowledge Base stores prior information that controls and coordinates the interaction between the modules.

Acquisition -> Enhancement -> Restoration -> Colour -> Wavelets
 -> Compression -> Morphology -> Segmentation
 -> Representation/Description -> Recognition
 (all guided by the Knowledge Base)
10Short answer5 marks

How do you measure distance between pixels? Discuss about region oriented segmentation.

Measuring Distance Between Pixels

For pixels pp at (x,y)(x,y) and qq at (s,t)(s,t), the common distance measures are:

  1. Euclidean distance DeD_e:
De(p,q)=(xs)2+(yt)2D_e(p,q)=\sqrt{(x-s)^2+(y-t)^2}

Points within DerD_e\le r form a disk of radius rr.

  1. City-block / Manhattan distance D4D_4 (moves along 4-neighbours):
D4(p,q)=xs+ytD_4(p,q)=|x-s|+|y-t|

Equidistant points form a diamond.

  1. Chessboard distance D8D_8 (moves along 8-neighbours):
D8(p,q)=max(xs,yt)D_8(p,q)=\max(|x-s|,\,|y-t|)

Equidistant points form a square.

A valid distance metric satisfies D(p,q)0D(p,q)\ge0 (=0 iff p=qp=q), D(p,q)=D(q,p)D(p,q)=D(q,p), and the triangle inequality D(p,q)D(p,z)+D(z,q)D(p,q)\le D(p,z)+D(z,q).

Region-Oriented Segmentation

Region-based methods group pixels directly into regions using a similarity (homogeneity) predicate PP, rather than detecting edges.

1. Region Growing

Start from one or more seed pixels and iteratively append neighbouring pixels that satisfy the predicate (e.g. intensity difference below a threshold). Growth stops when no neighbour qualifies. Pros: produces connected regions; cons: result depends on seed and threshold choice, sensitive to noise.

2. Region Splitting and Merging

Uses a quadtree:

  • Split: if a region is not homogeneous (P(Ri)=FALSEP(R_i)=\text{FALSE}), divide it into four quadrants; repeat recursively.
  • Merge: combine adjacent regions whose union is homogeneous (P(RiRj)=TRUEP(R_i\cup R_j)=\text{TRUE}).

The process stops when no region can be split or merged, yielding a final segmentation. Region methods guarantee connected, closed regions but can be computationally heavy.

11Short answer5 marks

Define clipping and contrast stretching. Compute Hadamard transform of the data sequence {1, 2, 0, 3}.

Clipping and Contrast Stretching

  • Clipping is a point operation that forces all intensities outside a chosen range [r1,r2][r_1,r_2] to fixed limits — values below r1r_1 become 00 (or r1r_1) and values above r2r_2 become L1L-1 (or r2r_2). It is a special case of contrast stretching with the slope set to 0 outside the band, used to suppress unwanted background/noise levels and isolate a feature band.

  • Contrast stretching expands a narrow range of input intensities to span the full available range [0,L1][0,L-1], increasing contrast. Using control points (r1,s1)(r_1,s_1) and (r2,s2)(r_2,s_2), the piecewise-linear transform stretches the mid-range while compressing the dark and bright ends. With s1=0,s2=L1s_1=0,\,s_2=L-1 it maps [r1,r2][0,L1][r_1,r_2]\to[0,L-1]:

s=L1r2r1(rr1).s = \frac{L-1}{r_2-r_1}\,(r-r_1).

Hadamard Transform of {1, 2, 0, 3}

The order-4 (natural-ordered) Hadamard matrix is H4=H2H2H_4=H_2\otimes H_2:

H4=[1111111111111111]H_4=\begin{bmatrix}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{bmatrix}

With input vector f=[1,2,0,3]T\mathbf{f}=[1,\,2,\,0,\,3]^T, the transform is F=H4f\mathbf{F}=H_4\,\mathbf{f}:

F0=1+2+0+3=6F_0 = 1+2+0+3 = 6 F1=12+03=4F_1 = 1-2+0-3 = -4 F2=1+203=0F_2 = 1+2-0-3 = 0 F3=120+3=2F_3 = 1-2-0+3 = 2 F=[6,  4,  0,  2]\boxed{\mathbf{F} = [\,6,\; -4,\; 0,\; 2\,]}

If the normalized form 1NH4f\frac{1}{N}H_4\mathbf{f} (with N=4N=4) is required:

F=[1.5,  1,  0,  0.5].\mathbf{F} = \left[\,1.5,\; -1,\; 0,\; 0.5\,\right].
12Short answer5 marks

Distinguish between forward and inverse transform. Define pixel coding, run length and bit plane.

Forward vs Inverse Transform

AspectForward TransformInverse Transform
DirectionSpatial domain \to transform domainTransform domain \to spatial domain
PurposeAnalysis — decompose image into frequency/basis coefficientsSynthesis — reconstruct the image from coefficients
Example (DFT)F(u,v)=xyf(x,y)ej2π(ux/M+vy/N)F(u,v)=\sum_x\sum_y f(x,y)e^{-j2\pi(ux/M+vy/N)}f(x,y)=1MNuvF(u,v)e+j2π(ux/M+vy/N)f(x,y)=\frac{1}{MN}\sum_u\sum_v F(u,v)e^{+j2\pi(ux/M+vy/N)}
KernelForward kernel ej()e^{-j(\cdot)}Conjugate kernel e+j()e^{+j(\cdot)} (with scaling)
Used inFiltering/compression: get coefficients to modifyRecover processed image after modification

In processing we apply the forward transform, modify coefficients (filter/quantize), then apply the inverse transform to obtain the result.

Pixel Coding, Run Length, Bit Plane

  • Pixel (PCM) coding: the basic representation where each pixel's intensity is coded independently using a fixed number of bits (e.g. 8 bits/pixel). It removes coding redundancy by assigning shorter codes to more frequent gray levels (variable-length / Huffman coding) instead of equal-length codes.

  • Run-length encoding (RLE): a lossless method that replaces sequences (runs) of identical pixels with a (value, run-length) pair. Highly effective for images with long uniform stretches (binary/fax images). Example: 0 0 0 0 1 1 → (0,4)(1,2).

  • Bit plane (slicing): an kk-bit image is decomposed into kk binary planes, one per bit position. The most significant bit-plane holds the major structure, while lower planes hold fine detail/noise. Coding only the high-order planes (often with RLE) achieves compression; bit-plane slicing also aids analysis and watermarking.

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