02

CSS 433 Computer Vision — Comprehensive Lecture & Exercise Guide

Instructor: Prof. Toshiaki Kondo (SIIT, Thammasat University)
Topic: Review of Acquisition Optics & Chapter 2: Digital Image Fundamentals

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Part 0: Review — The Camera Exposure Plane (Slide 2)

Slide 2 establishes the physical trade-offs in camera acquisition through a two-axis exposure coordinate plane:

• Vertical Axis (Aperture / F-number): Ranges from Large Aperture ($f/2,f/2.8,f/4,f/5.6$) at the top to Small Aperture ($f/8,f/11,f/16,f/22,f/32$) at the bottom.
• Horizontal Axis (Shutter Speed): Ranges from Fast Shutter ($1/1000\text{ s},1/500\text{ s},1/250\text{ s},1/125\text{ s},1/60\text{ s}$) on the left to Slow Shutter ($1/30\text{ s},1/15\text{ s},1/8\text{ s},1/4\text{ s},1/2\text{ s},1^{\prime \prime }$) on the right.

                         Large Aperture (f/2)
                         [Shallow DOF / More Light]
                                   ^
                     Quadrant A    |    Quadrant B
                   (Fast & Wide)   |   (Slow & Wide)
                                   |   ===> BRIGHT
       Fast Shutter <--------------+--------------> Slow Shutter
       [Frozen Motion]             |                [Motion Blur]
                     Quadrant D    |    Quadrant C
                   (Fast & Narrow) |   (Slow & Narrow)
                                   |
                                   v
                         Small Aperture (f/32)
                         [Deep DOF / Less Light]

Analysis of the Four Operating Quadrants:

1. Quadrant A (Large Aperture + Fast Shutter):
   • Optical Effect: Shallow Depth of Field (DOF) with Frozen Motion.
   • Visual Characteristic: Crisp foreground subject, heavily blurred background (bokeh), and high-speed motion captured without streak artifacts.
2. Quadrant B (Large Aperture + Slow Shutter):
   • Optical Effect: Maximum light collection (Bright / Overexposed) with severe Motion Blur and Shallow DOF.
3. Quadrant C (Small Aperture + Slow Shutter):
   • Optical Effect: Deep Depth of Field (Deep DOF) with Blurred Motion.
   • Visual Characteristic: Landscape photography look; both near and far objects are in focus, but moving elements (e.g., water, moving cars) are smoothed into motion trails.
4. Quadrant D (Small Aperture + Fast Shutter):
   • Optical Effect: Severely photon-starved (Dark / Underexposed) with Frozen Motion and Deep DOF.

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Part 1: Digitization — Spatial Sampling & Quantization (Slides 3–12)

Converting a continuous real-world scene $f(x,y)$ into a digital image $I[r,c]$ requires two distinct discretization steps:

Continuous Image f(x, y)
       │
       ▼  [Step 1: Spatial Sampling]  ---> Discretizes Coordinates (x, y) into (r, c)
Discrete Lattice Grid
       │
       ▼  [Step 2: Quantization]     ---> Discretizes Amplitude / Brightness into L levels
Digital Matrix I[r, c]

1. Spatial Sampling (Slides 4–6, 8–11)

• Definition: Discretization of the continuous spatial spatial domain into an $M\times N$ array of discrete grid elements (pixels or pels).
• Sensor Resolution: Determined by the physical sensor fabrication (e.g., Canon EOS 5DS: $8688\times 5792\approx 50.3\text{ Megapixels}$).
• Image as a Matrix: An image is a numerical 2D array: ${\mathbf{A}}_{2\times 2}=\left[\begin{array}{cc}0& 64\\ 128& 255\end{array}\right]$
  • A value of $0$ corresponds to pure black; $255$ represents maximum brightness (white).
• Coordinate Systems (Slide 6):
  • 0-indexed: Origin $(0,0)$ is at the top-left; rows $r\in [0,M-1]$, columns $c\in [0,N-1]$ (used in C/C++, Python, OpenCV).
  • 1-indexed: Origin $(1,1)$ is at the top-left; rows $r\in [1,M]$, columns $c\in [1,N]$ (used in MATLAB).
• Sampling Defect — Block Artifacts (Slide 11):
  • When an image is subsampled to an insufficient spatial resolution (e.g., downsampled from $1024\times 1024$ to $32\times 32$), the continuous contours break into visible square pixel blocks upon display.

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2. Quantization (Slides 7, 12)

• Definition: Discretization of continuous light amplitude (radiance) into a finite set of $L$ integer intensity levels.
• Hardware: Performed by an Analog-to-Digital Converter (ADC). In modern CMOS sensors, ADCs are placed on each column of the pixel array.
• Bit Depth ($b$): $L=2^b$
  • Monochromatic (Grayscale): 8 bits ($1\text{ Byte}$) $⟹2^8=256$ intensity levels, $I\in [0,255]$.
  • True-Color RGB: 3 channels $\times$ 8 bits = 24 bits $⟹2^{24}=16,777,216$ colors.
• Quantization Defect — False Contouring (Slide 12):
  • When the number of quantization levels is severely restricted (e.g., dropping from 256 levels to 16, 8, 4, or 2 levels), smooth, continuous luminance gradients degrade into distinct, step-like bands known as false contours.

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Fully Worked Exercise 1: Size of a Video Sequence (Slide 13)

Problem Statement:

Let a full-color video sequence have $30\text{ fps}$ (frames per second). A full-color image frame consists of 3 color channels (Red, Green, Blue). Each channel comprises $640\times 480$ pixels with 8-bit intensity levels.

1. Find the amount of uncompressed video data generated per second in both bits/sec and Bytes/sec.
2. How long can a standard single-layer DVD ($4.7\text{ GB}$ storage capacity) store this raw uncompressed video stream?

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Step-by-Step Solution:

Step 1: Calculate the data rate per second.

• Number of pixels per single color channel: $N_{\text{pixels}}=640\times 480=307,200\text{ pixels/frame}$
• Total bits per frame across 3 channels ($8\text{ bits/channel}$): $\text{Bits per frame}=640\times 480\times 3\times 8=7,372,800\text{ bits}$
• Total data rate in bits per second: ${\text{Data Rate}}_{\text{bits}}=640\times 480\times 3\times 8\times 30=221,184,000\text{ bits/sec}$
• Total data rate in Bytes per second (dividing by $8\text{ bits/Byte}$): ${\text{Data Rate}}_{\text{Bytes}}=640\times 480\times 3\times 30=27,648,000\text{ Bytes/sec}\approx 27.65\text{ MB/s}$

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Step 2: Calculate storage duration on a 4.7 GB DVD. Commercial storage media manufacturers define $1\text{ GB}=10^9\text{ Bytes}$.
Therefore, the DVD capacity is: $\text{Capacity}=4.7\text{ GB}=4,700,000,000\text{ Bytes}$

Dividing total capacity by the Byte rate per second: $T=\frac{4,700,000,000\text{ Bytes}}{27,648,000\text{ Bytes/sec}}\approx 169.9942\text{ seconds}$

Converting to minutes and seconds: $T\approx 169.99\text{ seconds}\approx 2\text{ minutes and }50\text{ seconds}$

(Note: If computed using binary gibibytes where $1\text{ GiB}=2^{30}\text{ Bytes}$, $4.7\times 1,073,741,824/27,648,000\approx 182.53\text{ s}$, but storage media specs universally follow the $10^9$ decimal standard shown in the slide’s handwritten notes).

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Part 2: The Sampling Theorem, Aliasing & Slit Animation (Slides 14–24, 44–46, 47–48)

1. The Nyquist-Shannon Sampling Theorem (Slide 16)

To accurately reconstruct an analog signal of highest frequency $f_{\text{input}}$ without aliasing distortion, the sampling frequency $f_{\text{sample}}$ must be at least twice the maximum signal frequency:

$f_{\text{sample}}\ge 2f_{\text{input}}⟺\frac{f_{\text{sample}}}{2}\ge f_{\text{input}}⟺f_{\text{Nyquist}}\ge f_{\text{input}}$

Where $f_{\text{Nyquist}}=\frac{f_{\text{sample}}}{2}$ is the Nyquist frequency.

Two Engineering Solutions to Prevent Aliasing:

1. Increase $f_{\text{sample}}$: Use a higher-density sensor array or higher temporal frame rate.
2. Apply an Anti-Aliasing Filter: Pass the input through an optical low-pass filter (OLPF) prior to sampling to eliminate frequencies above $f_{\text{Nyquist}}$.

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Fully Worked Exercise 2: The Sampling Theorem (Slide 17)

Problem Statement:

To represent the continuous signal $y$ without aliasing: $y=A\cos (Bx+C)+D$ What is the minimum sampling frequency? Express it using pertinent variables from $A,B,C$, and $D$.

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Step-by-Step Solution:

The standard mathematical form of a sinusoidal wave as a function of continuous coordinate $x$ is: $y=A\cos (2\pi f_{\text{input}}x+C)+D$ Comparing this to the given expression $y=A\cos (Bx+C)+D$:

• The angular frequency is $\omega =B$.
• The cyclical frequency of the input signal is: $2\pi f_{\text{input}}=B⟹f_{\text{input}}=\frac{B}{2\pi }$
• According to the Nyquist criterion, the minimum sampling frequency $f_{\text{sample, min}}$ must satisfy: $f_{\text{sample, min}}=2f_{\text{input}}=2\left(\frac{B}{2\pi }\right)=\frac{B}{\pi }$

Final Answer: $f_{\text{sample, min}}=\frac{B}{\pi }$ (Variables $A$ [amplitude], $C$ [phase angle], and $D$ [DC bias offset] affect only signal magnitude and offset; they have no bearing on frequency or required sampling rate).

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2. Mathematical Prediction of Aliasing Frequencies (Slides 18, 20, 48)

When a signal exceeds the Nyquist limit ($f_{\text{input}}>f_{\text{sample}}/2$), it folds back into the baseband $[0,f_{\text{sample}}/2]$ as an erroneous ghost frequency $f_{\text{aliasing}}$.

Slides 20 and 48 provide two analytical formulations to compute $f_{\text{aliasing}}$:

               Nyquist Limit (fs/2)
                      |
    0                 |                 fs               2fs
    +-----------------+-----------------+-----------------+-----> Frequency
    |    PASSBAND     |
    |  (No Aliasing)  |  Folds back     | Folds forward   |
    |                 |<--- (fs - r)    |<--- (r)         |

Method 1: The Modulo / Remainder Formulation (Slide 20, 48)

1. Calculate the remainder $r$ of the input frequency with respect to the sampling rate: $r=\text{mod}(f_{\text{input}},f_{\text{sample}})$
2. The alias frequency is the minimum distance to the nearest sampling harmonic (the smaller of $r$ and $f_{\text{sample}}-r$): $f_{\text{aliasing}}=\min (r,f_{\text{sample}}-r)$

Method 2: The Nearest Multiple Formulation (Slide 48)

1. Find the integer multiple index $n$ nearest to the ratio: $n=\text{round}\left(\frac{f_{\text{input}}}{f_{\text{sample}}}\right)$
2. Calculate the folded absolute difference: $f_{\text{aliasing}}=|f_{\text{input}}-n\cdot f_{\text{sample}}|$

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Fully Worked Exercise 3: Aliasing Errors (Slides 19–23)

Problem 1:

Given $f_{\text{sample}}=4\text{ Hz}$.
The Nyquist frequency is $f_{\text{Nyquist}}=4/2=2\text{ Hz}$.
Predict the alias frequency for $f_{\text{input}}=3\text{ Hz}$ and $f_{\text{input}}=4\text{ Hz}$.

• For $f_{\text{input}}=3\text{ Hz}$:
  • Method 1: $r=\text{mod}(3,4)=3$.
    $f_{\text{aliasing}}=\min (3,4-3)=\min (3,1)=1\text{ Hz}$.
  • Method 2: $n=\text{round}(3/4)=1$.
    $f_{\text{aliasing}}=|3-1\times 4|=|-1|=1\text{ Hz}$.
• For $f_{\text{input}}=4\text{ Hz}$:
  • Method 1: $r=\text{mod}(4,4)=0$.
    $f_{\text{aliasing}}=\min (0,4-0)=0\text{ Hz}$ (appears as a constant DC flatline).
  • Method 2: $n=\text{round}(4/4)=1$.
    $f_{\text{aliasing}}=|4-1\times 4|=0\text{ Hz}$.

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Problem 2:

Given $f_{\text{sample}}=8\text{ Hz}$.
The Nyquist frequency is $f_{\text{Nyquist}}=8/2=4\text{ Hz}$.
Predict $f_{\text{aliasing}}$ for inputs $f_{\text{input}}\in \{5,6,7,8,9,10,11,12,13,14\}\text{ Hz}$:

$f_{\text{input}}$ (Hz)	$r=\text{mod}(f_{\text{input}},8)$	$f_{\text{sample}}-r=8-r$	$n=\text{round}(f_{\text{in}}/8)$	$f_{\text{aliasing}} =	f_{\text{in}} - 8n	$	State
5	5	3	1	$	5 - 8	= \mathbf{3\text{ Hz}}$	Aliased
6	6	2	1	$	6 - 8	= \mathbf{2\text{ Hz}}$	Aliased
7	7	1	1	$	7 - 8	= \mathbf{1\text{ Hz}}$	Aliased
8	0	8	1	$	8 - 8	= \mathbf{0\text{ Hz}}$	DC Constant
9	1	7	1	$	9 - 8	= \mathbf{1\text{ Hz}}$	Aliased
10	2	6	1	$	10 - 8	= \mathbf{2\text{ Hz}}$	Aliased
11	3	5	1	$	11 - 8	= \mathbf{3\text{ Hz}}$	Aliased
12	4	4	1 or 2	$	12 - 8	= \mathbf{4\text{ Hz}}$	At Nyquist boundary
13	5	3	2	$	13 - 16	= \mathbf{3\text{ Hz}}$	Aliased
14	6	2	2	$	14 - 16	= \mathbf{2\text{ Hz}}$	Aliased

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Fully Worked Today’s Quiz: Part 1 (Slides 47–48)

Problem Statement:

Given $f_{\text{input}}=4700\text{ Hz}$ and $f_{\text{sample}}=1000\text{ Hz}$.
Find $f_{\text{aliasing}}$ using both Solution (i) and Solution (ii).

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Solution:

• Solution (i) — Modulo / Remainder Method: $r=\text{mod}(f_{\text{input}},f_{\text{sample}})=\text{mod}(4700,1000)=700\text{ Hz}$ $f_{\text{sample}}-r=1000-700=300\text{ Hz}$ $f_{\text{aliasing}}=\min (r,f_{\text{sample}}-r)=\min (700,300)=300\text{ Hz}$

• Solution (ii) — Nearest Multiple Method: $\frac{f_{\text{input}}}{f_{\text{sample}}}=\frac{4700}{1000}=4.7$ $n=\text{round}(4.7)=5$ $f_{\text{aliasing}}=|f_{\text{input}}-n\cdot f_{\text{sample}}|=|4700-5\times 1000|=|4700-5000|=|-300|=300\text{ Hz}$

Result: Both formulations yield $f_{\text{aliasing}}=300\text{ Hz}$.

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3. Slit Animation / Moiré Animation (Slides 24, 44–46)

While aliasing is typically an unwanted defect in computer vision, Moiré animation uses it systematically to generate motion illusions from static printed sheets:

1. Prepare $n$ frames representing a cyclic sequence of motion (e.g., $n=3$ frames of a shifting square, or $n=6$ frames of a galloping horse).
2. Interleave the frames: Construct a single composite image by interleaving vertical slices: column 1 from frame 1, column 2 from frame 2, …, column $n$ from frame $n$, then repeating cyclically.
3. Construct a sampling mask: Create an opaque physical barrier sheet with transparent vertical slits having a repetition period of $n$ columns (width of transparent slit = 1 column, opaque barrier = $n-1$ columns).
4. Slide the mask: As the barrier translates horizontally by $\Delta x=1\text{ pixel}$, it alternately unmasks frame 1, frame 2, …, frame $n$, producing smooth motion due to spatial subsampling.

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Part 3: Bit-Plane Decomposition & Run-Length Encoding (Slides 25–30, 47)

1. Bit-Plane Decomposition

Any 8-bit monochromatic image pixel can be represented as an 8-bit binary polynomial: $I(x,y)=b_7\cdot 2^7+b_6\cdot 2^6+b_5\cdot 2^5+b_4\cdot 2^4+b_3\cdot 2^3+b_2\cdot 2^2+b_1\cdot 2^1+b_0\cdot 2^0$ Decomposing an image creates 8 binary matrices (Bit-planes 0 to 7):

• Bit-plane 7 (MSB, weight 128): Contains the primary structural layout, global contrast, and boundaries.
• Bit-plane 0 (LSB, weight 1): Appears as visually imperceptible random noise, making it suitable for steganography and fragile watermarking.

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Fully Worked Exercise 4: Bit-Plane Decomposition (Slide 26)

Problem Statement:

Perform bit-plane decomposition on the $2\times 2$ grayscale image: $\mathbf{I}=\left[\begin{array}{cc}8& 57\\ 215& 120\end{array}\right]$ What will happen to this image if the least significant bit (LSB) plane is discarded?

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Step-by-Step Solution:

Step 1: Convert all matrix entries to 8-bit unsigned binary ($b_7{b}_6{b}_5{b}_4{b}_3{b}_2{b}_1{b}_0$).

• For 8: $8=0\cdot 128+0\cdot 64+0\cdot 32+0\cdot 16+1\cdot 8+0\cdot 4+0\cdot 2+0\cdot 1⟹00001000_2$
• For 57: $57=32+16+8+1⟹00111001_2$
• For 215: $215=128+64+16+4+2+1⟹11010111_2$ (Slide 27 shows conversions via base-8 octal: $215_{10}=327_8\to 011010111_2$, and base-16 hex: $215_{10}=\text{D}{7}_{16}\to 11010111_2$).
• For 120: $120=64+32+16+8⟹01111000_2$

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Step 2: Construct the 8 binary bit-plane matrices. Writing $\mathbf{I}={\sum }_{k=0}^7{\mathbf{P}}_k\cdot 2^k$:

${\mathbf{P}}_7(\text{MSB}\times 128)=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$

${\mathbf{P}}_6(\text{7th}\times 64)=\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]$

${\mathbf{P}}_5(\text{6th}\times 32)=\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]$

${\mathbf{P}}_4(\text{5th}\times 16)=\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$

${\mathbf{P}}_3(\text{4th}\times 8)=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

${\mathbf{P}}_2(\text{3rd}\times 4)=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$

${\mathbf{P}}_1(\text{2nd}\times 2)=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$

${\mathbf{P}}_0(\text{LSB}\times 1)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

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Step 3: Discard the LSB Plane (${\mathbf{P}}_0$). Setting the LSB to 0 (or subtracting ${\mathbf{P}}_0\times 1$):

• $8⟹00001000_2=8$
• $57⟹00111000_2=57-1=56$
• $215⟹11010110_2=215-1=214$
• $120⟹01111000_2=120$

The reconstructed image becomes: ${\mathbf{I}}_{\text{no LSB}}=\left[\begin{array}{cc}8& 56\\ 214& 120\end{array}\right]$ Conclusion: Discarding the LSB changes pixel intensities by at most $1$ unit. This change is imperceptible to the human eye.

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Fully Worked Today’s Quiz: Part 2 (Slide 47)

Problem Statement:

Perform bit-plane decomposition on the matrix: $\mathbf{Q}=\left[\begin{array}{cc}212& 105\\ 55& 176\end{array}\right]$

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Step-by-Step Solution:

Step 1: Convert each entry to 8-bit binary.

• 212:
  $212-128=84$
  $84-64=20$
  $20-16=4$
  $4-4=0$
  $212=11010100_2$
• 105:
  $105-64=41$
  $41-32=9$
  $9-8=1$
  $1-1=0$
  $105=01101001_2$
• 55:
  $55-32=23$
  $23-16=7$
  $7-4=3$
  $3-2=1$
  $1-1=0$
  $55=00110111_2$
• 176:
  $176-128=48$
  $48-32=16$
  $16-16=0$
  $176=10110000_2$

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Step 2: Construct the decomposition sum. $\mathbf{Q}={\mathbf{B}}_7\cdot 128+{\mathbf{B}}_6\cdot 64+{\mathbf{B}}_5\cdot 32+{\mathbf{B}}_4\cdot 16+{\mathbf{B}}_3\cdot 8+{\mathbf{B}}_2\cdot 4+{\mathbf{B}}_1\cdot 2+{\mathbf{B}}_0\cdot 1$

Matching bits for each plane:

${\mathbf{B}}_7(\text{MSB})=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

${\mathbf{B}}_6(\text{7th})=\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$

${\mathbf{B}}_5(\text{6th})=\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$

${\mathbf{B}}_4(\text{5th})=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

${\mathbf{B}}_3(\text{4th})=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

${\mathbf{B}}_2(\text{3rd})=\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$

${\mathbf{B}}_1(\text{2nd})=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$

${\mathbf{B}}_0(\text{LSB})=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

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2. Run-Length Encoding (RLE) & Compression Metrics (Slides 28–30)

Run-Length Encoding (RLE) is a lossless data compression technique that replaces consecutive repeated symbols (runs) with a single counter value.

Example from Slide 28:

Given binary sequence: $\text{"}\underset{9\text{ zeros}}{\underbrace{000000000}}\underset{8\text{ ones}}{\underbrace{11111111}}\underset{5\text{ zeros}}{\underbrace{00000}}\underset{10\text{ ones}}{\underbrace{1111111111}}\underset{3\text{ zeros}}{\underbrace{000}}\text{"}$ Alternative valid RLE formats:

1. Pair format: $(0,9)(1,8)(0,5)(1,10)(0,3)$
2. Alternating run-count format (assuming sequence starts with 0): $(9)(8)(5)(10)(3)$
3. Transition coordinate format: $(10,8)(23,10)$, storing the 1-based start positions and run lengths of the ones.

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Detailed Breakdown of Compression Ratio Example (Slide 30):

A binary image has dimensions: $M=1024\text{ columns},N=343\text{ rows}$ After RLE, the entire image is reduced to $12,166\text{ runs}$.

• Why are 11 bits necessary for each run?
  • The maximum possible length of a continuous run across a row is the full image width: $1024\text{ pixels}$.
  • A 10-bit integer can only index $2^{10}=1024$ states (values $0$ to $1023$).
  • To represent a maximum run of $1024$ without ambiguity (or values in range $[1,1024]$), the required bit allocation is: $b_{\text{run}}=\lceil {\log }_2(1024+1)\rceil =11\text{ bits}$
• Original uncompressed bit count: $\text{Original Bits}=1024\times 343\times 1\text{ bit}=351,232\text{ bits}$
• Compressed code bit count: $\text{Code Bits}=12,166\text{ runs}\times 11\text{ bits/run}=133,826\text{ bits}$
• Compression Ratio ($C_R$): $C_R=\frac{\text{Original Bits}}{\text{Code Bits}}=\frac{351,232}{133,826}\approx 2.62$
• Relative Data Redundancy ($R_D$): $R_D=1-\frac{1}{C_R}=1-\frac{1}{2.6246}\approx 1-0.381=0.62(62\%)$ (This indicates that $62\%$ of the original data was redundant).

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Part 4: Digital Watermarking (Slides 31–41)

Digital watermarking embeds verification data into an image for copyright management, tampering detection, and content authentication.

1. Visible Spatial-Domain Watermarking (Slides 33–34)

Blends a visible watermark pattern $w(x,y)$ over an image $f(x,y)$ using a linear transparency factor $\alpha \in [0,1]$: $f_w(x,y)=(1-\alpha )f(x,y)+\alpha w(x,y)$

• When $\alpha =0.3$, the watermark is semi-transparent ($70\%$ underlying image, $30\%$ watermark pattern).

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2. Invisible LSB Spatial-Domain Watermarking (Slides 35–39)

Hides information in the least significant bits of the carrier image.

The 2-LSB Insertion Formula (Slide 35):

$f_w=2^2\times \lfloor \frac{f}{2^2}\rfloor +\lfloor \frac{w}{2^6}\rfloor$ This operation works as follows:

• Dividing $f$ by $2^2=4$, taking the floor, and multiplying by $4$ clears (zeros out) the two least significant bits of $f$: $2^2\times \lfloor \frac{f}{2^2}\rfloor \equiv f\&11111100_2(\text{Bitwise AND with }252)$
• Dividing $w$ by $2^6=64$ and taking the floor isolates its two most significant bits ($b_7,b_6$) and shifts them to positions $b_1,b_0$: $\lfloor \frac{w}{2^6}\rfloor \equiv w\gg 6$
• The sum places the two MSBs of $w$ directly into the two vacated LSB positions of $f$.

Extraction Formula:

To retrieve the embedded watermark $\hat{w}$: $\hat{w}=(f_w\&00000011_2)\times 64$

Fragility and Compression Vulnerability (Slide 38):

Spatial LSB watermarks are fragile. When an image undergoes JPEG lossy compression, the Discrete Cosine Transform (DCT) combined with high-frequency quantization discards subtle spatial variations. The reconstructed LSBs are corrupted (RMS error $\approx 2.1\text{ bits}$), rendering the extracted watermark illegible.

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Fully Worked Exercise 5: Watermarking in 2 LSBs (Slides 36–37)

Problem Statement:

Using the equation $f_w=2^2\times \lfloor f/2^2\rfloor +\lfloor w/2^6\rfloor$, embed watermark $w$ into carrier image $f$: $f=\left[\begin{array}{cc}8& 57\\ 215& 120\end{array}\right],w=\left[\begin{array}{cc}206& 173\\ 68& 39\end{array}\right]$

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Step-by-Step Solution:

Step 1: Compute the modified carrier terms $4\times \lfloor f/4\rfloor$ (zeroing the 2 LSBs).

• $f(1,1)=8$:
  $\lfloor 8/4\rfloor =2⟹2\times 4=8$
  (Binary: $000010{00}_2$)
• $f(1,2)=57$:
  $\lfloor 57/4\rfloor =14⟹14\times 4=56$
  (Binary: $001110{00}_2$)
• $f(2,1)=215$:
  $\lfloor 215/4\rfloor =53⟹53\times 4=212$
  (Binary: $110101{00}_2$)
• $f(2,2)=120$:
  $\lfloor 120/4\rfloor =30⟹30\times 4=120$
  (Binary: $011110{00}_2$)

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Step 2: Compute the watermark terms $\lfloor w/64\rfloor$ (extracting the 2 MSBs).

• $w(1,1)=206$:
  $206=11001110_2⟹\lfloor 206/64\rfloor =3$
  (Binary: ${11}_2$)
• $w(1,2)=173$:
  $173=10101101_2⟹\lfloor 173/64\rfloor =2$
  (Binary: ${10}_2$)
• $w(2,1)=68$:
  $68=01000100_2⟹\lfloor 68/64\rfloor =1$
  (Binary: ${01}_2$)
• $w(2,2)=39$:
  $39=00100111_2⟹\lfloor 39/64\rfloor =0$
  (Binary: ${00}_2$)

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Step 3: Sum the two components to form $f_w$.

• Top-Left: $8+3=11$
• Top-Right: $56+2=58$
• Bottom-Left: $212+1=213$
• Bottom-Right: $120+0=120$

$f_w=\left[\begin{array}{cc}11& 58\\ 213& 120\end{array}\right]$ (This matches the professor’s handwritten result on Slide 37).

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Fully Worked Exercise 6: Spatial-Domain Watermarking (Slide 40)

Problem Statement:

An alphabet letter is embedded into the LSB plane of the following $8\times 8$ image. Retrieve and identify the letter.

$${\mathbf{I}}_{8\times 8}=\left[\begin{array}{cccccccc}4& 8& 12& 16& 2& 0& 10& 4\\ 12& 5& 17& 15& 15& 9& 7& 6\\ 14& 13& 9& 7& 17& 7& 7& 6\\ 6& 8& 4& 9& 13& 0& 12& 8\\ 10& 20& 6& 11& 11& 2& 12& 0\\ 12& 2& 14& 5& 7& 4& 10& 10\\ 8& 4& 8& 9& 17& 14& 8& 12\\ 0& 4& 2& 12& 6& 6& 10& 8\end{array}\right]$$

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Step-by-Step Solution:

To retrieve the LSB plane, compute $\text{LSB}(x,y)=I(x,y)(\mathrm{mod}2)$:

• If the pixel value is even, $\text{LSB}=0$.
• If the pixel value is odd, $\text{LSB}=1$.

Row-by-Row Parity Check:

• Row 1: $[4,8,12,16,2,0,10,4]⟹\text{all even}⟹[0,0,0,0,0,0,0,0]$
• Row 2: $[12,5,17,15,15,9,7,6]⟹[0,1,1,1,1,1,1,0]$
• Row 3: $[14,13,9,7,17,7,7,6]⟹[0,1,1,1,1,1,1,0]$
• Row 4: $[6,8,4,9,13,0,12,8]⟹[0,0,0,1,1,0,0,0]$
• Row 5: $[10,20,6,11,11,2,12,0]⟹[0,0,0,1,1,0,0,0]$
• Row 6: $[12,2,14,5,7,4,10,10]⟹[0,0,0,1,1,0,0,0]$
• Row 7: $[8,4,8,9,17,14,8,12]⟹[0,0,0,1,1,0,0,0]$
• Row 8: $[0,4,2,12,6,6,10,8]⟹\text{all even}⟹[0,0,0,0,0,0,0,0]$

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Reconstructed Binary Matrix & Pattern Visualization:

$$\mathbf{LSB}=\left[\begin{array}{cccccccc}0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 1& 1& 1& 1& 1& 0\\ 0& 1& 1& 1& 1& 1& 1& 0\\ 0& 0& 0& 1& 1& 0& 0& 0\\ 0& 0& 0& 1& 1& 0& 0& 0\\ 0& 0& 0& 1& 1& 0& 0& 0\\ 0& 0& 0& 1& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]$$

Rendering $1$ as $■$ and $0$ as $\cdot$:

.  .  .  .  .  .  .  .
.  █  █  █  █  █  █  .   <-- Horizontal bar of letter "T"
.  █  █  █  █  █  █  .   <-- Double-thickness horizontal bar
.  .  .  █  █  .  .  .   <-- Vertical stem of letter "T"
.  .  .  █  █  .  .  .   <-- Vertical stem
.  .  .  █  █  .  .  .   <-- Vertical stem
.  .  .  █  █  .  .  .   <-- Vertical stem
.  .  .  .  .  .  .  .

Final Answer: The recovered letter is T.