Every Fall 2025 quiz question, answered

Discuss logarithmic transformations for image enhancement.

Definition. The log transformation is a basic intensity transformation:

r = input gray level, s = output gray level, c = a constant, r ≥ 0. The 1 is added so that log 0 never occurs. To map 0…255 onto 0…255, c = 255 / log(256) = 105.89.

Shape. Draw this graph in the answer script:

The curve lies above the diagonal. It is steep for small r and flat for large r.

Properties.

• For low input values the range is expanded: dark regions are spread out and their details become visible.
• For high input values the range is compressed: bright regions are squeezed together.

Worked values (c = 105.89):

The input band 0–50 (20% of the range) becomes 0–181 (71%). The band 100–255 shrinks to 212–255.

Application. Use it when the input has a very large dynamic range that the display cannot show. The standard example is the Fourier spectrum, whose values run from 0 to about 1.5×10⁶. On a display only the bright centre shows; after the log transform the whole spectrum is visible. That is why it is called dynamic-range compression.

Inverse log, s = 10^r/c − 1, does the opposite: it expands the bright values and compresses the dark ones.

A medical X-ray image (8-bit) needs its bone structures in the range [120, 200] enhanced. Apply gray-level slicing with both approaches. (i) Write the transformation functions [4] (ii) Show the mapping for pixel values 100, 150 and 220 [4] (iii) Sketch the transformation curves [3].

Given: 8-bit, so L − 1 = 255. Band of interest A = 120, B = 200. Highlight level = 255.

(i) Transformation functions

Approach 1, without background (other levels suppressed):

Approach 2, with background (other levels preserved):

(ii) Mapping

r = 100 is below the band: approach 1 gives 0, approach 2 keeps 100. r = 150 is inside: both give 255. r = 220 is above: approach 1 gives 0, approach 2 keeps 220.

(iii) Curves

Result: approach 1 shows only the bones, white on black. Approach 2 shows the bones in white with the rest of the X-ray still visible.

How many shades of gray are possible in a 10-bit grayscale image? Show the numeric range. Write the expression for intensity resolution.

Intensity resolution = number of gray levels a pixel can take. With k bits per pixel:

Here k = 10, so L = 2¹⁰ = 1024 shades.

Range: 0 to L − 1 = 0 to 1023 (0 = black, 1023 = white).

How do you generate an image histogram? What does an image histogram say? [3+3]

Generate: (1) find the number of gray levels L = 2^k and make one bin per level 0 … L−1; (2) scan the image and, for each pixel with value r_k, add 1 to bin h(r_k); (3) plot n_k against r_k as bars. Σ h = total number of pixels.

Example (slide 7, 3-bit 4×4): h = [1, 1, 3, 5, 1, 2, 1, 2].

What it says: how many pixels have each intensity, i.e. the tonal distribution. From its shape we can tell whether the image is dark, bright, low- or high-contrast; judge the overall quality and the possibility of enhancement; check whether the image was scanned properly; and choose a threshold for object detection and segmentation. It gives only statistical information and no indication of pixel locations.

Explain the four basic types of intensity characteristics of an image and their corresponding histograms.

The histogram of a digital image with gray levels 0 … L−1 is h(r_k) = n_k, the number of pixels with gray level r_k. Its shape shows the four basic types:

1. Dark image: most pixels have low values, so the histogram is crowded on the left (dark) side. The image looks dark and details in the shadows are hidden.
2. Bright image: most pixels have high values, so the histogram is crowded on the right side. The image looks washed out.
3. Low-contrast image: the pixels occupy a narrow range of gray levels in the middle, so the histogram is narrow. The image looks dull and grayish.
4. High-contrast image: the pixels cover almost the whole range, and the histogram is spread fairly evenly. The image shows a lot of gray-level detail and looks sharp.

Histogram equalization tries to turn types 1–3 into type 4.

A 3-bit 8×8 image has histogram counts {{0:2, 1:3, 2:5, 3:6, 4:9, 5:12, 6:14, 7:13}}. (i) Sketch the normalized histogram (ii) compute and sketch the equalized histogram (iii) sketch the transformation function curve.

Given: 3-bit image, so L = 2³ = 8 and L − 1 = 7. Total pixels N = 64.

Step 1, histogram h(r_k) = n_k: count the pixels at each gray level.

Check: Σ n_k = 2 + 3 + 5 + 6 + 9 + 12 + 14 + 13 = 64.

Steps 2–5. PDF p(r_k) = n_k/N; CDF = running sum of the PDF; multiply by (L − 1) = 7; round off (.5 goes up).

Mapping the new gray levels into number of pixels:

Equalized histogram: add the n_k that land on the same s_k. Levels not listed get 0 pixels.

What is the histogram of an image? Two 20×20 images have equal numbers of black (0) and white (1) pixels. Are their histograms identical? Explain.

Histogram. For a digital image with gray levels in [0, L−1], the histogram is the discrete function

where n_k is the number of pixels with gray level r_k. It is drawn as bars of n_k against r_k. The normalized histogram is p(r_k) = n_k / MN.

The two images. Each has 20 × 20 = 400 pixels, and equal numbers of black and white, so 200 black and 200 white. For both images:

Yes, the histograms are identical, even if one image is half black and half white and the other is a checkerboard. A histogram records only how many pixels have each value, not where they are. It has no spatial information.

So different images can have the same histogram, and an image cannot be rebuilt from its histogram.

A 4×5, 3-bit image [1,2,4,7,0; 2,1,3,1,5; 5,2,1,4,1; 7,6,3,1,4]. (i) Draw the histogram (ii) sketch the normalized histogram (iii) compute and sketch the equalized histogram (iv) sketch the transformation function curve (v) show the equalized output image.

Given: 3-bit image, so L = 2³ = 8 and L − 1 = 7. Total pixels N = 20.

Step 1, histogram h(r_k) = n_k: count the pixels at each gray level.

Check: Σ n_k = 1 + 6 + 3 + 2 + 3 + 2 + 1 + 2 = 20.

Steps 2–5. PDF p(r_k) = n_k/N; CDF = running sum of the PDF; multiply by (L − 1) = 7; round off (.5 goes up).

Mapping the new gray levels into number of pixels:

Equalized histogram: add the n_k that land on the same s_k. Levels not listed get 0 pixels.

Equalized output image: replace every pixel r by its s.

What are the two principal approaches to image segmentation? Why is a Laplacian filter typically combined with a Gaussian smoothing filter (LoG operator) before performing edge detection in noisy environments?

Two approaches (based on two basic properties of intensity values):

1. Discontinuity: partition the image where the intensity changes abruptly. Methods: point, line and edge detection (gradient/Sobel, LoG, Canny).
2. Similarity: partition the image into regions that are alike by a predefined rule. Methods: thresholding, region growing, region splitting and merging.

Why Gaussian + Laplacian (LoG).

• The Laplacian is a 2nd derivative, so it is very sensitive to noise: every noisy pixel gives a strong response and false zero crossings.
• Gaussian smoothing first removes the noise, so the zero crossings of the Laplacian then mark real edges only.
• Both are linear, so they combine into one LoG mask ∇²G (cheaper), and its zero crossings give thin, well-localised edges.

Small proof (a noisy step, 2nd derivative f(x+1) + f(x−1) − 2f(x)):

Noisy: 8 sign changes, almost all false. Smoothed: the big one (7 → −7) sits at the real step, and the leftover small ones are weak, so a threshold on the jump drops them.

For the boundary in the grid figure (anticlockwise), compute the 8-directional chain code that is (i) invariant to the starting point; (ii) invariant to both the starting point and rotation.

Step 1 · chain code. Start at the top pixel (1,2) and walk anticlockwise round the boundary. Each move to the next boundary pixel gets the number of its direction (8-direction compass: 0 = right, then counterclockwise).

Step 2 · start-point invariance (shape number of slide 18). The boundary is a loop, so any digit can be the first. Write the code from every start (circular shifts) and pick the one that reads as the smallest number.

Step 3 · circular first difference (CFD). Slide 22 rule: CFD_i = (c_i+1 − c_i) mod 8; the last digit wraps round to the first. This removes rotation.

Step 4 · normalise (NCFD). Smallest circular shift of the CFD. This removes the start point too.

Give the equations of erosion and dilation; describe opening and closing. / Describe two basic morphological operations with equations, effects and applications. / Define dilation and erosion in binary images; how do they affect object size and shape?

Erosion A ⊖ B = {z | (B)_z ⊆ A}: a pixel stays 1 only where the SE fits inside the object. Effects: shrinks foreground objects (removes the outer layer), removes objects smaller than B, smooths the boundary by cutting spikes, can split touching objects, enlarges holes. Uses: removing isolated noise pixels, separating touching coins before counting, boundary extraction β(A) = A − (A ⊖ B).

Dilation A ⊕ B = {z | (B̂)_z ∩ A ≠ ∅}: a pixel becomes 1 where the SE hits the object. Effects: expands foreground objects (adds an outer layer), fills small holes, bridges gaps, joins nearby objects, smooths intrusions. Uses: repairing broken characters in low-resolution text, filling holes, edge detection (A ⊕ B) − A.

Size and shape: erosion makes objects smaller and thinner (size shrinks by the SE radius); dilation makes them larger and thicker. The SE's shape decides the direction: a vertical SE acts only vertically, a disk rounds corners.

Opening (A ⊖ B) ⊕ B removes small objects and thin bridges, keeps size. Closing (A ⊕ B) ⊖ B fills small holes and gaps, keeps size. Both are idempotent.

Construct a 6×6 binary image A and a 3×3 structuring element B. Using fundamental morphological operations, formulate an edge-detection algorithm for A with B, and demonstrate the procedure systematically with mathematical expressions.

Choose A = a 4×4 square in a 6×6 image; B = 3×3 square, origin at the centre.

Algorithm (inner boundary): 1. erode: A ⊖ B = {z | (B)_z ⊆ A}; 2. subtract: β(A) = A − (A ⊖ B) = A ∩ (A ⊖ B)^c.

Step 1: B fits only where the pixel and all 8 neighbours are 1, i.e. the inner 2×2. Step 2: removing those from A leaves the one-pixel ring.

Alternative (outer edge): (A ⊕ B) − A, dilation then subtraction, gives the ring just outside the square: