Smoothing Spatial Filters, from zero

g = (110 + 120 + 90 + 91 + 94 + 98 + 90 + 91 + 99) / 9 = 883/9 = 98.11 ≈ 98.1

Mask 1/16 [1 2 1; 2 4 2; 1 2 1]. Multiply each pixel by its weight:

(1×110 + 2×120 + 1×90 + 2×91 + 4×94 + 2×98 + 1×90 + 2×91 + 1×99)/16

= (110 + 240 + 90 + 182 + 376 + 196 + 90 + 182 + 99)/16 = 1565/16 = 97.81 ≈ 98

The mask moves from point to point; at each point (x,y) the response is calculated. Border pixels are skipped (discard method), so the output is 8×8.

First few positions, as on the slides: g = 1/9 × (0+0+0+0+0+0+0+0+90) = 10, then 20, 30, 30. At the hole: 1/9 × (0+0+0+0+90+90+90+90+90) = 50. Fully inside the box: 1/9 × (9×90) = 90.

Notice: the hard edge of the box became a ramp (0, 10, 20, 30 …), the dark hole was filled in (50 instead of 0), and the stray 90 spread into a faint 10-valued patch. That's blurring and noise reduction at the same time.

Step 1, sort ascending: 90, 90, 91, 94, 95, 98, 99, 110, 120. Step 2, replace the original value by the median: 95.

sorted: 10, 15, 20, 23, 27, 30, 31, 33, 90 → 5th value = 27. The 90 is replaced by 27.

The noisy image has random 0 (pepper) and 255 (salt) pixels. Below, both 3×3 filters are applied to the inner 5×5 pixels:

The median output is within 1 of the original everywhere. The mean output carries the noise along: a 255 raises its whole neighbourhood to about 70, and a 0 pulls it down.

aust.png (5×5) has random 0 and 255 pixels; zero padding. (i) Which noise affects aust.png? (ii) Apply a 3×3 mean and a 3×3 median filter at point (1,1). (iii) Is it possible to remove the noise using only a Max or a Min filter? Justify. [1+3+2]

(i) The image has random black (0) and white (255) pixels scattered among normal values: salt-and-pepper (impulse) noise.

(ii)

3×3 mean: (0 + 12 + 0 + 255 + 145 + 200 + 255 + 160 + 90) / 9 = 1117/9 = 124.11 ≈ 124

3×3 median: sorted: 0, 0, 12, 90, 145, 160, 200, 255, 255 → 5th value = 145

The mean (124) is dragged by the two 255s and the 0s; the median (145) is a real neighbouring value, and the extremes end up at the two ends of the sorted list.

(iii) No. A max filter picks the largest value, so it removes pepper (0) but spreads salt (255) to every neighbour of a white dot. A min filter removes salt but spreads pepper. Salt-and-pepper noise contains both at once, so either filter alone makes half of the noise worse. The median filter removes both, because both extremes are pushed to the ends of the sorted list.

(i) Smooth the 3×3 image [2 5 8; 5 1 3; 4 7 2] with a 3×3 average filter, zero padding; show the calculations. (ii) Apply a 3×3 median filter to the 4×4 image; ignore the border and put zero in the output border.

(i) Pad with a ring of zeros, then average every 3×3 window:

(ii) Only the 4 inner pixels have a full window:

(Part iii, the single-pass Laplacian, is on the deck 07 page.)

(i) Explain with an example how a spatial filter affects an image. (iii) Why does the median filter preserve more edge sharpness than the average filter?

(i) A spatial filter consists of a neighbourhood and a predefined operation. The mask is moved from pixel to pixel; at each position the response, the sum of products of mask coefficients and the pixels under them, becomes the new value of the centre pixel. Example: with the 1/9 mean mask, the pixel 94 in the slide-19 image becomes (110+120+90+91+94+98+90+91+99)/9 = 98.1. Applied everywhere, sharp features and noise are averaged out and the image looks smoother (blurred).

(iii) At an edge the 3×3 window holds pixels from both sides. The average mixes them, so it outputs an in-between value that did not exist in the image: the step becomes a ramp and the edge blurs. The median outputs one of the actual pixel values: whichever side has the majority (5 or more of the 9 pixels) wins, so pixels stay on their own side and the step stays a step. It also ignores a single extreme value instead of averaging it in. [Draw the 1-D step figure above.]

3-bit 4×4 image, Laplacian [0 1 0; 1 −4 1; 0 1 0], zero padding. (i) 3×3 mean filter at (2,2) (ii) 3×3 median at (0,0) (iii) Laplacian at (1,1) (iv) Why does the median preserve more edge sharpness than the mean?

(i)

3×3 mean: (3 + 6 + 2 + 4 + 4 + 1 + 5 + 7 + 0) / 9 = 32/9 = 3.56 ≈ 4

(ii)

3×3 median: sorted: 0, 0, 0, 0, 0, 1, 3, 6, 7 → 5th value = 0

Five of the nine values are padding zeros, so the median is 0 (this is why zero padding darkens corners).

(iii)

Laplacian response: (1)(1) + (1)(6) + (-4)(3) + (1)(6) + (1)(4) = 5

(iv) Same answer as SP22 (iii): the median picks a real value from the majority side of the edge, while the mean blends both sides into a new in-between value.

At point (3,2) of the reconstructed 7×7 image: Task 1, perform a 3×3 Max filter. Task 3, design a 3×3 spatial mask that averages the 4 neighbours of a point (x,y) but excludes the point itself.

Original image (rebuilt from bit planes; see the deck 04 page):

Task 1

3×3 max: sorted 0, 0, 0, 0, 0, 1, 1, 5, 5 → largest = 5

Task 2 (negative Laplacian, deck 07):

Laplacian response: (-1)(0) + (-1)(1) + (4)(1) + (-1)(0) + (-1)(0) = 3

Task 3. The 4 neighbours are up, down, left and right. Give each weight 1, the centre 0, the corners 0, and divide by 4 so the weights add to 1:

At (3,3): Task 1, 3×3 Max filter; Task 2, 3×3 positive Laplacian; Task 3, the 4-neighbour averaging mask.

Task 1

3×3 max: sorted 0, 0, 0, 0, 0, 0, 0, 1, 1 → largest = 1

Task 2

Laplacian response: (1)(0) + (1)(1) + (-4)(0) + (1)(1) + (1)(0) = 2

Task 3: the same mask 1/4 [0 1 0; 1 0 1; 0 1 0] as above.

(i) Explain how you can smooth an image. If you subtract the smoothed image from the original, what happens if you add the result back to the original? (ii) Group selfies of batch "Integer 43" suffer from salt-and-pepper noise. Apply a de-noising process: which filtering is most effective, and how does it improve the photos?

(i) Smooth by convolving the image with a low-pass mask whose coefficients add to 1, e.g. the 3×3 mean 1/9 [all 1] or the weighted 1/16 [1 2 1; 2 4 2; 1 2 1]. Each pixel becomes the (weighted) average of its neighbourhood, which blurs and reduces noise.

Original − smoothed = the detail (mask) image: it holds only edges and fine detail, since the smooth parts cancel out. Adding this detail back to the original, g = f + (f − f̄), sharpens the image: edges and fine detail are emphasized. This is unsharp masking (with a weight k > 1 on the detail, high-boost filtering; see deck 07). It is the same as the slide 52–54 mask 2δ − box.

(ii) Use a 3×3 median filter.

• Salt-and-pepper noise is random black (0) and white (255) dots. In each 3×3 window these extremes sort to the two ends of the list, so the median, a real neighbouring value, replaces them.
• Unlike the mean filter, the median does not smear the dots into gray blotches, and it keeps the edges of faces, eyes and hair sharp, because it never creates new in-between values.
• It introduces no new values and is very effective for impulse noise (slides 43–44). If the noise is dense, a second pass or a 5×5 median removes what's left, at the cost of a little extra blur.
• Result: clean skin and background, sharp facial features, and natural-looking group photos.

An image is corrupted by salt-and-pepper noise in the object part only (not the background). Give the best way to denoise it and explain.

Use a median filter (3×3), applied to the object region. Salt-and-pepper noise consists of isolated extreme values (0 and 255). The median of each window ignores them, since they sort to the ends, and replaces each noisy pixel with a real value from its neighbourhood. It keeps the object's edges sharp, unlike the mean, which would blur the object boundary and smear the dots.

Because the background is clean, the filter can be restricted to the object (e.g. a mask from thresholding), so the background stays untouched and no extra blur is added there. Part (ii), morphological removal, is on the morphology page.