Morphology I, from zero

Input 1 0 0 0 1 1 1 0 1 1, SE [1 1 1] with the origin at the centre. At each position take the minimum of the three values under the SE (superimpose and multiply: all three must be 1). The two end pixels are ignored.

Output: 0 0 0 0 1 0 0 0. Only the middle of the 1 1 1 run survives.

Only 5 pixels keep the whole cross inside the object. Every boundary pixel had at least one cross arm on background.

Count the number of coins in the given image by performing an erosion operation.

1. Binarize the image (threshold): coins become white blobs on a black background. Many coins touch each other, so they form joined blobs.
2. Erode with a disk SE large enough to break the narrow necks where coins touch. Each coin shrinks to a smaller, separate blob.
3. Use connected-component labeling to count the separate blobs: the count equals the number of coins.

Same input 1 0 0 0 1 1 1 0 1 1, SE [1 1 1]. Now take the maximum of the three values under the SE (border pixels ignored).

Output: 1 0 1 1 1 1 1 1. Only position 2, whose window 0 0 0 has no 1 at all, stays 0.

The dilated pixels (the slide's green cells) are A ⊕ B − A: 16 new pixels, one layer all round.

Erosion leaves only the two centres where the whole cross fits, (1,5) and (5,2). Dilation puts a cross back on each. The narrow link between the two blobs (row 3) and the blob corners are gone: they were thinner than the SE.

Dilation joins the two blobs across the gap. Erosion shrinks them back to their original outlines, but the new pixels (3,3) and (3,4) in the gap survive: closing bridged the gap without growing the objects.

The input map of Africa (white land) has black text "Northern Africa", "South Africa" written on it; the output is the solid white shape. Which operation was applied?

Closing. The black letters are small dark holes and gaps inside the white foreground. Dilation fills them (the SE touches the white on both sides of each stroke), and the following erosion restores the coastline, so the outline and the size of the continent are unchanged. Opening would not remove holes.

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:

F (5×6) and B = 3×1 vertical [1; 1; 1] with the origin at the centre. Compute (i) the reflection of B (ii) F ⊕ B (iii) F^c ⊖ B (iv) (F ⊕ B) − F (v) the opening of F by B (vi) the closing of F by B, with equations.

(i) B̂ = {w | w = −b, b ∈ B}. B = {(−1,0), (0,0), (1,0)} → B̂ = {(1,0), (0,0), (−1,0)} = B. B is symmetric about its origin, so B̂ = B.

(ii) F ⊕ B = {z | (B̂)_z ∩ F ≠ ∅}: a pixel becomes 1 if it, the pixel above or the pixel below is 1.

(iii) F^c ⊖ B = {z | (B)_z ⊆ F^c}: complement F, then keep a pixel only if it and both vertical neighbours are 1 in F^c.

(iv) (F ⊕ B) − F = the pixels added by the dilation (an outer edge above and below the object):

(v) F ∘ B = (F ⊖ B) ⊕ B:

(vi) F • B = (F ⊕ B) ⊖ B:

Notice: the opening kept only the columns where the object is at least 3 pixels tall (columns 2–3), so everything thinner than B vertically is gone. The closing kept the whole object; the two isolated pixels (0,5) and (4,0) disappeared because they sit on the image border, where the erosion needs the pixel outside the image, which counts as 0.

Images X and Y are given. (i) Design a structuring element B so that Y = X ⊕ B; mark the origin. (ii) Write the mathematical expression of the dilation. (iii) Erode Y with the same SE and show the eroded image. (iv) Explain erosion mathematically.

(i) Take an isolated pixel of X, e.g. (6,4). In Y it has become (5,4), (6,3), (6,2), (5,2), and (6,4) itself is gone. So B holds the offsets (−1,0), (0,−1), (0,−2), (−1,−2) relative to its origin, and the origin itself is not part of B (value 0). Every other pixel of X gives the same pattern, which confirms it.

Check: X ⊕ B reproduces Y exactly.

(ii) X ⊕ B = {z | (B̂)_z ∩ X ≠ ∅} = {x + b | x ∈ X, b ∈ B}.

(iii) Y ⊖ B = {z | (B)_z ⊆ Y}: z survives only if all four offsets from z land on 1s in Y.

Here Y ⊖ B gives back X exactly (this image is lucky; in general (X ⊕ B) ⊖ B is the closing, which can also fill gaps).

(iv) A ⊖ B = {z | (B)_z ⊆ A}: the set of all origin positions z where the SE translated by z lies completely inside A. Equivalently, g(x,y) = 1 if B fits A at (x,y), else 0.

Binary image of the Greek letter θ. Which morphological operation(s) turn θ into the digit 0? Four SEs are given: SE1 horizontal 1×2, SE2 cross, SE3 vertical 2×1 (origin + below), SE4 vertical 3×1. Use any as many times as needed; explain the choice and show the outputs.

What must go: the middle bar (row 6) is only 1 pixel tall. Every part of the ring is at least 2 pixels thick in the vertical direction. So an opening with SE3 (vertical, 2 pixels) removes the bar and keeps the ring: θ ∘ SE3 = (θ ⊖ SE3) ⊕ SE3.

Why not the others: SE1 is horizontal and the bar is long horizontally, so the bar survives. The cross (SE2) needs 3 pixels in both directions, which also deletes the thin ring corners. SE4 needs 3 pixels vertically, which deletes the 2-row top and bottom of the ring.

Step 1, erosion θ ⊖ SE3 (keep a pixel only if it and the pixel below are both 1). Step 2, dilation by SE3 (put back the pixel below every survivor).

Define dilation mathematically for a binary image. Explain the opening and closing operations. What happens when you apply closing to an image with small holes?

Dilation: A ⊕ B = {z | (B̂)_z ∩ A ≠ ∅}: the set of all displacements z such that the reflected SE B̂, translated to z, overlaps A in at least one pixel. It grows objects, fills small holes and bridges gaps.

Opening A ∘ B = (A ⊖ B) ⊕ B, an erosion followed by a dilation: removes objects and protrusions smaller than B, breaks thin bridges and smooths the outline, without changing the size of the remaining objects.

Closing A • B = (A ⊕ B) ⊖ B, a dilation followed by an erosion: fills small holes and gaps, joins nearby objects, and smooths indentations, again without changing the core size.

Closing on small holes: the dilation fills every hole smaller than B (the SE reaches the object on all sides of the hole); the following erosion shrinks the object back to its original outline, but the filled holes cannot reopen because the SE now fits there. So the small holes disappear and the object's size and boundary stay the same (e.g. the pizza image closed with a 22-pixel disc).

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.

What is a structuring element? What are the effects of erosion and dilation? Explain briefly with applications.

A structuring element is a small binary image (template) of any shape and size (square, cross, disk, line) with a defined origin. It is moved over the image like a filter mask, and at each position the fit or hit test between the SE and the image decides the output pixel.

Effects and applications of erosion and dilation: as in the answer above (erosion: shrink, remove noise, separate objects; dilation: grow, fill holes, bridge gaps, repair broken text).

A satellite image of the AUST campus road network has noise (small bright spots), gaps in roads and overlapping structures. (i) Which morphological operation first removes the small bright spots, and why? (ii) After noise removal the roads are fragmented with small gaps: which operation connects them, and how? (iii) Develop the most suitable operation to highlight road boundaries. (iv) Can highlighting boundaries help distinguish roads from the background terrain?

(i) Opening, A ∘ B = (A ⊖ B) ⊕ B, with a small disk/square SE larger than the spots. The erosion deletes every bright spot that cannot contain B; the dilation then grows the roads back to their original width, so only the noise is removed. (Dilation first would enlarge the spots instead.)

(ii) Closing, A • B = (A ⊕ B) ⊖ B, with an SE a bit longer than the gaps (a line SE along the road direction works best). The dilation joins the road fragments across each gap; the erosion shrinks the roads back to their width, but the filled gaps stay filled, so the network becomes connected without fattening.

(iii) Boundary extraction β(A) = A − (A ⊖ B) with a 3×3 SE: erode the cleaned road image and subtract it from itself, leaving a one-pixel outline of every road. (Or (A ⊕ B) − A for the outer edge.)

(iv) Yes. The boundaries outline the road shapes sharply, separating them from the surrounding terrain; they can be used to segment, measure and map the road network.

(i) Automated coin counting: coins are circular blobs, but some touch. Which morphological operations separate touching coins for accurate counting, and why? (ii) A microscope image has small insignificant particles and larger ones: which operations remove the small ones but keep the large ones? Explain mathematically, with applications.

(i) Binarize the image, then apply erosion A ⊖ B = {z | (B)_z ⊆ A} with a disk SE (repeated if needed). Touching coins are joined only by narrow necks; the disk cannot fit in a neck, so the necks disappear first and each coin becomes a separate, smaller blob. Then count the blobs with connected-component labeling (slides 43–44). Counting is not affected by the coins shrinking. (A dilation afterwards, i.e. an opening, would restore size, but could rejoin coins, so count after the erosion.)

(ii) Opening with an SE (disk) a little larger than the insignificant particles: A ∘ B = (A ⊖ B) ⊕ B. Erosion removes every particle that cannot contain B, i.e. all the small ones; the larger particles survive as shrunken cores, and the dilation grows them back to almost their original shape and size. Applications: cleaning cell or particle images before counting, removing dust in industrial inspection, removing noise in scanned documents.

An image is corrupted by salt-and-pepper noise in the object part only. Which morphological operation would you use to remove it? Explain the process.

Inside a white object, the visible noise is the pepper: small black dots, i.e. tiny holes in the foreground. Use closing A • B = (A ⊕ B) ⊖ B with a small (3×3) SE: the dilation fills each black dot because the SE touches white on all sides; the erosion then restores the object's outline. Any white specks left outside the object could then be removed with an opening (open–close filtering). The background is untouched because it has no noise.

Coins image: which morphological operation can be used to count the coins? Explain the process.

Erosion followed by connected-component labeling: binarize; erode with a disk to separate the touching coins (and remove small noise); label the connected components; the number of labels = number of coins.

(i) A text image has broken characters: which morphological operation joins them? (ii) Define opening and closing mathematically, with applications.

(i) Dilation with a small SE such as the 3×3 cross [0 1 0; 1 1 1; 0 1 0] (slide 59): any background pixel whose SE touches a character pixel becomes foreground, so the 1–2 pixel breaks are bridged and the characters become whole. A closing (dilation then erosion) joins the breaks the same way but returns the strokes to their original thickness.

(ii) Opening A ∘ B = (A ⊖ B) ⊕ B: removes small noise and thin bridges (cleaning scanned documents, separating objects). Closing A • B = (A ⊕ B) ⊖ B: fills small holes and gaps (repairing broken text, filling reflections in a thresholded object).