F R A M E S F O R C O L O U R M O R P H O L O G Y

Transcription

1 F R A M E S F O R C O L O U R M O R P H O L O G Y 3 In the last chapter we introduced the problem of false colours, and argued why we believe the problem with these colours is not that they are false (different from any of the input colours), but rather that they are unexpected or unintuitive. Furthermore, we argued that invariance to certain transformations played an important role in this (also see Fig. 3.1). In this chapter we will see what happens when one imposes certain invariances on product-order-based colour morphology. In addition to developing some group-invariant frames suitable for colour morphology, this chapter will also provide a much more compact statement of the main result of the previous chapter and enlarge its scope to include certain non-orthogonal transforms (saturation scaling). Some results are given, demonstrating how enforcing invariance to changes in hue or saturation can make the false colour problem much less objectionable. Additionally, it is shown that in both noise reduction and texture classification tasks our approach gives a (further) objective improvement over product orders. Our approach thus mitigates the false colour problem and leads to better results than a traditional product order-based approach. Illustrations explain why and when invariance leads to better results. 3.1 frame constructions We first give a quick recap of the relevant theory developed in the previous chapter. So let us assume that T is a group of linear transformations on the Hilbert space R K, and that ϕ 0 is an operator on R K that is not invariant to T. If {e k } k K is the original basis used for R K, then we con- Figure 3.1: The infimum of red and blue is black (in the RGB colour space). In contrast, the infimum of magenta and cyan is blue. If we desaturate red and blue, then their infimum suddenly becomes grey. To an observer (who does not know about RGB colour spaces), it is not immediately obvious why this should be so. Why is the infimum of red and blue not also some in-between (purplish) colour for example? 35

2 frames for colour morphology sider the set {f i } i I, with I = T K and f τ,k = τ e k (where τ is the adjoint of τ ). Under certain assumptions this set forms a frame that can be used to construct a T-invariant version of ϕ 0. As we saw before, the above choice of frame has the interesting property that the analysis operator essentially corresponds to taking the components of all transformed versions of a vector: (Fa) τ,k = (τ e k ) a = e k (τ a) = (τ a) k. As a consequence, (Fa) τ is considered equal to τ a. Similarly, if u R I, then u τ (with τ T) denotes the vector in R K described by the coefficients u τ,k, with k K. To construct an operator ϕ : R K R K that is invariant to T, based on an operator ϕ 0, we first define an operator ψ : R I R I such that ψ (u) τ = ϕ 0 (u τ ) for all τ T and u R I. So ψ (Fa) computes ϕ 0 (τ a) for all transformations τ T. We then simply define ϕ as F + ψ F, which Theorem 3.1 shows to be invariant to T under a mild condition on the norm on R I. Theorem 3.1. Assume ϕ 0 is an operator on R K, and that ψ : R I R I is defined by ψ (u) τ = ϕ 0 (u τ ) (for all τ T and u R I ). The operator ϕ : R K R K, defined as ϕ = F + ψ F, is then invariant to the transformation group T on the Hilbert space R K, provided the norm on R I is invariant to a permutation of T (in the sense that if P is a permutation operator mapping each index (τ,k) I to some index (p(τ ),k) I, then u = Pu for any choice of u R I ) 1. Proof. It should be clear that (Fτ a) σ = σ τ a = (Fa) σ τ for all τ,σ T and a R K. Thus, ψ (Fτ a) σ = ψ (Fa) σ τ. In other words, ψ (Fτ a) is a permuted version of ψ (Fa). Denote the effect of this permutation by the operator P, so that F τ = P F and ψ (Fτ a) = P ψ (Fa). Similarly, P 1 is used to denote the inverse permutation, and we naturally have F τ 1 = P 1 F. What remains, is to show that F + P = τ F +. By definition, F + u (with u R I ) is the least-squares solution a to Fa = u, while F + Pu is the least-squares solution b to Fb = Pu. As these linear systems are over-determined, these least-squares solutions are unique. It is thus sufficient to show that b must equal τ a. This follows directly from the invariance of the norm to any permutation. Due to this, the least-squares solution to F b = Pu must equal the least-squares solution to (P 1 F )b = u, or equivalently, Fτ 1 b = u. We can now see that b must indeed equal τ a, and thus F + P must equal τ F +. We thus have ϕ(τ a) = F + ψ (Fτ a) = F + P ψ (Fa) = τ F + ψ (Fa) = τ ϕ(a). This concludes the proof. 36

3 3.1 frame constructions Basis Hue invariant Figure 3.2: The top and bottom bars cycle through all possible hues. The middle two bars compute the infimum over a range of hues at each position (indicated by the white rectangles). The top-middle bar uses a product order on the RGB basis, giving a brighter and more saturated response at the primary colours than elsewhere. In contrast, the bottom-middle bar uses the hue-invariant frame, giving a similar result for all hues Hue invariance The previous chapter argued that the problem with false colours is not that they are different from the original colours, but rather that they do not make any sense. Why should the infimum of two colours that are both (perceptually) close to magenta be black or grey, while the infimum of two colours similarly close to green is green(ish)? Here we show how to construct a frame-dual pair that is invariant to hue rotations, resulting in a qualitatively similar result in both cases, see Fig Example 2.3 showed that the set of all hue rotations can be considered a transformation group T on the RGB colour space. The group T is (group) isomorphic to the two-dimensional rotation group SO(2). A rotation of θ degrees around the grey axis will be denoted by R д,θ, but instead of writing u Rд,θ we will use the more readable u θ. For simplicity, the RGB colour space is considered to be R 3 (R K with K = {1,2,3}), with an orthonormal basis corresponding to the red, green and blue components. The index set I can then be considered to be [0,2π ) {1,2,3} and f ϕ,k = R д,θ ek. Taking u,v R I, we now define the inner product on R I by using the inner product on R 3 : u v = 1 2π 2π 0 u θ v θ dθ. Clearly, this inner product is invariant to the group T. The set of vectors {f i } i I as defined above forms a tight frame with frame constants A = B = 1: 2π Fa 2 = Fa Fa = 1 2π 0 = 1 2π a 2 dθ 2π 0 = a 2. R д,θ a 2 dθ (rotations are orthogonal) 1 This condition is connected to the concept of a Haar measure [143]. 37

4 frames for colour morphology (a) v : (224, 120, 16) (b) v = F v (c) v w Figure 3.3: Illustration of taking the infimum in a hue-invariant representation. (a) Colours are often represented using their red, green and blue components. As established before, taking the infimum is not invariant to changes in hue in that representation. (b) The same colour as in (a), but now in the hue-invariant frame. The colour is shown using a polar plot of the inner product with the frame vectors. (c) The infimum of two colours (indicated using orange and magenta polar region plots) in a hue-invariant frame, with w = FR д, 60 v. This means that the frame is its own (canonical) dual. Also, since the frame itself is invariant to T (as illustrated in Fig. 3.3), it forms a T- invariant frame-dual pair (with itself), by Theorem The synthesis operator corresponding to the hue-invariant frame found above can be given explicitly. By the definitions above and the linearity of the inner product, we must have: F u a = u Fa = 1 2π 2π = 1 2π = ( 1 2π 0 2π 0 2π 0 u θ (R д,θ a) dθ (R д,θ u θ ) a dθ ) R д,θ u θ dθ a. So F u = 1 2π 2π R 0 д,θ u θ dθ. Note that the above is geared towards single colour values, rather than colour images. For treating colour images we simply apply the same technique per-pixel. So to filter a colour image using the rotationinvariant frame discussed here, we would first compute a greyscale image for every vector in the frame, then apply some greyscale morphological operator on each of these images, and then finally combine all the greyscale images (according to the canonical dual frame) Rotation Invariance Here we will see how to construct operators on RGB colour images that are invariant to all rotations of the colour space. This is particularly use- 38

5 3.1 frame constructions ful in the context of filters that should act on differences in colour regardless of the direction of that difference. For example, you might want to supress colours that are outliers in their neighbourhood, regardless of the direction (in the colour space) in which they are an outlier. The group T of all 3D rotations is SO(3). The image/range of the frame analysis operator can thus be considered to be R SO (3) 3. Elements of R SO (3) 3 should be interpreted as vectors whose components can be indexed by elements from I = SO(3) {1,2,3}. The next task is to define a suitable inner product on R SO (3) 3. We will build our inner product on top of the one for R 3. As we saw in Theorem 3.1, the inner product should be invariant with respect to permutations of SO(3). Take f (R) dr to be the integral of f (R) over SO (3) the entire group of rotations (the rotations represented by r), weighing all rotations equally 2. We assume that 1 dr is some finite SO (3) non-zero constant A. The inner product can then simply be defined as follows for u and v in R SO (3) 3 : u v = u R v R dr. SO (3) It can be verified that this inner product is invariant to the permutations alluded to in Theorem 3.1 (which makes sense, given that we weigh all rotations equally). Now we need to find F +. This proves particularly easy, since the frame is tight (A equals B in Eq. (1.1), see Section 1.3.2): Fa 2 = (Fa) (Fa) = (Fa) R 2 dr = Ra 2 dr = A a 2. SO (3) SO (3) So F + = 1 A F. The last step above is valid because rotations are orthogonal Saturation Invariance As shown in Section 2.4.2, the construction used above breaks down when the transformations have eigenvalues with non-unit magnitude. However, in some cases, we can still construct frames invariant to such transformations. We will illustrate this by constructing a frame invariant to scaling the saturation of a colour in the RGB colour space. The saturation, or colourfulness, is taken to be the distance to the grey axis, which is the line through both white and black. As group T we take the set of linear transformations {S s s R and 0 < s}, with S s defined by S s a = s a + (1 s) a g g gg for a R3 and g representing grey (it is not particularly important which shade of grey). These transformations only scale the component of a that is 2 Formally, we take the Haar integral [143]. 39

6 frames for colour morphology Grey axis Frame Original Frame Original: e 1 = (1,0). Grey vector: g = (1,1). Transformation: S s e 1 = s (1,0) + (1 s) 1 2 (1,1) = 1 2 (1 + s,1 s). (1+s,1 s) Frame: f 0,1 = lim s 0 = 1 (1,1), 2(1+s) 2 2 (1+s,1 s) f,1 = lim s = lim s (s, s) = 1 (1, 1),... 2(1+s) 2 2 s 2 2 Figure 3.4: Illustration of the process for constructing a saturation-invariant frame. If the original basis vector is scaled perpendicularly to the grey axis, it traces the horizontal line. If the operator ϕ 0 is invariant to scalar multiplication, then we can avoid vectors of arbitrarily large magnitude by normalizing them. This maps the horizontal line (of infinite extent) onto the quarter circle. For the frame we only take the limit vectors on either end; these are eigenvectors of all scalings. perpendicular to the grey axis, and can thus be interpreted as scaling the saturation. The next step is to assume that the operator ϕ 0 is already invariant to a uniform scaling of the colour space (so scaling all components equally). This is typically the case for morphological operators, and allows us to normalize all vectors. Now the trick is to realize that scaling (and normalizing) one of the original basis vectors by zero and by infinity (in the limit) produces two vectors that are eigenvectors of all saturation scalings: one corresponding to a purely achromatic part of the colour space and one corresponding to a purely chromatic part of the colour space (as illustrated in Fig. 3.4). So instead of creating a frame indexed by T {1,2,3}, we create a frame indexed by I = {0, } {1,2,3}. Noting that S s = S s for all s R, the six frame vectors {f i } i I are then defined as f 0,k = lim s 0 S s e k S s e k and S s e k f,k = lim s S s e k. Alternatively, using the concept of Haar measure [143], this construction can be seen as a natural generalization of the original construction. Although the Haar measure on the group of saturation scalings (and normalizations) is not finite, one can see that it causes the extreme vectors (those scaled by zero, or the limit cases of scaling by infinity) to have an infinitely higher measure in the frame than all other vectors. A frame containing those extreme vectors with measure one and the rest with measure zero naturally follows. 40

7 3.2 implementation details In practice, this means we get a frame consisting of a grey vector and three vectors that are perpendicular to that grey vector. These vectors are all eigenvectors of all S s T. In particular, for some S s T, the grey vector (f 0,1 = f 0,2 = f 0,3 ) has eigenvalue 1, while the other three vectors (f,1, f,2 and f,3 ) have eigenvalue s. Recalling that S s = S s, this means that (FS s a) 0 = (Fa) 0 and (FS s a) = s (Fa) for any S s T. Since we assumed that ϕ 0 is invariant to multiplication by a scalar, we can easily show that F + ψ F is invariant to T, withψ as in Theorem 3.1. It should be noted that the above approach is not terribly useful for much more general scalings. For example, illumination changes can be modelled by multiplying the red, green and blue channels with independent weights. The eigenvectors of such a transformation are in general only the red, green and blue vectors, thus the only possible frame would be the original basis. Clearly, a different method would be needed to combine invariance to illumination changes and other (non-linear) transformations, such as those discussed by Angulo [4, 3.4], with invariance to rotations. 3.2 implementation details The above considers frames containing infinitely many (even uncountably many) vectors, this can obviously not be implemented directly. In practice, the easiest solution is to use a finite set of vectors to approximate the frame. One method is to sample the group of transformations used to create the frame, and construct a frame (and canonical dual) based on this. Alternatively, we can also take a sample of (uniformly distributed) vectors in the ideal frame, and construct a discrete frame based on these. We can then take the Moore-Penrose pseudoinverse of the sampled frame to find its canonical dual frame. A practical implementation of an operator on a lifted image is fairly straightforward using Theorem 3.1: the original operator simply has to be called multiple times, on different, transformed, versions of the original (if we use a finite sample of the group of transformations). Alternatively, if frame is sampled rather than the group, a greyscale operator can be applied to all the (scalar) channels corresponding to the different frame vectors. In the (2D) examples shown here the sampled frames were about ten to fifty times as large as the original basis. Note that if we use this construction, memory can be saved by first computing the channel corresponding to the first frame vector, applying the operator and projecting back only that channel (this works because both F and F + are linear). The running time of our prototype frame-based operators is simply a (large) constant factor more than that of the operators based on bases. However, we expect that in practice many optimizations could be made to decrease the runtime impact of our method. For example, in addition to not always needing the absolute best quality possible, we expect that Code and online demo available via Bitly links MwNF3F and YarcHY (the first for the noise reduction and texture classification tasks, the other for the rest). 41

8 frames for colour morphology R R R R G G G G (a) Original (b) Basis (c) Vector (d) Frame Figure 3.5: The original (a) is filtered using a median filter applied independently to the channels using the original basis (b), a vector median filter (c), and a median filtered applied independently to the channels in a rotation-invariant frame (d). In each case, both a plot of the red (R) and green (G) channels, as well as the actual (colour) image are shown. The median filter has a (centered) support of five pixels. The spike in the red channel in (a) is assumed to be noise. the choice of frame vectors could be optimized for specific images to further reduce the number of directions to sample (in the degenerate case of greyscale images encoded as RGB images one or two directions would suffice for example). Finally, some frame-based filters can be computed directly on the original representation (see Chapter 5). 3.3 results Aptoula and Lefèvre [13] evaluated different orders for use with mathematical morphology. We use the same noise reduction and texture classification tasks to demonstrate the objective benefits of group-invariant frames. In our comparison we consider the RGB basis, the hue-invariant frame and the (fully) rotation-invariant frame. However, we first explore two more theoretical (and illustrative) examples of historical significance Median filters Astola et al. [19] gave two examples to illustrate the problem with marginal processing (/product-order-based operators). In the first example (Fig. 3.5), the per-channel median filter does not pick up on the fact that yellow is a different colour from both red and green, and thus finds that just before the switch to red, there are three out of five pixels that have a non-zero red component, instead of seeing one yellow pixel, two green pixels and two red pixels. In contrast, the other two filters indeed give much less importance to the yellow pixel, resulting in a more natural transition from red to green. The second example by Astola et al. (see Fig. 3.6) shows how processing channels independently can result in an unnatural bias towards filtering only along the axes. In particular, when processing the channels independently there is a clear flattening of the result in the direction of 42

9 3.3 results (a) Original (b) Basis (c) Vector (d) Frame Figure 3.6: The original (a) is filtered using a median filter applied independently to the channels in the original basis (b), a vector median filter (c), and a median filter applied to the channels in a rotationinvariant frame (d). The 1D signals are plotted as strings of points in the 2D value(/colour) space. the axes, and a simple 45 rotation of the input results in a completely different output. In contrast, neither the vector median filter, nor the median filter using the rotation-invariant frame, shows such a directional bias. Astola et al. [19] suggested solving both these issues by creating a vector median filter. This was based on minimizing the sum of the distances to all vectors. As can be seen in Figs. 3.5 and 3.6, our method gives very similar results 3. However, in contrast to the vector median filter, our approach generalizes easily to any operator (not just the median filter), and simply follows logically from enforcing certain constraints Noise reduction Aptoula and Lefèvre [13] used a noise reduction task to compare the objective utility of different ordering schemes. This section replicates their results as they relate to product orders, while showing a (slight) improvement for hue and rotation invariant frames. For this, Gaussian noise was added to an image (Lena), the noisy image was lifted to one of three different frames, filtered, and then projected back to the original colour space. Only the RGB colour space was used, as this makes the most sense with this particular test (after all, the noise is added in this space), and led to best performance in the original experiments. Also, the aim is primarily to demonstrate the potential impact of using groupinvariant frames. 3 It should be noted that the original vector median filter always picked the result from one of the input values. We have chosen not to do this, as this forces an arbitrary decision to be made in ambiguous cases (like the one shown in Fig. 3.5). 43

10 frames for colour morphology ρ RGB Hue Rot. Lex Normalised MSE ( 100) RMSE remaining (%) Figure 3.7: The Lena image, with image values in the range [0, 255], was corrupted with Gaussian noise (σ = 32), with different correlation coefficients (ρ). The OCCO operator was applied to the corrupted images in the original RGB basis, the hue-invariant frame and the rotationinvariant frame. For comparison purposes, the same filter was also applied using a lexicographical order (ordering first on red, then on green, then on blue). The values in the top part of the table show the normalised mean squared error (multiplied by 100), while the bottom part uses the RMSE remaining. As can be seen, using a groupinvariant frame can give a clear improvement. The noise was Gaussian, with the covariance matrix *1 Σ = σ 2...ρ,ρ ρ 1 ρ ρ+ / ρ //. 1- Here ρ denotes the correlation coefficient. To generate Gaussian ran1 dom numbers with the above covariance matrix: first find a matrix Σ such that Σ 2 (Σ 2 )T = Σ; the correlated noise is then generated by gen1 erating normally distributed samples and multiplying them by Σ 2. The noise reduction filter was the OCCO filter. This filter is built from a structural opening γb and a structural closing ϕb, both using the same structuring element b. In this case a 3 3 cross was used (the origin and its four nearest neighbours). Both filters operate on L = Fun(E,C) using a product order, with C the (RGB) colour space. Following Aptoula and Lefèvre [13], the operator is defined by (a L) OCCOb (a) = 1 (γb (ϕb (a)) + ϕb (γb (a))). 2 (3.1) The OCCO operator was lifted to the group-invariant frames by taking OCCOb (u)τ = OCCOb (uτ ). An image was then filtered by lifting it to M = Fun(E,C T ), applying the (lifted) OCCO operator, and then projecting back to L = Fun(E,C) using the canonical dual frame. The results of the noise reduction task are summarized in Fig The error metric used originally was the normalised mean squared error 44

11 3.3 results Figure 3.8: The effect of rotation invariance on the OCCO filter. Each diagram shows part of a one-dimensional signal in which each value is a 2D vector. The signals are plotted as in Fig The error-free signals are not shown, but assumed to be linear ramps of the form u[i] = (1/2, 1/2)+i (cos(a), sin(a)). Shown in grey are the perturbed signals, where the value at i = 0 is offset at a right angle to the original signal. Shown in black are the OCCO filtered signals (the structuring element has a width of 3). The top row shows the results for using the OCCO filter directly on the 2D basis, the bottom row shows the results for the rotation-invariant frame (in 2D). (MSE), which is just the squared L 2 -norm of the error in the filtered image divided by the squared L 2 -norm of the original image (see [13] for details). Unfortunately, this error metric is quite difficult to interpret when comparing results for different images. Therefore, we also use the root mean squared error (RMSE) remaining: the RMSE in the filtered image divided by the RMSE in the noisy image. For example, looking at the table in Fig. 3.7 we see that noise is reduced to approximately 40% (or less) of its original strength. The group-invariant frames clearly lead to better noise reduction 4. This makes sense: the OCCO filter essentially removes bumps. When using the RGB basis it only removes bumps in the red, green and blue directions. So yellow noise is less likely to be suppressed than red or green noise. Using the hue-invariant frame remedies this, but still not covers all possible directions (for example the direction (1, 1, 0)). This could explain why the rotation-invariant frame (whose frame vectors point in all directions) performs best. In fact, as can be seen in Fig. 3.8, the OCCO filter on a 2D basis performs perfectly for signals that are aligned with one of the basis directions, but quite badly for signals that are even slightly misaligned. In contrast, the OCCO filter on the rotation-invariant frame gives a much more consistent result, effectively the average behaviour of the filter on the basis, averaged over all rotations of the space (this is better than the RGB basis result in most cases). The effect of correlation is also interesting. As can be seen in Fig. 3.7, increasing correlation decreases the benefit of using group-invariant frames. This likely has to do with the fact that in many images the main 4 The experiment was repeated using a set of 20 images, with similar results. 45

12 frames for colour morphology ρ RGB Rot (40.3) 36.5 (36.5) (40.2) 38.0 (39.2) RMSE remaining (%) Figure 3.9: For the rotation-invariant frame, inverting the green channel of Lena leads to improved performance when filtering images corrupted with correlated noise. Presumably this effect is a result of having less of the data aligned with the noise. The performance for the RGB basis remains essentially the same (as the filter is self-dual and applied per channel). For comparison purposes, the values from Fig. 3.7 are shown between parentheses. variation is along (or close to) the grey axis. As the correlated noise is strongest along the grey axis, noise and data values line up. As points on a line can only be ordered in two ways, rotational invariance does not help here. This is corroborated by repeating the experiment using a version of Lena where the green channel has been inverted. In that case the predominant gradient directions are no longer aligned with the grey axis, and the effect is less pronounced (see Fig. 3.9). This shows the ability of group-invariant frames to do better than a basis, while gracefully degrading to similar performance when it cannot do better Texture classification In this task a number of textures are grouped into classes. Part of the textures serve as training data and the rest is used for testing. The textures are taken from Outex13 [147]. Nearest-neighbour classification is used, based on the Euclidean distance between feature vectors. Again we limit ourselves to the RGB colour space (and derived group-invariant frames), as using different colour spaces had hardly any effect on the performance in the original tests using a product order [13]. The feature vectors are based on morphological covariance. To compute the morphological covariance of an image (for a specific shift), a point-wise infimum is computed of the image and a shifted version of itself. The result is then summed over all positions (but independently for each channel ), and divided by the sum of the pixel values of the original image. This is done for a fixed set of shifts, and when using group-invariant frames, for a finite (fixed) subset of the frame. Shifts in four different directions (0, 45, 90 and 135 ), and in each direction by L distances of 1 to 49 pixels (in step sizes of two), were used in all cases. In total this gives 25 feature values per direction and 100 per channel. In the RGB case 3 channels were used, for the groupinvariant frames subsets of 150 frame vectors were used to create

13 3.3 results RGB Hue Rot Classification rate (%) Figure 3.10: The classification rates for a simple nearest-neighbour scheme based on features computed using different frames. Clearly, the group-invariant frames again lead to better performance than the original RGB basis. On the left a couple of the textures used in the test are shown. channels. (Note that in this task we do not need to project back to the original colour space.) This follows the description given by Aptoula and Lefèvre [13]. The results summarized in Fig again show improved performance using the group-invariant frames. However, there are some complications. First of all, the classification results are better than in the original work, even using the RGB basis. This is probably due to slight differences in implementation. Secondly, we need some tricks to make sure that the features can be computed in a sensible manner when using the group-invariant frames. Specifically, for each direction we make sure that the projection of the entire RGB cube is non-negative. Otherwise some directions give much less stable results than others (due to division by numbers close to zero). Still, that the group-invariant frames lead to improved performance is not surprising. As an extreme case, consider a texture that simply oscillates between pure red and green. When analysing this using just the original RGB basis such a texture is indistinguishable from a texture that oscillates between black and yellow (a combination of red and green), as we effectively have no information about the relative phase of the red and blue components. On the other hand, using the rotationinvariant frame we also look at the projection onto the vector ( 1,0,1), giving excellent discrimination. This is illustrated in Fig Further examples The effect of saturation invariance is illustrated in Fig Without saturation invariance, the infimum of pure red and blue is black, while the infimum of desaturated red and blue is grey. With a saturation-invariant frame, the infimum of pure red and blue is grey as well. Conceptually, this makes a lot more sense; both colours are pretty light. Using a hue and saturation-invariant frame provides an interesting alternative to filtering in the HSL colour space. Instead of representing saturation and hue using a magnitude and an angle (which is hard to order sensibly) we effectively represent them as a function of the angle. 47

14 frames for colour morphology (a) (b) (c) (d) (e) (f) Figure 3.11: The two signals in (a) and (b) are plotted using the original basis in (c) and (d), respectively. In (c) to (f), the signals are plotted against time (top) and in the 2D value space (bottom), like in Fig The basis vectors used for the middle row are shown as arrows in the bottom row. Figure 3.12: Three different pairs of colours and their infima using different frames. From left to right: the original basis (same as in Fig. 3.1), a saturation-invariant frame and a saturation and hue-invariant frame. Gamma correction (srgb) was used; this avoids big differences in lightness between the saturated and desaturated colours. Figures 3.13 and 3.14 show some results on more natural images. Figure 3.13 shows the result of applying a dilation using the original basis, as well as using a hue-invariant frame and a rotation-invariant frame. Figure 3.14 does the same for the OCCO operator. Since the OCCO operator is a self-dual operator 5, it is an excellent candidate for use with a rotation-invariant frame. In fact, the OCCO operator can be derived from taking the opening of a closing as ϕ 0 in Theorem 3.1, with T the group consisting of the identity mapping and the inversion mapping. 3.4 summary This chapter provided a more compact statement of the main result of the previous chapter. Furthermore, it was shown that this result can be extended to certain non-orthogonal transformations. Using several examples this chapter illustrated how using the newly created frames can lead to more intuitive and higher quality results. 5 A self-dual operator is taken to be invariant to inverting the image. 48

15 3.4 summary Figure 3.13: Dilations (by a square) of the originals (top row, both pixels) using different frames, from left to right: the RGB basis, a hue-invariant frame and a rotation-invariant frame. Rows three and five are zoomed versions of rows two and four. A pink haze appears between orange and blue patches on the colour chart when using the RGB basis. The same kind of effect is quite visible all over the parrot. Using a hue-invariant frame solves these problems (middle column). Using the rotation-invariant frame (right column) is similar to averaging a dilation and an erosion, but otherwise shows no significant colour artefacts. (The parrot image is based on a photograph by Luc Viatour / used under the CC BY 2.0 license.) 49

16 frames for colour morphology Figure 3.14: Similar to Fig. 3.13, except that now the self-dual OCCO operator is used, with a square structuring element. Again colour artefacts are visible in the left column, especially between the orange and blue patches on the colour chart (the transition region becomes green). Using a hue-invariant frame (middle column), and especially using a rotation-invariant frame (right column), eliminates these artefacts. Similarly, in the left column the back of the neck of the parrot is suddenly bright green, even though it is originally blueish, and there is only yellow and dark green in the neighbourhood. Again this artefact is largely gone in the middle and right columns. 50