On basis images for the digital image representation
V.N. Gorbachev, L.A. Denisov, E.M. Kaynarova, I.K. Metelev, E.S. Yakovleva
OOn basis images for the digital image representation
Gorbachev V.N., Denisov L.A., Kaynarova E.M., Metelev I.K. , Yakovleva E.S. January 24, 2019 High School of Printing and Mediatechnology of St. Petersburg State University of Industrial Technology andDesign, St. Petersburg, Russia St. Petersburg State University, St. Petersburg, Russia a r X i v : . [ c s . MM ] J a n bstract Digital array orthogonal transformations that can be presented as a decomposition over basis items orbasis images are considered. The orthogonal transform provides digital data scattering, a process of pixelenergy redistributing, that is illustrated with the help of basis images. Data scattering plays importantrole for applications as image coding and watermarking. We established a simple quantum analogues ofbasis images. They are representations of quantum operators that describe transition of single particlebetween its states.Considering basis images as items of a matrix, we introduced a block matrix that is suitable for orthogonaltransforms of multi-dimensional arrays such as block vector, components of which are matrices. Wepresent an orthogonal transform that produces correlation between arrays. Due to correlation new featureof data scattering was found. A presented detection algorithm is an example of how it can be used infrequency domain watermarking. ontents
A digital image has various representations and some of them are required by applications. Many usefulrepresentations are produced by orthogonal transforms that are powerful tools of image processing. Wellknown examples are JPEG and JPEG2000 lossy compression formats based on DCT (Discrete CosineTransform) and DWT (Discrete Wavelet Transform). For the image compression problem block basedDCT and DWT techniques are developed [1] and generalized to non-separable transforms [2].Orthogonal transform produces scattering of digital data, a process that redistributes pixel energy oftransformed image. It is useful for protection the hiding data in steganography, when a message isembedded into image. The hidden data is scattered among all digital cover image and becomes morerobust to lossy data compression and some statistical attacks [3].The orthogonal transform of images may be considered as a decomposition over matrices known asbasis matrices [4]. Being some kind of grayscale images, the basis matrices look attractive and they areoften reproduced by textbooks [5]. We will also call these matrices basis images.In this paper we study basis images. We focus on the following questions: color and wavelet basisimages, orthogonal transform by matrix of basis images and their quantum analogues. For color imagesthe solution is directly achieved by considering three-dimensional orthogonal transform but for waveletsthe solution is not so simple. The reason is that in practice, DWT is calculated by algorithms usingsignal processing techniques instead of orthogonal transforms. Nevertheless these algorithms can be usedto calculate wavelet basis images. So it was found for various wavelets that the basis has a block structuresimilar to DWT coefficients [6].Basis images may be considered as items of a matrix. We introduced such a matrix, it is orthogonal andsuitable for transforms of multi-dimensional arrays such as a block vectors consisting of matrices. In thiscase there is a large number of degrees of freedom that may be correlated by the transformation. Thecorrelation results in new features of the orthogonal transform for data scattering. he orthogonal transform
The orthogonal transform can scatter digital data.
The orthogonal matrix
A square matrix U of real items is orthogonal if [7] U U T = 1 or U T U = 1 . (1)Columns of this matrix u m and rows u Tn are orthonormal vectors (cid:104) u m u n (cid:105) = δ mn , (cid:104) u Tm u Tn (cid:105) = δ mn , (2)where (cid:104) xy (cid:105) denotes scalar product of two vectors. Scattering.
We will study data scattering that can be illustrated by orthogonal transform of vec-tors.Let us assume that f = { f k } , k = 1 , . . . N is a vector and U is an N × N orthogonal matrix. Taking intoaccount that f = U U T f , we find orthogonal transform of vector ff = U g.g = U T f, (3)where vector g = { g p } , p = 1 , . . . N is often called a representation of f . In matrix form these equationslook as follows f k = (cid:88) p U kp g p ,g p = (cid:88) k U kp f k . (4)As a result two points concerning data scattering can be made.1. Every item of f transforms into all items of g with the weight U kp f k , where p = 1 , , . . . .
2. To get f k , we need to know all items of g .Let us assume that the data are hidden in f k , for example, by steganography and is distributed amongall the digital space of g by an orthogonal transform. The data can be extracted, we need all space of g in spite of every point the data have. Formally the problem is to find f k for given g p and the orthogonal asis images f k (cid:28) { g , g , . . . } . (5)It is clear that due to symmetry, the vector f may be replaced with g . The considered features are truefor the orthogonal transforms of matrices and other multi-dimensional arrays.Data scattering can be directly demonstrated by the orthogonal transform of a set of basic vectors.Let us consider a set of vectors and each of them has a nonzero component e k = { δ kn } , k, n = 1 , . . . , N .The vectors are known to be unit vectors and form a standard basis [7] e = ... , e = ... , . . . e N = ... . An orthogonal matrix U transforms the standard basis into another basis consisting of the columns of Uu k = U e k . (6)This equation shows that a single nonzero item of e k distributes among a column u k e k = ... ... (cid:28) u k ... u kk ... u N = u k . Since the column has at least two non zero items this transformation can be considered as scattering.Scattering may results in energy concentration, a process that is important for applications. Thearray energy, defined as the sum of all components squared, is preserved under orthogonal transforms.Due to scattering, the energy can be distributed into a small amount of components, that is a base ofcoding in the image compression field. It depends on the orthogonal matrix, regardless of whether theenergy would be concentrated or not. It is known that DCT, WHT (Walsh Hadamard Transform) andKLT (Karhunen Loeve Transform) can concentrate the image energy, if image is not random, but DST(Discrete Sine Transform) can’t do it [5].
The orthogonal transform of a matrix and three-dimensional array provides decomposition over thegrayscale and color basis images.
Representation of matrix . Let F = { F mn } be a real rectangular M × N matrix, that corresponds toa grayscale image. We introduce two orthogonal matrices U = { U mn } and V = { V pk } of M × M and N × N . Then taking into account, that F = U U T F V V T , we find F = U GV T ,G = U T F V. (7) asis images G is a M × N matrix.Let us assume that F is an image in a spatial domain (that is the image as we see it). Matrix G is usuallycalled a frequency representation of F or an image in frequency domain. The frequency domain imagemay look senseless, however the orthogonal transform is reversible and the original image can always beretrieved.Using the matrix form of (7), for example, F xy = (cid:88) kp U xk G kp ( V T ) py = (cid:88) kp ( u k ⊗ v p ) xy G kp , we get a decomposition over tensor products of rows and columns of the matrices U and V . Here andlater we assume U = V and M = N that is a more interesting case. Then the decomposition producedby the orthogonal transformation takes the form F = (cid:88) k,p ( u k ⊗ u p ) G kp ,G = (cid:88) x,y ( u Tx ⊗ u Ty ) F xy . (8)We introduce the matrices a kp = u k ⊗ u p , (9) d xy = u Tx ⊗ u Ty , that we call basis images. There are N basis images of size N × N , every image pixel is a product oftwo items of the orthogonal matrix Ua kp ( x, y ) = U xk U yp . Color basis images.
A color RGB image is a three-dimensional array and similar to matrices it providesa decomposition over basis images.Let T = { T mnz } be a three-dimensional array of M × N × Z . The orthogonal transform of T can beachieved by three orthogonal matrices U , V and W . The matrices have size M × M , N × N and Z × Z respectively. Similarly to (8) the array T can be presented as follows T = (cid:88) kps ( u k ⊗ v p ⊗ w s ) τ kps , (10)where w s , s = 1 , . . . , Z is a column of the matrix W . Tensor products t kps = u k ⊗ v p ⊗ w s = a kp ⊗ w s produce a basis, the basis items are t kps ( m, n, q ) = a kp ( m, n ) w q . (11)In general a three-dimensional array can not be a color image. The color RGB image is described bythree matrices R , G and B of equal dimensions, say M × N . Matrices are concatenated in a N × N × array C = cat (3 , R, G, B ) , roperties of basis images cat is concatenation. Here we use notation of MATLAB, it means that C mn = R mn , C mn = G mn and C mn = B mn .Let us assume that in (10) W is a matrix of size × and introduce color basis images t kps = cat (3 , r kps , g kps , b kps ) . Using (11) we find the color channels r kps = a kp w s , g kps = a kp w s and b kps = a kp w s . Full basis has M · N color items. As a result we get the decomposition of RGB images over basis color images C = (cid:88) kps cat (3 , r kps , g kps , b kps ) τ kps . Being tensor products of orthogonal matrix columns and rows the basis images have properties that followfrom orthogonality, and they have a simple analogue came from quantum mechanics.
Properties.
Now let us consider the basis images a kp , if U = V , properties of d xy are the same.1. The matrix product of two basis images is another basis image a kp · a mn = a kn δ pm .
2. The scalar product (cid:104) a kp , a mn (cid:105) = δ km δ pn , (12)where the scalar product of matrices is (cid:104) A, B (cid:105) = (cid:80) mn A mn B mn .3. The sum of diagonal elements, trace (cid:88) k a kk = 1 , (cid:88) x a kp ( x, x ) = δ kp . (13)It follows that (cid:80) k a kk ( xy ) = δ xy .Analysing these properties we came to the conclusion that basis images are orthonormal. This observationallows us to consider the orthogonal transform (8) as a standard decomposition over the orthonormalbasis. Is is obvious that the first equation takes the form F = (cid:88) k,p a kp G kp , (14)where G kp = (cid:104) F, a kp (cid:105) . Generation of basis images.
There are at least two ways to get basis images. The first is to useits definitions. In this case the orthogonal matrix has to be given. The second way follows from orthog-onal transform of the basis images. roperties of basis images F = a kp be in equation (14). Then we find the basis imagerepresentation of the form G k,p = δ ka δ pb . It means that the matrix G has one nonzero pixel, it is equalto 1 and its position is ( a, b ) . So, the orthogonal transform of a basis image is a binary matrix of unitbrightness. We denote such unit matrix as e ab = { δ ka δ pb } , (15)where k, p = 1 , . . . , N . Then the next relations are true a ab = U e ab U T , (16) d ab = U T e ab U. These equations are two-dimensional analogue of (6) and they have a simple meaning. So together withthe unit vectors e k the unit matrices e ab form a standard basis and the orthogonal transform of the basisis a set of basis images a ab .Indeed, with the help of the standard basis any matrix can be presented in the following form G = (cid:88) kp G kp e kp . Then we get the decomposition given by (14), using the orthogonal transform and taking into account (16).
Example.
WHT basis images. The × orthogonal WHT matrix known also as Hadamard matrixconsists of plus 1 and minus 1 H = 1 √ (cid:20) − (cid:21) . (17)In optics this matrix describes so called beam splitter, a linear optical element often used in exper-iments to split the beam into two parts. Four basis images a kp , denoted as tensor product of columns,have the following form a = 12 (cid:20) (cid:21) , a = 12 (cid:20) − − (cid:21) ,a = 12 (cid:20) − − (cid:21) , a = 12 (cid:20) − − (cid:21) . The determinant of every matrix equals to 0 and the matrices are non invertable. The matrices can begenerated from a unit matrix by WHT: H : e = (cid:20) (cid:21) (cid:28) (cid:20) (cid:21) = a . (18)This equation illustrates relations between the basis images and the standard two-dimensional basis. Butwhat is more interesting, the equation demonstrates scattering of digital data (5). So, a nonzero pixel ofthe unit matrix transforms into a basis images of a matrix with only nonzero pixels. ♦ As a result, basis images can be produced by transformation of unit matrices.
The quantum analogue.
The presented features allow us to consider basis images as a represen-tation of quantum operators. These operators describe transitions of a physical system between its states asis wavelet images {| k (cid:105)} and {| q (cid:105)} are two basis of a single particle Hilbert space (cid:88) k | k (cid:105)(cid:104) k | = 1 , (cid:88) q | q (cid:105)(cid:104) q | = 1 , where k ∈ Z = { , , . . . } , q ∈ Q = { x, y, . . . } . Let the overlapping integrals be real (cid:104) k | q (cid:105) ∗ = (cid:104) q | k (cid:105) . (19)Then we find a real matrix ˜ U qk = (cid:104) q | k (cid:105) that is orthogonal because Z and Q are complete basis.The following operator | k (cid:105)(cid:104) p | = ˆ a kp , (20)where k, p ∈ Z , describes transition from the state or level | p (cid:105) into level | k (cid:105) . If k = p , this operator isknown as projection operator.Using Q , the introduced operator (20) can be presents as a real matrix (cid:104) x | ˆ a kp | y (cid:105) = a kp ( xy ) , where x, y ∈ Q . It is not difficult to understand, that these matrices are basis images, considered above.Using Z we can present any single particle operator ˆ F as follows ˆ F = (cid:88) kp | k (cid:105)(cid:104) p |(cid:104) k | ˆ F | p (cid:105) . Operator F can be written as a matrix using Q and (19), then the right part of this equation takes theform (14). As result we find that some of representations of single particle operators can be consideredas basis grayscale images. Basis images can be generated by DWT. In calculation the DWT techniques do not use matrix methodsand the basis wavelet images can be achieved by transform of standard basis.
Wavelet coefficients.
The DWT coefficients have a block structure due to orthogonal matrix U . Incase of single level transform this matrix consists of two parts L and H known as low and high frequencyblocks. Let G be a frequency representation of a N × N grayscale image F = U GU T . Applying theMATLAB notation, we write DWT as follows G = dwt ( F ) = (cid:20) cA cHcV cD (cid:21) , (21) F = idwt ( cA, cH, cV, cD ) . Here the introduced blocks cA , cH , cV and cD — are approximation coefficients, horizontal, vertical anddiagonal details or LL , LH , HL and HH frequency bands. asis wavelet images G can be considered as a three-dimensional array G = { G kpz } of size N/ × N/ × . Index z = 1 , , , labels the cA , cH , cV and cD blocks, for example, G kp = cA kp . Block structure of basis and basis images.
To calculate basis images we use equation (16) a kp = U e kp U T . According to (21) indexes ( k, p ) belong to one of the blocks cA , cH , cV or cD . Let ( k, p ) ∈ cD , so thereis a set of basis items E ( kpD ) = idwt ( O, O, O, e kp ) , (22)where O — is a N/ × N/ matrix of zeros. Here the upper indexes are in brackets to label number of thematrices instead of indicating the pixel position. In other words, we perform an orthogonal transformationof the unit block matrix E ( kpD ) (cid:28) (cid:20) O OO e kp (cid:21) . The total number of basis images of E D = { E ( kpD ) } is N / , every image is a N × N matrix.It is important to note that the equation (16) gives solution by Matlab functions dwt and idwt . Thereason is that in practice the DWT calculations are often based on the filter function techniques [9]. Thesetechniques were developed for signal processing without referring to the orthogonal matrix U . Usuallywavelets are introduced numerically or by recurrent equations so the calculation of U is a problem (except,for example, the Haar wavelet).Using the block coefficients cD and E kpD we can achieve an approximation of original image D = (cid:88) kp cD kp E ( kpD ) . This image has diagonal details only.The wavelet coefficient structure results in basis of four blocks. The blocks refer to cA , cH , cV and cD similarly to (22) (cid:110) { E A } . { E H } , { E V } , { E D } (cid:111) . Every block has N / basis N × N images. As a result the representation over the wavelet basis imageslooks as follows F = (cid:88) kp (cid:16) cA kp E ( kpA ) + cH kp E ( kpH ) + cV kp E ( kpV ) + cD kp E ( kpD ) (cid:17) . Indeed, the considered above function dwt can produce another basis. For this case in accordance with(21) every basis images has a block structure J ( xy ) = dwt ( e xy ) = (cid:20) J ( xyA ) J ( xyH ) J ( xyV ) J ( xyD ) (cid:21) , block matrix Basis images may be items of a matrix that can be orthogonal.
A matrix of basis images.
Consider a square N × N matrix, which elements are basis images b = { b mn } ,b mn = a ( nm ) . (23)Elements of b do not commute. The introduced matrix is a four-dimensional array, consisting of ( N × N ) × ( N × N ) elements K kpxy = a kp ( x, y ) = U xk U yp . The matrix b has the following important feature: bb = 1 . (24)So considering the matrix elements we find ( bb ) mn = (cid:88) k b mk b kn = (cid:88) k a km a nk = δ mn (cid:88) k a kk = δ mn . Indeed, matrix β , which elements are basis images, β mn = a mn , doesn’t have the property given by (24).In this case ββ = N β . The biorthogonal decomposition.
The equation (24) tells that the matrix b has rows orthogonalto columns (cid:104) b Tm b n (cid:105) = δ mn . (25)However, the rows are not orthogonal vectors themselves and similarly to columns. An orthonormal basisis obtained from rows and columns. The basis is known to be biorthogonal or biorthonormal [8] and itcan be used to represent digital arrays.Let us consider a vector f = { f k } , k = 1 , . . . , N . Using (24), we find f = bbf = bg,g = bf, (26)where the introduced vector g = { g p } , p = 1 , . . . , N is a representation of f . To focus on the particularfeature of transform (26), we introduce decomposition of vectors f and g over columns of matrix bf = (cid:88) k b k g k , In contrast to orthogonal transform, the coefficients g k are denoted by rows but not by columns g k = (cid:104) b Tk , f (cid:105) . That is a biorthogonal decomposition.The biorthogonal decompositions are applied in the wavelet field. So, to perform dwt (21) and inversetransform idwt , we need two different wavelets. An example is the Cohen-Daubechies-Feauveau wavelet he block based representation
Orthogonality.
Is the matrix b orthogonal? The answer is not clear because b is a four-dimensionalarray. However, we can refer to the array primitives and consider rows and columns consisting of therows and columns of the basis images. Let r k be a block row. It has items b k , b k , . . . , b kN of basisimages a k , a k , . . . , a Nk . Selecting a row x of every basis image, we get a row r kx . This may be done fora column as well. Introduced rows and columns will be orthonormal vectors. This is a reason to considerthe block matrix b as an orthogonal matrix.Indeed, this result follows from the definition of the transposing operation. In case of block matrix Z itcan be presented as follows Z T = (cid:20) Z Z Z Z (cid:21) T = (cid:20) Z T Z T Z T Z T (cid:21) . The orthogonal matrix of basis images provides a block based representation of multi-dimensional arrays.
Representation.
The block based representation follows from the equations (26), if they are written inthe matrix form f k = (cid:88) k a pk g p ,g p = (cid:88) k a kp f k . (27)Here f and g are two block vectors f = ( f , f , . . . , f N ) ,g = ( g , g , . . . , g N ) . items of which may be chosen as vectors, matrices etc.Let us assume that f k = { f k ( x, y ) } and g p = { g p ( x, y ) } are N × Q matrices. Restrictions on Q will beestablished later. For this case the equations (27) take the following form f k ( x, y ) = (cid:88) p,z a pk ( x, z ) g p ( z, y ) ,g p ( x, y ) = (cid:88) k,z a kp ( x, z ) f k ( z, y ) . (28)It is important to notice that index y plays minor role in these equations and from the formal point ofview it is unnecessary. It means that f k and g p have to be not less than one-dimensional arrays. Thenfor considered matrices we find the following condition Q ≥ .The unnecessary index indicates that there is more space to which matrices a kp do not belong. From thephysical point of view we have two systems, for example, atoms and light. Both systems are described byits observations that can be represented by matrices that, however, affect its Hilbert spaces. To describeelements of different spaces, e.g. two matrices A and B a tensor product is introduced A ⊗ B . on separability and scattering The block based representation leads to new features of data scattering and has a quantum analogue.
Correlation.
Formally, the block based representation (28) looks as one-dimensional transform (4)and we find properties given by (7) for data scattering. However, due to large number of degrees offreedom, scattering obtains new features.Let us assume that both arrays f k and g p are block matrices consisting of other matrices. They arefour-dimensional arrays that we specify by four indexes ( x, y, α, β ) . Let, in contrast to g p , the array f k be dependant on the last pair of indexes only f k = δ xy ψ k ( α, β ) ,g p = g p ( x, y, α, β ) . Under these conditions the equations (28) take the following form ⊗ ψ k = (cid:88) p ( a pk ⊗ g p ,g p = (cid:88) k a kp ⊗ ψ k , (29)where ⊗ ψ k = f k .An important fact follows that the array g p is non-separable. We will use the term separable as divisibility,when the variables are factorized. For example, the function F ( x, y ) = cos( x ) cos y is separable over x and y and the function Φ( x, y ) = cos( x + y ) is not. In our case we focus on two pairs of variables, a pair x, y , that describe basis images a pk ( x, y ) , and pair α, β . From this point of view, the array f k is separablein contrast to g p , that is a non-separable array, because it is a sum of products g p ( x, y, α, β ) = a p ( x, y ) ψ ( α, β ) + a p ( x, y ) ψ ( α, β ) + . . . (30)Non separability is a kind of correlation. Now this is a correlation between the matrices from differentspaces, the basis images and the matrices ψ k . Scattering.
Due to the property of the scalar product of basis matrices (12), we find that (cid:88) xy a kp ( x, y ) g p ( x, y, α, β ) = ψ k ( α, β ) or (cid:104) a kp , g p (cid:105) = ψ k . (31)For data scattering this result tells us the following. The component ψ k scatters into every g p with itsweight a kp and it may be established from every item g p ψ k → { g , g , . . . , } ,ψ k ← g p . This is a new property and it is usually impossible. The property arises from non separability producedby orthogonal transform of the block matrix b . The transform results in correlation between the set of on separability and scattering A quantum analogue.
The block-based representation can be introduced for a three particle quantumsystem.Let us consider a three particle operator given by ˆ c = (cid:88) k,p | k (cid:105)(cid:104) p | ⊗ | p (cid:105)(cid:104) k | ⊗ , where k, p ∈ Z = { , , . . . } and {| k (cid:105)} is a single particle basis. The operator ˆ c is Hermitian and unitary ˆ c = ˆ c † , ˆ c ˆ c = 1 . Let us note that two particle operator ˆ b = (cid:88) k,p | k (cid:105)(cid:104) p | ⊗ | p (cid:105)(cid:104) k | = ˆ b † is a quantum analogue of the matrix b , given by (23).Let us introduce three particle operators ˆ f and ˆ g that are equal up to orthogonal transform given by ˆ c ˆ f = ˆ c ˆ g, ˆ g = ˆ c ˆ f . These equations can be written in a block form. Introducing the matrix elements over particle 1 foroperators ˆ f and ˆ g way we get two operators of particle 2 and 3, which we denote as (cid:104) k | ˆ f | m (cid:105) = ˆ f k , (cid:104) p | ˆ g | m (cid:105) = ˆ g p , where k, p, m ∈ Z . Then we have the block representation ˆ f k = (cid:88) p (ˆ a pk ⊗ g p , ˆ g p = (cid:88) k (ˆ a kp ⊗
1) ˆ f k . Let ˆ f k be the operator of particle 3 only, ˆ f k = 1 ⊗ ˆ ψ k , then we find ˆ f k being a two particle non-separableoperator ˆ g p = ˆ a p ⊗ ˆ ψ + ˆ a p ⊗ ˆ ψ . . . . This equation is a quantum analogue of (31) found for digital data scattering. It is obvious, that Sp { ˆ a kp ˆ g p } = ˆ ψ k , where the average refers to particle 2. steganographic scheme The block-based representation may be useful for frequency domain steganographic technique.
Scheme.
Let the digital data f be images in a spatial domain and g be its representation in a fre-quency domain. Any standard frequency embedding scheme has the following steps. • Transform data into the frequency domain f → g and embed a message M using an algorithm g → g M = emb ( g, M, K ) , where K is a set of parameters with a possible secrete key. • Transform data into the spatial domain g M → f M and send it to a receiver via the communicationchannel. • Extract the embedded message using detection algorithm f M → M = det ( f M , K ) .The scheme includes transformations f → g → g M → f M → g M → M. Indeed, the transform g M → f M can scatter the embedded data among the spatial domain. Scatteringmay result in more robust of hidden data to degradation due to various transformations. An example is aJPEG lossy compression, that stores image in a graphical format. By decreasing the image redundancy,the lossy compression introduces changes into embedded data that exploits the redundancy. So, there is atrade between the compression and the quality of the extracted information. The higher the compressionlevel is, the worse the quality is. Data scattering in the spatial domain.
Let us consider data scattering in the block based rep-resentation (27), assuming k, p = 1 , . Let a message be embedded into g → g M . Then two spatialitems will be changed f → f M = a f + a g M ,f → f M = a f + a g M . (32)To extract the message, we need g M or two items f M and f M g M = a f M + a f M . The equation is a basis for the detection algorithm det ( f M , f M , K ) → M. Data scattering means that all spatial items were changed after embedding and all items are required fordetection. Any frequency domain watermarking technique has these properties regardless of whether ituse the block based representation or not. However, the representation leads to new features appearance.
Embedding.
Let us assume that both vectors f and g have two components. Also f k = 1 ⊗ ψ k ,where ψ k is an image in the spatial domain, k = 1 , . In accordance to (29), the frequency representation g consists of pair of four-dimensional arrays g = a ⊗ ψ + a ⊗ ψ ,g = a ⊗ ψ + a ⊗ ψ . onclusions ψ k with messages g → g M = a ⊗ M + a ⊗ M ,g → g M = a ⊗ M + a ⊗ M , (33)where four matrices M , . . . , M are introduced messages. The main feature of this algorithm is store theability to the structure of the array that holds a set of tensor products including basis images. In thespatial domain we have f M = a ⊗ M + a ⊗ M ,f M = a ⊗ M + a ⊗ M . This allows us to exploit the equation (31) for detection. Then the embedded messages can be extracted,if the component f M or the component f M is given f M → M = (cid:104) a , f M (cid:105) . For this case the detection algorithm works as follows det ( f M , K ) → M , M ,det ( f M , K ) → M , M . Let us note that it differs from the standard algorithm (33) that needs two spatial items instead of one.Moreover there is a difference between this and the spatial domain embedding. We assume that messageis embedded into spatial components ψ k → ψ kM , k = 1 , . It is clear that two messages can be embed-ded only. In the frequency domain there are four messages that may be embedded. But what is moreimportant, these four messages can be distinguished. This fact plays a key role in detection and arisesfrom coupling the messages and basis images to be orthogonal and hence to be well distinguished.Taking into account the considered quantum analogues, we admit that the presented scheme can beextended to quantum mechanic fields.
1. Orthogonal transform provides decomposition over basis items or basis images that have a simplequantum analogue. So, they are a representation of single particle operators that describe transitionsof a particle between its states.2. Grayscale, color and wavelet basis images can be introduced for decomposition of two- and three-dimensional arrays.3. Basis images can be achieved by orthogonal transform of a standard basis that is a set of unit vectors,unit matrices and etc. This fact illustrates digital data scattering, a process of redistributing pixelenergy.4. Due to scattering, energy can be concentrated in small amount of items or, in contrast, be spread.Both cases are interesting for applications. For example, in lossy compression scattering allows toextract the image redundancy, in watermarking it can increase the robustness of a watermark. onclusions ibliographyibliography