Fast 2D discrete cosine transform on compressed image
in restricted quadtree and shading format
Kuo-Liang Chung
a,∗, Wen-Ming Yan
b,1aDepartment of Information Management, Institute of Computer Science and Information Engineering,
National Taiwan University of Science and Technology, No. 43, Section 4, Keelung Road, Taipei, Taiwan 10672
bDepartment of Computer Science and Information Engineering, National Taiwan University, No. 1, Section 4,
Roosevelt Road, Taipei, Taiwan 106
Received 15 January 2001; received in revised form 6 March 2001 Communicated by A.A. Bertossi
Abstract
Given a compressed image in the restricted quadtree and shading format, this paper presents a fast algorithm for computing 2D discrete cosine transform (DCT) on the compressed grey image directly without the need to decompress the compressed image. The proposed new DCT algorithm takes O(K2log K+ N2) time where the decompressed image is of size N× N and K denotes the number of nodes in the restricted quadtree. Since commonly K < N , the proposed algorithm is faster than the indirect method by decompressing the compressed image first, then applying the conventional DCT algorithm on the decompressed image. The indirect method takes O(N2log N ) time.2002 Elsevier Science B.V. All rights reserved.
Keywords: Algorithms; Discrete cosine transform; Gouraud shading; Quadtree; Compressed image
1. Introduction
Given a binary image of size N× N (= 2n× 2n), based on the quadtree spatial data structure [10], the binary image can be represented in a compressed form. Many efficient algorithms [11] have been developed on the compressed binary image in quadtree format directly without the need to decompress it. However, in the literature, only few algorithms on the compressed grey image domain in the quadtree format were presented. In [2,9], the
N× N grey image is considered. According the restricted quadtree format, which is a complete quadtree, the grey
image is partitioned into 2s × 2s (= S × S) squares, s n, where each square is of size 2n−s × 2n−s and each pixel within the square has the same grey level. Using the subsampling technique, the subsampled image with size 2s× 2s is called the compressed image in the restricted quadtree format. Anguh [2] and Philips [9] presented two different algorithms for Fourier transform, although the proposed two algorithms have the same time complexity, O(S2log S+ N2).
*Corresponding author. Supported by NSC89-2213-E011-062.
E-mail addresses: [email protected] (K.-L. Chung), [email protected] (W.-M. Yan).
1Supported by NSC87-2119-M002-006.
0020-0190/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved. PII: S 0 0 2 0 - 0 1 9 0 ( 0 1 ) 0 0 1 8 8 - 0
The experimental results [3] reveal that combining the quadtree structure and Gouraud shading technique [5]
has better compression effect. That is, given the same grey image of size N× N, we have K S when comparing
the compressed image with size K× K in the restricted quadtree and shading format and the compressed image
with size S× S in the restricted quadtree format. In addition, the decompressed image in [3] can avoid the blocking effect, i.e., has better image quality. In this research, we focus on the 2D discrete cosine transform (DCT) [1, 12,7,6,13,4,8,14] and the compressed image domain in the restricted quadtree and shading format which is more
general than the previous compressed domain [2,9]. The DCT on the original N× N image can be performed in
O(N2log N ) time. The motivation of this research is to design an efficient 2D DCT algorithm on the compressed image in the restricted quadtree and shading format directly without the need to decompress it.
On the compressed image in the restricted quadtree and shading format, this paper presents a novel and fast 2D DCT algorithm and the proposed algorithm takes O(K2log K+ N2) time. Since commonly K < N , the proposed
DCT algorithm is faster than the indirect method by decompressing the compressed image first, then applying the
conventional DCT algorithm [1] on the decompressed image. The indirect method takes O(N2log N ) time. The
main contribution of this paper is that the compression effect in the compressed image implies that the proposed DCT algorithm has a computation-saving effect when compared to the indirect method. However, the compressed image in the restricted quadtree and shading format considered in this paper is a special case of the one in quadtree and shading format [3]. In [3], the constructed quadtree may be incomplete and it is still an open problem to design an efficient algorithm for DCT transform on such a compressed image.
2. The compressed image in restricted quadtree and shading format
Following the shading concept used in [3], the given image is first augmented by duplicating the original last
column and the original last row to become of size (N+ 1) × (N + 1). Then the augmented image is partitioned
into 2k× 2k= K × K overlapping homogeneous squares, i.e., subimages, as few as possible, where each square is of size (2n−k+ 1) × (2n−k+ 1) = (M + 1) × (M + 1). Here, N = K × M. For each square, we only save the grey levels of its four corners. Each corner’s grey level of one square is shared by the neighboring three corners in the neighboring squares. Thus, any homogeneous square shares its top edge with the northern homogeneous square’s bottom edge, and so on. The formal definition of a homogeneous square [3] will be given in next paragraph. Using Gouraud shading method [5], the grey levels of the pixels within one square can be interpolated using the the four corners’ grey levels of that square. When compared to the compressed image in the restricted quadtree format [2, 9], the purpose of making the blocks overlap can alleviate the blockiness effect [3] when employing the shading technique and leads to better image quality. Throughout this paper, the subsampled image in the restricted quadtree and shading format is called the subsampled image without confusion.
It is known that each subimage has four corners, namely, the left-bottom corner, the right-bottom corner, the left-upper corner, and the right-upper corner. The grey levels of the pixels within one square can be interpolated using the the four corners’ grey levels at positions (x1, y1), (x2, y1), (x1, y2), and (x2, y2); their grey levels are g1, g2, g3, and g4, respectively. Here, x2= x1+ M and y2= y1+ M. Let
g5= g1+ g2− g1 x2− x1 (x− x1) and g6= g3+ g4− g3 x2− x1 (x− x1),
thus the estimated grey level of the pixel at (x, y) within the subimage is calculated as
f (x, y)= g5+ g6− g5 y2− y1
(y− y1). (1)
Given a specified error tolerance ε, if|g(x, y) − f (x, y)| ε holds for all the estimated pixels at position (x, y) in the subimage, x1 x < x2and y1 y < y2, then the subimage is homogeneous. Here, g(x, y) denotes the original
grey level at position (x, y). From (1), we have
where a0= g1, a1= g2− g1 x2− x1 , a2= g3− g1 y2− y1 , a3= g4− g3− g2+ g1 (x2− x1)(y2− y1) .
Using the above restricted quadtree and shading approach to compress the grey image with size N× N, we
obtain the compressed image with size K× K. Among these K × K squares, for each square, only four corners’
grey levels are required to be stored. Totally there are 4K2 grey levels to be stored and these 4K2 grey levels will be used as the input of our proposed DCT algorithm. Following the same assumption as in [2,9], we omit the preprocessing time for preparing the subsampled image.
3. The proposed algorithm
For exposition, we let 0 kx, ky< 2k, x1= kxM, x2= kxM+ M, y1= kyM, and y2= kyM+ M. According to
the shading method mentioned in (1), the grey level f (x1+lx, y1+ly), 0 lx, ly< M, at position (x1+lx, y1+ly)
is interpolated using the four corners’ grey levels, g1 = g(x1, y1), g2 = g(x2, y1), g3= g(x1, y2), and g4= g(x2, y2). Since x2− x1= y2− y1= M, by (2), if we let a0(kx, ky) = g1= g(x1, y1), a1(kx, ky)= g2− g1 x2− x1 = g(x2, y1)− g(x1, y1) M , a2(kx, ky) = g3− g1 y2− y1= g(x1, y2)− g(x1, y1) M , (3) a3(kx, ky) = g4− g3− g2+ g1 (y2− y1)(x2− x1)= g4− g3− g2+ g1 M2 , then we have f (kxM+ lx, kyM+ ly) = f (x1+ lx, y1+ ly) = a0(kx, ky)+ a1(kx, ky)lx+ a2(kx, ky)ly+ a3(kx, ky)lxly (4)
for 0 lx, ly< M. Thus, the approximate image with grey levels f (x, y)’s is used to help the computation of the
2D DCT. From (3) and (4), f (kxM+ lx, kyM+ ly) can be written as f (kxM+ lx, kyM+ ly) = c0(kx, ky)+ c1(kx, ky) lx− M− 1 2 + c2(kx, ky) ly− M− 1 2 + c3(kx, ky) lx− M− 1 2 ly− M− 1 2 , where c0(kx, ky)= a0(kx, ky)+ a1(kx, ky)+ a2(kx, ky)M − 1 2 + a3(kx, ky) M− 1 2 2 , c1(kx, ky)= a1(kx, ky)+ a3(kx, ky) M− 1 2 , c2(kx, ky)= a2(kx, ky)+ a3(kx, ky) M− 1 2 , (5) c3(kx, ky)= a3(kx, ky).
The concerning N× N 2D DCT is defined by F (u, v) =N2e(u)e(v)F (u, v), where
e(k)= 1 √ 2 k= 0, 1 1 k N − 1,
and F (u, v)= K −1 kx=0 K −1 ky=0 M −1 lx=0 M −1 ly=0
f (kxM+ lx, kyM+ ly) cosπ(2kxM+ 2lx+ 1)u
2N cos
π(2kyM+ 2ly+ 1)v
2N .
From the trigonometric formula, we have cosπ(2kxM+ 2lx+ 1)u 2N = cosπ(2kx+ 1)u 2K cos π(2lx+ 1 − M)u 2N − sin π(2kx+ 1)u 2K sin π(2lx+ 1 − M)u 2N .
Similarly, it is easy to derive cosπ(2kyM+ 2ly+ 1)v 2N = cosπ(2ky+ 1)v 2K cos π(2ly+ 1 − M)v 2N − sin π(2ky+ 1)v 2K sin π(2ly+ 1 − M)v 2N . Let h1(kx, ky)= M −1 lx=0 M −1 ly=0 f (kxM+ lx, kyM+ ly) cos π(2lx+ 1 − M)u 2N cos π(2ly+ 1 − M)v 2N , h2(kx, ky)= M −1 lx=0 M −1 ly=0 f (kxM+ lx, kyM+ ly) cos π(2lx+ 1 − M)u 2N sin π(2ly+ 1 − M)v 2N , h3(kx, ky)= M −1 lx=0 M −1 ly=0 f (kxM+ lx, kyM+ ly) sin π(2lx+ 1 − M)u 2N cos π(2ly+ 1 − M)v 2N , h4(kx, ky)= M −1 lx=0 M −1 ly=0 f (kxM+ lx, kyM+ ly) sin π(2lx+ 1 − M)u 2N sin π(2ly+ 1 − M)v 2N , then we have F (u, v) = K −1 kx=0 K −1 ky=0 h1(kx, ky) cos π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K − K −1 kx=0 K −1 ky=0 h2(kx, ky) cos π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K − K −1 kx=0 K −1 ky=0 h3(kx, ky) sin π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K + K −1 kx=0 K −1 ky=0 h4(kx, ky) sin π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K .
From (5), it is known that
f (kxM+ lx, kyM+ ly) = c0(kx, ky)+ c1(kx, ky)
lx−M− 1 2 + c2(kx, ky) ly− M− 1 2 + c3(kx, ky) lx− M− 1 2 ly− M− 1 2 , so we have
h1(kx, ky)= c0(kx, ky)S0(u, v)+ c1(kx, ky)S1(u, v)+ c2(kx, ky)S2(u, v)+ c3(kx, ky)S3(u, v),
where S0(u, v) = M −1 lx=0 cosπ(2lx+ 1 − M)u 2N M −1 ly=0 cosπ(2ly+ 1 − M)v 2N , S1(u, v) = M −1 lx=0 lx− M− 1 2 cosπ(2lx+ 1 − M)u 2N M −1 ly=0 cosπ(2ly+ 1 − M)v 2N , S2(u, v) = M −1 lx=0 cosπ(2lx+ 1 − M)u 2N M −1 ly=0 ly− M− 1 2 cosπ(2ly+ 1 − M)v 2N , S3(u, v) = M −1 lx=0 lx− M− 1 2 cosπ(2lx+ 1 − M)u 2N M −1 ly=0 ly− M− 1 2 cosπ(2ly+ 1 − M)v 2N .
In what follows, we want to show that the closed forms of S1(u, v), S2(u, v), and S3(u, v) are zeros and want to
derive the condensed forms of h1(kx, ky), h2(kx, ky), h3(x, ky), and h4(kx, ky). Let
Ac(t)= M −1 x=0 cosπ(2x+ 1 − M)t 2N and Bc(t)= M −1 x=0 x−M− 1 2 cosπ(2x+ 1 − M)t 2N .
Lemma 1. Ac(t)= sin(πMt/2N)/sin(πt/2N) and Bc(t)= 0. Proof. See Appendix A. ✷
Plugging the two closed forms in Lemma 1 into S0(u, v), S1(u, v), S2(u, v), and S3(u, v), it yields to S0(u, v)= Ac(u)Ac(v), S1(u, v)= Bc(u)Ac(v)= 0, S2(u, v)= Ac(u)Bc(v)= 0, and S3(u, v)= Bc(u)Bc(v)= 0. Thus, we
have h1(kx, ky)= c0(kx, ky)Ac(u)Ac(v). By the same arguments, let
As(t)= M −1 x=0 sinπ(2x+ 1 − M)t 2N and Bs(t)= M −1 x=0 x−M− 1 2 sinπ(2x+ 1 − M)t 2N ,
we have the following result.
Lemma 2. As(t)= 0 and the closed form of Bs(t) is equal to
−N π π M 2N sin π t 2Ncos π Mt 2N − π 2Ncos π t 2Nsin π Mt 2N sin2 π t2N .
Therefore, we have h2(kx, ky)= c2(kx, ky)Ac(u)Bs(v), h3(kx, ky)= c1(kx, ky)Bs(u)Ac(v), and h4(kx, ky)= c3(kx, ky)Bs(u)Bs(v).
From the above derivations, we have
K −1 kx=0 K −1 ky=0 h1(kx, ky) cos π(2kx+ 1)u 2K cos π(2ky+ 1)u 2K = Ac(u)Ac(v) K −1 kx=0 K −1 ky=0 c0(kx, ky) cos π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K , K −1 kx=0 K −1 ky=0 h2(kx, ky) cos π(2kx+ 1)u 2K sin π(2ky+ 1)u 2K = Ac(u)Bs(v) K −1 kx=0 K −1 ky=0 c2(kx, ky) cos π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K , K −1 kx=0 K −1 ky=0 h3(kx, ky) sin π(2kx+ 1)u 2K cos π(2ky+ 1)u 2K = Bs(u)Ac(v) K −1 kx=0 K −1 ky=0 c1(kx, ky) sin π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K , K −1 kx=0 K −1 ky=0 h4(kx, ky) sin π(2kx+ 1)u 2K sin π(2ky+ 1)u 2K = Bs(u)Bs(v) K −1 kx=0 K −1 ky=0 c3(kx, ky) sin π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K . Consequently, we have F (u, v) = Ac(u)Ac(v) K −1 kx=0 K −1 ky=0 c0(kx, ky) cos π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K − Ac(u)Bs(v) K −1 kx=0 K −1 ky=0 c2(kx, ky) cos π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K − Bs(u)Ac(v) K −1 kx=0 K −1 ky=0 c1(kx, ky) sin π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K + Bs(u)Bs(v) K −1 kx=0 K −1 ky=0 c3(kx, ky) sin π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K .
C0(u, v) = K −1 kx=0 K −1 ky=0 c0(kx, ky) cos π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K , C1(u, v) = K −1 kx=0 K −1 ky=0 c1(kx, ky) sin π(2kx+ 1)u 2K cos π(2ky+ 1)v 2K , C2(u, v) = K −1 kx=0 K −1 ky=0 c2(kx, ky) cos π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K , C3(u, v) = K −1 kx=0 K −1 ky=0 c3(kx, ky) sin π(2kx+ 1)u 2K sin π(2ky+ 1)v 2K ,
the above four equalities concern one K× K 2D DCT, one K × K discrete sine and cosine transform (DSCT), one
K× K discrete cosine and sine transform (DCST), and one K × K discrete sine transform (DST), respectively.
All the related K× K DCT, DSCT, DCST, and DST can be computed in O(K2log K) time [1]. Originally, for
C0(u, v), we have the range limitation: 0 u < K and 0 v < K. Because of C0(K, v)= C0(u, K)= 0, the
feasible range of u and v could be extended to 0 u, v v. For C1(u, v), we have the range limitation: 0 < u K
and 0 v < K. From C1(0, v)= C1(u, K)= 0, the feasible range of u and v could be extended to 0 u, v K.
Similarly, from C2(K, v)= C2(u, 0)= 0, and C3(0, v)= C3(u, 0)= 0, for C2(u, v) and C3(u, v), the feasible
range of u and v could be extended to 0 u, v K.
However, we have to extend the feasible range of u and v from 0 to N − 1. It is easy to verify that
C0(2K−u, v) = −C0(u, v), C0(u, 2K−v) = −C0(u, v), C1(2K−u, v) = C1(u, v), C1(u, 2K−v) = −C1(u, v), C2(2K−u, v) = −C2(u, v), C2(u, 2K−v) = C2(u, v), C3(2K−u, v) = C3(u, v), and C3(u, 2K−v) = C3(u, v).
Using the above eight equalities, the feasible range of u and v can be extended to 0 to 2K− 1. Similarly, it can be verified that Ci(u+ 2K, v) = −Ci(u, v) and Ci(u, v+ 2K) = −Ci(u, v) for i= 0, 1, 2, 3 and the feasible
range of u and v can be extended to 0 to 4K− 1. By the same arguments, from Ci(u+ 4K, v) = Ci(u, v) and
Ci(u, v+ 4K) = Ci(u, v) for i= 0, 1, 2, 3, the feasible range of u and v can be extended to 0 to N − 1. Then from
the simplified forms of h1(kx, ky), h2(kx, ky), h3(kx, ky), and h4(kx, ky), it yields to
F (u, v)= Ac(u)Ac(v)C0(u, v)− Ac(u)Bs(v)C2(u, v)− Bs(u)Ac(v)C1(u, v)+ Bs(u)Bs(v)C3(u, v). (6)
It takes O(N2) time for computing (6). In addition, it also takes O(N2) time for computing F (u, v)=N2e(u)e(v) F (u, v) for 0 u, v N − 1. Consequently, we have the main result.
Theorem 1. On the compressed image in the restricted quadtree and shading format, the proposed 2D DCT
algorithm takes O(K2log K+ N2) time, where the decompressed grey image is of size N× N and K denotes the number of nodes in the restricted quadtree.
The compressed image domain considered in this paper is more general than the previous domain [2,9] since in the previous restricted quadtree format with size S× S, each quadrant is of constant grey value. The experimental
results in [3] reveal K S. This better compression effect implies that the proposed DCT algorithm has a
computation-saving effect. Since commonly K < N , the proposed DCT algorithm is faster than the indirect method by decompressing the compressed image first, then applying the conventional DCT algorithm on the decompressed image. The indirect method takes O(N2log N ) time. However, in [3], the constructed quadtree may be incomplete, so it is still an open problem to design an efficient DCT algorithm on such a compressed image.
Acknowledgements
The authors appreciate the referees and the Editor Professor Alan Bertossi for their valuable comments that lead to the improved version of the paper.
Appendix A. The proof of Lemma 1
Since Ac(t)= M −1 x=0 cosπ(2x+ 1 − M)t 2N , we have 2 sin π t 2NAc(t)= M −1 x=0 2 cosπ(2x+ 1 − M)t 2N sin π t 2N = M −1 x=0 sinπ(2x+ 2 − M)t 2N − sin π(2x− M)t 2N = sinπ Mt 2N − sin π(−Mt) 2N = 2 sin π Mt 2N .
Therefore, the closed form of Ac(t) is given by Ac(t)= sinπ Mt2N sin2Nπ t . From Bc(t) = M −1 x=0 x−M− 1 2 cosπ(2x+ 1 − M)t 2N ,
let y= M − 1 − x, then we have
Bc(t) = M −1 y=0 M− 1 − y −M− 1 2 cosπ(2(M− 1 − y) + 1 − M)t 2N = M −1 y=0 M− 1 2 − y cosπ(−2y − 1 + M)t 2N = M −1 x=0 M− 1 2 − x cosπ(−2x − 1 + M)t 2N = − M −1 x=0 x−M− 1 2 cosπ(2x+ 1 − M)t 2N = −Bc(t). Thus, it yields to Bc(t)= 0.
Appendix B. The proof of Lemma 2
From As(t)= M −1 x=0 sinπ(2x+ 1 − M)t 2N ,
let y= M − 1 − x, then we have As(t)= M −1 y=0 sinπ(2(M− 1 − y) + 1 − M)t 2N = M −1 y=0 sinπ(−2y − 1 + M)t 2N = M −1 x=0 sinπ(−2x − 1 + M)t 2N = − M −1 x=0 sinπ(2x+ 1 − M)t 2N = −As(t). It implies that As(t)= 0. From Ac(t)= − M −1 x=0 π(2x+ 1 − M) 2N sin π(2x+ 1 − M)t 2N = − π NBs(t),
the closed form of Bs(t) is given by Bs(t)= −N πA c(t)= − N π π M 2N sin π t 2Ncos π Mt 2N − π 2Ncos π t 2Nsin π Mt 2N sin2 π t2N . References
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