1 /*
2 Copyright (c) 2014 Timur Gafarov
3 
4 Boost Software License - Version 1.0 - August 17th, 2003
5 
6 Permission is hereby granted, free of charge, to any person or organization
7 obtaining a copy of the software and accompanying documentation covered by
8 this license (the "Software") to use, reproduce, display, distribute,
9 execute, and transmit the Software, and to prepare derivative works of the
10 Software, and to permit third-parties to whom the Software is furnished to
11 do so, all subject to the following:
12 
13 The copyright notices in the Software and this entire statement, including
14 the above license grant, this restriction and the following disclaimer,
15 must be included in all copies of the Software, in whole or in part, and
16 all derivative works of the Software, unless such copies or derivative
17 works are solely in the form of machine-executable object code generated by
18 a source language processor.
19 
20 THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
21 IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
22 FITNESS FOR A PARTICULAR PURPOSE, TITLE AND NON-INFRINGEMENT. IN NO EVENT
23 SHALL THE COPYRIGHT HOLDERS OR ANYONE DISTRIBUTING THE SOFTWARE BE LIABLE
24 FOR ANY DAMAGES OR OTHER LIABILITY, WHETHER IN CONTRACT, TORT OR OTHERWISE,
25 ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER
26 DEALINGS IN THE SOFTWARE.
27 */
28 
29 // dimage is actually stripped out part of dlib - just to support reading PNG and JPEG
30 module dimage.idct; //dlib.image.io.idct
31 
32 import std.math;
33 
34 /*
35  * Inverse discrete cosine transform (DCT) for 64x64 blocks
36  */
37 
38 enum blockSize = 64; // A DCT block is 8x8.
39 
40 enum w1 = 2841; // 2048*sqrt(2)*cos(1*pi/16)
41 enum w2 = 2676; // 2048*sqrt(2)*cos(2*pi/16)
42 enum w3 = 2408; // 2048*sqrt(2)*cos(3*pi/16)
43 enum w5 = 1609; // 2048*sqrt(2)*cos(5*pi/16)
44 enum w6 = 1108; // 2048*sqrt(2)*cos(6*pi/16)
45 enum w7 = 565; // 2048*sqrt(2)*cos(7*pi/16)
46 
47 enum w1pw7 = w1 + w7;
48 enum w1mw7 = w1 - w7;
49 enum w2pw6 = w2 + w6;
50 enum w2mw6 = w2 - w6;
51 enum w3pw5 = w3 + w5;
52 enum w3mw5 = w3 - w5;
53 
54 enum r2 = 181; // 256/sqrt(2)
55 
56 // idct performs a 2-D Inverse Discrete Cosine Transformation.
57 //
58 // The input coefficients should already have been multiplied by the
59 // appropriate quantization table. We use fixed-point computation, with the
60 // number of bits for the fractional component varying over the intermediate
61 // stages.
62 //
63 // For more on the actual algorithm, see Z. Wang, "Fast algorithms for the
64 // discrete W transform and for the discrete Fourier transform", IEEE Trans. on
65 // ASSP, Vol. ASSP- 32, pp. 803-816, Aug. 1984.
66 void idct64(int* src)
67 {
68     // Horizontal 1-D IDCT.
69     for (uint y = 0; y < 8; y++)
70     {
71         int y8 = y * 8;
72         // If all the AC components are zero, then the IDCT is trivial.
73         if (src[y8+1] == 0 && src[y8+2] == 0 && src[y8+3] == 0 &&
74             src[y8+4] == 0 && src[y8+5] == 0 && src[y8+6] == 0 && src[y8+7] == 0)
75         {
76             int dc = src[y8+0] << 3;
77             src[y8+0] = dc;
78             src[y8+1] = dc;
79             src[y8+2] = dc;
80             src[y8+3] = dc;
81             src[y8+4] = dc;
82             src[y8+5] = dc;
83             src[y8+6] = dc;
84             src[y8+7] = dc;
85             continue;
86         }
87 
88         // Prescale.
89         int x0 = (src[y8+0] << 11) + 128;
90         int x1 = src[y8+4] << 11;
91         int x2 = src[y8+6];
92         int x3 = src[y8+2];
93         int x4 = src[y8+1];
94         int x5 = src[y8+7];
95         int x6 = src[y8+5];
96         int x7 = src[y8+3];
97 
98         // Stage 1.
99         int x8 = w7 * (x4 + x5);
100         x4 = x8 + w1mw7*x4;
101         x5 = x8 - w1pw7*x5;
102         x8 = w3 * (x6 + x7);
103         x6 = x8 - w3mw5*x6;
104         x7 = x8 - w3pw5*x7;
105 
106         // Stage 2.
107         x8 = x0 + x1;
108         x0 -= x1;
109         x1 = w6 * (x3 + x2);
110         x2 = x1 - w2pw6*x2;
111         x3 = x1 + w2mw6*x3;
112         x1 = x4 + x6;
113         x4 -= x6;
114         x6 = x5 + x7;
115         x5 -= x7;
116 
117         // Stage 3.
118         x7 = x8 + x3;
119         x8 -= x3;
120         x3 = x0 + x2;
121         x0 -= x2;
122         x2 = (r2*(x4+x5) + 128) >> 8;
123         x4 = (r2*(x4-x5) + 128) >> 8;
124 
125         // Stage 4.
126         src[y8+0] = (x7 + x1) >> 8;
127         src[y8+1] = (x3 + x2) >> 8;
128         src[y8+2] = (x0 + x4) >> 8;
129         src[y8+3] = (x8 + x6) >> 8;
130         src[y8+4] = (x8 - x6) >> 8;
131         src[y8+5] = (x0 - x4) >> 8;
132         src[y8+6] = (x3 - x2) >> 8;
133         src[y8+7] = (x7 - x1) >> 8;
134     }
135 
136     // Vertical 1-D IDCT.
137     for (uint x = 0; x < 8; x++)
138     {
139         // Similar to the horizontal 1-D IDCT case, if all the AC components are zero, then the IDCT is trivial.
140         // However, after performing the horizontal 1-D IDCT, there are typically non-zero AC components, so
141         // we do not bother to check for the all-zero case.
142 
143         // Prescale.
144         int y0 = (src[8*0+x] << 8) + 8192;
145         int y1 = src[8*4+x] << 8;
146         int y2 = src[8*6+x];
147         int y3 = src[8*2+x];
148         int y4 = src[8*1+x];
149         int y5 = src[8*7+x];
150         int y6 = src[8*5+x];
151         int y7 = src[8*3+x];
152 
153         // Stage 1.
154         int y8 = w7*(y4+y5) + 4;
155         y4 = (y8 + w1mw7*y4) >> 3;
156         y5 = (y8 - w1pw7*y5) >> 3;
157         y8 = w3*(y6+y7) + 4;
158         y6 = (y8 - w3mw5*y6) >> 3;
159         y7 = (y8 - w3pw5*y7) >> 3;
160 
161         // Stage 2.
162         y8 = y0 + y1;
163         y0 -= y1;
164         y1 = w6*(y3+y2) + 4;
165         y2 = (y1 - w2pw6*y2) >> 3;
166         y3 = (y1 + w2mw6*y3) >> 3;
167         y1 = y4 + y6;
168         y4 -= y6;
169         y6 = y5 + y7;
170         y5 -= y7;
171 
172         // Stage 3.
173         y7 = y8 + y3;
174         y8 -= y3;
175         y3 = y0 + y2;
176         y0 -= y2;
177         y2 = (r2*(y4+y5) + 128) >> 8;
178         y4 = (r2*(y4-y5) + 128) >> 8;
179 
180         // Stage 4.
181         src[8*0+x] = (y7 + y1) >> 14;
182         src[8*1+x] = (y3 + y2) >> 14;
183         src[8*2+x] = (y0 + y4) >> 14;
184         src[8*3+x] = (y8 + y6) >> 14;
185         src[8*4+x] = (y8 - y6) >> 14;
186         src[8*5+x] = (y0 - y4) >> 14;
187         src[8*6+x] = (y3 - y2) >> 14;
188         src[8*7+x] = (y7 - y1) >> 14;
189     }
190 }
191 
192 /*
193 void idct(float* inMat, float* outMat)
194 {
195     uint i, j, u, v;
196     float s;
197 
198     for (i = 0; i < 8; i++)
199     for (j = 0; j < 8; j++)
200     {
201         s = 0;
202 
203         for (u = 0; u < 8; u++)
204         for (v = 0; v < 8; v++)
205         {
206             s += inMat[u * 8 + v]
207                 * cos((2 * i + 1) * u * PI / 16.0f)
208                 * cos((2 * j + 1) * v * PI / 16.0f)
209                 * ((u == 0) ? 1 / sqrt(2.0f) : 1.0f)
210                 * ((v == 0) ? 1 / sqrt(2.0f) : 1.0f);
211         }
212 
213         outMat[i * 8 + j] = s / 4.0f;
214     }
215 }
216 */
217