1// Copyright 2009 The Go Authors. All rights reserved.
2// Use of this source code is governed by a BSD-style
3// license that can be found in the LICENSE file.
4
5package jpeg
6
7// This is a Go translation of idct.c from
8//
9// http://standards.iso.org/ittf/PubliclyAvailableStandards/ISO_IEC_13818-4_2004_Conformance_Testing/Video/verifier/mpeg2decode_960109.tar.gz
10//
11// which carries the following notice:
12
13/* Copyright (C) 1996, MPEG Software Simulation Group. All Rights Reserved. */
14
15/*
16 * Disclaimer of Warranty
17 *
18 * These software programs are available to the user without any license fee or
19 * royalty on an "as is" basis.  The MPEG Software Simulation Group disclaims
20 * any and all warranties, whether express, implied, or statuary, including any
21 * implied warranties or merchantability or of fitness for a particular
22 * purpose.  In no event shall the copyright-holder be liable for any
23 * incidental, punitive, or consequential damages of any kind whatsoever
24 * arising from the use of these programs.
25 *
26 * This disclaimer of warranty extends to the user of these programs and user's
27 * customers, employees, agents, transferees, successors, and assigns.
28 *
29 * The MPEG Software Simulation Group does not represent or warrant that the
30 * programs furnished hereunder are free of infringement of any third-party
31 * patents.
32 *
33 * Commercial implementations of MPEG-1 and MPEG-2 video, including shareware,
34 * are subject to royalty fees to patent holders.  Many of these patents are
35 * general enough such that they are unavoidable regardless of implementation
36 * design.
37 *
38 */
39
40const blockSize = 64 // A DCT block is 8x8.
41
42type block [blockSize]int32
43
44const (
45	w1 = 2841 // 2048*sqrt(2)*cos(1*pi/16)
46	w2 = 2676 // 2048*sqrt(2)*cos(2*pi/16)
47	w3 = 2408 // 2048*sqrt(2)*cos(3*pi/16)
48	w5 = 1609 // 2048*sqrt(2)*cos(5*pi/16)
49	w6 = 1108 // 2048*sqrt(2)*cos(6*pi/16)
50	w7 = 565  // 2048*sqrt(2)*cos(7*pi/16)
51
52	w1pw7 = w1 + w7
53	w1mw7 = w1 - w7
54	w2pw6 = w2 + w6
55	w2mw6 = w2 - w6
56	w3pw5 = w3 + w5
57	w3mw5 = w3 - w5
58
59	r2 = 181 // 256/sqrt(2)
60)
61
62// idct performs a 2-D Inverse Discrete Cosine Transformation.
63//
64// The input coefficients should already have been multiplied by the
65// appropriate quantization table. We use fixed-point computation, with the
66// number of bits for the fractional component varying over the intermediate
67// stages.
68//
69// For more on the actual algorithm, see Z. Wang, "Fast algorithms for the
70// discrete W transform and for the discrete Fourier transform", IEEE Trans. on
71// ASSP, Vol. ASSP- 32, pp. 803-816, Aug. 1984.
72func idct(src *block) {
73	// Horizontal 1-D IDCT.
74	for y := 0; y < 8; y++ {
75		y8 := y * 8
76		// If all the AC components are zero, then the IDCT is trivial.
77		if src[y8+1] == 0 && src[y8+2] == 0 && src[y8+3] == 0 &&
78			src[y8+4] == 0 && src[y8+5] == 0 && src[y8+6] == 0 && src[y8+7] == 0 {
79			dc := src[y8+0] << 3
80			src[y8+0] = dc
81			src[y8+1] = dc
82			src[y8+2] = dc
83			src[y8+3] = dc
84			src[y8+4] = dc
85			src[y8+5] = dc
86			src[y8+6] = dc
87			src[y8+7] = dc
88			continue
89		}
90
91		// Prescale.
92		x0 := (src[y8+0] << 11) + 128
93		x1 := src[y8+4] << 11
94		x2 := src[y8+6]
95		x3 := src[y8+2]
96		x4 := src[y8+1]
97		x5 := src[y8+7]
98		x6 := src[y8+5]
99		x7 := src[y8+3]
100
101		// Stage 1.
102		x8 := w7 * (x4 + x5)
103		x4 = x8 + w1mw7*x4
104		x5 = x8 - w1pw7*x5
105		x8 = w3 * (x6 + x7)
106		x6 = x8 - w3mw5*x6
107		x7 = x8 - w3pw5*x7
108
109		// Stage 2.
110		x8 = x0 + x1
111		x0 -= x1
112		x1 = w6 * (x3 + x2)
113		x2 = x1 - w2pw6*x2
114		x3 = x1 + w2mw6*x3
115		x1 = x4 + x6
116		x4 -= x6
117		x6 = x5 + x7
118		x5 -= x7
119
120		// Stage 3.
121		x7 = x8 + x3
122		x8 -= x3
123		x3 = x0 + x2
124		x0 -= x2
125		x2 = (r2*(x4+x5) + 128) >> 8
126		x4 = (r2*(x4-x5) + 128) >> 8
127
128		// Stage 4.
129		src[y8+0] = (x7 + x1) >> 8
130		src[y8+1] = (x3 + x2) >> 8
131		src[y8+2] = (x0 + x4) >> 8
132		src[y8+3] = (x8 + x6) >> 8
133		src[y8+4] = (x8 - x6) >> 8
134		src[y8+5] = (x0 - x4) >> 8
135		src[y8+6] = (x3 - x2) >> 8
136		src[y8+7] = (x7 - x1) >> 8
137	}
138
139	// Vertical 1-D IDCT.
140	for x := 0; x < 8; x++ {
141		// Similar to the horizontal 1-D IDCT case, if all the AC components are zero, then the IDCT is trivial.
142		// However, after performing the horizontal 1-D IDCT, there are typically non-zero AC components, so
143		// we do not bother to check for the all-zero case.
144
145		// Prescale.
146		y0 := (src[8*0+x] << 8) + 8192
147		y1 := src[8*4+x] << 8
148		y2 := src[8*6+x]
149		y3 := src[8*2+x]
150		y4 := src[8*1+x]
151		y5 := src[8*7+x]
152		y6 := src[8*5+x]
153		y7 := src[8*3+x]
154
155		// Stage 1.
156		y8 := w7*(y4+y5) + 4
157		y4 = (y8 + w1mw7*y4) >> 3
158		y5 = (y8 - w1pw7*y5) >> 3
159		y8 = w3*(y6+y7) + 4
160		y6 = (y8 - w3mw5*y6) >> 3
161		y7 = (y8 - w3pw5*y7) >> 3
162
163		// Stage 2.
164		y8 = y0 + y1
165		y0 -= y1
166		y1 = w6*(y3+y2) + 4
167		y2 = (y1 - w2pw6*y2) >> 3
168		y3 = (y1 + w2mw6*y3) >> 3
169		y1 = y4 + y6
170		y4 -= y6
171		y6 = y5 + y7
172		y5 -= y7
173
174		// Stage 3.
175		y7 = y8 + y3
176		y8 -= y3
177		y3 = y0 + y2
178		y0 -= y2
179		y2 = (r2*(y4+y5) + 128) >> 8
180		y4 = (r2*(y4-y5) + 128) >> 8
181
182		// Stage 4.
183		src[8*0+x] = (y7 + y1) >> 14
184		src[8*1+x] = (y3 + y2) >> 14
185		src[8*2+x] = (y0 + y4) >> 14
186		src[8*3+x] = (y8 + y6) >> 14
187		src[8*4+x] = (y8 - y6) >> 14
188		src[8*5+x] = (y0 - y4) >> 14
189		src[8*6+x] = (y3 - y2) >> 14
190		src[8*7+x] = (y7 - y1) >> 14
191	}
192}
193