1 /*
2 * CDDL HEADER START
3 *
4 * The contents of this file are subject to the terms of the
5 * Common Development and Distribution License, Version 1.0 only
6 * (the "License"). You may not use this file except in compliance
7 * with the License.
8 *
9 * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
10 * or http://www.opensolaris.org/os/licensing.
11 * See the License for the specific language governing permissions
12 * and limitations under the License.
13 *
14 * When distributing Covered Code, include this CDDL HEADER in each
15 * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
16 * If applicable, add the following below this CDDL HEADER, with the
17 * fields enclosed by brackets "[]" replaced with your own identifying
18 * information: Portions Copyright [yyyy] [name of copyright owner]
19 *
20 * CDDL HEADER END
21 */
22 /*
23 * Copyright 2005 Sun Microsystems, Inc. All rights reserved.
24 * Use is subject to license terms.
25 */
26
27 /* Copyright (c) 1984, 1986, 1987, 1988, 1989 AT&T */
28 /* All Rights Reserved */
29
30 #pragma ident "%Z%%M% %I% %E% SMI"
31
32
33 #include <unistd.h>
34 #include <stdlib.h>
35 #include <stdio.h>
36
37 #define BYTE 8
38 #define QW 1 /* width of bas-q digit in bits */
39
40 /*
41 * this stuff should be local and hidden; it was made
42 * accessible outside for dirty reasons: 20% faster spell
43 */
44 #include "huff.h"
45 struct huff huffcode;
46
47 /*
48 * Infinite Huffman code
49 *
50 * Let the messages be exponentially distributed with ratio r:
51 * P {message k} = r^k*(1-r), k = 0, 1, ...
52 * Let the messages be coded in base q, and suppose
53 * r^n = 1/q
54 * If each decade(base q) contains n codes, then
55 * the messages assigned to each decade will be q times
56 * as probable as the next. Moreover the code for the tail of
57 * the distribution after truncating one decade should look
58 * just like the original, but longer by one leading digit q-1.
59 * q(z+n) = z + (q-1)q^w
60 * where z is first code of decade, w is width of code, in shortest
61 * full decade. Examples, base 2:
62 * r^1 = 1/2 r^5 = 1/2
63 * 0 0110
64 * 10 0111
65 * 110 1000
66 * 1110 1001
67 * ... 1010
68 * 10110
69 * w = 1, z = 0 w = 4, z = 0110
70 * Rewriting slightly
71 * (q-1)z + q*n = (q-1)q^w
72 * whence z is a multiple of q and n is a multiple of q-1. Let
73 * z = cq, n = d(q-1)
74 * We pick w to be the least integer such that
75 * d = n/(q-1) <= q^(w-1)
76 * Then solve for c
77 * c = q^(w-1) - d
78 * If c is not zero, the first decade may be preceded by
79 * even shorter (w-1)-digit codes 0, 1, ..., c-1. Thus
80 * the example code with r^5 = 1/2 becomes
81 * 000
82 * 001
83 * 010
84 * 0110
85 * 0111
86 * 1000
87 * 1001
88 * 1010
89 * 10110
90 * ...
91 * 110110
92 * ...
93 * The expected number of base-q digits in a codeword is then
94 * w - 1 + r^c/(1-r^n)
95 * The present routines require q to be a power of 2
96 */
97 /*
98 * There is a lot of hanky-panky with left justification against
99 * sign instead of simple left justification because
100 * unsigned long is not available
101 */
102 #define L (BYTE*(sizeof (long))-1) /* length of signless long */
103 #define MASK (~((unsigned long)1<<L)) /* mask out sign */
104
105 /*
106 * decode the prefix of word y (which is left justified against sign)
107 * place mesage number into place pointed to by kp
108 * return length (in bits) of decoded prefix or 0 if code is out of
109 * range
110 */
111 int
decode(long y,long * pk)112 decode(long y, long *pk)
113 {
114 int l;
115 long v;
116 if (y < cs) {
117 *pk = y >> (long)(L+QW-w);
118 return (w-QW);
119 }
120 for (l = w, v = v0; y >= qcs;
121 y = ((unsigned long)y << QW) & MASK, v += n)
122 if ((l += QW) > L)
123 return (0);
124 *pk = v + (y>>(long)(L-w));
125 return (l);
126 }
127
128 /*
129 * encode message k and put result (right justified) into
130 * place pointed to by py.
131 * return length (in bits) of result,
132 * or 0 if code is too long
133 */
134
135 int
encode(long k,long * py)136 encode(long k, long *py)
137 {
138 int l;
139 long y;
140 if (k < c) {
141 *py = k;
142 return (w-QW);
143 }
144 for (k -= c, y = 1, l = w; k >= n; k -= n, y <<= QW)
145 if ((l += QW) > L)
146 return (0);
147 *py = ((y-1)<<w) + cq + k;
148 return (l);
149 }
150
151
152 /*
153 * Initialization code, given expected value of k
154 * E(k) = r/(1-r) = a
155 * and given base width b
156 * return expected length of coded messages
157 */
158 static struct qlog {
159 long p;
160 double u;
161 } z;
162
163 static struct qlog
qlog(double x,double y,long p,double u)164 qlog(double x, double y, long p, double u) /* find smallest p so x^p<=y */
165 {
166
167 if (u/x <= y) {
168 z.p = 0;
169 z.u = 1;
170 } else {
171 z = qlog(x, y, p+p, u*u);
172 if (u*z.u/x > y) {
173 z.p += p;
174 z.u *= u;
175 }
176 }
177 return (z);
178 }
179
180 double
huff(float a)181 huff(float a)
182 {
183 int i, q;
184 long d, j;
185 double r = a/(1.0 + a);
186 double rc, rq;
187
188 for (i = 0, q = 1, rq = r; i < QW; i++, q *= 2, rq *= rq)
189 continue;
190 rq /= r; /* rq = r^(q-1) */
191 (void) qlog(rq, 1./q, 1L, rq);
192 d = z.p;
193 n = d*(q-1);
194 if (n != d * (q - 1))
195 abort(); /* time to make n long */
196 for (w = QW, j = 1; j < d; w += QW, j *= q)
197 continue;
198 c = j - d;
199 cq = c*q;
200 cs = cq<<(L-w);
201 qcs = (((long)(q-1)<<w) + cq) << (L-QW-w);
202 v0 = c - cq;
203 for (i = 0, rc = 1; i < c; i++, rc *= r) /* rc = r^c */
204 continue;
205 return (w + QW*(rc/(1-z.u) - 1));
206 }
207
208 void
whuff(void)209 whuff(void)
210 {
211 (void) fwrite((char *) & huffcode, sizeof (huffcode), 1, stdout);
212 }
213
214 int
rhuff(FILE * f)215 rhuff(FILE *f)
216 {
217 return (read(fileno(f), (char *)&huffcode, sizeof (huffcode)) ==
218 sizeof (huffcode));
219 }
220