game_timing_state() now has access to the game_ui. This means that
[sgt/puzzles] / misc.c
CommitLineData
4efb3868 1/*
2 * misc.c: Miscellaneous helpful functions.
3 */
4
5#include <assert.h>
6#include <stdlib.h>
74476385 7#include <string.h>
8#include <stdio.h>
4efb3868 9
10#include "puzzles.h"
11
077f3cbe 12void free_cfg(config_item *cfg)
13{
14 config_item *i;
15
16 for (i = cfg; i->type != C_END; i++)
17 if (i->type == C_STRING)
18 sfree(i->sval);
19 sfree(cfg);
20}
74476385 21
22/*
23 * The Mines (among others) game descriptions contain the location of every
24 * mine, and can therefore be used to cheat.
25 *
26 * It would be pointless to attempt to _prevent_ this form of
27 * cheating by encrypting the description, since Mines is
28 * open-source so anyone can find out the encryption key. However,
29 * I think it is worth doing a bit of gentle obfuscation to prevent
30 * _accidental_ spoilers: if you happened to note that the game ID
31 * starts with an F, for example, you might be unable to put the
32 * knowledge of those mines out of your mind while playing. So,
33 * just as discussions of film endings are rot13ed to avoid
34 * spoiling it for people who don't want to be told, we apply a
35 * keyless, reversible, but visually completely obfuscatory masking
36 * function to the mine bitmap.
37 */
38void obfuscate_bitmap(unsigned char *bmp, int bits, int decode)
39{
40 int bytes, firsthalf, secondhalf;
41 struct step {
42 unsigned char *seedstart;
43 int seedlen;
44 unsigned char *targetstart;
45 int targetlen;
46 } steps[2];
47 int i, j;
48
49 /*
50 * My obfuscation algorithm is similar in concept to the OAEP
51 * encoding used in some forms of RSA. Here's a specification
52 * of it:
53 *
54 * + We have a `masking function' which constructs a stream of
55 * pseudorandom bytes from a seed of some number of input
56 * bytes.
57 *
58 * + We pad out our input bit stream to a whole number of
59 * bytes by adding up to 7 zero bits on the end. (In fact
60 * the bitmap passed as input to this function will already
61 * have had this done in practice.)
62 *
63 * + We divide the _byte_ stream exactly in half, rounding the
64 * half-way position _down_. So an 81-bit input string, for
65 * example, rounds up to 88 bits or 11 bytes, and then
66 * dividing by two gives 5 bytes in the first half and 6 in
67 * the second half.
68 *
69 * + We generate a mask from the second half of the bytes, and
70 * XOR it over the first half.
71 *
72 * + We generate a mask from the (encoded) first half of the
73 * bytes, and XOR it over the second half. Any null bits at
74 * the end which were added as padding are cleared back to
75 * zero even if this operation would have made them nonzero.
76 *
77 * To de-obfuscate, the steps are precisely the same except
78 * that the final two are reversed.
79 *
80 * Finally, our masking function. Given an input seed string of
81 * bytes, the output mask consists of concatenating the SHA-1
82 * hashes of the seed string and successive decimal integers,
83 * starting from 0.
84 */
85
86 bytes = (bits + 7) / 8;
87 firsthalf = bytes / 2;
88 secondhalf = bytes - firsthalf;
89
90 steps[decode ? 1 : 0].seedstart = bmp + firsthalf;
91 steps[decode ? 1 : 0].seedlen = secondhalf;
92 steps[decode ? 1 : 0].targetstart = bmp;
93 steps[decode ? 1 : 0].targetlen = firsthalf;
94
95 steps[decode ? 0 : 1].seedstart = bmp;
96 steps[decode ? 0 : 1].seedlen = firsthalf;
97 steps[decode ? 0 : 1].targetstart = bmp + firsthalf;
98 steps[decode ? 0 : 1].targetlen = secondhalf;
99
100 for (i = 0; i < 2; i++) {
101 SHA_State base, final;
102 unsigned char digest[20];
103 char numberbuf[80];
104 int digestpos = 20, counter = 0;
105
106 SHA_Init(&base);
107 SHA_Bytes(&base, steps[i].seedstart, steps[i].seedlen);
108
109 for (j = 0; j < steps[i].targetlen; j++) {
110 if (digestpos >= 20) {
111 sprintf(numberbuf, "%d", counter++);
112 final = base;
113 SHA_Bytes(&final, numberbuf, strlen(numberbuf));
114 SHA_Final(&final, digest);
115 digestpos = 0;
116 }
117 steps[i].targetstart[j] ^= digest[digestpos++];
118 }
119
120 /*
121 * Mask off the pad bits in the final byte after both steps.
122 */
123 if (bits % 8)
124 bmp[bits / 8] &= 0xFF & (0xFF00 >> (bits % 8));
125 }
126}
127
128/* err, yeah, these two pretty much rely on unsigned char being 8 bits.
129 * Platforms where this is not the case probably have bigger problems
130 * than just making these two work, though... */
131char *bin2hex(const unsigned char *in, int inlen)
132{
133 char *ret = snewn(inlen*2 + 1, char), *p = ret;
134 int i;
135
136 for (i = 0; i < inlen*2; i++) {
137 int v = in[i/2];
138 if (i % 2 == 0) v >>= 4;
139 *p++ = "0123456789abcdef"[v & 0xF];
140 }
141 *p = '\0';
142 return ret;
143}
144
145unsigned char *hex2bin(const char *in, int outlen)
146{
147 unsigned char *ret = snewn(outlen, unsigned char);
148 int i;
149
150 debug(("hex2bin: in '%s'", in));
151
152 memset(ret, 0, outlen*sizeof(unsigned char));
153 for (i = 0; i < outlen*2; i++) {
154 int c = in[i];
155 int v;
156
157 assert(c != 0);
158 if (c >= '0' && c <= '9')
159 v = c - '0';
160 else if (c >= 'a' && c <= 'f')
161 v = c - 'a' + 10;
162 else if (c >= 'A' && c <= 'F')
163 v = c - 'A' + 10;
164 else
165 v = 0;
166
167 ret[i / 2] |= v << (4 * (1 - (i % 2)));
168 }
169 return ret;
170}
171
937a9eff 172void game_mkhighlight(frontend *fe, float *ret,
173 int background, int highlight, int lowlight)
174{
175 float max;
176 int i;
177
178 frontend_default_colour(fe, &ret[background * 3]);
179
180 /*
181 * Drop the background colour so that the highlight is
182 * noticeably brighter than it while still being under 1.
183 */
184 max = ret[background*3];
185 for (i = 1; i < 3; i++)
186 if (ret[background*3+i] > max)
187 max = ret[background*3+i];
188 if (max * 1.2F > 1.0F) {
189 for (i = 0; i < 3; i++)
190 ret[background*3+i] /= (max * 1.2F);
191 }
192
193 for (i = 0; i < 3; i++) {
194 ret[highlight * 3 + i] = ret[background * 3 + i] * 1.2F;
195 ret[lowlight * 3 + i] = ret[background * 3 + i] * 0.8F;
196 }
197}
198
74476385 199/* vim: set shiftwidth=4 tabstop=8: */