diff --git a/auxiliary/obfusc.c b/auxiliary/obfusc.c index c62189c..1194276 100644 --- a/auxiliary/obfusc.c +++ b/auxiliary/obfusc.c @@ -30,13 +30,82 @@ #include "puzzles.h" +static bool self_tests(void) +{ + bool ok = true; + + #define FAILED (ok = false, "failed") + + /* + * A few simple test vectors for the obfuscator. + * + * First test: the 28-bit stream 1234567. This divides up + * into 1234 and 567[0]. The SHA of 56 70 30 (appending + * "0") is 15ce8ab946640340bbb99f3f48fd2c45d1a31d30. Thus, + * we XOR the 16-bit string 15CE into the input 1234 to get + * 07FA. Next, we SHA that with "0": the SHA of 07 FA 30 is + * 3370135c5e3da4fed937adc004a79533962b6391. So we XOR the + * 12-bit string 337 into the input 567 to get 650. Thus + * our output is 07FA650. + */ + { + unsigned char bmp1[] = "\x12\x34\x56\x70"; + obfuscate_bitmap(bmp1, 28, false); + printf("test 1 encode: %s\n", + memcmp(bmp1, "\x07\xfa\x65\x00", 4) ? FAILED : "passed"); + obfuscate_bitmap(bmp1, 28, true); + printf("test 1 decode: %s\n", + memcmp(bmp1, "\x12\x34\x56\x70", 4) ? FAILED : "passed"); + } + /* + * Second test: a long string to make sure we switch from + * one SHA to the next correctly. My input string this time + * is simply fifty bytes of zeroes. + */ + { + unsigned char bmp2[50]; + unsigned char bmp2a[50]; + memset(bmp2, 0, 50); + memset(bmp2a, 0, 50); + obfuscate_bitmap(bmp2, 50 * 8, false); + /* + * SHA of twenty-five zero bytes plus "0" is + * b202c07b990c01f6ff2d544707f60e506019b671. SHA of + * twenty-five zero bytes plus "1" is + * fcb1d8b5a2f6b592fe6780b36aa9d65dd7aa6db9. Thus our + * first half becomes + * b202c07b990c01f6ff2d544707f60e506019b671fcb1d8b5a2. + * + * SHA of that lot plus "0" is + * 10b0af913db85d37ca27f52a9f78bba3a80030db. SHA of the + * same string plus "1" is + * 3d01d8df78e76d382b8106f480135a1bc751d725. So the + * second half becomes + * 10b0af913db85d37ca27f52a9f78bba3a80030db3d01d8df78. + */ + printf("test 2 encode: %s\n", + memcmp(bmp2, "\xb2\x02\xc0\x7b\x99\x0c\x01\xf6\xff\x2d\x54" + "\x47\x07\xf6\x0e\x50\x60\x19\xb6\x71\xfc\xb1\xd8" + "\xb5\xa2\x10\xb0\xaf\x91\x3d\xb8\x5d\x37\xca\x27" + "\xf5\x2a\x9f\x78\xbb\xa3\xa8\x00\x30\xdb\x3d\x01" + "\xd8\xdf\x78", 50) ? FAILED : "passed"); + obfuscate_bitmap(bmp2, 50 * 8, true); + printf("test 2 decode: %s\n", + memcmp(bmp2, bmp2a, 50) ? FAILED : "passed"); + } + + #undef FAILED + + return ok; +} + int main(int argc, char **argv) { enum { BINARY, DEFAULT, HEX } outputmode = DEFAULT; char *inhex = NULL; unsigned char *data; int datalen; - enum { UNKNOWN, DECODE, ENCODE } mode = UNKNOWN; + enum { UNKNOWN, DECODE, ENCODE, SELFTEST } mode = UNKNOWN; bool doing_opts = true; while (--argc > 0) { @@ -56,6 +125,9 @@ int main(int argc, char **argv) case 'd': mode = DECODE; break; + case 't': + mode = SELFTEST; + break; case 'b': outputmode = BINARY; break; @@ -81,9 +153,14 @@ int main(int argc, char **argv) if (mode == UNKNOWN) { fprintf(stderr, "usage: obfusc < -e | -d > [ -b | -h ] [hex data]\n"); + fprintf(stderr, " or: obfusc -t to run self-tests\n"); return 0; } + if (mode == SELFTEST) { + return self_tests() ? 0 : 1; + } + if (outputmode == DEFAULT) outputmode = (mode == DECODE ? BINARY : HEX); diff --git a/mines.c b/mines.c index c9deb82..a945dc4 100644 --- a/mines.c +++ b/mines.c @@ -1902,72 +1902,7 @@ static char *describe_layout(bool *grid, int area, int x, int y, static bool *new_mine_layout(int w, int h, int n, int x, int y, bool unique, random_state *rs, char **game_desc) { - bool *grid; - -#ifdef TEST_OBFUSCATION - static int tested_obfuscation = false; - if (!tested_obfuscation) { - /* - * A few simple test vectors for the obfuscator. - * - * First test: the 28-bit stream 1234567. This divides up - * into 1234 and 567[0]. The SHA of 56 70 30 (appending - * "0") is 15ce8ab946640340bbb99f3f48fd2c45d1a31d30. Thus, - * we XOR the 16-bit string 15CE into the input 1234 to get - * 07FA. Next, we SHA that with "0": the SHA of 07 FA 30 is - * 3370135c5e3da4fed937adc004a79533962b6391. So we XOR the - * 12-bit string 337 into the input 567 to get 650. Thus - * our output is 07FA650. - */ - { - unsigned char bmp1[] = "\x12\x34\x56\x70"; - obfuscate_bitmap(bmp1, 28, false); - printf("test 1 encode: %s\n", - memcmp(bmp1, "\x07\xfa\x65\x00", 4) ? "failed" : "passed"); - obfuscate_bitmap(bmp1, 28, true); - printf("test 1 decode: %s\n", - memcmp(bmp1, "\x12\x34\x56\x70", 4) ? "failed" : "passed"); - } - /* - * Second test: a long string to make sure we switch from - * one SHA to the next correctly. My input string this time - * is simply fifty bytes of zeroes. - */ - { - unsigned char bmp2[50]; - unsigned char bmp2a[50]; - memset(bmp2, 0, 50); - memset(bmp2a, 0, 50); - obfuscate_bitmap(bmp2, 50 * 8, false); - /* - * SHA of twenty-five zero bytes plus "0" is - * b202c07b990c01f6ff2d544707f60e506019b671. SHA of - * twenty-five zero bytes plus "1" is - * fcb1d8b5a2f6b592fe6780b36aa9d65dd7aa6db9. Thus our - * first half becomes - * b202c07b990c01f6ff2d544707f60e506019b671fcb1d8b5a2. - * - * SHA of that lot plus "0" is - * 10b0af913db85d37ca27f52a9f78bba3a80030db. SHA of the - * same string plus "1" is - * 3d01d8df78e76d382b8106f480135a1bc751d725. So the - * second half becomes - * 10b0af913db85d37ca27f52a9f78bba3a80030db3d01d8df78. - */ - printf("test 2 encode: %s\n", - memcmp(bmp2, "\xb2\x02\xc0\x7b\x99\x0c\x01\xf6\xff\x2d\x54" - "\x47\x07\xf6\x0e\x50\x60\x19\xb6\x71\xfc\xb1\xd8" - "\xb5\xa2\x10\xb0\xaf\x91\x3d\xb8\x5d\x37\xca\x27" - "\xf5\x2a\x9f\x78\xbb\xa3\xa8\x00\x30\xdb\x3d\x01" - "\xd8\xdf\x78", 50) ? "failed" : "passed"); - obfuscate_bitmap(bmp2, 50 * 8, true); - printf("test 2 decode: %s\n", - memcmp(bmp2, bmp2a, 50) ? "failed" : "passed"); - } - } -#endif - - grid = minegen(w, h, n, x, y, unique, rs); + bool *grid = minegen(w, h, n, x, y, unique, rs); if (game_desc) *game_desc = describe_layout(grid, w * h, x, y, true);