+ for (i = cfg; i->type != C_END; i++)
+ if (i->type == C_STRING)
+ sfree(i->sval);
+ sfree(cfg);
+}
+
+/*
+ * The Mines (among others) game descriptions contain the location of every
+ * mine, and can therefore be used to cheat.
+ *
+ * It would be pointless to attempt to _prevent_ this form of
+ * cheating by encrypting the description, since Mines is
+ * open-source so anyone can find out the encryption key. However,
+ * I think it is worth doing a bit of gentle obfuscation to prevent
+ * _accidental_ spoilers: if you happened to note that the game ID
+ * starts with an F, for example, you might be unable to put the
+ * knowledge of those mines out of your mind while playing. So,
+ * just as discussions of film endings are rot13ed to avoid
+ * spoiling it for people who don't want to be told, we apply a
+ * keyless, reversible, but visually completely obfuscatory masking
+ * function to the mine bitmap.
+ */
+void obfuscate_bitmap(unsigned char *bmp, int bits, int decode)
+{
+ int bytes, firsthalf, secondhalf;
+ struct step {
+ unsigned char *seedstart;
+ int seedlen;
+ unsigned char *targetstart;
+ int targetlen;
+ } steps[2];
+ int i, j;
+
+ /*
+ * My obfuscation algorithm is similar in concept to the OAEP
+ * encoding used in some forms of RSA. Here's a specification
+ * of it:
+ *
+ * + We have a `masking function' which constructs a stream of
+ * pseudorandom bytes from a seed of some number of input
+ * bytes.
+ *
+ * + We pad out our input bit stream to a whole number of
+ * bytes by adding up to 7 zero bits on the end. (In fact
+ * the bitmap passed as input to this function will already
+ * have had this done in practice.)
+ *
+ * + We divide the _byte_ stream exactly in half, rounding the
+ * half-way position _down_. So an 81-bit input string, for
+ * example, rounds up to 88 bits or 11 bytes, and then
+ * dividing by two gives 5 bytes in the first half and 6 in
+ * the second half.
+ *
+ * + We generate a mask from the second half of the bytes, and
+ * XOR it over the first half.
+ *
+ * + We generate a mask from the (encoded) first half of the
+ * bytes, and XOR it over the second half. Any null bits at
+ * the end which were added as padding are cleared back to
+ * zero even if this operation would have made them nonzero.
+ *
+ * To de-obfuscate, the steps are precisely the same except
+ * that the final two are reversed.
+ *
+ * Finally, our masking function. Given an input seed string of
+ * bytes, the output mask consists of concatenating the SHA-1
+ * hashes of the seed string and successive decimal integers,
+ * starting from 0.
+ */
+
+ bytes = (bits + 7) / 8;
+ firsthalf = bytes / 2;
+ secondhalf = bytes - firsthalf;
+
+ steps[decode ? 1 : 0].seedstart = bmp + firsthalf;
+ steps[decode ? 1 : 0].seedlen = secondhalf;
+ steps[decode ? 1 : 0].targetstart = bmp;
+ steps[decode ? 1 : 0].targetlen = firsthalf;
+
+ steps[decode ? 0 : 1].seedstart = bmp;
+ steps[decode ? 0 : 1].seedlen = firsthalf;
+ steps[decode ? 0 : 1].targetstart = bmp + firsthalf;
+ steps[decode ? 0 : 1].targetlen = secondhalf;
+
+ for (i = 0; i < 2; i++) {
+ SHA_State base, final;
+ unsigned char digest[20];
+ char numberbuf[80];
+ int digestpos = 20, counter = 0;
+
+ SHA_Init(&base);
+ SHA_Bytes(&base, steps[i].seedstart, steps[i].seedlen);