Journal TheRaven64's Journal: Decrypting the OpenBSD Theme Song 5
Converting them to decimal, we get:
100001 = 32 + 1 = 33
1010101 = 64 + 16 + 4 + 1 = 85
As ASCII codes, these are ! and U. Not particularly meaningful, but it gives us a hint. Considering the song's subject some connection to money could be a good guess. Considering OpenBSD's focus on cryptography, it seems like it might be encrypted in some way, but presumably some way that's known to be insecure enough that someone with only two characters and a knowledge of the context can decrypt it. A Caesar Cypher is an obvious bet. Since it's in binary, a power of two seems like a nice bet for an easy-to-guess key. We want one that leaves both characters in the letters region of the character set (65-90, 97-122). Picking 32, we get:
1000001 = 64 + 1 = 65
1110101 = 64 + 32 + 16 + 4 + 1 = 117
This corresponds to the letters A and u. Since Au is the chemical element for gold, this is probably the answer.
Of course, with only a two-letter cyphertext and no knowledge of the algorithm or key, we can't be sure, and the 'real' decryption could be anything, but it seems likely that the correct answer is gold considering the subject of the song. Assuming it is a Caesar Cypher, we know that the distance between the two characters must be 52, so we can write a simple program that will output them all:
#include <stdio.h>
int main(void)
{
for(int i=0 ; i<128 ; i++)
{
printf("%d: %c%c\n", i - 33, (char)i, (char)(i + 52) & 127);
}
return 0;
}
The only results where both are in the letter range are:
32: Au
33: Bv
34: Cw
35: Dx
36: Ey
37: Fz
Of these, only Au is an atomic symbol. The others might have some other meaning, but Au still seems like the best bet. No other powers of two give us a value in the letter range, although 16 gives 1e, which might mean something to someone (decimal 30? ASCII code for record separator?).
In summary, the title for the new OpenBSD theme song could be anything, but is probably Gold. Also, I am definitely a geek.
decoded... (Score:1)
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