I think the answer is:
Uranus
I looked for commonalities between the words to get a substitute started. It looks like:
t
and y
mean themselves. They have not been changed.
This gives a starting point.
The letter o is a common letter and I see it at the beginning of the phrase. Used in to
. This also helps with u
, giving the words our
and you
.
Judging by the number of letters, in the first sentence, I've deemed it to state:
Welcome to our home
Working, from this assumption, I have a partial cypher.
welcome to our home
9| you =re here to co61uer u3 tur6 home 6ow
th93 3oc9ety 93 h97hly =,0=6ce,
the t=(le o| eleme6t3 co6t=963 =ct969um
Now, I know a few more letters. I can assume:
=
is a
. And the third sentence says: "This society is highly advanced"
This gives me more to work with.
I run it again.
welcome to our home
i| you are here to con1uer us turn home now
this society is highly advanced
the ta(le o| elements contains actinium
Clearly, we're near the end. Add the last few letters and:
welcome to our home
if you are here to conquer us turn home now
this society is highly advanced
the table of elements contains actinium
Now....
Where can we find actinium?
This is where it gets a little shaky, and I'm not certain.
if you are here to conquer us turn home now
this society is highly advanced
A society with nuclear bombs could be considered highly advanced.
actinium is found as a byproduct of uranium. The catch, though, is that I'm basing the answer purely off the relationship between the words Uranium, and Uranus.
I used the following to decrypt, making it as I went:
void Main()
{
string phrase = @"./4)-5/ t- -+2 8-5/
9| y-+ =2/ 8/2/ t- )-61+/2 +3 t+26 8-5/ 6-.
t893 3-)9/ty 93 89784y =,0=6)/,
t8/ t=(4/ -| /4/5/6t3 )-6t=963 =)t969+5";
Dictionary Cypher = new Dictionary()
{
{ '.', 'w' },
{ '/', 'e' },
{ '4', 'l' },
{ ')', 'c' },
{ '-', 'o' },
{ '5', 'm' },
{ 't', 't' },
{ '+', 'u' },
{ '2', 'r' },
{ '8', 'h' },
{ 'y', 'y' },
{ '=' , 'a' },
{ '9' , 'i' },
{ '3' , 's' },
{ '7' , 'g' },
{ ',' , 'd' },
{ '6' , 'n' },
{ '0' , 'v' },
{ '|' , 'f' },
{ '(' , 'b' },
{ '1' , 'q' },
};
GetDecoded(phrase, Cypher).Dump();
}
public string GetDecoded(string phrase, Dictionary cypher)
{
StringBuilder result = new StringBuilder();
for (int i = 0; i < phrase.Length; i++)
{
try
{
result.Append(cypher.Single(a => a.Key == phrase[i]).Value);
}
catch
{
result.Append(phrase[i]);
}
}
return result.ToString();
}