116313333134261542156511321455121126113213151461123615112516114142
OK, so recently I made up a way of sending coded messages and I was just wondering how difficult it would be for someone to crack. So let's say, for example, that you are in the army and you caught a suspected spy and you found a slip of paper on him with those numbers on it, exactly as they are above.
The first thing that's apparent is that
the numbers are all in the range 1..6.
Since
the number of letters in the alphabet is comparable to 6 squared,
it's natural to consider
pairs of digits,
especially as
the total number of digits is even. It's also a multiple of 3, but it doesn't look as if there's much repetition in the digit-triples, so that isn't a very promising line of attack.
An obvious guess is that
letters are mapped into pairs of digits in some consistent way (there's already a pretty obvious guess, but let's be a little methodical here).
We then notice that
there are quite a lot of 11 and 15, which would work nicely if those were A (first letter) and E (fifth letter).
So the obvious guess is
that at least 11...16 are a..f.
Looking at what that gives us
and considering whether then 21..26 are g..l, etc.,
we notice
ba.a.cedwhich could be "balanced" if 26 is L
and now
we just do the obvious thing, interpreting a digit pair ab as letter 6(a-1)+b,
and get
a_complete_and_balanced_breakfastwhere each_is a (different, as it happens) "out-of-range" digit pair.
Seems like we've cracked it.
(Is there an egg joke somewhere around here?)
In answer to the question of how difficult it is to crack,
I think it took somewhere between three and five minutes to go through the process above, though I confess my very first thought on looking at the string of digits was pretty much the actual answer.
