I'm currently reading Gödel Escher Bach. His first formal system he introduces is the MIU system. It follows four rules.
I
to the U
?The only rule that can reduce the number of I
the the 3rd rule, so we should focus on that.
To remove all I
, we need to fullfill count(I) mod 3 = 0
, which means that the number of I
must be divisible by 3.
count
represents the number of I
, mod
is the mathematical operation modulo.
With the second rule, we can duplicate the starting I
.
But doubling a number that is not divisible by 3 will never make it divisible by 3.
Proof sketch:Suppose you have a number
n
that is not divisible by 3, son mod 3 != 0
, which means thatn = x * 3 + y
, wherex
can be any number andy
is 1 or 2. When doubling the number we getn * 2 = x * 6 + y * 2
.y * 2
is now 2 or 4, both numbers are still not divisible by 2.x * 6
is divisible by 3, so adding 2 or 4 will make it not divisible by 3.
Reducing the number of I
by 3 obviously won't help, we can only use this rule in a useful way when we already have a number that is divisible by 3.
All other rules don't affect the number of I
.
So the first exercise given by Douglas Hofstadter in Gödel Escher Bach is impossible to solve.
When you don't believe this proof, simply try it out and check if you're able to find a solution. You won't.
The code for this page can be found at https://github.com/TN1ck/MIU.
- Tom Nick