The Mu Puzzle
22 Sep 2015
Imagine strings that only contains the letters
M
,I
andU
. Every valid string begins with anM
. You can generate new strings by successively applying one of four rules: (1) you can add anU
to any string that ends in anI
; (2) you can double everything from anM
to the end of the string; (3) any occurrence ofIII
can be replaced by a singleU
; and (4) you can simply remove any occurrences ofUU
. The puzzle is thus: starting from the stringMI
, what is the sequence of rules one must apply to produce the stringMU
? ^{1}
Let’s first summarize the rules in a nice table, so that we can be clear about their semantics:
Rule  Example  

1  xI → xIU 
MII → MIIU 
2  Mx → Mxx 
MUI → MUIUI 
3  xIIIy → xUy 
MUIIIU → MUUU 
4  xUUy → xy 
MIUUUI → MIUI 
Assume x
and y
denote any arbitrary (sub)string. Let’s play the game a little bit. Suppose we start with the string MUI
:
MUI { axiom }
MUIU { by applying rule 1 }
MUIUUIU { rule 2 }
MUIUUIUUIU { 2 }
MUIIUUIU { 4 }
MUIIIU { 4 }
MUU { 3 }
M { 4 }
It seems we’ve reached a point where there’s no rule we can apply (well, except rule 2, but that would not change the string M
). Now, the puzzle asks us: what’s the sequence of steps that transforms a MI
into a MU
?
Why a BFS and not a DFS? Suppose you’ve chosen to use DFS. Now, when would a long sequence of rules be long enough that you should begin to search for alternatives?
We can attempt to solve this problem with a program that does a breadthfirst search, by recursively applying every rule that is applicable (i.e. valid) to a reached string and checking if it is equal to MU
. We would, however, quickly reach the conclusion that the program doesn’t stop after a considerable amount of time/steps.
In these situations, one might begin to wonder: “regardless of the sequence of rules one apply, would it be possible to reach MU
starting from a MI
?”
The Unbearable Impossibility of MU
Should the reader begin to be convinced that maybe this puzzle doesn’t have a solution, how would one proceed to mathematically prove it?
One common strategy, particularly when loops or recursion is part of the problem, is to (1) establish a proposition that asserts something about the underlying formal system and (2) show that if the proposition holds for a certain string, it would also hold for all strings that are the result of applying the rules. Such kind of proposition is called an invariant, for obvious reasons:
… (an) invariant is a property, held by a class of mathematical objects, which remains unchanged when transformations of a certain type are applied to the objects.
What would be a good invariant in this case? Well, starting from MI
one needs to get rid of the I
and create a U
. Creating U
’s is straightforward due to rule 1 (or rule 3), but getting rid of I
’s is trickier — rule 3 requires three I
’s to get rid of them, and the only way to create more I
’s is through the slightly chaotic rule 2, provided we start with any I
in the first place. Getting rid of extra U
’s is also easy due to rule 4.
Thinking backwards, one would need to reach a state where three I
’s exist (or any multiple of 3). Starting from MI
, that has only one I
, can we create 3 of them (or 6, 9…)? Here’s a hint to the property we were looking for: the number of I
’s. Let’s state the invariant as such:
Regardless of what we do, the number of
I
’s in a string is not a multiple of 3.
MI
has only one I
, so the invariant holds in the beginning. If you recall proof by induction, this is the base case. Now, regardless of the string, does applying rule 1 preserve the invariant? Yes it does, particularly because it doesn’t change the number of I
’s. Same goes for rule 4.
In doubt, recall the notion of divisor and note that 2 and 3 are prime numbers.
Rule 2 doubles the number of I
’s present in a string. However, if $n$ is not a multiple of 3, 2$n$ is still not a multiple of 3 either. Rule 3, on the other hand, reduces the number of I
’s by 3; but similarly, if $n$ is not a multiple of 3, $n$3 isn’t either.
Hence, we are forced to accept that, regardless the sequence of rules we try, by beginning with MI
we would never be able to reach a state where producing a MU
is a possibility. ∎

I first learned about this puzzle in the amazing Douglas Hofstadter’s book “Gödel, Escher, Bach: An Eternal Golden Braid”, and I’ve been using it in the last years as a way to introduce formal proofs via invariant preservation. As always, one can find more information about this puzzle in Wikipedia. ↩