TheMU puzzle is a puzzle stated byDouglas Hofstadter and found inGödel, Escher, Bach involving a simple formal system called "MIU". Hofstadter's motivation is to contrast reasoning within a formal system (i.e., deriving theorems) against reasoning about the formal system itself. MIU is an example of aPost canonical system and can be reformulated as astring rewriting system.
Suppose there are the symbolsM
,I
, andU
which can be combined to produce strings of symbols. The MU puzzle asks one to start with the "axiomatic" stringMI
and transform it into the stringMU
using in each step one of the following transformation rules:[1][2]
Nr. | Formal rule[note 1] | Informal explanation | Example | ||||
1. | xI | → | xIU | Add aU to the end of any string ending inI | MI | to | MIU |
2. | M x | → | M xx | Double the string after theM | MIU | to | MIUIU |
3. | xIII y | → | xU y | Replace anyIII with aU | MUIIIU | to | MUUU |
4. | xUU y | → | xy | Remove anyUU | MUUU | to | MU |
The puzzle cannot be solved: it is impossible to change the stringMI
intoMU
by repeatedly applying the given rules. In other words, MU is not a theorem of the MIU formal system. To prove this, one must step "outside" the formal system itself.
In order to prove assertions like this, it is often beneficial to look for aninvariant; that is, some quantity or property that doesn't change while applying the rules.
In this case, one can look at the total number ofI
in a string. Only the second and third rules change this number. In particular, rule two will double it while rule three will reduce it by 3. Now, theinvariant property is that, in any string produced when starting withMI
, the number ofI
is notdivisible by 3:
I
s is 1 which is not divisible by 3.Thus, the goal ofMU
with zeroI
cannot be achieved because 0is divisible by 3.
In the language ofmodular arithmetic, the numbern ofI
obeys the congruence
wherea counts how often the second rule is applied.
More generally, an arbitrarily given stringx can be derived fromMI
by theabove four rulesif, and only if,x respects the three following properties:
M
and any number ofI
andU
,M
, andI
inx is not divisible by 3.Only if: No rule moves theM
, changes the number ofM
, or introduces any character out ofM
,I
,U
. Therefore, everyx derived fromMI
respects properties 1 and 2. As shownbefore, it also respects property 3.
If: Ifx respects properties 1 to 3, let and be the number ofI
andU
inx, respectively, and let.By property 3, the number cannot be divisible by 3, hence, cannot be, either. That is,. Let such that and.[note 2] Beginning from the axiomMI
, applying the second rule times obtainsMIII
...I
withI
. Since is divisible by 3, by construction of, applying the third rule times will obtainMIII
...IU
...U
, with exactlyI
, followed by some number ofU
. TheU
count can always be made even, by applying the first rule once, if necessary. Applying the fourth rule sufficiently often, allU
can then be deleted, thus obtainingMIII
...I
withI
. Applying the third rule to reduce triplets ofI
into aU
in the right spots will obtainx. Altogether,x has been derived fromMI
.
To illustrate the construction in theIf part of the proof, the stringMIIUII
, which respects properties 1 to 3, leads to,,,; it can be hence derived as follows:
MI
2→MII
2→MIIII
2→MIIIIIIII
2→MIIIIIIIIIIIIIIII
3→MIIIIIIIUIIIIII
3→MIIIIIIIUUIII
1→MIIIIIIIUUIIIU
3→MIIIIIIIUUUU
4→MIIIIIIIUU
4→MIIIIIII
3→MIIUII
.Chapter XIX ofGödel, Escher, Bach gives a mapping of the MIU system to arithmetic, as follows. First, every MIU-string can be translated to an integer by mapping the lettersM
,I
, andU
to the numbers 3, 1, and 0, respectively. (For example, the stringMIUIU
would be mapped to 31010.)
Second, the single axiom of the MIU system, namely the stringMI
, becomes the number 31.
Third, the four formal rules given above become the following:
Nr. | Formal rule[note 3] | Example | ||||||
1. | k × 10 + 1 | → | 10 × (k × 10 + 1) | 31 | → | 310 | (k = 3) | |
2. | 3 × 10m +n | → | 10m × (3 × 10m +n) +n | 310 | → | 31010 | (m = 2,n = 10) | |
3. | k × 10m+3 + 111 × 10m +n | → | k × 10m + 1 +n | 3111011 | → | 30011 | (k = 3,m = 3,n = 11) | |
4. | k × 10m+2 +n | → | k × 10m +n | 30011 | → | 311 | (k = 3,m = 2,n = 11) |
(NB: The rendering of the first rule above differs superficially from that in the book, where it is written as "[i]f we have made 10m + 1, then we can make 10 × (10m + 1)". Here, however, the variablem was reserved for use in exponents of 10 only, and therefore it was replaced byk in the first rule. Also, in this rendering, the arrangement of factors in this rule was made consistent with that of the other three rules.)
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The MIU system illustrates several important concepts in logic by means of analogy.
It can be interpreted as an analogy for aformal system — an encapsulation of mathematical and logical concepts using symbols. The MI string is akin to a singleaxiom, and the four transformation rules are akin torules of inference.
The MU string and the impossibility of its derivation is then analogous to a statement of mathematical logic which cannot beproven or disproven by the formal system.
It also demonstrates the contrast between interpretation on the "syntactic" level of symbols and on the "semantic" level of meanings. On the syntactic level, there is no knowledge of the MU puzzle's insolubility. The system does notrefer to anything: it is simply a game involving meaningless strings. Working within the system, an algorithm could successively generate every valid string of symbols in an attempt to generate MU, and though it would never succeed, it would search forever, never deducing that the quest was futile. For a human player, however, after a number of attempts, one soon begins to suspect that the puzzle may be impossible. One then "jumps out of the system" and starts to reasonabout the system, rather than working within it. Eventually, one realises that the system is in some wayabout divisibility by three. This is the "semantic" level of the system — a level of meaning that the system naturally attains. On this level, the MU puzzle can be seen to be impossible.
The inability of the MIU system to express or deduce facts about itself, such as the inability to derive MU, is a consequence of its simplicity. However, more complex formal systems, such as systems of mathematical logic, may possess this ability. This is the key idea behindGödel's Incompleteness Theorem.
In her textbook,Discrete Mathematics with Applications,Susanna S. Epp uses the MU puzzle to introduce the concept ofrecursive definitions, and begins the relevant chapter with a quote fromGEB.[3]