MU puzzle explained

The MU puzzle is a puzzle stated by Douglas Hofstadter and found in Gödel, 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 a Post canonical system and can be reformulated as a string rewriting system.

The puzzle

Suppose there are the symbols,, and which can be combined to produce strings of symbols. The MU puzzle asks one to start with the "axiomatic" string and transform it into the string using in each step one of the following transformation rules:[1] [2]

Nr.          COLSPAN=3 Formal rule[3] Informal explanationCOLSPAN=3 Example
1.ALIGN=RIGHT x → x Add a to the end of any string ending in ALIGN=RIGHT to
2.ALIGN=RIGHT x → xxDouble the string after the ALIGN=RIGHT to
3.ALIGN=RIGHT xy → xyReplace any with a ALIGN=RIGHT to
4.ALIGN=RIGHT xy → xyRemove any ALIGN=RIGHT to

Solution

The puzzle cannot be solved: it is impossible to change the string into 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 an invariant; that is, some quantity or property that doesn't change while applying the rules.

In this case, one can look at the total number of 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, the invariant property is that the number of is not divisible by 3:

Thus, the goal of with zero cannot be achieved because 0 is divisible by 3.

In the language of modular arithmetic, the number n of obeys the congruence

n\equiv2a\not\equiv0\pmod3.

where a counts how often the second rule is applied.

A decidable criterion for derivability

More generally, an arbitrarily given string x can be derived from by the above four rules if, and only if, x respects the three following properties:

  1. x is only composed with one and any number of and,
  2. x begins with, and
  3. the number of in x is not divisible by 3.

Proof

Only if: No rule moves the, changes the number of, or introduces any character out of,, . Therefore, every x derived from respects properties 1 and 2. As shown before, it also respects property 3.

If: If x respects properties 1 to 3, let

NI

and

NU

be the number of and in x, respectively, and let

N=NI+3NU

.By property 3, the number

NI

cannot be divisible by 3, hence,

N

cannot be, either. That is,

N\equiv1orN\equiv2\pmod3

. Let

n\in\N

such that

2n>N

and

2n\equivN\pmod3

.[4] Beginning from the axiom, applying the second rule

n

times obtains ... with

2n

. Since

2n-N

is divisible by 3, by construction of

n

, applying the third rule
2n-N
3
times will obtain ......, with exactly

N

, followed by some number of . The count can always be made even, by applying the first rule once, if necessary. Applying the fourth rule sufficiently often, all can then be deleted, thus obtaining ... with

NI+3NU

. Applying the third rule to reduce triplets of into a in the right spots will obtain x. Altogether, x has been derived from .

Example

To illustrate the construction in the If part of the proof, the string, which respects properties 1 to 3, leads to

NI=4

,

NU=1

,

N=7

,

n=4

; it can be hence derived as follows:

.

Arithmetization

Chapter XIX of Gödel, 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 letters,, and to the numbers 3, 1, and 0, respectively. (For example, the string would be mapped to 31010.)

Second, the single axiom of the MIU system, namely the string, becomes the number 31.

Third, the four formal rules given above become the following:

Nr.          COLSPAN=3 ALIGN=CENTER Formal rule[5] COLSPAN=4 ALIGN=CENTER Example 
1.ALIGN=RIGHT k × 10 + 1  →  10 × (k × 10 + 1)          ALIGN=RIGHT 31  →  310ALIGN=LEFT  (k = 3)
2.ALIGN=RIGHT 3 × 10m + n  →  10m × (3 × 10m + n) + nALIGN=RIGHT 310  →  31010ALIGN=LEFT  (m = 2, n = 10)
3.ALIGN=RIGHT k × 10m+3 + 111 × 10m + n  →  k × 10m + 1 + nALIGN=RIGHT 3111011  →  30011ALIGN=LEFT  (k = 3, m = 3, n = 11)
4.ALIGN=RIGHT k × 10m+2 + n  →  k × 10m + nALIGN=RIGHT 30011  →  311ALIGN=LEFT  (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 variable m was reserved for use in exponents of 10 only, and therefore it was replaced by k 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.)

Relationship to logic

The MIU system illustrates several important concepts in logic by means of analogy.

It can be interpreted as an analogy for a formal system — an encapsulation of mathematical and logical concepts using symbols. The MI string is akin to a single axiom, and the four transformation rules are akin to rules of inference.

The MU string and the impossibility of its derivation is then analogous to a statement of mathematical logic which cannot be proven 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 not refer 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 reason about the system, rather than working within it. Eventually, one realises that the system is in some way about 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 behind Gödel's Incompleteness Theorem.

Pedagogical uses

In her textbook, Discrete Mathematics with Applications, Susanna S. Epp uses the MU puzzle to introduce the concept of recursive definitions, and begins the relevant chapter with a quote from GEB.[6]

See also

External links

Notes and References

  1. Justin Curry / Curran Kelleher . 2007 . Gödel, Escher, Bach: A Mental Space Odyssey . MIT OpenCourseWare.
  2. Here: Chapter I.
  3. Here, x and y are variables, standing for strings of symbols. A rule may be applied only to the whole string, not to an arbitrary substring.
  4. Such an

    n

    always exists, since the powers of 2 alternatingly evaluate to 1 and 2, modulo 3.
  5. Here, k and m stand for arbitrary natural numbers, and n is any natural number smaller than 10m. Each rule of the form "x → y" should be read as "if we have made x we can make y." As the Example column illustrates, a rule may be applied only to an entire MIU-number, not to an arbitrary part of its decimal representation.
  6. Discrete Mathematics with Applications, Third Edition, Brooks/Cole, 2004. Chapter 8.4, "General Recursive Definitions," p. 501.

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