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| The '''MU puzzle''' is a puzzle stated by [[Douglas Hofstadter]] and found in ''[[Gödel, Escher, Bach]]''. As stated, it is an example of a [[Post canonical system]] and can be reformulated as a [[string rewriting system]].
| | Hello and welcome. My title is Irwin and I totally dig that name. To collect coins is 1 of the things I adore most. My day occupation is a librarian. For a while she's been in South Dakota.<br><br>Feel free to surf to my web page - [http://www.pinaydiaries.com/user/LConsiden std testing at home] |
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| ==The puzzle==
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| {{Quotation|Has the dog Buddha-nature? MU!|[[Zen Koan]]<ref>{{cite web | url=http://ocw.mit.edu/high-school/courses/godel-escher-bach/lecture-notes/geb_v3_5.pdf | title=The MU–Puzzle | accessdate=29 July 2013}}</ref>||}}
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| Suppose there are the symbols <code>M</code>, <code>I</code>, and <code>U</code> which can be combined to produce strings of symbols called "words". The MU puzzle asks one to start with the "axiomatic" word <code>MI</code> and transform it into the word <code>MU</code> using in each step one of the following transformation rules:
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| # Add a <code>U</code> to the end of any string ending in <code>I</code>. For example: <code>MI</code> to <code>MIU</code>.
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| # Double the string after the <code>M</code> (that is, change <code>Mx</code>, to <code>Mxx</code>). For example: <code>MIU</code> to <code>MIUIU</code>.
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| # Replace any <code>III</code> with a <code>U</code>. For example: <code>MUIIIU</code> to <code>MUUU</code>.
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| # Remove any <code>UU</code>. For example: <code>MUUU</code> to <code>MU</code>.
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| Using these four rules is it possible to change <code>MI</code> into <code>MU</code> in a finite number of steps?
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| The production rules can be written in a more schematic way. Suppose <code>x</code> and <code>y</code> behave as variables (standing for strings of symbols). Then the [[Formal grammar|production rules]] can be written as:
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| # <code>xI → xIU</code>
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| # <code>Mx → Mxx</code>
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| # <code>xIIIy → xUy</code>
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| # <code>xUUy → xy</code>
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| Is it possible to obtain the word <code>MU</code> using these rules?
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| <ref>
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| {{cite AV media
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| | people = Justin Curry / Curran Kelleher
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| | year = 2007
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| | title = ''Gödel, Escher, Bach: A Mental Space Odyssey''
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| | url = http://www.youtube.com/watch?v=GSh0XY67CE4
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| | location = [[MIT OpenCourseWare]]
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| }}</ref>
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| ==Solution==
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| The puzzle's solution is no. It is impossible to change the string <code>MI</code> into <code>MU</code> by repeatedly applying the given rules.
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| In order to prove assertions like this, it is often beneficial to look for an [[Invariant (mathematics)|invariant]], that is some quantity or property that doesn't change while applying the rules.
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| In this case, one can look at the total number of <code>I</code> 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 <code>I</code> is not [[divisible]] by 3:
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| * In the beginning, the number of <code>I</code>s is 1 which is not divisible by 3.
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| * Doubling a number that is not divisible by 3 does not make it divisible by 3.
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| * Subtracting 3 from a number that is not divisible by 3 does not make it divisible by 3 either.
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| Thus, the goal of <code>MU</code> with zero <code>I</code> cannot be achieved because 0 ''is'' divisible by 3.
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| In the language of [[modular arithmetic]], the number <math>n</math> of <code>I</code> obeys the congruence
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| :<math>n \equiv 2^a \not\equiv 0 \pmod 3.\,</math>
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| where <math>a</math> counts how often the second rule is applied.<ref name="mu_solution">{{cite web | url=http://www.cems.uwe.ac.uk/~irjohnso/coursenotes/lrc/lect/lect5b21.html | title=Solution to MIU puzzle | accessdate=29 July 2013}}</ref>
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| ==Relationship to provability==
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| {{Refimprove section|date=July 2013}}
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| The result that MU cannot be obtained with these rules demonstrates the notion of [[Independence (mathematical logic)|independence]] in mathematical logic. The MIU system can be viewed as a formal logic in which a string is considered provable if it can be derived by application of the rules starting from MI. In this interpretation, the question is phrased as "Is MU provable in the MIU logic?".
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| Finding an invariant of the inference rules is a common method for establishing independence results.
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| ==See also==
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| *[[Invariant (computer science)]]
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| *[[Unrestricted grammar]]
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| ==References==
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| {{Reflist}}
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| {{Douglas Hofstadter}}
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| {{DEFAULTSORT:Mu Puzzle}}
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| [[Category:Logic puzzles]]
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Hello and welcome. My title is Irwin and I totally dig that name. To collect coins is 1 of the things I adore most. My day occupation is a librarian. For a while she's been in South Dakota.
Feel free to surf to my web page - std testing at home