Presentation on theme: "The MU Puzzle “Gödel, Escher, Bach – An Eternal Golden Braid” By Douglas Hofstadter “Gödel, Escher, Bach – An Eternal Golden Braid” By Douglas Hofstadter."— Presentation transcript:
The MU Puzzle “Gödel, Escher, Bach – An Eternal Golden Braid” By Douglas Hofstadter “Gödel, Escher, Bach – An Eternal Golden Braid” By Douglas Hofstadter
Kurt Gödel Gödel is one of the most famous logicians of all time. He is best known for his "Incompleteness Theorem," which proved that in any axiom system, there are statements that can be neither proved nor disproved. Much later in his life, he showed that the axiom of choice and continuum hypothesis are consistent with the axioms of set theory, which helped lead the way for Cohen’s proof of the independence of these propositions.
Maurits Cornelis Escher M.C. Escher is a very well known artist. His lithographs are extremely popular and thought provoking. These mathematically inspired woodcuts, lithographs and mezzotints, feature impossible constructions, explorations of infinity, architecture, and tessellations. Some of the most famous of this lithographs are Drawing Hands, Relativity, Waterfall, Metamorphosis, Ascending and Descending.
Escher’s Artworks Ascending and Descending Waterfall
Escher’s Artworks Hand with Reflecting Sphere Relativity
Johann Sebastian Bach Johann Sebastian Bach was a German composer, organist, violist, and violinist whose ecclesiastical and secular works for choir, orchestra, and solo instruments drew together the strands of the Baroque period and brought it to its ultimate maturity. Bach’s most important contribution is his work on Baroque fugue. He is now regarded as the supreme composer of the Baroque, and as one of the greatest of all time.
The MU Puzzle "Can you produce MU?" We start with String MI. We have to generate string MU. At any stage one or more rules rules can be applied to obtain alternate strings and the choice of the rule to be applied depend upon the judgement of the solver. But the string can be changed only within the rules. This is called “The Requirement of Formality”.
Rules to be Followed A U can be added at the end of any string whose last letter is I. For Example : MI to MIU xI xIU String after a M can be doubled. For Example : MIU to MIUIU Mx Mxx A III appearing anywhere in the string can be replaced by U. For example : MIUIII to MIUU xIIIY xUy A UU appearing anywhere inside a string can be dropped. For Example: MIUU to MI xUUy xy I
Decision Procedures Imagine a genie (or a super computer) who has all the time in the world, and who enjoys using it to produce theorems of the MIU-system, in a rather methodical way. Here, for instance, is a possible way the genie might go about it Step 1: Apply every applicable rule to the axiom MI. This yields two new theorems MIU, MII. Step 2: Apply every applicable rule to the theorems produced in step 1. This yields three new theorems: MIIU, MIUIU, MIIII. Step 3: Apply every applicable rule to the theorems produced in step 2. This yields five new theorems: MIIIIU, MIIUIIU, MIUIUIUIU, MIIIIIIII, MUI. And so on …. This method produces every single theorem sooner or later, because the rules are applied in every conceivable order. All of the lengthening-shortening alternations which we mentioned above eventually get carried out. However, it is not clear how long to wait for a given string
A systematically constructed "tree" of all the theorems of the MIU-system. The N th level down contains those theorems whose derivations contain exactly N steps. The encircled numbers tell which rule was employed. Is MU anywhere in this tree?
Solution The answer to the MU Puzzle is No. It is impossible to change the string MI into MU by repeatedly applying the given rules. Explanation : The number of I contained in the strings produced in this case is not divisible by 3. This is because only 2 nd and 3 rd rule changes the number of I. The 2 nd one doubles it while the 3 rd one reduces it by 3. In the beginning, the number of Is is 1 which is not divisible by 3. Doubling a number that is not divisible by 3 does not make it divisible by 3. Subtracting 3 from a number that is not divisible by 3 does not make it divisible by 3 either. Thus, the goal of MU with zero I cannot be achieved because 0 is divisible by 3.
AI and the MU Puzzle We can read M-I-U as M = machine I = intelligence U = understanding In other words, the MU puzzle is really a metaphor for some basic questions about artificial (that is, machine-based) intelligence. Minds can understand things. Minds also exhibit the properties of intelligence and can derive understanding from intelligence. Computers, on the other hand, do not currently understand things, but can and do behave logically, and therefore can be imbued with a certain degree of intelligence. Can we produce MU? (Starting with MI) -- This question essentially explores the difference between Machine Intelligence and Understanding.
Continued.. When enough intelligence is applied, the computer may eventually exhibit output that is practically indistinguishable from otherwise mind-like (that is, human) understanding. But would such a display be actual or real understanding? By using the decision procedures like the one described earlier might take ever to answer such which might otherwise be trivial for a human mind