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- 1. Hilbert’s 10th problem Thorsten Altenkirch Hilbert’s 10th problem – p.1/31 David Hilbert (1862-1943) 1900 Paris International Congress Hilbert proposed 23 outstanding problems in Mathematics Hilbert’s 10th pro Hilbert’s problems 1a Is there a transfinite number between that of a denumerable set and the numbers of the continuum? Independent, Cohen 1963 1b Can the continuum of numbers be considered a well ordered set? Yes, Zermelo 1904 using the Axiom of Choice which is independent, Fraenkel 1925 2. Can it be proven that the axioms of logic are consistent? No, Gödel 1931 8. Prove the Riemann hypothesis. Still open 10. Does there exist a universal algorithm for solving Diophantine equations? Topic today Hilbert’s 10th problem – p.3/31 Diophantine equations Example Are there solutions for , yes, , no In general? Hilbert’s 10th pro
- 2. Relatively Prime Every number can be (uniquely) represented as a product of primes, e.g. Def.: Two numbers are relatively prime, iff the lists of primes is disjoint. and are relatively prime. and are not relatively prime. Prop.: has integer solutions, iff and are relatively prime. Hilbert’s 10th problem – p.5/31 Hilbert 10th, revisited Consider Diophantine equations made up from and . Dioph Is there a computer program which decides Dioph ? Given an equation in Dioph the program would answe yes if there is a solution no if there is no solution. Hilbert’s 10th pro Undecidability Turing, 1930: The problem Halt to decide whether a given pro- gram (Turing machine) halts is undecidable, i.e it cannot be solved by any program (Turing machine). Hilbert’s 10th problem – p.7/31 Reduction To show that a problem is undecidable, we construct a reduction Halt , that is a computer program which translates instances of the halting problem into instances of . Why does this work ? Hilbert’s 10th pro
- 3. Yuri Matiyasevich (1947) 1971 Matiyasevich shows that Halt Dioph Hence, the answer to Hilbert’s 10th is negative Hilbert’s 10th problem – p.9/31 Julia Roberts Halt ExpDioph Dioph Hilbert’s 10th prob ExpDioph Consider Diophantine equations made up from and and ExpDioph We will show: ExpDioph is undecidable. By reducing it to the Halting problem for register machines. Hilbert’s 10th problem – p.11/31 Eliminating Prop: A conjunction of 2 equations can be reduced to one. How? Use . This also works for equations (and variables). Hilbert’s 10th prob
- 4. Eliminating Prop: A disjunction of 2 equations can be reduced to one. How? Use . This also works for equations (and variables). Hilbert’s 10th problem – p.13/31 Eliminating negative numbers Dioph Dioph Dioph Dioph How? Hilbert’s 10th prob Dioph Dioph Hint Lagrange: Every natural number can be written as the sum of four squares. ! ! # % $ # % # # % # % ' # $ # # # # ' Hilbert’s 10th problem – p.15/31 Rest of the talk We are going to show that Halt ExpDioph . Here Halt is the Halting problem for Register machines. Hence we have shown that ExpDioph is undecidable. We believe Matiyasevich that ExpDioph Dioph or read his paper. Hilbert’s 10th prob
- 5. Register machines registers: with values in . Program: . . . What are the possible instructions ? Hilbert’s 10th problem – p.17/31 Instructions INC (and DEC ) increments (decrements) register by one. GOTO Goto line . IF GOTO Goto line , if is HALT Ends the program. Why don’t we have or ? Hilbert’s 10th prob IF GOTO DEC GOTO Hilbert’s 10th problem – p.19/31 IF GOTO DEC INC INC IF GOTO DEC INC GOTO Hilbert’s 10th prob
- 6. The Halting problem Given a register machine started with all registers 0, will the machine stop? We are going to constract a set of equations in ExpDioph which has a solution iff the machine stops. Hilbert’s 10th problem – p.21/31 The dominance relation Def.: the th bit of the th bit of , because , because We will see: is definable in ExpDioph Hilbert’s 10th prob Important Variables the largest integer, the number of steps until HALT Values of register Sequence number for instruction is executed at time otherwise. Hilbert’s 10th problem – p.23/31 First equations But these are not equations? Hilbert’s 10th prob
- 7. Hilbert’s 10th problem – p.25/31 More equations Exactly one instruction is executed at any time. The program starts with the first instruction. The last instruction is HALT. Initially all registers are . Hilbert’s 10th prob GOTO Also for INC DEC we add Hilbert’s 10th problem – p.27/31 INC,DEC INC DEC Hilbert’s 10th prob
- 8. IF GOTO The next step is either or To test : Hilbert’s 10th problem – p.29/31 Back to Theorem (Kummer,Lucas): is odd Hence we replace by Hilbert’s 10th prob What about ? Hilbert’s 10th problem – p.31/31