Lecture 4: Unsolvable Problems

Lecture 4: Unsolvable Problems 虞台文大同大學資工所智慧型多媒體研究室

Content • Enumerable and Decidable Sets • Algorithm Solvability • Problem Reduction • Diagonalization Method

Lecture 4: Unsolvable Problems Enumerable and Decidable Sets 大同大學資工所智慧型多媒體研究室

Set Enumerability and Generability A * is enumerable (generable) if  computable functiong: N* such that If g is not totally defined, then D(g)={0, 1, …, k 1}.

= = = = Enumerable (Generable) The Generation Process A Generate

   = = = Enumerable (Generable) The Enumeration Process A Enumerate 

   = = = Enumerable (Generable) The Enumeration Process Is such a process always terminable? A Enumerate 

   = = = Enumerable (Generable) The Enumeration Process A Enumerate

   = = = Enumerable (Generable) The Labeling Process A Labeling 2

Exercises • Write two algorithms to enumerate (generate) the set of all even numbers of integer in different sequences. • What is difference between ‘countable’ (studied in discrete math) and `enumerable’?

The Countability of Real Number Is R countable? Is (0,1)R countable?

… 0 1 2 3 4 … d00 d01 d02 d03 d04 0. f0 … d10 d11 d12 d13 d14 0. f1 … d20 d21 d22 d23 d24 0. f2 … d30 d31 d32 d33 d34 0. f3 … d40 d41 d42 d43 d44 0. f4 … … … … … … … … Prove (0,1)R is uncountable Assume that (0, 1) is countable. Suppose that it is counted as:

… 0 1 2 3 4 … d01 d02 d03 d04 0. f0 … d10 d12 d13 d14 0. f1 … d20 d21 d23 d24 0. f2 … d30 d31 d32 d34 0. f3 … d40 d41 d42 d43 0. f4 … … … … … … … Prove (0,1)R is uncountable Assume that (0, 1) is countable. Suppose that it is counted as: Consider the following number: d00 d11 d22 d33 d44 …

… 0 1 2 3 4 … d00 d01 d02 d03 d04 0. f0 … d10 d11 d12 d13 d14 0. f1 … d20 d21 d22 d23 d24 0. f2 … d30 d31 d32 d33 d34 0. f3 … d40 d41 d42 d43 d44 0. f4 … … … … … … … … 0. h0 h1 h2 h3 h4 v=fk … … … … … … … Prove (0,1)R is uncountable Assume that (0, 1) is countable. Suppose that it is counted as: Consider the following number: …

… 0 1 2 3 4 … d00 d01 d02 d03 d04 0. f0 … d10 d11 d12 d13 d14 0. f1 … d20 d21 d22 d23 d24 0. f2 … d30 d31 d32 d33 d34 0. f3 … d40 d41 d42 d43 d44 0. f4 … … … … … … … … … … … … … … … Prove (0,1)R is uncountable Assume that (0, 1) is countable. Suppose that it is counted as: Consider the following number: … 0. h0 h1 h2 h3 h4 v=fk

A Semidecidability A * is semidecidable if there exists an algorithm which, when applying to any   A, can decide that  is in A, i.e., there exists an membership algorithm for A. True False or Never Halt

Theorem 1 Ais enumerable iff A is semidecidable. “” Pf) Ais enumerable  g: N* such that g(N)=A. Hence, given any A, we generate strings . . . . . . . . g(1)=1 g(0)=0 Then, we will finally reach g(k) = k = . This allows us to concludes that A.

We have Want Theorem 1 Ais enumerable iff A is semidecidable. “” Pf) Thinking first. What do we know? There exists an membership algorithm  for A on TM such that A is semidecidable How to prove? Find g such that

1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 We have Want Theorem 1 Ais enumerable iff A is semidecidable. “” Pf) Numbering the following black cells Preliminaries

1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 7 3 4 5 6 0 1 2 8 9 10 Enumerable We have Want Theorem 1 Ais enumerable iff A is semidecidable. “” Pf) Numbering the following black cells Preliminaries 6  = {0, 1} 0 * = {, 0, 1, 00, 01, 10, 11, 000, …} 1 5 2 3 4 7

We have Want Theorem 1 Ais enumerable iff A is semidecidable. “” Pf) Step 1. Set n = k + 1. Step 2. List the first n strings Step 3. Hand simulates n steps of  on TM for each The algorithm to generate g(k):

- 0 n1 0 1 2 3 4 5 6 7 8 9 1 2 3 n Steps 4 5 6 7 8 9 10 We have Want Theorem 1 Ais enumerable iff A is semidecidable. “” Pf) Step 1. Set n = k + 1. Step 2. List the first n strings Step 3. Hand simulates n steps of  on TM for each The algorithm to generate g(k): Terminating Record

- 0 n1 0 1 2 3 4 5 6 7 8 9 1 2 3 n Steps 4 5 6 7 8 9 10 We have Want Theorem 1 Ais enumerable iff A is semidecidable. “” Pf) Step 1. Set n = k + 1. Step 2. List the first n strings Step 3. Hand simulates n steps of  on TM for each The algorithm to generate g(k): Step 4. Let m=#terminating computations. Step 5. if m < k + 1 n=n + 1; goto Step 2; else return the kth string. k

Theorem 2 Ais enumerable iff A is a domain of some computable functions. Pf) Exercise

Theorem 2 Ais enumerable iff A is a domain of some computable functions. How about if A is not enumerable?

Discussion Enumeration Program Computability (Some reformation may be required) Enumeration Membership Determination (Semidecidability)

Complement Set

Decidable Set A * is decidable if there exists an computable function, say, D such that

Ais decidable iff and are enumerable. Ais decidable iff and are semidecidable. Theorem 3

Ais decidable iff and are 0 1 2 3 4 . . . D: . . . t f t f t . . . g(0) g(1) g(2) enumerable. semidecidable. Theorem 3 Pf) “” • Fact: * can be enumerated as 0, 1, 2, … • Since A is decidable,  a computable function D st. • The computable function g to enumerate A can be: Show that A is enumerable as follows: g: Ais enumerable.

Ais decidable iff and are Show that is enumerable as follows: 0 1 2 3 4 . . . D: . . . t f f t t is enumerable. . . . h(0) h(1) enumerable. semidecidable. Theorem 3 Pf) “” • Fact: * can be enumerated as 0, 1, 2, … • Since A is decidable,  a computable function D st. • The computable function h to enumerate can be: h:

Ais decidable iff and are enumerable. semidecidable. Theorem 3 Pf) “” • Suppose 1 semidecides A, 2 semidecides . • For any *, input it to 1 and 2 simultaneously. • If 1 terminates, return true. • If 2 terminates, return false.

Lecture 4: Unsolvable Problems Algorithm Solvability 大同大學資工所智慧型多媒體研究室

Enumeration, Decidability and Solvability Enumeration: (Semidecidability) Decidability: To deal with the existence of algorithm (program) to solve problems. Solvability: That is, given a problem, whether there exists a program to solve (any instance of) the problem.

Example Problem (PE: Program Equivalence): PE q = “Are program 1and program 2equivalent?” Any Algorithm?

string to represent 1 string to represent 2 Solvable? Computable? Example Problem (PE: Program Equivalence): PE q = “Are program 1and program 2equivalent?” Decidable?

string to represent 1 string to represent 2 Solvable? Computable? Problem Reformulation Example Problem (PE: Program Equivalence): PE q = “Are program 1and program 2equivalent?” Decidable? Problem PE is solvable iff A is decidable (recursive).

Definitions • Semisolvable: • Solvable: • Totally Unsolvable: if q is true, it answers ‘yes’. if q is false, it may not give answer (never halt).  algorithm if q is true, it answers ‘yes’. if q is false, it answers ‘no’.  algorithm not semisolvale.

Problem and Complement Problem

Problem and Complement Problem Example: Turing Machine Version

Problem and Complement Problem Example: Register Machine Version

P is solvable iff and are semisolvable. Theorem 4 Pf) “” algorithm P such that P is solvable if q is true, it answers ‘yes’. if q is false, it answers ‘no’. To show that P is semisolvable, we need to find a algorithm ’ such that ?

P is solvable iff and are semisolvable. ’: START q false true P  TRUE HALT Theorem 4 Pf) “” algorithm P such that P is solvable if q is true, it answers ‘yes’. if q is false, it answers ‘no’. To show that P is semisolvable, we need to find a algorithm ’ such that It can be

P is solvable iff and are semisolvable. To show that is semisolvable, we need to find a algorithm ” such that ”: START q true false P  TRUE HALT Theorem 4 Pf) “” algorithm P such that P is solvable if q is true, it answers ‘yes’. if q is false, it answers ‘no’. It can be

P is solvable iff and are semisolvable. Theorem 4 Pf) “” To run ’ and ” alternatively, if ’ answers ‘yes’, then ‘yes’; if ” answers ‘yes’, then ‘no’.

P is solvable iff and are semisolvable. Corollary P is semisolvable but not solvable is totally unsolvable.

More simple, The Halting Problem • Register Machine Version • Turing Machine Version

Head S1 S2 … Sk1 Sk Sk+1 … Sn1 Sn Blank Tape Blank Tape START  HALT Theorem 5  The HP (halting problem) is not solvable. 會停嗎? We will prove a simpler version.

Head S1 S2 … Sk1 Sk Sk+1 … Sn1 Sn Blank Tape Blank Tape START  HALT Theorem 5  The HP (halting problem) is not solvable. 會停嗎?

Lecture 4: Unsolvable Problems