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::ICS 804:: Theory of Computation - Ibrahim Otieno - [email protected] +254-0722-429297PowerPoint Presentation

::ICS 804:: Theory of Computation - Ibrahim Otieno - [email protected] +254-0722-429297

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Theory of Computation

- Ibrahim Otieno -

+254-0722-429297

SCI/ICT Building Rm. G15

Course Outline

- Mathematical Preliminaries
- Turing Machines
- Recursion Theory
- Markov Algorithms
- Register Machines
- Regular Languages and finite-state automata
- Aspects of Computability

Last Week: Markov Algorithms

- Sequential computation and Markov Algorithms
- Language Acceptors/Recognizers
- Number-theoretic functions
- Labeled Markov algorithms
- Markov-Computable functions and partial recursive functions
- Efficiency

Course Outline

- Mathematical Preliminaries
- Turing Machines
- Additional Varieties of Turing Machines

- Recursion Theory
- Markov Algorithms
- Register Machines
- Regular Languages and finite-state automata
- Aspects of Computability

Register Machines

- Register machines
- Register machines and formal languages
- Model-independent characterization of computational feasibility

Register Machines

- More up-to-date way of thinking about computation
- A register machine (RM) consists of a finite number of registers, instruction set and a CU.
- We write R0, R1, . . . for the registers.
- Each register contains a natural number: [Ri] denotes the content of the ith register.
- Note: there is no bound on the size of numbers stored in our registers
- The CU capable of executing certain instructions that manipulate register contents.

Example

- Register machine M with two registers: R1 and R2
- Register: variable storage
- Initial: R1 contains natural number n

Example

- R2 initializes to 0

Example

- M checks whether R1=R2
- If yes, then M halts. If no, R2’s value is incremented

Some Pseudocode

Input R1 = n

Output R2 = n

Algorithm

Start;

R2:=0;

Loop: R1=R2? goto end;

R2++;

goto loop;

End: Halt

Exercise

- What does this register machine compute?
- Try starting with R1 and R2 both containing 10

Exercise

- What does this register machine compute?
- successor function

Additional Instructions

- *Ri++: increments register Rj where j is the natural number currently stored in register Ri
- Ri = *Rj? goto L: branch to instruction labeled L if the value in Ri is the same as the value in Rk, where k is the value currently stored in register Rj
- Also: *Ri =Rj? goto L
*Ri = *Rj? goto L

*Ri := 0

- This is called indirect addressing (cf. Assembler)

Formal Definition

- A register machine M is any pair <,>
- = {R1, R2, …, Rm} with m1 is a possibly infinite set of registers
- =<t1,t2,…,tt> with t2 is a finite, nonempty sequence of instructions such that:
- Instruction t1 is the start instruction
- Each of t2,…,tt is either the halt instruction or an instruction of the form:
(a) Ri:=0 or *Ri:=0

(b) Ri++ or *Ri++

(c) Ri=Rj? goto L or *Ri=Rj? goto L or

Ri=*Rj? goto L or *Ri=*Rj? goto L

whereRi and Rj are members of while L is an identifier with exactly one of instructions t1,t2,…,tt

- tt is the Halt instruction

M computes k-aryf

- M’s registers R1,R2,…,Rk contain arguments of f
- Ultimately, register Rk+1 contains f(n1,n2,…,nk)
- If f(n1,n2,…,nk) is undefined, then M’s computation never terminates

Register-Machine-Computability

- f is said to be register-machine-computable if there exists a register machine that computes f

An Equivalence Result

- Let h be Turing-computable number-theoretic function.
- Then h is register-machine-computable.
- Then h is a partial recursive function as well.
- If f is register-machine-computable, then f is Turing-computable.

Language Acceptance/recognition

- Adds
- Input tape and output tape
- read(Ri) and write(Ri) instructions

- Initially, “input word” w on tape followed by terminating 0 (sentinel)
- If w L, then 1 written to output tape
- If w L, then 0 written to output tape (recogn)

Example

- L = {w*|na(w) = nb(w)}
- 1s represent as and 2s represent bs
- Register-machine-acceptable language
also: palindromes

Time Analysis feasibility

- Time analysis of register-machine M: count each primitive operation
- Uniform Cost Assumption: size of register contents make no difference
- M in previous example carries out 4 steps for each symbol in word w
- Before loop, there are 2 steps, after end-of-input sentinel 0 is encountered, five steps are executed
- With |w| = n, then there are 4n+5 execution steps
- M computes in linear time

Feasibility (recap) feasibility

- Cobham–Edmonds thesis regarding computational feasibility: the problem of determining whether a given string is a member of a given language L is feasible if and only if L is in P

Complexity class P: class of all languages accepted in polynomial time by some (single-tape) Turing Machine

Problems with the Thesis feasibility

- Computable by what? Focus on single-tape Turing machine
- Which model of computation is the favored one?
- Does it make a difference?

Problems with the Thesis feasibility

e.g. L = {w*|na(w) = nb(w)}

- Register machine: O(n) steps
- Single Tape Turing Machine: O(n2) steps
- Multi-Tape Turing Machine: O(n) steps
- Markov Machine: O(n2) steps

Polynomial Relatedness feasibility

- Suppose that multitape MTM accepts language L in O(n) steps. Then there exists MRM that accepts L in time O([timeTM(n)]2) under the uniform cost assumption.
- Suppose that MRM accepts language L in O(n) steps under the assumption of uniform cost. Then there exists multitape MTM that accepts L in time O([timeTM(n)]3).

Complexity (recap: MA) feasibility

- Given Turing machine M accepting L, there exists a Markov algorithm AS that accepts L in O(timeM(n)) steps
- Given Markov algorithm AS accepting L, there exists a Turing machine M that accepts L in O([timeAS(n)]4) steps

Time and Space (recap: TM) feasibility

- Corollary Suppose that language L is accepted by k-tape Turing machine M. Then L is accepted by a single-tape Turing machine M1 such that timeM1(n) is O([timeM(n)]2)
- Corollary Suppose that language L is accepted by k-tape Turing machine M. Then L is accepted by a single-tape Turing machine M´ such that spaceM´(n) is O(spaceM(n)).

Results Regarding Polynomial Relatedness of Models feasibility

Markov

Algorithm

Model

Theorem 4.3

Multitape

Turing Machine model

Corollary 2.1

Theorem 5.3

Register Machine

Model

Single-tape Turing Machine model

Reformulation feasibility

- In invoking the Cobham-Edmonds thesis, it seems to make no difference which model we take
Reformulation:

- A language-acceptance problem is feasible provided that the language in question is accepted in polynomially bounded time

More generally feasibility

Invariance principle for models of sequential Computation

Let Model be some machine model of sequential computation. Then Model will be said to be a reasonable model of sequential computation if it is polynomially related to the multitape Turing machine model

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