How to resolve the algorithm Execute a Markov algorithm step by step in the SETL programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Execute a Markov algorithm step by step in the SETL programming language

Table of Contents

Problem Statement

Create an interpreter for a Markov Algorithm. Rules have the syntax: There is one rule per line. If there is a   .   (period)   present before the   ,   then this is a terminating rule in which case the interpreter must halt execution. A ruleset consists of a sequence of rules, with optional comments.

Rulesets Use the following tests on entries:

Sample text of: Should generate the output:

A test of the terminating rule Sample text of: Should generate:

This tests for correct substitution order and may trap simple regexp based replacement routines if special regexp characters are not escaped. Sample text of: Should generate:

This tests for correct order of scanning of rules, and may trap replacement routines that scan in the wrong order.   It implements a general unary multiplication engine.   (Note that the input expression must be placed within underscores in this implementation.) Sample text of: should generate the output:

A simple Turing machine, implementing a three-state busy beaver. The tape consists of 0s and 1s,   the states are A, B, C and H (for Halt), and the head position is indicated by writing the state letter before the character where the head is. All parts of the initial tape the machine operates on have to be given in the input. Besides demonstrating that the Markov algorithm is Turing-complete, it also made me catch a bug in the C++ implementation which wasn't caught by the first four rulesets. This ruleset should turn into

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Execute a Markov algorithm step by step in the SETL programming language

Source code in the setl programming language

program markov_algorithm;
    magic := false;
    if command_line(1) = om then
        print("error: no ruleset file given");
        stop;
    elseif command_line(2) = om then
        print("error: no input string given");
        stop;
    end if;

    rules := read_file(command_line(1));

    input := command_line(2);
    loop do
        loop for [pat, repl, trm] in rules do
            if pat in input then
                input(pat) := repl;
                if trm then
                    quit;
                else
                    continue loop do;
                end if;
            end if;
        end loop;
        quit;
    end loop;
    print(input);

    proc read_file(file_name);
        if (rulefile := open(file_name, "r")) = om then
            print("error: cannot open ruleset file");
            stop;
        end if;
    
        rules := [];
        loop doing
            line := getline(rulefile);
        while line /= om do
            rule := parse_rule(line);
            if rule /= om then rules with:= rule; end if;
        end loop;
        return rules;
    end proc;
    
    proc parse_rule(rule);
        if rule(1) = "#" then return om; end if; $ comment
        if " -> " notin rule then return om; end if; $ not a rule
        
        [s, e] := mark(rule, " -> ");
        pattern := rule(..s-1);
        repl := rule(e+1..);
        
        whitespace := "\t\r\n ";
        span(pattern, whitespace);
        rspan(pattern, whitespace);
        span(repl, whitespace);
        rspan(repl, whitespace);
        
        trm := match(repl, ".") /= "";
        return [pattern, repl, trm];
    end proc;
end program;

  

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