9.3 KiB
title | id | challengeType |
---|---|---|
Execute a Markov algorithm | 59e09e6d412c5939baa02d16 | 5 |
Description
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:
Ruleset 1:This rules file is extracted from Wikipedia: http://en.wikipedia.org/wiki/Markov_AlgorithmA -> apple B -> bag S -> shop T -> the the shop -> my brother a never used -> .terminating rule
Sample text of:
I bought a B of As from T S.
Should generate the output:
I bought a bag of apples from my brother.
A test of the terminating rule
Slightly modified from the rules on WikipediaA -> apple B -> bag S -> .shop T -> the the shop -> my brother a never used -> .terminating rule
Sample text of:
I bought a B of As from T S.
Should generate:
I bought a bag of apples from T shop.
This tests for correct substitution order and may trap simple regexp based replacement routines if special regexp characters are not escaped.
BNF Syntax testing rulesA -> apple WWWW -> with Bgage -> ->.* B -> bag ->.* -> money W -> WW S -> .shop T -> the the shop -> my brother a never used -> .terminating rule
Sample text of:
I bought a B of As W my Bgage from T S.
Should generate:
I bought a bag of apples with my money from T shop.
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.)
## Unary Multiplication Engine, for testing Markov Algorithm implementations ## By Donal Fellows. Unary addition engine_+1 -> _1+ 1+1 -> 11+ Pass for converting from the splitting of multiplication into ordinary addition1! -> !1 ,! -> !+ _! -> _ Unary multiplication by duplicating left side, right side times1*1 -> x,@y 1x -> xX X, -> 1,1 X1 -> 1X _x -> _X ,x -> ,X y1 -> 1y y_ -> _ Next phase of applying1@1 -> x,@y 1@_ -> @_ ,@_ -> !_ ++ -> + Termination cleanup for addition_1 -> 1 1+_ -> 1 _+_ ->
Sample text of:
_1111*11111_
should generate the output:
11111111111111111111
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.
Turing machine: three-state busy beaver # state A, symbol 0 => write 1, move right, new state BA0 -> 1B state A, symbol 1 => write 1, move left, new state C0A1 -> C01 1A1 -> C11 state B, symbol 0 => write 1, move left, new state A0B0 -> A01 1B0 -> A11 state B, symbol 1 => write 1, move right, new state BB1 -> 1B state C, symbol 0 => write 1, move left, new state B0C0 -> B01 1C0 -> B11 state C, symbol 1 => write 1, move left, halt0C1 -> H01 1C1 -> H11
This ruleset should turn
000000A000000
into
00011H1111000
Instructions
Tests
tests:
- text: <code>markov</code> is a function.
testString: assert(typeof markov === 'function', '<code>markov</code> is a function.');
- text: <code>markov(["A -> apple","B -> bag","S -> shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")</code> should return "I bought a bag of apples from my brother.".
testString: assert.deepEqual(markov(rules[0],tests[0]),outputs[0],'<code>markov(["A -> apple","B -> bag","S -> shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")</code> should return "I bought a bag of apples from my brother.".');
- text: <code>markov(["A -> apple","B -> bag","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")</code> should return "I bought a bag of apples from T shop.".
testString: assert.deepEqual(markov(rules[1],tests[1]),outputs[1],'<code>markov(["A -> apple","B -> bag","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")</code> should return "I bought a bag of apples from T shop.".');
- text: <code>markov(["A -> apple","WWWW -> with","Bgage -> ->.*","B -> bag","->.* -> money","W -> WW","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As W my Bgage from T S.")</code> should return "I bought a bag of apples with my money from T shop.".
testString: assert.deepEqual(markov(rules[2],tests[2]),outputs[2],'<code>markov(["A -> apple","WWWW -> with","Bgage -> ->.*","B -> bag","->.* -> money","W -> WW","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As W my Bgage from T S.")</code> should return "I bought a bag of apples with my money from T shop.".');
- text: <code>markov(["_+1 -> _1+","1+1 -> 11+","1! -> !1",",! -> !+","_! -> _","1*1 -> x,@y","1x -> xX","X, -> 1,1","X1 -> 1X","_x -> _X",",x -> ,X","y1 -> 1y","y_ -> _","1@1 -> x,@y","1@_ -> @_",",@_ -> !_","++ -> +","_1 -> 1","1+_ -> 1","_+_ -> "],"_1111*11111_")</code> should return "11111111111111111111".
testString: assert.deepEqual(markov(rules[3],tests[3]),outputs[3],'<code>markov(["_+1 -> _1+","1+1 -> 11+","1! -> !1",",! -> !+","_! -> _","1*1 -> x,@y","1x -> xX","X, -> 1,1","X1 -> 1X","_x -> _X",",x -> ,X","y1 -> 1y","y_ -> _","1@1 -> x,@y","1@_ -> @_",",@_ -> !_","++ -> +","_1 -> 1","1+_ -> 1","_+_ -> "],"_1111*11111_")</code> should return "11111111111111111111".');
- text: <code>markov(["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"],"")</code> should return "00011H1111000".
testString: assert.deepEqual(markov(rules[4],tests[4]),outputs[4],'<code>markov(["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"],"")</code> should return "00011H1111000".');
Challenge Seed
function markov (rules,test) {
// Good luck!
}
Solution
function markov(rules,test) {
let pattern = new RegExp("^([^#]*?)\\s+->\\s+(\\.?)(.*)");
let origTest = test;
let captures = [];
rules.forEach(function(rule){
let m = pattern.exec(rule);
for (let j = 0; j < m.length; j++)
m[j] = m[j + 1];
captures.push(m);
});
test = origTest;
let copy = test;
for (let j = 0; j < captures.length; j++) {
let c = captures[j];
test = test.replace(c[0], c[2]);
if (c[1]==".")
break;
if (test!=copy) {
j = -1;
copy = test;
}
}
return test;
}
// tail:
let rules=[["A -> apple","B -> bag","S -> shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
["A -> apple","B -> bag","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
["A -> apple","WWWW -> with","Bgage -> ->.*","B -> bag","->.* -> money","W -> WW","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
["_+1 -> _1+","1+1 -> 11+","1! -> !1",",! -> !+","_! -> _","1*1 -> x,@y","1x -> xX","X, -> 1,1","X1 -> 1X","_x -> _X",",x -> ,X","y1 -> 1y","y_ -> _","1@1 -> x,@y","1@_ -> @_",",@_ -> !_","++ -> +","_1 -> 1","1+_ -> 1","_+_ -> "],
["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"]];
let tests=["I bought a B of As from T S.",
"I bought a B of As from T S.",
"I bought a B of As W my Bgage from T S.",
"_1111*11111_",
"000000A000000"];
let outputs=["I bought a bag of apples from my brother.",
"I bought a bag of apples from T shop.",
"I bought a bag of apples with my money from T shop.",
"11111111111111111111",
"00011H1111000"];