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---
id: 59e09e6d412c5939baa02d16
title: 执行马尔可夫算法
challengeType: 5
videoUrl: ''
dashedName: execute-a-markov-algorithm
---
# --description--
任务:
为[马尔可夫算法](<https://en.wikipedia.org/wiki/Markov algorithm> "wp马尔可夫算法")创建解释器。
规则的语法如下:
<pre>[ruleset] ::= (([comment] | [rule]) [newline]+)*
[comment] ::= # {[any character]}
[rule] ::= [pattern] [whitespace] -> [whitespace] [.] [replacement]
[whitespace] ::= ([tab] | [space]) [[whitespace]]
</pre>
每行有一条规则。
如果在 \[replacement] 之前有一个 `.`(句点),那么这就是一个终止规则。在这种情况下,解释器必须停止执行。
规则集由一系列规则组成,可能包含一些注释。
规则集
我们会对你提交的代码进行如下测试:
**规则集 1**
<pre># 此条规则来自 Wikipedia
# <code>http://en.wikipedia.org/wiki/Markov_Algorithm</code>
A -> apple
B -> bag
S -> shop
T -> the
the shop -> my brother
(终止规则) -> .
</pre>
对于这段文本:
`I bought a B of As from T S.`
应该输出:
`I bought a bag of apples from my brother.`
**规则集 2**
终止规则的测试
<pre># 基于 Wikipedia 的规则稍做修改
A -> apple
B -> bag
S -> .shop
T -> the
the shop -> my brother
(终止规则) -> .
</pre>
对于这段文本:
`I bought a B of As from T S.`
应该输出:
`I bought a bag of apples from T shop.`
**规则集 3**
这条不仅可以用来测试替换顺序是否正确,还可以测试你的代码中对正则表达式的处理是否完备。如果你的代码没有对正则表达式进行正确的转义处理,那在替换的时候就会出现问题。
<pre># BNF Syntax testing rules
A -> apple
WWWW -> with
Bgage -> ->.*
B -> bag
->.* -> money
W -> WW
S -> .shop
T -> the
the shop -> my brother
(终止规则) -> .
</pre>
对于这段文本:
`I bought a B of As W my Bgage from T S.`
应该输出:
`I bought a bag of apples with my money from T shop.`
**规则集 4**
这条是用来测试规则扫描的顺序是否正确,并可能捕获以错误顺序扫描的替换例程。这里我们选取了通用的一元乘法引擎(请注意,在此实现中,输入的表达式必须放在两个下划线之间)。
<pre> ##一元乘法引擎,用于测试马尔可夫算法实现
### by Donal Fellows
# 一元加法引擎
_+1 -> _1+
1+1 -> 11+
# 将乘法转换为普通加法
1! -> !1
,! -> !+
_! -> _
# 一元乘法,左侧为被乘数,右侧为乘数
1*1 -> x,@y
1x -> xX
X, -> 1,1
X1 -> 1X
_x -> _X
,x -> ,X
y1 -> 1y
y_ -> _
# 下一阶段
1@1 -> x,@y
1@_ -> @_
,@_ -> !_
++ -> +
# 加法的终止条件
_1 -> 1
1+_ -> 1
_+_ ->
</pre>
对于这段文本:
`_1111*11111_`
应该输出:
`11111111111111111111`
**规则集5**
一台简单的[图灵机](http://en.wikipedia.org/wiki/Turing_machine)包含三个状态的["忙碌海狸"](http://en.wikipedia.org/wiki/Busy_beaver)。纸带由 0 和 1 组成,状态为 A、B、C 和代表终止Halt的 H。通过在字符前写状态字母来的方式来指示读写头的位置。机器运行时需要的初始纸带必须通过输入在一开始全部给出。
这一规则集除了可以证明 Markov 算法是图灵完备的,它还帮我找出了使用 C++ 完成此题中的一个错误,而且这个错误没有被前四个规则集抓到。
<pre># 图灵机:三个状态的"忙碌海狸"
# 状态 A符号 0 => 写入 1向右移动新状态 B
A0 -> 1B
# 状态 A符号 1 => 写入 1向左移动新状态 C
0A1 -> C01
1A1 -> C11
# 状态 B符号 0 => 写入 1向左移动新状态 A
0B0 -> A01
1B0 -> A11
# 状态 B符号 1 => 写入 1向右移动新状态 B
B1 - > 1B
# 状态 C符号 0 => 写入 1向左移动新状态 B
0C0 - > B01
1C0 - > B11
# 状态 C符号 1 => 写入 1向左移动停止
0C1 - > H01
1C1 - > H11
</pre>
这个规则集应将这段输入:
`000000A000000`
转换成:
`00011H1111000`
# --hints--
`markov` 应是一个函数。
```js
assert(typeof markov === 'function');
```
`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.")` 应返回 "I bought a bag of apples from my brother."。
```js
assert.deepEqual(markov(rules[0], tests[0]), outputs[0]);
```
`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.")` 应返回 "I bought a bag of apples from T shop."。
```js
assert.deepEqual(markov(rules[1], tests[1]), outputs[1]);
```
`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.")` 应返回 "I bought a bag of apples with my money from T shop."。
```js
assert.deepEqual(markov(rules[2], tests[2]), outputs[2]);
```
`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_")` 应返回 "11111111111111111111"。
```js
assert.deepEqual(markov(rules[3], tests[3]), outputs[3]);
```
`markov(["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"],"")` 应返回 "00011H1111000"。
```js
assert.deepEqual(markov(rules[4], tests[4]), outputs[4]);
```
# --seed--
## --seed-contents--
```js
function markov(rules,test) {
}
```
# --solutions--
```js
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"];
```