📄 4.10.14.htm
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<title>4.14的解答</title>
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<center><font class="title2"><b>练习4.14</b></font></center><br>
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解：<br>
(1) 构造其拓广文法G'的产生式为
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<tr><td>(0) S' → S</td><td>(1) S → A</td></tr>
<tr><td>(2) A → BA</td><td>(3) A → ε</td></tr>
<tr><td>(4) B → aB</td><td>(5) B → b</td></tr>
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构造其LR(0)项目集规范族和goto函数(识别活前缀的DFA)如下:<br>
I<sub>0</sub> = { [S'→·S, $], [S→·A, $], [A→·BA, $], [A→·, $],<br>
<b> <b> [B→·aB, a/b/$], [B→·b, a/b/$]}
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I<sub>1</sub> = {[S'→S·, $]}<br>
I<sub>2</sub> = {[S→A·, $]}<br>
I<sub>3</sub> = {[A→B·A, $], [A→·BA, $], [A→·, $],<br>
<b> <b> [B→·aB, a/b/$], [B→·b, a/b/$]}<br>
I<sub>4</sub> = {[B→b·, a/b/$]}<br>
I<sub>5</sub> = {[B→a·B, a/b/$], [B→·aB, a/b/$], <br>
<b> <b> [B→·b, a/b/$]}<br>
I<sub>6</sub> = {[A→BA·, $]}<br>
I<sub>7</sub> = {[B→aB·, a/b/$]}<br>
<center><img src="images/ex4.141.gif"></center><br>
该文法的LR(1)项目集规范族中没有冲突，所以该文法是LR(1)文法。<br>
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(2)构造LR(1)分析表如下：<br>
<center><img src="images/ex4.142.gif" ></center><br>
以上分析表无多的定义入口，所以该文法为LR(1)文法。<br>
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(3)对于输入串abab，其分析过程如下：
<center><img src="images/ex4.143.gif" ></center><br>
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