📄 4.10.13.htm
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<title>4.13的解答</title>
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<center><font class="title2"><b>练习4.13</b></font></center><br>
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解：<br>
该文法的拓广文法G'为
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<tr><td>(0) S' → S</td><td>(1) S → (SR</td></tr>
<tr><td>(2) S → a</td><td>(3) R → ,SR</td></tr>
<tr><td>(4) R → )</td><td></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→·(SR, S→·a}<br>
I<sub>1</sub> = {S'→S·}<br>
I<sub>2</sub> = {S→(·SR, S→·(SR, S→·a}<br>
I<sub>3</sub> = {S→a·}<br>
I<sub>4</sub> = {S→(S·R, R→·,SR, R→·)}<br>
I<sub>5</sub> = {S→(SR·}<br>
I<sub>6</sub> = {R→)·}<br>
I<sub>7</sub> = {R→,·SR, S→·(SR, S→·a}<br>
I<sub>8</sub> = {R→,S·R, R→·,SR, R→·)}<br>
I<sub>9</sub> = {R→,SR·}<br>
<center><img src="images/ex4.131.gif"></center><br>
每个LR(0)项目集中没有冲突。因此，此文法是LR(0)文法。其分析表如下：<br>
<center><img src="images/ex4.132.gif"></center><br>
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