📄 4.10.11(a).htm
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<title>4.11(a)的解答</title>
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<center><font class="title2"><b>练习4.11(a)</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 → Sab</td></tr>
<tr><td>(2) S → bR</td><td>(3) R → S</td></tr>
<tr><td>(4) R → a</td><td></td></tr>
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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→·Sab, S→·bR}<br>
I<sub>1</sub> = {S'→S·, S→S·ab}<br>
I<sub>2</sub> = {S→b·R, R→·S, R→·a, S→·Sab, S→·bR}<br>
I<sub>3</sub> = {S→Sa·b}<br>
I<sub>4</sub> = {S→bR·}<br>
I<sub>5</sub> = {R→S·, S→S·ab}<br>
I<sub>6</sub> = {R→a·}<br>
I<sub>7</sub> = {S→Sab·}<br>
<center><img src="images/ex4.11a.gif"></center><br>
求<b>FOLLOW</b>集：<br>
<b>FOLLOW</b>(S')={＄}<br>
<b> FOLLOW</b>(R)=<b>FOLLOW</b>(S)={a,＄}<br>
在I<sub>5</sub>中，出现移进－归约冲突，且<b>FOLLOW</b>(R)∩{a}={a}<br>
因此，此文法不是SLR(1)文法。<br>
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