L={ a^nb^m: n <= m+3, n,m>=0} 的 CFG
[英]CFG for L={ a^n b^m : n <= m+3 , n,m>=0}
问题描述
I want to find Context Free Grammar for L={ a^nb^m: n <= m+3, n,m>=0}
我想找到 L={ a^nb^m: n <= m+3, n,m>=0} 的上下文无关语法
What I have so far到目前为止我所拥有的
S -> AAAB
A -> a | ε
B -> aBb | Bb | ε
Does this make any sense?这有道理吗？
1 个解决方案
解决方案1
0 已采纳 2022-09-15 11:18:49
First of all this CFG should work properly But the CFG below is more readable:首先，这个 CFG 应该可以正常工作但下面的 CFG 更具可读性：
S -> aSb | A | B
A -> a | aa | aaa | ε
B -> bB | ε
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