[英]Is my transition function correct (String matching with finite automata)
I am learning String matching with finite automata from CLRS . 我正在从CLRS学习具有有限自动机的字符串匹配。 I am solving some exercise problems .For the exercise problem 32.3-1 ,
我正在解决一些运动问题。对于运动问题32.3-1,
Construct the string-matching automaton for the pattern P = aabab and illustrate its operation on the text string T = aaababaabaababaab.
构造模式P = aabab的字符串匹配自动机,并说明其在文本字符串T = aaababaabaababaab上的操作。
the following is my transition function , 以下是我的转换函数,
states a b
0 1 0
1 2 0
2 2 3
3 4 3
4 4 5
5 ? ?
Is my transition function correct ? 我的转换函数正确吗? And how do i fill the last row ?
那我怎么填最后一行呢? Any help
任何帮助
I assume you are creating a Finite Automata which accepts a string containing the pattern aabab
. 我假设您正在创建一个有限自动机,该自动机接受包含模式
aabab
的字符串。
There are two mistakes in your finite automata, 有限自动机有两个错误,
on state 3
and state 4, 在状态
3
和状态4,
For state 3
, if the input is b
, you have to go back to state 0
. 对于状态
3
,如果输入为b
,则必须返回状态0
。 For example the pattern aabb
will force you back to state 0
. 例如,
aabb
模式将迫使您返回状态0
。 Here you have to start all over again from state 0
. 在这里,您必须从状态
0
重新开始。
For state 4
, if the input is a
, you have to go back to state 2
because you have the pattern aa
. 对于状态
4
,如果输入为a
,则必须返回状态2
因为您具有模式aa
。 For example the pattern aabaa
will force you back to state 2
. 例如,模式
aabaa
将迫使您返回状态2
。
The corrected Finite Automaton is given below, 修正后的有限自动机如下所示,
states a b
0 1 0
1 2 0
2 2 3
3 4 0
4 2 5
5 5 5
Here 5 is your Accepting state. 这里5是您的接受状态。 You will reach this state only when you have found the required pattern in a string.
仅当在字符串中找到所需的模式时,您才会达到此状态。 Once a pattern is found no matter what the string remains in the accepting state.
一旦找到模式,无论什么字符串保持接受状态。 Hence for both inputs
a
and b
on state 5
remains on 5
itself. 因此,对于状态
5
输入a
和b
,状态5
本身都保持不变。
The transition function is that of an fa accepting a string with sub string ' aabab '. 过渡函数是fa接受带有子字符串“ aabab ”的字符串的函数 。 If you are going back to state
1
for a
and 0
for b
, then the transition function accepts strings ending with the sub string ' aabab '. 如果你想回到状态
1
的a
和0
为b
,那么转换函数接受与子串“aabab的 ”结尾的字符串。 Given that only state 5 is the accepting state. 假定只有状态5是接受状态。
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