L = {ww^rx | w，x在哪里属于{a，b}^*}是一种常规语言？
我已经知道l = {wxw^r | w，x属于{a，b}^*}是常规的，因为事实证明它是以相同的符号开始和结束的模式如何说l = {ww^rx | w，x属于{a，b}*}是使用DFA设计的常规语言。 请帮助我理解这一点！
I have understood that L={wxw^r|w,x belongs to {a,b}^* } is regular because it turns out to be the pattern of starting and ending with same symbol but I am not getting the proper explanation that how to say L={ww^rx|w,x belongs to {a,b}*} is regular language using DFA design.
Please help me in understanding this!
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这是一个诀窍。您指定的语言l是正则表达式
（a + b）*的语言，即A和B的任何字符串。诀窍是对于任何字符串y = s1.s2.s3 ... sk其中si in {a，b}，我们可以写y = wxw ^r其中w是空字符串，x = y。基本上，诀窍是我们可以始终选择w作为空字符串，在这种情况下，我们将留下l = {x | x in {a，b}^*}，显然是常规的。另一种思考的方式是：您能找到不在L中的A和B的字符串吗？即使您将W是空字符串，它也不在L中吗？This is a trick question. The language L as you have specified is the language of the regular expression
(a + b)*, that is, any string of a's and b's. The trick is that for any stringy = s1.s2.s3...skwheresi in {a, b}, we can writey = wxw^Rwherewis the empty string andx = y. Basically, the trick is that we can always choosewto be the empty string, and in that case we are left withL = {x | x in {a, b}^*}, clearly regular. Another way of thinking about it is this: can you find any string of a's and b's that is not in L? Is it not in L even if you take w to be the empty string?