Question

为{w | w∈{0,1，＃} ∗，w = b（n）R＃b（n + 1），n≥1， b（x）将x转换为没有前导0的二进制数}

b（n）R表示二进制字符串反转。

我尝试制作一种可以描述这种语言的CFG，然后转换为PDA，但我真的不知道如何开始。我在想与b（n + 1）二进制数相对应的0和1的数目之间存在某种关系？

一些例子：

For n=1, the recognized string is "1#10"  
For n=2, the recognized string is "01#11"  
For n=3, the recognized string is "11#100"  
For n=4, the recognized string is "001#101"

Answer 1

如果我们从1开始，我们知道RHS上的+1将涉及进位，因此我们可以记录相反的值并保持在有进位的状态。一旦丢失进位，就无法取回，只能记住我们看到的数字。所以：

q    S    s    q'    S'
q0   Z0   0    q1    1Z0   starts with 0, no carry, just copy
q0   Z0   1    q2    0Z0   starts with 1, some carry, copy backwards

q1   x    0    q1    0x    no more carry, just copy input
q1   x    1    q1    1x    to stack so we can read it off backwards
q1   x    #    q3    x

q2   x    0    q1    1x    still have carry, keep carrying as long
q2   x    1    q2    0x    as we keep seeing 1
q2   x    #    q4    #     (go write an extra 1 of we carried all the way)

q3   0x   0    q3    x     read back the stack contents, backwards
q3   1x   1    q3    x     
q3   Z0   -    q5    Z0    

q4   x    1    q3    x     if the LHS is 1^n, write the extra 1 on RHS

q5                         accepting state reachable on empty stack

构建满足{w | w∈{0,1，＃} ∗，w = b（n）R＃b（n + 1），n≥1，b（x）将x转换为无前导0的二进制数

1 个答案: