Construct a PDA that satisfies {w | w∈{0,1,#}∗,w=b(n)R#b(n+1),n≥1, b(x) converts x to binary with no leading 0}

Question

Construct a PDA for the language {w | w∈{0,1,#}∗,w=b(n)R#b(n+1),n≥1, b(x) converts x to binary with no leading 0}

b(n)R means the binary string reversed.

I tried making a CFG that can describe this language and then converting to PDA, but I don't really know how to start. I was thinking there is some relationship between the number of 0 and 1s that correspond to the b(n+1) binary number?

Some Examples:

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"

Try ignoring the fact that a binary string is a number, and think about it just as a string of characters. What string algorithm would you use to find the next larger binary string? (Hint: it's very simple.) How might you simultaneously build up a number and the next number? You can build two numbers in parallel if you build one of them backwards... — rici, May 20 '19 at 01:39

score 0 · Accepted Answer · answered May 22 '19 at 13:37

If we start with a 1, we know there is going to be carry involved from the +1 on the RHS, so we can record the inverse and stay in a state where we have the carry. Once we lose the carry we can't get it back and can just remember the digits we're seeing. So:

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

Construct a PDA that satisfies {w | w∈{0,1,#}∗,w=b(n)R#b(n+1),n≥1, b(x) converts x to binary with no leading 0}

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