Multiplying by three is the same as shift + add. Essentially you have a binary adder with carry, or rather two carries: the input bit is also a carry. So you have an FSM with three states for the three possible carry values 0, 1, 2. 

    00111
 +   00111
  --------
     10101

Let's start with a recursive regex with groups for each of the states. 

    qr/ (?(DEFINE)
          ( 00 (?1) | 11 (?2) | )               # group 1:  carry = 0
          ( 01 (?1) | 10 (?3) )                 # group 2:  carry = 1
          ( 00 (?2) | 11 (?3) )                 # group 3:  carry = 2
        )

        \A(?1)\Z /x;

[download]
From here, we can inline the group 2 in g1 and g3: 
    qr/ (?(DEFINE)
          ( 00 (?1) | 11 01 (?1) | 11 10 (?3) | )
          ( )
          ( 00 01 (?1) | 00 10 (?3) | 11 (?3) )
        )

        \A(?1)\Z /x;

[download]
Now it should be obvious that the group 3 i.e. carry=2 state is entered with a 11 10 and left with 00 01. Rewriting the regex with repeats instead of recursion: 
my $Regex_Pattern =

    qr/ \A ( 00 | 11 01 | 11 10 ( 00 10 | 11 )* 00 01 )* \Z /x;

[download]

ps. I'm assuming that the input string must have 2*n digits.

In reply to Re: Regular Expreso 2 by oiskuu
in thread Regular Expresso 2 by choroba

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