泵引理证明与最小泵长度：L=(0+1)1*0的相关技术问询

Great question! Let's break this down step by step, since it's easy to mix up the pumping length with the shortest string in a language.

Is the minimal pumping length of ( L=(0+1)1^*0 ) equal to 2? #

No, it's not. Here's why:

• The pumping length ( p ) for a regular language isn't tied to the shortest string in the language. Instead, ( p ) is the smallest number such that every string in ( L ) with length ≥ ( p ) can be split into ( xyz ) (per the pumping lemma rules) and pumped to produce another string in ( L ).
• The shortest strings in ( L ) are 00 and 10 (both length 2). For either of these, any split ( xyz ) with ( |y| ≥ 1 ) will result in ( xy^0z = xz ), which has length ≤1. Since ( L ) has no strings shorter than 2, ( xz \notin L ). This violates the pumping lemma's requirement, so ( p=2 ) can't be the pumping length.

What's the actual minimal pumping length for ( L )? #

The minimal pumping length ( p ) is 3.

All strings in ( L ) with length ≥3 follow the pattern: [0 or 1] + [one or more 1s] + 0 (e.g., 010, 110, 0110). For any such string, we can always find a valid split that satisfies the pumping lemma.

How to split strings into ( xyz ) for valid pumping? #

The key is to pick ( y ) from the middle sequence of 1s (since repeating 1s won't break the language's structure). Here are concrete examples:

• For ( s = 010 ) (length 3):
  • Split ( x = 0 ), ( y = 1 ), ( z = 0 )
  • Pumping gives ( xy^kz = 01^k0 ):
    • ( k=0 ): 00 (valid, since it's in ( L ))
    • ( k=1 ): 010 (original string, valid)
    • ( k=2 ): 0110 (valid, fits ( L )'s pattern)
• For ( s = 1110 ) (length 4):
  • Split ( x = 1 ), ( y = 1 ), ( z = 10 )
  • Pumping gives ( xy^kz = 11^k10 = 11^{k+1}0 ), which is always in ( L ) (starts with 1, ends with 0, middle is all 1s)

In general:

• Let ( s = c + 1^m + 0 ) where ( c \in {0,1} ) and ( m ≥1 ) (since ( |s|≥3 ))
• Choose ( x = c ), ( y = 1 ), ( z = 1^{m-1}0 )
• This split satisfies all pumping lemma rules:
  1. ( |xy| = 2 ≤ 3 ) (meets ( |xy| ≤ p ))
  2. ( |y| =1 ≥1 ) (non-empty ( y ))
  3. ( xy^kz = c1^{k+m-1}0 ), which always fits ( L )'s definition.

A quick refresher on the pumping lemma for regular languages #

Remember:

• The pumping lemma is a necessary condition for regularity (all regular languages satisfy it, but not all languages that satisfy it are regular).
• Since ( L ) is clearly regular (we have a valid regular expression for it), it must satisfy the pumping lemma. Our choice of ( p=3 ) works because every string in ( L ) of length ≥3 can be pumped as shown.

内容的提问来源于stack exchange，提问作者Jan Lovšin