EQUAL language

1
Lecture 28

• EQUAL language
• Designing a CFG
• Proving the CFG is correct

2
EQUAL language

• Designing a CFG

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EQUAL

• EQUAL is the set of strings with an equal number
  of as and bs
• Strings in EQUAL include
• aabbab
• bbbaaa
• abba
• Strings in a,b not in EQUAL include
• aaa
• bbb
• aab
• ababa

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Designing a CFG for EQUAL

• Think recursively
• Base Case
• What is the shortest possible string in EQUAL?
• Answer l
• Production Rule
• S -- l

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Recursive Case

• Recursive Case
• Now consider a longer string x in EQUAL
• Since x has length 0, x must have a first
  character
• This must be a or b
• Two possibilities for what x looks like
• x ay
• What must be true about relative number of as
  and bs in y?
• y must have one extra b
• x bz
• What must be true about relative number of as
  and bs in z?
• z must have one extra a

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Case 1 xay

• x ay where y has one extra b
• What must y look like?
• Some examples
• b
• babba
• aabbbab
• aaabbbb
• Is there a general pattern?
• In particular, are there substrings in y which
  belong to EQUAL?

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Decomposing y

• y has one extra b
• Possible examples
• b, babba, aabbbab, aaabbbb
• Common pattern
• In each string, there is some prefix with exactly
  one more b than a
• That is, y ubv where
• ub is the prefix with an extra b
• u and v both have an equal number of as and bs
• Decompose the 4 strings above into u, b, v
• lbl, aabbbab, lbabba, aaabbbbl

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Implication

• Case 1 xay
• y has one extra b
• Case 1 refined xaubv
• u, v belong to EQUAL
• Production rule for this case
• S -- aSbS
• The first S can derive the string u
• The second S can derive the string v

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Case 2 xbz

• Case 2 xbz
• z has one extra a
• Case 2 refined xbuav
• u, v belong to EQUAL
• Production rule for this case
• S -- bSaS
• The first S can derive the string u
• The second S can derive the string v

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Final Grammar

• EG (V, S, S, P)
• V S
• S a,b
• S S
• P S -- l aSbS bSaS

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EQUAL language

• Proving CFG is correct

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Is our grammar correct?

• How do we prove our grammar is correct?
• Informal
• Test some strings
• Review logic behind program (CFG) design
• Formal
• First, show every string derived by EG belongs to
  EQUAL
• That is, show L(EG) is a subset of EQUAL
• Second, show every string in EQUAL can be derived
  by EG
• That is, show EQUAL is a subset of L(EG)
• Both proofs will be inductive proofs
• Inductive proofs and recursive algorithms go well
  together

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L(EG) subset of EQUAL

• Let x be an arbitrary string in L(EG)
• What does this mean?
• S EG x
• Follows from definition of x in L(EG)
• We will prove the following
• If S 1EG x, then x is in EQUAL
• If S 2EG x, then x is in EQUAL
• If S 3EG x, then x is in EQUAL
• If S 4EG x, then x is in EQUAL
• ...

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L(EG) subset of EQUAL

• Statement to be proven
• For all n 1, if S nEG x, then x is in EQUAL
• Prove this by induction on n
• Base Case
• n 1
• The only string x such that S 1EG x is the
  string l
• Follows from inspection of EG
• The string l belongs to EQUAL

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Inductive Case

• Inductive Hypothesis
• For 1 jEG x, then x is in
  EQUAL
• Note, this is a strong induction hypothesis
• Traditional inductive hypothesis would take form
• For some n 1, if S nEG x, then x is in
  EQUAL
• The difference is we assume the basic hypothesis
  for all integers between 1 and n, not just n
• Statement to be Proven in Inductive Case
• If S n1EG x, then x is in EQUAL

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Proving Inductive Case

• If S n1EG x, then x is in EQUAL
• Let x be an arbitrary string such that S n1EG
  x
• There are three possible first derivation steps
• Case 1 S l nEG x
• Case 2 S aSbS nEG x
• Case 3 S bSaS nEG x
• These 3 cases follow from looking at grammar EG
• Case 1 is not possible
• n was assumed to be at least 1 which means x is
  derived in at least 2 steps

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Case 2 S aSbS nEG x

• This means x has the form aubv where
• S
• S
• This follows because in n steps, we go from aSbS
  to x
• Thus, the first S can take at most n-1 steps to
  generate u
• Likewise for the second S generating v
• Apply the inductive hypothesis
• u and v belong to EQUAL
• Note we needed the strong inductive hypothesis
• Conclude x belongs to EQUAL
• x aubv where u and v belong to EQUAL
• Clearly the number of as in x equals the number
  of bs in x

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Case 3 S bSaS nEG x

• This means x has the form buav where
• S
• S
• This follows because in n steps, we go from bSaS
  to x
• Thus, the first S can take at most n-1 steps to
  generate u
• Likewise for the second S generating v
• Apply the inductive hypothesis
• u and v belong to EQUAL
• Note we needed the strong inductive hypothesis
• Conclude x belongs to EQUAL
• x buav where u and v belong to EQUAL
• Clearly the number of as in x equals the number
  of bs in x

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L(EG) subset of EQUAL

• Wrapping up inductive case
• In all possible derivations of x, we have shown
  that x belongs to EQUAL
• Thus, we have proven the inductive case
• Conclusion
• By the principle of mathematical induction, we
  have shown that L(EG) is a subset of EQUAL

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EQUAL subset of L(EG)

• Let x be an arbitrary string in EQUAL
• What does this mean?
• x has an equal number of as and bs
• Follows from definition of x in EQUAL
• We will prove the following
• If x 0 and x is in EQUAL, then x is in L(G)
• If x 1 and x is in EQUAL, then x is in L(G)
• If x 2 and x is in EQUAL, then x is in L(G)
• If x 3 and x is in EQUAL, then x is in L(G)
• ...

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EQUAL subset of L(EG)

• Statement to be proven
• For all n 0, if x n and x is in EQUAL,
  then x is in L(EG)
• Prove this by induction on n
• Base Case
• n 0
• The only string x such that x0 and x is in
  EQUAL is the string l
• Follows from definition of EQUAL
• The string l belongs to L(EG)
• Follows from production S -- l

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Inductive Case

• Inductive Hypothesis
• For 0 then x is in L(EG)
• Again, this is a strong induction hypothesis
• Statement to be Proven in Inductive Case
• If x n1 and x is in EQUAL, then x is in
  L(EG)
• note n 0

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Proving Inductive Case

• If xn1 and x is in EQUAL, then x is in L(EG)
• Let x be an arbitrary string such that xn1
  and x is in L(EG)
• There are two possibilities for the first
  character in x
• Case 1 first character in x is a
• Case 2 first character in x is b
• These 2 cases follow from looking at alphabet S
• Case 1 x ay where y has one extra b
• Case 2 x bz where z has one extra a

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Case 1 x ay

• This means x has the form aubv where
• u is in EQUAL and has length
• v is in EQUAL and has length
• Proving this statement true
• Consider all the prefixes of string y
• length 0 l
• length 1 y1
• length 2 y1y2
• length n y1y2 yn y

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Case 1 x ay

• Consider all the prefixes of string y
• length 0 l
• length 1 y1
• length 2 y1y2
• length n y1y2 yn y
• The first prefix l has the same number of as as
  bs
• The last prefix y has one extra b
• The relative number of as and bs changes in the
  length i prefix differs by only one from the
  length i-1 prefix
• Thus, there must be a first prefix t of y where t
  has one extra b
• Furthermore, the last character of t must be b
• Otherwise, t would not be the FIRST prefix of y
  with one extra b
• Break t into u and b and let the remainder of y
  be v
• The statement follows

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Case 1 x aubv

• x aubv
• u is in EQUAL and has length
• v is in EQUAL and has length
• Apply the induction hypothesis
• u and v belong to L(EG)
• This means S EG u and S EG v.
• Conclude x is in L(EG) by constructing a
  derivation
• S aSbS EG aubS EG aubv
• The first derivation step follows from production
  S -- aSbS
• The second series of derivation steps follows
  from S EG u
• The third series of derivation steps follows from
  S EG v

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Case 2 x buav

• x buav
• u is in EQUAL and has length
• v is in EQUAL and has length
• Follows from similar proof as in case 1
• Apply the induction hypothesis
• u and v belong to L(EG)
• This means S EG u and S EG v.
• Conclude x is in L(EG) by constructing a
  derivation
• S bSaS EG buaS EG buav
• The first derivation step follows from production
  S -- bSaS
• The second series of derivation steps follows
  from S EG u
• The third series of derivation steps follows from
  S EG v

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EQUAL subset of L(EG)

• Wrapping up inductive case
• For all possible first characters of x, we have
  shown that x belongs to L(EG)
• Thus, we have proven the inductive case
• Conclusion
• By the principle of mathematical induction, we
  have shown that EQUAL is a subset of L(EG)