CS 3240 – Chapter 6
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6.1: Simplifying Grammars
 Substitution
 Removing useless variables
 Removing λ
 Removing unit productions
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6.2: Normal Forms
 Chomsky Normal (CNF)
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6.3: A Membership Algorithm
 CYK Algorithm
 An example of bottom-up parsing
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Variables in CFGs can often be eliminated
 If they are not recursive, you can substitute their
rules throughout
 See next slide
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A ➞ a | aaA | abBc
B ➞ abbA | b
Just substitute directly for B:
A ➞ a | aaA | ababbAc | abbc
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A variable is useless if:
 It can’t be reached from the start state, or
 It never leads to a terminal string
▪ Due to endless recursion
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Both problems can be detected by a
dependency graph
 See next slide
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S ➞ aSb | A | λ
A ➞ aA
A is useless (non-terminating):
S ➞ aSb | λ
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S➞A
A ➞ aA | λ
B ➞ bA
B is useless (non-reachable):
S➞A
A ➞ aA | λ
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Simplify the following:
S ➞ aS | A | C
A➞a
B ➞ aa
C ➞ aCb
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Simplify the following grammar:
S ➞ AB | AC
A ➞ aAb | bAa | a
B ➞ bbA | aaB | AB
C ➞ abCa | aDb
D ➞ bD | aC
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Any variable that can eventually terminate in
the empty string is said to be nullable
 Note: a variable may be indirectly nullable
 In general: if A ➞ V1V2…Vn, and all the Vi are
nullable, then A is also nullable
▪ See Theorem 6.3
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Consider the following grammar:
S ➞ a | Xb | aYa
X ➞Y | λ
Y➞b|X
Which variables are nullable?
How can we substitute the effect of λ before removing it?
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First find all nullable variables
Then substitute (A + λ) for every nullable
variable A, and expand
 Then remove λ everywhere from the grammar
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What’s left is equivalent to the original
grammar
 except the empty string may be lost
 we won’t worry about that
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Consider the following grammar (again):
S ➞ a | Xb | aYa
X ➞Y | λ
Y➞b|X
How can we substitute the effect of λ before removing it?
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S → aSbS | bSaS | λ
S → aSa | bSb | X
X → aYb | bYa
Y → aY | bY | λ
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Unit Productions often occur in chains
 A➞B➞C
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Must maintain the effect of B and C when
substituting for A throughout
Procedure:
 Find all unit chains
 Rebuild grammar by:
▪ Keeping all non-unit productions
▪ Keeping only the effect of all unit productions/chains
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S ➞ A | bb
A➞B|b
B➞S|a
Note that S ⇒* {A,B}, A⇒* {B,S}, B ⇒* {S,A}
Giving:
S ➞ bb | b | a // Added non-unit part of A and B
A ➞ b | a | bb // Added non-unit part of B and S
B ➞ a | bb | b // Added non-unit part of S and A
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1) Determine all variables reachable by unit
rules for each variable
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2) Keep all non-unit rules
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3) Substitute non-unit rules in place of each
variable reachable by unit productions
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S ➞ aA
A ➞ BB
B ➞ aBb | λ
Now remove nulls and see what happens….
(Also see the solution for #15 in 6.1)
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S ➞ AB
A➞B
B ➞ aB | BB | λ
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Do things in the following recommended
order:
 Remove nulls
 Remove unit productions
 Remove useless variables
 Simplify by substitution as desired
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Very important for our purposes
All CNF rules are of one of the following two
forms:
 A➞c
 A ➞ XY
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(a single terminal)
(exactly two variables)
Must begin the transformation after
simplifying the grammar (removing λ, all unit
productions, useless variables, etc.)
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Convert to CNF:
S ➞ bA | aB
A ➞ bAA | aS | a
B ➞ aBB | bS | b
(NOTE: already has no nulls or units)
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Convert the following grammar to CNF:
S ➞ abAB
A ➞ bAB | λ
B ➞ BAa | A | λ
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Convert the following grammar to CNF:
S ➞ aS | bS | B
B ➞ bb | C | λ
C ➞ cC | λ
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Single terminal character followed by zero or
more variables (cV*, c ∈ Σ )
V→a
V → aBCD…
λ allowed only in S → λ
Sometimes need to make up new variable
names
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S → AB
A → aA | bB | b
B→b
Substitute for A in first rule (i.e., add B to each rule for A):
S → aAB | bBB | bB
The other rules are okay
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S → abSb |aa
Add rules to generate a and b:
S → aBSB |aA
A→a
B→b
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The “parsing” problem
 How do we know if a string is generated by a
given grammar?
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Bottom-up parsing (CYK Algorithm)
 An Example of Dynamic Programming
 Requires Chomsky Normal Form (CNF)
 Start by considering A ➞ c rules
 Build up the parse tree from there
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S ➞ XY
X ➞ XA | a | b
Y ➞ AY | a
A➞a
Does this grammar generate “baaa”?
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CNF yields binary trees.
(Can you find a third parse tree?)
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S ➞ XY
X ➞ XA | a | b
Y ➞ AY | a
A➞a
Stage 3:
baa ⇐ ba a ⇐ (S,X)(X,Y,A) = SX, SY, SA,
XX, XY, XA
= S, X
baa ⇐ b aa ⇐ X(S,X,Y) = XS, XX, XY = S
Stage 1:
b⇐X
a ⇐ X, Y, A
aaa ⇐ aa a ⇐ (S,X,Y)(X,Y,A) = _________
aaa ⇐ a aa ⇐ (X,Y,A)(S,X,Y) = _________
Stage 2:
ba ⇐ X(X,Y,A) = XX, XY, XA ⇐ S, X
aa ⇐ (X,Y,A)(X,Y,A) = XX, XY, XA,
YX, YY, YA,
AX, AY, AA
⇐ S, X, Y
Stage 4: (you finish…)
baaa:
b aaa ⇐ __________
ba aa ⇐ __________
baa a ⇐ __________
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Does the following grammar generate “abbaab”?
S ➞ SAB | λ
A ➞ aA | λ
B ➞ bB | λ
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