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Compiler first set_followset_brief

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shubham509

The document discusses calculating the first sets of grammar rules. It provides examples of determining the first sets for various grammar rules through multiple steps. In the first example, it calculates the first sets for a grammar involving rules for E, Prefix, and Tail over multiple steps, updating the values as more information is obtained. In a second example, it similarly determines the first sets for a grammar involving rules for S, B, and C. It performs the calculation in iterative steps, with the first sets of some rules depending on the results of earlier steps.

Engineering
Related topics:
Computer Science Basics • Compiler Design
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First Set • First (=A) = First(A) , if  ∉ First(A) First(A) – {} ∪First(), if  ∈ First(A) We will use some examples for this operation… Red : A Blue : 
First Set (2) EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue : 
First Set (2) • First (EPrefix(E))=First(Prefix(E)) ? = {Waiting operation} – We don’t know First(Prefix) at this moment. – So we keep the operation of First(Prefix(E)) until we know First(Prefix). • First (EV Tail) = First(V) = {V} • First (PrefixF) = First(F) = {F} • First (Prefix )= First() - {} ∪First() = {} • First (Tail+E) = First(+) = {+} • First (Tail ) = First() - {} ∪First() = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step 1:
First Set (2) • First (EPrefix(E))=First(Prefix(E)) ? = {Waiting operation} – We don’t know First(Prefix) at this moment. – So we keep the operation of First(Prefix(E)) until we know First(Prefix). • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, } Step 1:
First Set (2) • First (EPrefix(E))=First(Prefix(E)) ? = {Waiting operation} – In this step, we know First(Prefix). – So we replace the First(Prefix) with {F, }. • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  {F, } Step 2: Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, }
First Set (2) • First (EPrefix(E))=First(Prefix(E)) = {F, } = {F, } - {} ∪ First((E)) = {F, (} • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step 2: Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, }
First Set (2) • First (EPrefix(E))= {F, (} • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, } Step 2 {F, (} ∪ {V} ={F, (, V} {F, } {+, } Step 2:
First Set (2) • First (EPrefix(E))= {F, (} • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, } Step 2 {F, (} ∪ {V} ={F, (, V} {F, } {+, } Step 3 {F, (, V} {F, } {+, } {(} {)} {V} {F} {+} If no more change… The first set of a terminal symbol is itself Red : A Blue :  Step 3:
Another Example….
First Set (2) S  aSe S  B B  bBe B  C C  cCe C  d G0 Red : A Blue : 
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = First(a) ={a} • First (SB) = First(B) • First (B  bBe) = First(b)={b} • First (B  C) = First(C) • First (C  cCe) = First(c) ={c} • First (C  d) = First(d)={d} Red : A Blue :  Step 1:
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = First(B) • First (B  bBe) = {b} • First (B  C) = First(C) • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 1:
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = First(B) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = First(C) • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2:
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = First(C) • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 2: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C)
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = First(C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 2: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C)
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d} Step 2:
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) = {b}∪ {c, d} • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 3: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d}
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b, c, d} • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 3: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d}
S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b, c, d} • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 3: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d} Step 3 {a}∪ {b}∪{c, d} = {a,b,c,d} {b}∪{c, d} = {b,c,d} {c, d} {a} {b} {c} {d} If no more change… The first set of a terminal symbol is itself
Another Example….
First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue : 
First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = First(ABc) • First (Aa) = First(a) • First (A  ) = First()∪First() • First (B  b) = First(b) • First (B  ) = First()∪First() Step 1:
First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = First(ABc) • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 1: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, }
First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = First(ABc) = {a, } = {a, } - {} ∪ First(Bc) = {a} ∪ First(Bc) • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 2: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, } Step 2 {a} ∪ First(Bc) {a, } {b, }
First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = {a} ∪ First(Bc) = {a} ∪{b, } = {a} ∪{b, } - {} ∪First(c) = {a} ∪{b,c} • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 3: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, } Step 2 {a} ∪ First(Bc) {a, } {b, } Step 3 {a} ∪ {b, c}= {a,b,c} {a, } {b, }
First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = {a,b,c} • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 3: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, } Step 2 {a} ∪ First(Bc) {a, } {b, } If no more change… The first set of a terminal symbol is itself
Follow Set • Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() • We will use some examples for this operation… E…A
E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() • Step 0: EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0 {}  
E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step 1: • From “EPrefix(E) “, We can get… • Follow (Prefix) = Follow (Prefix) ∪First(() = Follow (Prefix) ∪{(} and • Follow (E) = Follow (E) ∪ First()) = Follow (E) ∪ {)} • From “EVTail “, We can get… • Follow (Tail) = Follow (Tail) ∪ First() = Follow (Tail) ∪ {} = Follow (Tail) ∪ Follow (E) • From “Tail+E “, We can get… • Follow (E) = Follow (E) ∪ First() = Follow (E) ∪ {} = Follow (E) ∪ Follow (Tail) EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0    Step 1 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {}
E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0    Step 1 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {} Step 2 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {, }} Step 2: • From “EPrefix(E) “, We can get… • Follow (Prefix) = Follow (Prefix) ∪First(() = Follow (Prefix) ∪{(} and • Follow (E) = Follow (E) ∪ First()) = Follow (E) ∪ {)} • From “EVTail “, We can get… • Follow (Tail) = Follow (Tail) ∪ First() = Follow (Tail) ∪ {} = Follow (Tail) ∪ Follow (E) • From “Tail+E “, We can get… • Follow (E) = Follow (E) ∪ First() = Follow (E) ∪ {} = Follow (E) ∪ Follow (Tail)
E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step 3: EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0    Step 1 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {} Step 2 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {, }} Step 3 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {, }}
Another Example….
S  aSe S  B B  bBe B  C C  cCe C  d G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First()
Step 0: S  aSe S  B B  bBe B  C C  cCe C  d G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0 {}  
Step 1: S  aSe S  B B  bBe B  C C  cCe C  d G0 • S  aSe  Follow (S) = Follow (S) ∪ {e} • S  B  Follow (B ) = Follow (B) ∪ Follow (S) • B  bBe  Follow (B) = Follow (B) ∪ {e} • B  C  Follow (C)  = Follow (C) ∪ Follow (B) • C  cCe  Follow (C) = Follow (C) ∪ {e} • C  d => No Non-Terminal E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0    Step 1 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} = {e}
Step 2: S  aSe S  B B  bBe B  C C  cCe C  d G0 • S  aSe  Follow (S) = Follow (S) ∪ {e} • S  B  Follow (B ) = Follow (B) ∪ Follow (S) • B  bBe  Follow (B) = Follow (B) ∪ {e} • B  C  Follow (C)  = Follow (C) ∪ Follow (B) • C  cCe  Follow (C) = Follow (C) ∪ {e} • C  d => No Non-Terminal E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0    Step 1 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} = {e} Step 2 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} ={, e}
Step 3: S  aSe S  B B  bBe B  C C  cCe C  d G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0    Step 1 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} = {e} Step 2 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} ={, e} Step 3 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} ={, e}
Another Example….
S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First()
Step 0: S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S A B Step 0 {}  
Step 1: • S  ABc  Follow (A) = Follow (A) ∪ First (Bc) = Follow (A) ∪ {b,c} • S  ABc  Follow (B) = Follow (B) ∪ {c} • A  a  No Non-Terminal • A   => No Non-Terminal • B  b => No Non-Terminal • B  => No Non-Terminal S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S A B Step 0 {}   Step 1 {} Follow (A) ∪ {b,c} = {b,c} Follow (B) ∪ {c} = {c}
Step 2: • S  ABc  Follow (A) = Follow (A) ∪ First (Bc) = Follow (A) ∪ {b,c} • S  ABc  Follow (B) = Follow (B) ∪ {c} • A  a  No Non-Terminal • A   => No Non-Terminal • B  b => No Non-Terminal • B  => No Non-Terminal S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S A B Step 0 {}   Step 1 {} Follow (A) ∪ {b,c} = {b,c} Follow (B) ∪ {c} = {c} Step 2 {} Follow (A) ∪ {b,c} = {b,c} Follow (B) ∪ {c} = {c}

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Compiler first set_followset_brief

  • 1. First Set • First (=A) = First(A) , if  ∉ First(A) First(A) – {} ∪First(), if  ∈ First(A) We will use some examples for this operation… Red : A Blue : 
  • 2. First Set (2) EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue : 
  • 3. First Set (2) • First (EPrefix(E))=First(Prefix(E)) ? = {Waiting operation} – We don’t know First(Prefix) at this moment. – So we keep the operation of First(Prefix(E)) until we know First(Prefix). • First (EV Tail) = First(V) = {V} • First (PrefixF) = First(F) = {F} • First (Prefix )= First() - {} ∪First() = {} • First (Tail+E) = First(+) = {+} • First (Tail ) = First() - {} ∪First() = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step 1:
  • 4. First Set (2) • First (EPrefix(E))=First(Prefix(E)) ? = {Waiting operation} – We don’t know First(Prefix) at this moment. – So we keep the operation of First(Prefix(E)) until we know First(Prefix). • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, } Step 1:
  • 5. First Set (2) • First (EPrefix(E))=First(Prefix(E)) ? = {Waiting operation} – In this step, we know First(Prefix). – So we replace the First(Prefix) with {F, }. • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  {F, } Step 2: Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, }
  • 6. First Set (2) • First (EPrefix(E))=First(Prefix(E)) = {F, } = {F, } - {} ∪ First((E)) = {F, (} • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step 2: Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, }
  • 7. First Set (2) • First (EPrefix(E))= {F, (} • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Red : A Blue :  Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, } Step 2 {F, (} ∪ {V} ={F, (, V} {F, } {+, } Step 2:
  • 8. First Set (2) • First (EPrefix(E))= {F, (} • First (EV Tail) = {V} • First (PrefixF) = {F} • First (Prefix )= {} • First (Tail+E) = {+} • First (Tail ) = {} EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step First Set E Prefix Tail ( ) V F + Step 1 First(Prefix(E))∪{V} {F, } {+, } Step 2 {F, (} ∪ {V} ={F, (, V} {F, } {+, } Step 3 {F, (, V} {F, } {+, } {(} {)} {V} {F} {+} If no more change… The first set of a terminal symbol is itself Red : A Blue :  Step 3:
  • 10. First Set (2) S  aSe S  B B  bBe B  C C  cCe C  d G0 Red : A Blue : 
  • 11. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = First(a) ={a} • First (SB) = First(B) • First (B  bBe) = First(b)={b} • First (B  C) = First(C) • First (C  cCe) = First(c) ={c} • First (C  d) = First(d)={d} Red : A Blue :  Step 1:
  • 12. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = First(B) • First (B  bBe) = {b} • First (B  C) = First(C) • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 1:
  • 13. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = First(B) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = First(C) • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2:
  • 14. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = First(C) • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 2: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C)
  • 15. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = First(C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 2: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C)
  • 16. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d} Step 2:
  • 17. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b}∪First(C) = {b}∪ {c, d} • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 3: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d}
  • 18. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b, c, d} • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 3: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d}
  • 19. S  aSe S  B B  bBe B  C C  cCe C  d G0 First Set (2) • First (SaSe) = {a} • First (SB) = {b, c, d} • First (B  bBe) = {b} • First (B  C) = {c, d} • First (C  cCe) = {c} • First (C  d) = {d} Red : A Blue :  Step 3: Step First Set S B C a b c d Step 1 {a}∪First(B) {b}∪First(C) {c, d} Step 2 {a}∪ {b}∪First(C) {b}∪{c, d} = {b,c,d} {c, d} Step 3 {a}∪ {b}∪{c, d} = {a,b,c,d} {b}∪{c, d} = {b,c,d} {c, d} {a} {b} {c} {d} If no more change… The first set of a terminal symbol is itself
  • 21. First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue : 
  • 22. First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = First(ABc) • First (Aa) = First(a) • First (A  ) = First()∪First() • First (B  b) = First(b) • First (B  ) = First()∪First() Step 1:
  • 23. First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = First(ABc) • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 1: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, }
  • 24. First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = First(ABc) = {a, } = {a, } - {} ∪ First(Bc) = {a} ∪ First(Bc) • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 2: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, } Step 2 {a} ∪ First(Bc) {a, } {b, }
  • 25. First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = {a} ∪ First(Bc) = {a} ∪{b, } = {a} ∪{b, } - {} ∪First(c) = {a} ∪{b,c} • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 3: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, } Step 2 {a} ∪ First(Bc) {a, } {b, } Step 3 {a} ∪ {b, c}= {a,b,c} {a, } {b, }
  • 26. First Set (2) S  ABc A  a A   B  b B   G0 Red : A Blue :  • First (SABc) = {a,b,c} • First (Aa) = {a} • First (A  ) = {} • First (B  b) = {b} • First (B  ) = {} Step 3: Step First Set S A B a b c Step 1 First(ABc) {a, } {b, } Step 2 {a} ∪ First(Bc) {a, } {b, } If no more change… The first set of a terminal symbol is itself
  • 27. Follow Set • Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() • We will use some examples for this operation… E…A
  • 28. E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() • Step 0: EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0 {}  
  • 29. E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step 1: • From “EPrefix(E) “, We can get… • Follow (Prefix) = Follow (Prefix) ∪First(() = Follow (Prefix) ∪{(} and • Follow (E) = Follow (E) ∪ First()) = Follow (E) ∪ {)} • From “EVTail “, We can get… • Follow (Tail) = Follow (Tail) ∪ First() = Follow (Tail) ∪ {} = Follow (Tail) ∪ Follow (E) • From “Tail+E “, We can get… • Follow (E) = Follow (E) ∪ First() = Follow (E) ∪ {} = Follow (E) ∪ Follow (Tail) EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0    Step 1 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {}
  • 30. E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0    Step 1 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {} Step 2 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {, }} Step 2: • From “EPrefix(E) “, We can get… • Follow (Prefix) = Follow (Prefix) ∪First(() = Follow (Prefix) ∪{(} and • Follow (E) = Follow (E) ∪ First()) = Follow (E) ∪ {)} • From “EVTail “, We can get… • Follow (Tail) = Follow (Tail) ∪ First() = Follow (Tail) ∪ {} = Follow (Tail) ∪ Follow (E) • From “Tail+E “, We can get… • Follow (E) = Follow (E) ∪ First() = Follow (E) ∪ {} = Follow (E) ∪ Follow (Tail)
  • 31. E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step 3: EPrefix(E) EV Tail PrefixF Prefix Tail+E Tail  G0 Step Follow Set E Prefix Tail Step 0    Step 1 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {} Step 2 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {, }} Step 3 Follow(E) ∪ {)} ∪ Follow (Tail) = {, )} Follow (Prefix) ∪{(} ={(} Follow (Tail) ∪ Follow (E) = {, }}
  • 33. S  aSe S  B B  bBe B  C C  cCe C  d G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First()
  • 34. Step 0: S  aSe S  B B  bBe B  C C  cCe C  d G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0 {}  
  • 35. Step 1: S  aSe S  B B  bBe B  C C  cCe C  d G0 • S  aSe  Follow (S) = Follow (S) ∪ {e} • S  B  Follow (B ) = Follow (B) ∪ Follow (S) • B  bBe  Follow (B) = Follow (B) ∪ {e} • B  C  Follow (C)  = Follow (C) ∪ Follow (B) • C  cCe  Follow (C) = Follow (C) ∪ {e} • C  d => No Non-Terminal E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0    Step 1 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} = {e}
  • 36. Step 2: S  aSe S  B B  bBe B  C C  cCe C  d G0 • S  aSe  Follow (S) = Follow (S) ∪ {e} • S  B  Follow (B ) = Follow (B) ∪ Follow (S) • B  bBe  Follow (B) = Follow (B) ∪ {e} • B  C  Follow (C)  = Follow (C) ∪ Follow (B) • C  cCe  Follow (C) = Follow (C) ∪ {e} • C  d => No Non-Terminal E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0    Step 1 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} = {e} Step 2 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} ={, e}
  • 37. Step 3: S  aSe S  B B  bBe B  C C  cCe C  d G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S B C Step 0    Step 1 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} = {e} Step 2 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} ={, e} Step 3 Follow(S) ∪ {e} = {, e} Follow (B) ∪Follow(S) ∪ {e} ={, e} Follow (C) ∪ Follow(B) ∪ {e} ={, e}
  • 39. S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First()
  • 40. Step 0: S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S A B Step 0 {}  
  • 41. Step 1: • S  ABc  Follow (A) = Follow (A) ∪ First (Bc) = Follow (A) ∪ {b,c} • S  ABc  Follow (B) = Follow (B) ∪ {c} • A  a  No Non-Terminal • A   => No Non-Terminal • B  b => No Non-Terminal • B  => No Non-Terminal S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S A B Step 0 {}   Step 1 {} Follow (A) ∪ {b,c} = {b,c} Follow (B) ∪ {c} = {c}
  • 42. Step 2: • S  ABc  Follow (A) = Follow (A) ∪ First (Bc) = Follow (A) ∪ {b,c} • S  ABc  Follow (B) = Follow (B) ∪ {c} • A  a  No Non-Terminal • A   => No Non-Terminal • B  b => No Non-Terminal • B  => No Non-Terminal S  ABc A  a A   B  b B   G0 E…A Follow (A) = Follow(A) ∪ First() , if  ∉ First() Follow(A) ∪First() – {} ∪Follow(E) , if  ∈ First() Step Follow Set S A B Step 0 {}   Step 1 {} Follow (A) ∪ {b,c} = {b,c} Follow (B) ∪ {c} = {c} Step 2 {} Follow (A) ∪ {b,c} = {b,c} Follow (B) ∪ {c} = {c}