Lecture 15


  1. \(S'\rightarrow\vdash S\dashv\)
  2. \(S\rightarrow b S d\)
  3. \(S\rightarrow p S q\)
  4. \(S\rightarrow C\)
  5. \(C\rightarrow lC\)
  6. \(C\rightarrow\epsilon\)


Iteration 0 1 2 3
\(S'\) false false false false
\(S\) false false true true
\(C\) false true true true


Iteration 0 1 2 3
\(S'\) \(\{\}\) \(\{\vdash\}\) \(\{\vdash\}\) \(\{\vdash\}\)
\(S\) \(\{\}\) \(\{b,p\}\) \(\{b,p,l\}\) \(\{b,p,l\}\)
\(C\) \(\{\}\) \(\{l\}\) \(\{l\}\) \(\{l\}\)


Iteration 0 1 2
\(S\) \(\{\}\) \(\{\dashv, d, q\}\) \(\{\dashv, d, q\}\)
\(C\) \(\{\}\) \(\{\dashv, d, q\}\) \(\{\dashv, d, q\}\)


\(Predict(A,a)=\{A\rightarrow\beta |\beta\Rightarrow ^*\alpha\gamma\}\cup\{A\rightarrow\beta | Nullable(\beta), a\in Follow(A)\}\)

\(\vdash\) \(\dashv\) \(b\) \(d\) \(p\) \(q\) \(l\)
\(S'\) 1
\(S\) 4 2 4 3 4 4
\(C\) 6 6 6 5

A grammar is \(LL(1)\) if


  1. \(E\rightarrow E+T|T\)
  2. \(T\rightarrow T*F|F\)
  3. \(F\rightarrow id\)
    This grammar is not \(LL(1)\)
    Why? - left recursion

\(E\Rightarrow E+T\Rightarrow T+T\Rightarrow F+T\Rightarrow id+T\)
\(E\Rightarrow T\Rightarrow F\Rightarrow id\)
id:same first symbol

Thus, left-recursion is always not \(LL(1)\)

Make it right-recursive

  1. \(E\rightarrow E+T|T\)
  2. \(T\rightarrow F*T|F\)
  3. \(F\rightarrow id\)

This grammar is still not \(LL(1)\)
Why? \(T+E\) and \(T\) generate the same first symbols.


  1. \(E\rightarrow TE'\)
  2. \(E'\rightarrow \epsilon|+E\)
  3. \(T\rightarrow FT'\)
  4. \(T'\rightarrow \epsilon|*T\)
  5. \(F\rightarrow id\)

This grammar is \(LL(1)\)

\(LL(1)\) conflicts with left-associativity.

Bottom-Up Parsing

go from \(w\) to \(S\)

Stack stores partially-reduced input read so far.

\(w\Leftarrow\alpha_k\Leftarrow\alpha_{k-1}\Leftarrow ... \Leftarrow\alpha_1\Leftarrow\ S\)

Invariant: stack and unread input = \(\alpha_i\) (or \(w\) or \(S\))


  1. \(S'\rightarrow\vdash S\dashv\)
  2. \(S\rightarrow AyB\)
  3. \(A\rightarrow ab\)
  4. \(A\rightarrow cd\)
  5. \(B\rightarrow Z\)
  6. \(B\rightarrow wx\)
Stack Read Input Unread Input Actions
\(\epsilon\) \(\vdash abywx\dashv\) Shift \(\vdash\)
\(\vdash\) \(\vdash\) \(abywx\dashv\) Shift \(a\)
\(\vdash a\) \(\vdash a\) \(bywx\dashv\) Shift \(b\)
\(\vdash ab\) \(\vdash ab\) \(ywx\dashv\) Reduce \(A\rightarrow ab\); pop \(b,a\), push \(A\)
\(\vdash A\) \(\vdash ab\) \(\vdash ywx\dashv\) Shift \(y\)
\(\vdash Ay\) \(\vdash aby\) \(wx\dashv\) Shift \(w\)
\(\vdash Ayw\) \(\vdash abw\) \(x\dashv\) Shift \(x\)
\(\vdash Aywx\) \(\vdash abywx\) \(\dashv\) Reduce \(B\rightarrow wx\); pop \(x,w\), push \(B\)
\(\vdash AyB\) \(\vdash abywx\) \(\dashv\) Reduce \(S\rightarrow AyB\); pop \(B,y,A\), push \(S\)
\(\vdash S\) \(\vdash abywx\) \(\dashv\) Shift \(\dashv\)
\(\vdash S \dashv\) \(\vdash abywx\dashv\) \(\epsilon\) Reduce \(S'\rightarrow\vdash S\dashv\), pop \(S\), push \(S'\)
\(\vdash S \dashv\) \(\vdash abywx\dashv\) \(\epsilon\) Accept

Choice at each step:

  1. Shift a character from input to stack
  2. Reduce the top of stack is the RHS of a grammar rule - replace with LHS.

Accept if stack contains \(S'\) when input is \(\epsilon\)
(Equivalent \(\vdash S\dashv\) on empty input)
(Equivalent accept when shift \(\dashv\))

How do we know whether to shift or reduce?