Bottom-up Parsing

Bottom-up parsing is

Revert to the (unambiguous) natural grammar for our example

E --> T + E | T
T --> int * T | int | (E)
Consider the string int * int + int

Bottom-up parsing reduces a string to the start symbol by inverting productions

int * int + int                int * T + int               T --> int
int * T + int                  T + int                     T --> int * T
T + int                        T + T                       T --> int
T + T                          T + E                       E --> T
T + E                          E                           E --> T + E
Reading the productions bottom-to-top, these are the productions.
Reading them top-to-bottom, these are reductions

What's the point? The productions read backwards, trace a rightmost derivation, i.e., we are producing the right-most non-terminal at each step

Important fact #1 about bottom-up parsing
Bottom-up parsing traces a right-most derivation in reverse


int * int + int
 |     |     |
 |   	 T     |
 \   /       |
   T         |
	 |         T
	  \       /
		 \     /
		  \   /

Shift-reduce parsing : strategy used by bottom-up parsers

Important fact #1 has an interesting consequence:

Idea: split string into two substrings

Example : int * int + int
Assume an oracle tells us whether to shift of reduce

|int * int  + int                     shift
int |* int + int                      shift
int * |int + int                      shift
int * int| + int                      shift
int *   T| + int                      reduce
T | + int                             reduce
T +| int                              shift
T + int|                              shift
T +   T|                              reduce
T +   E|                              reduce
E|                                    reduce
Left string can be implemented by a stack, because we only reduce the suffix of the left string. Top of the stack is |

shift pushes a terminal on the stack

reduce pops off of the stack (production rhs) and pushes a non-terminal on the stack (production lhs)

In a given state, more than one action (shift or reduce) may lead to a valid parse.

If it is legal to shift or reduce, there is a shift-reduce conflict. Need some techniques to remove them

If it is legal to reduce using two productions, there is a reduce-reduce conflict. Are always bad and usually indicate a serious problem with the grammar.

In either case, the parser does not know what to do, and we either need to rewrite the grammar, or need to give the parser a hint on what to do in this situation.

Deciding when to shift and when to reduce

Example : int * int + int
Assume an oracle tells us whether to shift of reduce

|int * int  + int                     shift
int |* int + int                      shift

At this point, we could reduce.

T |* int + int                        reduce

But this would be a fatal mistake because there is no way we can reduce to E (there is no production that begins with T *).

Intuition: want to reduce only if the result can still be reduced to the start symbol (E)

Assume a right-most derivation:

S -->* \alpha X \omega --> \alpha \beta \omega
Then \alpha \beta is a handle of \alpha \beta \omega

Handles formalize the intuition: A handle is a reduction that allows further reductions back to the start symbol.

We only want to reduce at handles.

Note: we have said what a handle is, not how to find handles.

Important fact #2: In shift-reduce parsing, handles appear only at the top of the stack, never inside. Proof by induction

  1. True initially, stack is empty
  2. Immediately after reducing a handle
    • rightmost non-terminal on top of stack
    • next handle must be to right of rightmost non-terminal, because this is a right-most derivation.
    • i.e., sequence of shift moves reaches the next handle
Handles are never to the left of the rightmost non-terminal, and are always at the top of the stack.

Hence shift-reduce parsers can only shift or reduce at the top of the stack. However, how to recognize handles, i.e., when to shift and when to reduce?

Recognizing handles

For an arbitrary grammar, no efficient algorithms known to recognize handles.

Heuristics to identify handles. On certain types of CFGs, heuristics are always correct.

Venn diagram: All CFGs \superset Unambiguous CFGs \superset LR(k) CFGs \superset LALR(k) CFGs \superset SLR(k) \superset LR(0)

LR(k) are quite general, but most practical grammers are LALR(k). SLR(k) are simplifications over LALR(k) [simple LR grammars]. There are grammars that are SLR(k) but not LALR(k) and so on.

\alpha is a viable prefix if there is a valid right-most derivation of the string: S' --> ... --> \alpha\omega --> ... --> string. e.g., \epsilon is a viable prefix

In other words, \alpha is a viable prefix if there is an \omega such that \alpha|\omega is a state of a shift-reduce parser. Here \alpha is the stack and \omega is the rest of the input. It can lookahead at \omega, but the parser does not know the whole think. The parser knows the whole stack.

What does this mean? A viable prefix does not extend past the right end of the handle. It's a viable prefix because it is a prefix of the handle. As long as a parser has viable prefixes on the stack, no parsing error has been detected.

An item is a production with a "." somewhere on the RHS in the production: e.g., the items for T --> (E) are: T --> .(E), T --> (.E), T --> (E.), T --> (E)..

The only item for an \epsilon production, X --> \epsilon is X --> .. Items are often called "LR(0) items".

The problem in recognizing viable prefixes is that the stack has only bits and pieces of the RHS of productions

These bits and pieces are always prefixes of RHS of some production(s).

Consider the input (int) for the grammar:

E --> T + E | T
T --> int * T | int | (E)

The stack may have many prefixes on RHS's:

Prefix1 Prefix2 Prefix3 ... Prefix(n-1) Prefix(n)
Let Prefix(i) be a prefix of RHS of Xi --> \alpha_i Recursively, Prefix(k+1) ... Prefix(n-1) Prefix(n) eventually reduces to the missing part of \alpha(k)

Important fact #3 about bottom-up parsing: For any grammar, the set of viable prefixes is a regular language. The regular language represents the language formed by concatenating 0 or more prefixes of the productions (items).

For example, the language of viable prefixes for the example grammar:

S --> \epsilon | [S]
\epsilon | "["* | "["*S | "["*S"]"

The language of viable prefixes for the example grammar:

S --> \epsilon | [S] | S.S
X = \epsilon | "[" | "["* | "["*S | "["*S"]"
Y = "["*S.(X | Y)*
Z = X + Y

Conversely, In a string is parse-able through the bottom-up parser, then every potential state of the stack should always be a viable prefix.

Consider the string (int * int):

Alternatively, we can represent this as a "stack of items"
T --> (.E)
E --> .T
T --> int * .T
says that Reading backwards, the LHS of every item becomes the RHS of the predecessor production.

In other words, every viable prefix can be represented as a stack of items, where the (n+1)th item is a production for a non-terminal that follows the "." in the nth item.

To recognize viable prefixes, we must

Recognizing Viable Prefixes

  1. Add a dummy production S' --> S to G
  2. We will construct an NFA that will behave as follows:
    NFA(stack) = yes if stack is a viable prefix
               = no otherwise
  3. The NFA will read the input (stack) bottom-to-top
  4. The NFA states are the items of G
  5. For item E --> \alpha.X\beta, add transition from (E --> \alpha.X\beta) --X--> (E --> \alphaX.\beta)
  6. For item E --> \alpha.X\beta and production X --> \gamma, add (E --> \alpha.X\beta) --\epsilon--> (X --> .\gamma)
  7. Every state is an accepting state (i.e., if the entire stack is consumed, the stack is a viable prefix)
  8. Start state is (S' --> .S)

Valid Items

Item X --> \beta.\gamma is valid for a viable prefix \alpha\beta if
S' -->* \alpha X \omega --> \alpha \beta \gamma \omega
by a right-most derivation.

After parsing \alpha \beta, the valid items are the possible tops of the stack of items

An item I is valid for a viable prefix \alpha if the DFA recognizing viable prefixes terminates on input \alpha in a state s containing I.

The items in s describe what the top of the item stack might be after reading input \alpha

An item is often valid for many prefixes. e.g., The item T --> (.E) is valid for prefixes

We can see this by looking at the DFA, which will keep looping into the same DFA state for each open paren. Need to show the NFA and DFA construction for our example grammar, and the valid items for these string prefixes. Will need a laptop and a projector!

Simple LR (SLR) parsing

SLR = "Simple LR": improves on LR(0) shift/reduce heuristics so fewer state have conflicts

Idea: Assume

If there are conflicts under these rules, the grammar is not SLR

SLR Parsing algorithm

  1. Let M be DFA for viable prefixes of G
  2. Let |x1...xn$ be initial configuration
  3. Repeat until configuration is S|$
    • Let \alpha|\omega be current configuration
    • Run M on current stack \alpha
    • If Mrejects \alpha, report parsing error
      • Stack \alpha is not a viable prefix
    • If M accepts \alpha with items I, let a be next input
    • Shift if X --> \beta.a\gamma \in I
    • Reduce if X --> \beta. \in I and a \in Follow(X)
    • Report parsing error if neither applies

If there is a conflict in the last step, grammar is not SLR(k). k is the amount of lookahead (in practice, k = 1)

SLR Improvements

Action table: For each state si and terminal a

SLR Improvements

  • Let I = w$ be initial input
  • Let j = 0
  • Let DFA state 1 have item S'-->.S
  • Let stack = <dummy, 1>
  • repeat
      case action[top_state(stack),I[j]] of
      • shift k: push < I[j++], k >
      • reduce X-->A:
        • pop |A| pairs
        • push <X,goto[top_state(stack),X]>
      • accept: halt normally
      • error: halt and report error
  • Note that the algorithm uses only the DFA states and the input
    • The stack symbols are never used!
  • However, we still need the symbols for semantic actions (e.g., code generation)
  • Some common constructs are not SLR(1)
  • LR(1) is more powerful
    • Build lookahead into the items (fine-grained disambiguation for each item, rather than just checking the follow set)
    • An LR(1) item is a pair: LR(0) item, lookahead
    • [T --> .int *T,$] means
      • After seeing T--> int * T, reduce if lookahead is $
    • More accurate than just using follow sets
    • Take a look at the LALR(1) automaton for your parser! (uses these items, but has a slight optimization over it)
      • LALR automaton is composed of states containing LR(1) items, with the added optimization that if two states differ only in lookahead then we combine those states. If the resulting states (after this minimization), have no conflict in the grammar, then that grammar is LALR also.

SLR Examples

S' --> S
S --> Sa
S --> b
SLR parsers do not mind left-recursive grammars

The first state in the corresponding DFA will look like (\epsilon closure of the NFA state): (state 1)

S' --> .S
S --> .Sa
S --> .b
If we see a b in this state, we get another DFA state: (state 2)
S --> b.
Alternatively, if we see a S in this state, we get another DFA state: (state 3)
S' --> S.
S --> S.a
From this state, if we see a, we get (state 4)
S --> Sa.

The only state with a shift-reduce conflict is state3. Here, if we look at follow of S', we have only "$", and hence we can resolve this conflict by one lookahead. Hence this is an SLR grammar

Another example grammar

S' --> S
S --> SaS
S --> b
Looking at the corresponding DFA: (state 1)
S' --> .S
S' --> .SaS
S --> .b
One possibility is that we see b in this state to get: (state 2)
S --> b.
Another possibility is that we see S in this state to get: (state 3)
S' --> S.
S' --> S.aS
If we get a in this state, we get (state 4)
S' --> Sa.S
S --> .SaS
S --> .b
(notice that we formed an \epsilon-closure of the first item to add more items

From here, if we get S, we get the following state: (state 5)

S --> SaS.
S --> S.aS
From here, if we get a again, we go back to state 4! If we get b, we go to state 2

The only states that have conflicts are: state3 (resolved by follow as follow(S') = $) and state5 (has a shift/reduce conflict because a \in follow(S))

Thus this is not an SLR(1) grammar