1. Bottom-Up Parsing Algorithms LR(k) parsing L: scan input Left to right R: produce Rightmost derivation k tokens of lookahead LR(0) zero tokens of look-ahead SLR Simple LR: like LR(0), but uses FOLLOW sets to build more “precise” parsing tables LR(0) is a toy, so we focus on SLR Reading: Section 4.7

2. Problem: when to shift, when to reduce? Recall our favorite grammar: E  T + E | T T  int * T | int | (E) The step T * int + int  int * int + int is not part of any rightmost derivation Hence, reducing first int to T was a mistake How to know when to reduce and when to shift?

3. What we need for LR parsing LR(0) states describe states in which the parser can be Note: LR(0) states are used by both LR(0) and SLR parsers Parsing tables transitions between LR(0) states, actions to take when transiting: shift, reduce, accept, error How to construct LR(0) states? How to construct parsing tables? How to drive the parser?

4. LR(0) state = set of LR(0) items An LR(0) item [X   ] says that the parser is looking for an X it has an  on top of the stack expects to find input string derived from  Notes: [X . a  ] means that if a is on the input, it can be shifted (resulting in  a. . That is: a is a correct token to see on the input, and shifting a would not “over-shift” (still a viable prefix). [X .] means that we could reduce  to X

5. T  int. * T T  int. 5 LR(0) states E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

6. Naïve SLR Parsing Algorithm 1. Let M be LR(0) state machine for G each state contains a set I of LR(0) items 2. Let |x 1 …x n $ be initial configuration 3. Repeat until configuration is S’|$ Let  |  be current configuration Run M on current stack  If M rejects , report parsing error If M accepts , let a be next input Shift if [X  . a  ]  Items Reduce if [X    Items and a  Follow(  )   a    X  a  Report parsing error if neither applies

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

8. T  int. * T T  int. 5 LR(0) states E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

9. SLR Example ConfigurationDFA Halt StateAction | int * int $1

10. T  int. * T T  int. 5 Configuration | int * int $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

11. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5

12. T  int. * T T  int. 5 Configuration int | * int $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

13. T  int. * T T  int. 5 Configuration int | * int $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3 T  (. E) E .T E .T + E T .(E) T .int * T T .int 4 E  T +. E E .T E .T + E T .(E) T .int * T T .int 6 7T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E. T  (E).10 T  int * T.11 T

14. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8

15. T  int. * T T  int. 5 Configuration int * | int $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

16. T  int. * T T  int. 5 Configuration int * | int $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3 T  (. E) E .T E .T + E T .(E) T .int * T T .int 4 E  T +. E E .T E .T + E T .(E) T .int * T T .int 6 7T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E. T  (E).10 T  int * T.11 T

17. T  int. * T T  int. 5 Configuration int * | int $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

18. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8shift int * int | $5

19. T  int. * T T  int. 5 Configuration int * int | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

20. T  int. * T T  int. 5 Configuration int * int | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

21. T  int. * T T  int. 5 Configuration int * int | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

22. T  int. * T T  int. 5 Configuration int * int | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

23. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8shift int * int | $ 5 $  Follow(T)reduce T  int int * | T $8

24. T  int. * T T  int. 5 Configuration int * | T $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

25. T  int. * T T  int. 5 Configuration int * | T $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

26. T  int. * T T  int. 5 Configuration int * | T $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

27. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8shift int * int | $ 5 $  Follow(T)reduce T  int int * | T $8shift int * T | $11

28. T  int. * T T  int. 5 Configuration int * T | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10 T  int * T.11 T

29. T  int. * T T  int. 5 Configuration int * T | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10 T  int * T.11 T

30. T  int. * T T  int. 5 Configuration int * T | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10 T  int * T.11 T

31. T  int. * T T  int. 5 Configuration int * T | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10 T  int * T.11 T

32. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8shift int * int | $ 5 $  Follow(T)reduce T  int int * | T $8shift int * T | $ 11 $  Follow(T)reduce T  int * T | T $1

33. T  int. * T T  int. 5 Configuration | T $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

34. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8shift int * int | $ 5 $  Follow(T)reduce T  int int * | T $8shift int * T | $ 11 $  Follow(T)reduce T  int * T | T $1shift T | $3

35. T  int. * T T  int. 5 Configuration T | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2 E  T. E  T. + E 3 T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

36. T  int. * T T  int. 5 Configuration T | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2 E  T. E  T. + E 3 T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

37. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8shift int * int | $ 5 $  Follow(T)reduce T  int int * | T $8shift int * T | $ 11 $  Follow(T)reduce T  int * T | T $1shift T | $ 3 $  Follow(E)reduce E  T | E $1

38. T  int. * T T  int. 5 Configuration | E $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1S’  E.2E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

39. SLR Example ConfigurationDFA Halt StateAction | int * int $1shift int | * int $5 * not in Follow(T)shift int * | int $8shift int * int | $ 5 $  Follow(T)reduce T  int int * | T $8shift int * T | $ 11 $  Follow(T)reduce T  int * T | T $1shift T | $ 3 $  Follow(E)reduce E  T | E $1shift E | $2ACCEPTt

40. T  int. * T T  int. 5 Configuration E | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2 E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

41. T  int. * T T  int. 5 Configuration E | $ E T ( int * ) E E T ( ( T  ( S’ . E E . T E .T + E T .(E) T .int * T T .int 1 S’  E.2 E  T. E  T. + E 3T  (. E) E .T E .T + E T .(E) T .int * T T .int 4E  T +. E E .T E .T + E T .(E) T .int * T T .int 67T  (E.) 8 T  int *.T T .(E) T .int * T T .int 9E  T + E.T  (E).10T  int * T.11 T

42. Notes Can also use one more state: it accepts in state “S’  E $. ” i.e., it accepts in configuration E$|, not in E|$. Rerunning the automaton at each step is wasteful Most of the work is repeated

43. An Improvement Remember the state of the automaton on each prefix of the stack Change stack to contain pairs  DFA State, Symbol 

44. An Improvement (Cont.) For a stack  state 1, sym 1 ...  state n, sym n  state n is the final state of the DFA on sym 1 … sym n Detail: bottom of stack is  start,any  where any is any dummy input start is the start state of the DFA

45. Goto Table Define Goto[i,A] = j if state i  A state j Goto is just the transition function of the DFA One of two parsing tables

46. Refined Parser Moves Shift x Push  a, x  on the stack a is current input x is a DFA state Reduce X   As before Accept Error

47. Action Table For each state s i and terminal a If s i has item X  .a  and Goto[i,a] = j then Action[i,a] = shift j If s i has item X  . and a  Follow(X) and X  S’ then Action[i,a] = reduce X   If s i has item S’  S. then action[i,$] = accept Otherwise, action[i,a] = error

48. SLR Parsing Algorithm Let I = w$ be initial input Let J = 1 Let DFA state 1 have item S’ .S Let stack =  1, dummy  repeat case action[top_state(stack),I J ] of shift k: push  k, I J  J++ reduce X  A: pop |A| pairs, replace I J-|A| to I J-1 with X J = J - |A| accept: halt normally error: halt and report error

49. Notes on SLR Parsing Algorithm 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

50. Constructing SLR states LR(0) state machine encodes all strings that are valid on the stack each valid string is a configuration, and hence corresponds to a state of the LR(0) state machine each state tells us what to do (shift or reduce?)

51. Example SLR Parse Table int*+()$ET 1s5s4s2s3 2acc 3s6r2 4s5s4s7s3 5s8r4 6s5s4s9s3 7s10 8s5s4s11 9r1 10r5 11r3 1: E  T + E 2: E  T 3: T  int * T 4: T  int 5: T  (E)

52. Example SLR Parse StackIJAct <1,?><1,?>int * int $1s5 2s8 3s5 4r4 int * T $3s11 4r3 <1,?><1,?>T $1s3 2r2 <1,?><1,?>E $1s2 2acc

53. Another Example int * (int + int) * int $