-Mandakinee Singh (11CS10026)

 What is parsing? ◦ Discovering the derivation of a string: If one exists. ◦ Harder than generating strings.  Two major approaches ◦ Top-down parsing ◦ Bottom-up parsing

 A parser is top-down if it discovers a parse tree top to bottom. ◦ A top-down parse corresponds to a preorder traversal(preorder expansion) of the parse tree. ◦ A leftmost derivation is applied at each derivation step.  Start at the root of the parse tree and grow toward leaves.  Pick a production & try to match the input.  Bad “pick”  may need to backtrack.

 Top Down Parser –LL(1) Grammar  LL(1) parsers ◦ Left-to-right input ◦ Leftmost derivation ◦ 1 symbol of look-ahead Preorder Expansion: The Leftmost non terminal production will occur first. Grammars that this can handle are called LL(1) grammars

 Start with the root of the parse tree ◦ Root of the tree: node labeled with the start symbol.  Algorithm: ◦ Repeat until the fringe of the parse tree matches input string. ◦ Declare a pointer which will represent the current position of the parser in string. ◦ Start scanning character by character from left to right from the parse tree and match it with input string.

 If the scanned symbol is: ◦ Terminal: Increase the pointer by one. ◦ Non-Terminal: Go for a production. Add a child node for each symbol of chosen production.  If a terminal symbol is added that doesn’t match, backtrack.  Find the next node to be expanded (a non-terminal)  Repeat The process.  Done when: ◦ Leaves of parse tree match input string (success) ◦ All productions exhausted in backtracking (failure)

 Grammar E E+T(rule 1) | E-T(2) | T(3) TT*F(4) | T/F (5)| F(6) Fnumber(7) | Id(8) Input String:x-2*y

 Problem: ◦ Can’t match next terminal ◦ We guessed wrong at step 2 RuleSentential formInput string - E E T x + T F T 1 E + T  x - 2 * y 3 T + T  x – 2 * y 6 F + T  x – 2 * y 8 + T x  – 2 * y - + T x  – 2 * y  x - 2 * y

Go for next production. RuleSentential formInput string - E 1 E + T  x - 2 * y 3 T + T  x – 2 * y 6 F + T  x – 2 * y 8 + T x  – 2 * y ? + T x  – 2 * y  x - 2 * y Undo all these productions

 Problem: ◦ More input to read ◦ Another cause of backtracking RuleSentential formInput string - E E E x - T F T 2 E - T  x - 2 * y 3 T - T  x – 2 * y 6 F - T  x – 2 * y 8 - T x  – 2 * y - - T x –  2 * y  x - 2 * y 6 - F x –  2 * y 7 - x – 2  * y F 2

All terminals matches- we are done. RuleSentential formInput string - E E E x - T F T 2 E - T  x - 2 * y 3 T - T  x – 2 * y 6 For - T  x – 2 * y 8 - T x  – 2 * y - - T x –  2 * y  x - 2 * y 4 - T * F x –  2 * y 6 - F * F x –  2 * y * F x – 2  * y F - - * F x – 2 *  y 8 - * x – 2 * y  T * F y

 If we see it carefully then there is one more possibility  Problem: Termination ◦ Wrong choice leads to infinite expansion (More importantly: without consuming any input!) ◦ May not be as obvious as this ◦ Our grammar is left recursive RuleSentential formInput string - E 2 E + T  x - 2 * y 2 E + T + T  x – 2 * y 2 E + T + T + T  x – 2 * y 2 E + T + T + T + T  x – 2 * y  x - 2 * y

 Formally, A grammar is left recursive if  a non-terminal A such that A → A  | b (for some set of symbols  ) A → AA  A → AAA  ……………… A → AAAAAAAAA  A → bAAAAAA……AAAAAAA   How to remove it: A → b A’ A’ →   A’|  

 Up Next: Predictive Parser