Lexical Analysis. Amitabha Sanyal. ( as) Department of Computer Science and Engineering, Indian Institute of Technology, Bombay

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1 Lexicl Anlysis Amith Snyl ( s) Deprtment of Computer Science nd Engineering, Indin Institute of Technology, Bomy Septemer 27

2 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp The input progrm s you see it. min () { int i,sum; sum = ; for (i=; i<=; i++); sum = sum + i; printf("%d\n",sum); } Amith Snyl IIT Bomy

3 College of Engineering, Pune Lexicl Anlysis: 4/6 Recp The sme progrm s the compiler sees it (initilly). min for () { (i=; i<=; int i++); printf("%d\n",sum); } i,sum; sum = sum sum = + ; i; The lnk spce chrcter The return chrcter Amith Snyl IIT Bomy

4 College of Engineering, Pune Lexicl Anlysis: 4/6 Recp The sme progrm s the compiler sees it (initilly). min for () { (i=; i<=; int i++); i,sum; sum = sum sum = + printf("%d\n",sum); } How do you mke the compiler see wht you see? ; i; Amith Snyl IIT Bomy

5 College of Engineering, Pune Lexicl Anlysis: 6/6 Recp - Discovering the structure of the progrm Step :. Brek up this string into words the smllest logicl units. min ; for ( ) { ( i = ; sum = sum = i <= ; sum + int ; printf ( "%d\n", sum ) ; } We get sequence of lexemes or tokens. i; i, sum i ++ ) ; Amith Snyl IIT Bomy

6 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step :. Clen up remove the nd the chrcters. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } Steps. nd. re interleved. Amith Snyl IIT Bomy

7 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step :. Clen up remove the nd the chrcters. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } This is lexicl nlysis or scnning. Amith Snyl IIT Bomy

8 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } Amith Snyl IIT Bomy

9 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef Amith Snyl IIT Bomy

10 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt Amith Snyl IIT Bomy

11 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt identifier Amith Snyl IIT Bomy

12 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt identifier min Amith Snyl IIT Bomy

13 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt identifier ( ) min Amith Snyl IIT Bomy

14 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt identifier ( ) { vdecl slist } min Amith Snyl IIT Bomy

15 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt identifier min ( ) { vdecl slist } type vrlist ; Amith Snyl IIT Bomy

16 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt identifier min ( ) { vdecl slist } type vrlist ; int Amith Snyl IIT Bomy

17 College of Engineering, Pune Lexicl Anlysis: /6 Recp - Discovering the structure of the progrm Step 2: Now group the lexemes to form lrger structures. min ( ) { int i, sum ; sum = ; for ( i = ; i <= ; i ++ ) ; sum = sum + i ; printf ( "%d\n", sum ) ; } fundef fnme prms compound-stmt identifier ( ) { vdecl slist } min type vrlist ;... int... Amith Snyl IIT Bomy

18 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp - Discovering the structure of the progrm fundef fnme prms compound-stmt identifier ( ) { vdecl slist } min type vrlist ;... int vrlist, vr vr identifier identifier sum i This is syntx nlysis or prsing. Amith Snyl IIT Bomy

19 College of Engineering, Pune Lexicl Anlysis: 4/6 Lexemes, Tokens nd Ptterns Definition: Lexicl nlysis is the opertion of dividing the input progrm into sequence of lexemes (tokens). Amith Snyl IIT Bomy

20 College of Engineering, Pune Lexicl Anlysis: 4/6 Lexemes, Tokens nd Ptterns Definition: Lexicl nlysis is the opertion of dividing the input progrm into sequence of lexemes (tokens). Distinguish etween Amith Snyl IIT Bomy

21 College of Engineering, Pune Lexicl Anlysis: 4/6 Lexemes, Tokens nd Ptterns Definition: Lexicl nlysis is the opertion of dividing the input progrm into sequence of lexemes (tokens). Distinguish etween lexemes smllest logicl units (words) of progrm. Exmples i, sum, for,, ++, "%d\n", <=. Amith Snyl IIT Bomy

22 College of Engineering, Pune Lexicl Anlysis: 4/6 Lexemes, Tokens nd Ptterns Definition: Lexicl nlysis is the opertion of dividing the input progrm into sequence of lexemes (tokens). Distinguish etween lexemes smllest logicl units (words) of progrm. Exmples i, sum, for,, ++, "%d\n", <=. tokens sets of similr lexemes. Exmples identifier = {i, sum, uffer,... } int constnt = {,,... } ddop = {+, -} Amith Snyl IIT Bomy

23 College of Engineering, Pune Lexicl Anlysis: 6/6 Lexemes, Tokens nd Ptterns Things tht re not counted s lexemes Amith Snyl IIT Bomy

24 College of Engineering, Pune Lexicl Anlysis: 6/6 Lexemes, Tokens nd Ptterns Things tht re not counted s lexemes white spces t, lnks nd newlines Amith Snyl IIT Bomy

25 College of Engineering, Pune Lexicl Anlysis: 6/6 Lexemes, Tokens nd Ptterns Things tht re not counted s lexemes white spces t, lnks nd newlines comments Amith Snyl IIT Bomy

26 College of Engineering, Pune Lexicl Anlysis: 6/6 Lexemes, Tokens nd Ptterns Things tht re not counted s lexemes white spces t, lnks nd newlines comments These too hve to e detected nd ignored. Amith Snyl IIT Bomy

27 College of Engineering, Pune Lexicl Anlysis: /6 Lexemes, Tokens nd Ptterns The lexicl nlyser: detects the next lexeme ctegorises it into the right token psses to the syntx nlyser the token nme for further syntx nlysis the lexeme itself, in some form, for stges eyond syntx nlysis Amith Snyl IIT Bomy

28 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp - Lexemes, Tokens nd Ptterns How does one descrie the lexemes tht mke up the token identifier. Vrints in different lnguges. Amith Snyl IIT Bomy

29 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp - Lexemes, Tokens nd Ptterns How does one descrie the lexemes tht mke up the token identifier. Vrints in different lnguges. String of lphnumeric chrcters. The first chrcter is n lphet. Amith Snyl IIT Bomy

30 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp - Lexemes, Tokens nd Ptterns How does one descrie the lexemes tht mke up the token identifier. Vrints in different lnguges. String of lphnumeric chrcters. The first chrcter is n lphet. string of lphnumeric chrcters in which the the first chrcter is n lphet. It hs length of t most 3. Amith Snyl IIT Bomy

31 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp - Lexemes, Tokens nd Ptterns How does one descrie the lexemes tht mke up the token identifier. Vrints in different lnguges. String of lphnumeric chrcters. The first chrcter is n lphet. string of lphnumeric chrcters in which the the first chrcter is n lphet. It hs length of t most 3. string of lphet or numeric or underline chrcters in which the the first chrcter is n lphet or n underline. It hs length of t most 3. Any chrcters fter the 3st chrcter re ignored. Amith Snyl IIT Bomy

32 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp - Lexemes, Tokens nd Ptterns How does one descrie the lexemes tht mke up the token identifier. Vrints in different lnguges. String of lphnumeric chrcters. The first chrcter is n lphet. string of lphnumeric chrcters in which the the first chrcter is n lphet. It hs length of t most 3. string of lphet or numeric or underline chrcters in which the the first chrcter is n lphet or n underline. It hs length of t most 3. Any chrcters fter the 3st chrcter re ignored. Such descriptions re clled ptterns. The description my e informl or forml. Amith Snyl IIT Bomy

33 College of Engineering, Pune Lexicl Anlysis: 22/6 Recp - Bsic concepts nd issues A pttern is used to specify tokens precisely uild recognizer from such specifictions Amith Snyl IIT Bomy

34 College of Engineering, Pune Lexicl Anlysis: 24/6 Exmple tokens in Jv. Identifier: A Jvletter followed y zero or more Jvletterordigits. A Jvletter includes the chrcters -z, A-Z, _ nd $. 2. Constnts: 2. Integer Constnts Octl, Hex nd Deciml 4 yte nd yte representtion 2.2 Floting point constnts flot - ends with f doule 2.3 Boolen constnts true nd flse 2.4 Chrcter constnts, \u34, \t 2.5 String constnts "", "\"", "A string". 2.6 Null constnt null. 3. Delimiters: (, ), {, },, ], ;,. nd, 4. Opertors: =, >, <... >>>= 5. Keywords: strct, oolen... voltile, while. Amith Snyl IIT Bomy

35 College of Engineering, Pune Lexicl Anlysis: 26/6 Recp - Bsic concepts nd issues Where does lexicl nlyser fit into the rest of the compiler? The front end of most compilers is prser driven. When the prser needs the next token, it invokes the Lexicl Anlyser. Insted of nlysing the entire input string, the lexicl nlyser sees enough of the input string to return single token. The ctions of the lexicl nlyser nd prser re intertwined. prser input progrm lexicl nlyser rest of the compiler Amith Snyl IIT Bomy

36 College of Engineering, Pune Lexicl Anlysis: 2/6 Recp - Token Attriutes Aprt from the token itself, the lexicl nlyser lso psses other informtions regrding the token. These items of informtion re clled token ttriutes EXAMPLE lexeme <token, token ttriute> 3 < const, 3> A <identifier, A> if <if, > = <ssignop, > > <gt, > ; <semicolon, > Amith Snyl IIT Bomy

37 College of Engineering, Pune Lexicl Anlysis: 3/6 Creting Lexicl Anlyzer Two pproches: Amith Snyl IIT Bomy

38 College of Engineering, Pune Lexicl Anlysis: 3/6 Creting Lexicl Anlyzer Two pproches:. Hnd code This is only of historicl interest now. Possily more efficient. Amith Snyl IIT Bomy

39 College of Engineering, Pune Lexicl Anlysis: 3/6 Creting Lexicl Anlyzer Two pproches:. Hnd code This is only of historicl interest now. Possily more efficient. 2. Use genertor To generte the lexicl nlyser from forml description. The genertion process is fster. Less prone to errors. Amith Snyl IIT Bomy

40 College of Engineering, Pune Lexicl Anlysis: 32/6 Automtic Genertion of Lexicl Anlysers Inputs to the lexicl nlyser genertor: A specifiction of the tokens of the source lnguge, consisting of: regulr expression descriing ech token, nd code frgment descriing the ction to e performed, on identifying ech token. The generted lexicl nlyser consists of: A deterministic finite utomton (DFA) constructed from the token specifiction. A code frgment ( driver routine) which cn trverse ny DFA. Code for the ction specifictions. Amith Snyl IIT Bomy

41 College of Engineering, Pune Lexicl Anlysis: 34/6 Automtic Genertion of Lexicl Anlysers specifiction regulr expression ction routines processed copied input progrm DFA ction routines generted lexicl nlyser Driver routine tokens Amith Snyl IIT Bomy

42 College of Engineering, Pune Lexicl Anlysis: 36/6 Exmple of Lexicl Anlyser Genertion Suppose lnguge hs two tokens Pttern Action * { printf( "Token found");} c+ { printf( "Token 2 found");} From the description, construct structure clled deterministic finite utomton (DFA). c 2 3 c Amith Snyl IIT Bomy

43 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: IIT Bomy

44 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: IIT Bomy

45 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: IIT Bomy

46 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: IIT Bomy

47 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found IIT Bomy

48 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found IIT Bomy

49 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found IIT Bomy

50 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found IIT Bomy

51 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found IIT Bomy

52 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found Token found IIT Bomy

53 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found Token found IIT Bomy

54 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found Token found IIT Bomy

55 College of Engineering, Pune Lexicl Anlysis: 3/6 Exmple of Lexicl Anlyser Genertion Now consider the following together: The DFA c The ctions void ction(); { switch(stte) 2:{printf("Token found"); rek;} 3:{printf("Token 2 found"); rek;} } Amith Snyl 3 c 2 The driver routine void nexttoken () {stte = ; c = nextchr(); while (vlid(nextsttestte,c])) {stte = nextsttestte,c]; c = nextchr();} if (!finl(stte)) {error; return;} else {unput(c);ction();return;}} The input nd output Input: dcc Output: Token found Token found Token 2 found IIT Bomy

56 College of Engineering, Pune Lexicl Anlysis: 4/6 Exmple of Lexicl Anlyser Genertion In summry: The DFA, the driver routine nd the ction routines tken together, constitute the lexicl nlyser. ctions re supplied s prt of specifiction. driver routine is common to ll generted lexicl nlyzers The only issue how re the ptterns, specified y regulr expressions, converted to DFA. In two steps: Convert regulr expression into NFA. Convert NFA to DFA. Amith Snyl IIT Bomy

57 College of Engineering, Pune Lexicl Anlysis: 42/6 Exmple of Lexicl Anlyser Genertion Consider lnguge with the following tokens: egin representing the lexeme egin integer Exmples:, -5, 25 identifier Exmples:, A, mx Amith Snyl IIT Bomy

58 College of Engineering, Pune Lexicl Anlysis: 44/6 Converting Regulr Expressions to NFA In two prts; First convert the regulr expression corresponding to ech token into NFA. Invrint: A single finl stte corresponding to ech token. Join the NFAs otined for ll the tokens. Amith Snyl IIT Bomy

59 College of Engineering, Pune Lexicl Anlysis: 46/6 Converting Regulr Expressions to DFA RE for ɛ RE for ε ε Amith Snyl IIT Bomy

60 College of Engineering, Pune Lexicl Anlysis: 4/6 Converting Regulr Expressions to NFA r r2 RE for r r 2 r r r 2 r 2 Amith Snyl IIT Bomy

61 College of Engineering, Pune Lexicl Anlysis: 5/6 Converting Regulr Expressions to NFA r RE for r r 2 r r 2 r r2 ε r 2 ε ε ε Amith Snyl IIT Bomy

62 College of Engineering, Pune Lexicl Anlysis: 52/6 Converting Regulr Expressions to NFA r RE for r ε r r* ε ε ε Amith Snyl IIT Bomy

63 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA Amith Snyl IIT Bomy

64 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA 2 e g 3 4 i 5 n 6 Amith Snyl IIT Bomy

65 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA 2 e g 3 4 i 5 n digit 9 digit digit Amith Snyl IIT Bomy

66 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA 2 e g 3 4 i 5 n digit 9 digit letter letter digit digit Amith Snyl IIT Bomy

67 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit Amith Snyl IIT Bomy

68 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit,,7, Amith Snyl IIT Bomy

69 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit,,7, 2, Amith Snyl IIT Bomy

70 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit,,7, - 2, 6 Amith Snyl IIT Bomy

71 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit,,7, 2, digit Amith Snyl IIT Bomy

72 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit,,7, 2, digit Amith Snyl -{} IIT Bomy

73 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit Amith Snyl,,7, -{} 2, digit e -{e} 3, 2 IIT Bomy

74 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit Amith Snyl,,7, -{} 2, digit 7 digit e -{e} 3, 2 IIT Bomy

75 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit Amith Snyl,,7, -{} 2, digit 7 digit e -{e} 3, 2 letter digit IIT Bomy

76 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit Amith Snyl,,7, -{} 2, digit 7 digit e -{e} 3, 2 letter digit 4, g 3 -{g} IIT Bomy

77 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit Amith Snyl,,7, -{} 2, digit 7 digit e -{e} 3, 2 letter digit 4, g 3 -{g} digit IIT Bomy

78 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit Amith Snyl,,7, -{} 2, digit 7 digit e -{e} 3, 2 letter digit 4, g 3 -{g} -{i} digit i 5, 4 IIT Bomy

79 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit,,7, 2, 6 e -{e} 3, digit 7 4, g 3 -{g} -{i} digit i 5, 4 -{n} n 6, 5 -{} digit letter digit Amith Snyl IIT Bomy

80 College of Engineering, Pune Lexicl Anlysis: 54/6 Converting NFA to DFA ɛ 2 e g 3 4 i 5 n 6 ɛ 7 - digit 9 digit ɛ digit letter letter digit,,7, 2, 6 e -{e} 3, digit 7 4, g 3 -{g} -{i} digit i 5, 4 -{n} n 6, 5 letter digit -{} digit letter digit Amith Snyl IIT Bomy

81 College of Engineering, Pune Lexicl Anlysis: 56/6 LEXICAL ERRORS Primrily of two kinds:. Lexemes whose length exceed the ound specified y the lnguge. In Fortrn, n identifier more thn 7 chrcters long is lexicl error. Most lnguges hve ound on the precision of numeric constnts. A constnt whose length exceeds this ound is lexicl error. 2. Illegl chrcters in the progrm. The chrcters ~, & occuring in Pscl progrm (ut not within string or comment) re lexicl errors. 3. Unterminted strings or comments. Amith Snyl IIT Bomy

82 College of Engineering, Pune Lexicl Anlysis: 5/6 Hndling Lexicl Errors issuing n error messge, the ction tken on detection of n error re:. Issue n pproprite error messge. 2. Error of the first type the entire lexeme is red nd then truncted to the specified length. Error of the second type skip illegl chrcter. pss the chrcter to the prser which hs etter knowledge of the context in which error hs occurred. more possiilities of recovery - replcement insted of deletion. Error of the third type wit till end of file n issue error messge. Amith Snyl IIT Bomy

83 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES The DFA constructed for ( ɛ)( ) Amith Snyl IIT Bomy

84 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES The DFA constructed for ( ɛ)( ). There is nother DFA for the sme regulr expression with lesser numer of sttes A B C D Amith Snyl IIT Bomy

85 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES The DFA constructed for ( ɛ)( ). There is nother DFA for the sme regulr expression with lesser numer of sttes A B C D For typicl lnguge, the numer of sttes of the DFA is in order of hundreds. Amith Snyl IIT Bomy

86 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES The DFA constructed for ( ɛ)( ). There is nother DFA for the sme regulr expression with lesser numer of sttes A B C D For typicl lnguge, the numer of sttes of the DFA is in order of hundreds. Therefore we should try to minimize the numer of sttes. Amith Snyl IIT Bomy

87 College of Engineering, Pune Lexicl Anlysis: 62/6 MINIMIZING THE NUMBER OF STATES The second DFA hs een otined y merging sttes nd 3 of the first DFA A B C D Amith Snyl IIT Bomy

88 College of Engineering, Pune Lexicl Anlysis: 62/6 MINIMIZING THE NUMBER OF STATES The second DFA hs een otined y merging sttes nd 3 of the first DFA. Under wht conditions cn this merging tke plce? A B C D Amith Snyl IIT Bomy

89 College of Engineering, Pune Lexicl Anlysis: 64/6 MINIMIZING THE NUMBER OF STATES A B C D The string tkes oth sttes nd 3 to finl stte. Amith Snyl IIT Bomy

90 College of Engineering, Pune Lexicl Anlysis: 64/6 MINIMIZING THE NUMBER OF STATES A B C D The string tkes oth sttes nd 3 to finl stte. The string tkes oth sttes nd 3 to non-finl stte. Amith Snyl IIT Bomy

91 College of Engineering, Pune Lexicl Anlysis: 64/6 MINIMIZING THE NUMBER OF STATES A B C D The string tkes oth sttes nd 3 to finl stte. The string tkes oth sttes nd 3 to non-finl stte. The string ɛ tkes oth sttes nd 3 to non-finl stte. Amith Snyl IIT Bomy

92 College of Engineering, Pune Lexicl Anlysis: 64/6 MINIMIZING THE NUMBER OF STATES A B C D The string tkes oth sttes nd 3 to finl stte. The string tkes oth sttes nd 3 to non-finl stte. The string ɛ tkes oth sttes nd 3 to non-finl stte. The string tkes oth sttes nd 3 to finl stte. Amith Snyl IIT Bomy

93 College of Engineering, Pune Lexicl Anlysis: 64/6 MINIMIZING THE NUMBER OF STATES A B C D The string tkes oth sttes nd 3 to finl stte. The string tkes oth sttes nd 3 to non-finl stte. The string ɛ tkes oth sttes nd 3 to non-finl stte. The string tkes oth sttes nd 3 to finl stte. Conclusion: Any string tht tkes stte to finl stte lso tkes 3 to finl stte. Conversely, ny string tht tkes stte to non-finl stte lso tkes 3 to non-finl stte. Amith Snyl IIT Bomy

94 College of Engineering, Pune Lexicl Anlysis: 66/6 MINIMIZING THE NUMBER OF STATES Sttes nd 3 re sid to e indistinguishle. Minimimiztion strtegy: Find indistinguishle sttes. Merge them. A B C D Question: How does one find indistingushle sttes? Amith Snyl IIT Bomy

95 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES Key ide: Amith Snyl IIT Bomy

96 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES Key ide: Initilly ssume ll sttes to e indistinguishle. Put them in single set. Amith Snyl IIT Bomy

97 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES Key ide: The string ɛ distinguishes etween finl sttes nd non-finl sttes. Crete two prtitions. Amith Snyl IIT Bomy

98 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES Key ide: tkes 4 to red prtition nd retins other lue sttes in lue prtition. Put 4 in seprte prtition. Amith Snyl IIT Bomy

99 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES Key ide: The string distinguishes 3 from other sttes in the lue prtition. Amith Snyl IIT Bomy

100 College of Engineering, Pune Lexicl Anlysis: 6/6 MINIMIZING THE NUMBER OF STATES Key ide: A B C D No other prtition possile. Merge ll sttes in the sme prtition. Amith Snyl IIT Bomy

101 College of Engineering, Pune Lexicl Anlysis: 7/6 Summry of the Method. Construct n initil prtition π = {S F,F,..., F n,}, where F = F F 2... F n s, nd ech F i is the set of finl sttes for some token i. 2. for ech set G in π do prtition G into susets such tht two sttes s nd t of G re in the sme suset if nd only if for ll input symols, sttes s nd t hve trnsitions onto sttes in the sme set of π; replce G in π new y the set of ll susets formed 3. If π new = π, let π finl := π nd continue with step 4. Otherwise repet step 2 with π := π new. 4. Merge sttes in the sme set of the prtition. 5. Remove ny ded sttes. Amith Snyl IIT Bomy

102 College of Engineering, Pune Lexicl Anlysis: 72/6 EFFICIENT REPRESENTATION OF DFA A nive method to represent DFA uses two dimensionl rry. c c c x 3 3 For typicl lnguge: the numer of DFA sttes is in the order of hundreds (sometimes ), the numer of input symols is greter thn. It is desirle to find spce-efficient representtion of the DFA. Amith Snyl IIT Bomy

103 College of Engineering, Pune Lexicl Anlysis: 74/6 The Four Arrys Scheme Key Oservtion For DFA tht we hve seen erlier, the sttes mrked with # ehve like stte on ll symols except for one symol. # - -{} # - digit 6 7 digit e -{e} letter digit # # # g 2 3 i 4 -{g} -{i} digit -{n} n 5 letter digit Therefore informtion out stte cn lso e used for these sttes. Amith Snyl IIT Bomy

104 College of Engineering, Pune Lexicl Anlysis: 76/6 Four Arrys Representtion of DFA Symols nd their numering z DEFAULT BASE NEXT CHECK Amith Snyl IIT Bomy

105 College of Engineering, Pune Lexicl Anlysis: 7/6 Four Arrys Representtion of DFA Symols nd their numering z DEFAULT BASE NEXT CHECK Amith Snyl IIT Bomy

106 College of Engineering, Pune Lexicl Anlysis: /6 Four Arrys Representtion of DFA Symols nd their numering z DEFAULT BASE NEXT CHECK Amith Snyl IIT Bomy

107 College of Engineering, Pune Lexicl Anlysis: 2/6 Four Arrys Representtion of DFA Symols nd their numering z DEFAULT BASE NEXT CHECK Amith Snyl IIT Bomy

108 College of Engineering, Pune Lexicl Anlysis: 4/6 Four Arrys Representtion of DFA If s is stte nd is the numeric representtion of symol, then. BASEs] gives the se loction for the informtion stored out stte s. 2. NEXT BASEs]+] gives the next stte for s nd symol, only if CHECKBASEs]+] = s. 3. If CHECKBASEs]+] s, then the next stte informtion is ssocited with DEFAULT s]. function nextstte(s,); egin if CHECKBASEs] + ] = s then NEXT BASEs]+] else return(nextstte(default s],)) end Amith Snyl IIT Bomy

109 College of Engineering, Pune Lexicl Anlysis: 6/6 Four Arrys Representtion of DFA All the entries for stte hve een stored in the rry NEXT. The CHECK rry shows tht the entries re vlid for stte. Stte hs trnsition on e(4), which is different from the corresponding trnsition on stte. This differing entry is stored in NEXT 37]. Therefore BASE] is set to 37 4 = 33. By similr resoning BASE] is set to 36. To find nextstte, ], we first refer to NEXT 33 + ], But since CHECK33 + ] is not we hve to refer to DEFAULT ] which is. So the correct next stte is found from NEXT BASE] + ] =. To fill up the four rrys, we hve to use heuristic method. One possiility, which works well in prctice, is to find for given stte, the lowest BASE, so tht the specil entries cn e filled without conflicting with existing entries. Amith Snyl IIT Bomy

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