COMS W4115 Monday, March 9, 2009 Solutions to Homework #1

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1 1 COMS W4115 Mondy, Mrch 9, 2009 Solutions to Homework #1 1. Regulr expressions. Lex regulr expression for ll lowercse English words with the five vowels in order. Throughout we ssume tht ech input line consists of single word, tht the five vowels re, e, i, o, u, nd tht in order mens tht the word is of the form uvewixoyuz, where u, v, w, x, y, nd z re ny strings of lowercse letters (tht my contin vowels). One Lex regulr expression definition is: L W [-z]* ^{L}{L}e{L}i{L}o{L}u{L}$ Lex progrm: %{ %} L W {W} #include <stdio.h> #include <string.h> int mxlen = 0; chr mxstr[100]; [-z]* {L}{L}e{L}i{L}o{L}u{L} {if (yyleng > mxlen){ mxlen = yyleng; strcpy(mxstr, yytext);}} \n. ; int min() { yylex(); if (mxlen == 0) printf("no word ws found\n"); else printf("longest word is %s\n", mxstr); } First longest word in /usr/dict/words: dventitious. Lex regulr expression for ll lowercse English words with eginning nd ending with the sustring d:

2 2 W ^d([-z]*d)?$ First longest word in /usr/dict/words: d c. Lex regulr expression for ll lowercse English words with exctly one vowel: C [-df-hj-np-tv-z]* W ^{C}[eiou]{C}$ First longest word in /usr/dict/words: dystrophy Note: Techniclly we should lso include y s vowel. d. Lex regulr expression for ll lowercse English words in which the letters re in strictly incresing lphetic order: W ^??c?d?e?f?g?h?i?j?k?l?m?n?o?p?q?r?s?t?u?v?w?x?y?z?$ First longest word in /usr/dict/words: lmost The lex progrms for ()-(d) re similr to tht for (). 2. Let L e the lnguge { x x is ny string of 's nd 's not contining the sustring }.. A regulr expression for L: **. From your regulr expression construct nondeterministic finite utomton (NFA) for L using the McNughton-Ymd- Thompson lgorithm:

3 c. From your NFA construct n equivlent deterministic finite utomton (DFA). E A B C D Here A = the set of nondeterministic sttes {1}, B = {2}, C = {3,4,6,7,9}, D = {8,7,9}, E={5,4,6,7,9}. In ddition, there is ded stte F to which there is trnsition on ny unspecified input, nd there is trnsition from F to F on nd. d. Minimize the numer sttes in your DFA. A B G D Here stte G is the merger of the two equivlent sttes C nd E of the DFA in (c).

4 4 e. Show how your regulr expression genertes, nd how ech of your finite utomt recognizes : Let R e the regulr expression *. The first in R genertes n, then the first genertes, then * genertes two s, nd the finl genertes n. Thus R genertes the string. The NFA in () mkes the following sequence of stte trnsitions: The DFA in (c) mkes the following sequence of stte trnsitions: A B C E E D The DFA in (d) mkes the following sequence of stte trnsitions: A B G G G D 3. Sequences of oolen expressions. Construct grmmr tht genertes newline-terminted sequences of oolen expressions whose vlue is true. The expressions cn contin the logicl constnts TRUE nd FALSE, nd the oolen opertors AND, OR, nd NOT. The expressions cn contin prentheses. For exmple, your grmmr should generte the expression TRUE OR FALSE, ut not TRUE AND FALSE.

5 5 %left OR %left AND %right NOT lines lines true NL true true AND true true OR flse flse OR true true OR true NOT flse (true) TRUE flse flse OR flse flse AND true true AND flse flse AND flse NOT true (flse) FALSE. Show the prse tree ccording to your grmmr for the input TRUE AND NOT FALSE OR FALSE: lines lines true NL true OR flse true AND true FALSE TRUE NOT flse FALSE

6 6 c. Implement n interpreter tht tkes s input lines of oolen expressions nd produces s output the truth vlue of ech expression. You cn use lex nd ycc or their equivlents to implement your interpreter. Print the source code for your interpreter nd the sequences of commnds you used to crete it nd test it. The progrm elow is ptterned fter the desk clcultor in ALSU, Figs nd 4.61, pp ool.l lexicl nlyzer [ ] { } AND {return AND;} FALSE {return FALSE;} NOT {return NOT;} OR {return OR;} TRUE {return TRUE;} \n. {return yytext[ 0];} ool.y trnsltor %{ #include <ctype.h> #include <stdio.h> %} %token AND FALSE NOT OR TRUE %left OR %left AND %right NOT lines : lines lines expr '\n' { $2?printf("true\n"):printf("flse\n"); } '\n' /* empty */ error '\n' { yyerror("reenter lst line:"); yyerrok; } ; expr : expr OR expr { $$ = $1 $ 3; } expr AND expr { $$ = $1 && $3; } NOT expr { $$ =!$2; } '(' expr ')' { $$ = $2; } TRUE { $$ = 1; } FALSE { $$ = 0; } ; #include "lex.yy.c" Compile ool >lex ool.l >ycc ool.y >gcc y.t.c ly ll -o ool

7 7 d. Run your interpreter on the inputs TRUE OR FALSE TRUE AND FALSE TRUE AND (TRUE OR FALSE) TRUE OR FALSE result: true TRUE AND FALSE result: flse TRUE AND (TRUE OR FALSE) result: true 4. Let L e the lnguge consisting of ll strings of 's nd 's hving the sme numer of 's s 's.. Construct n LL(1) grmmr for L. S AS BS A AA B BB. Construct leftmost derivtion for the string. S => AS // S AS => S // A => BS // S BS => S // B => A S // S AS => S // A => // S

8 8 c. Construc t predictive prsing tle for your grmmr. Input Symol Nonterminl $ S S AS S BS S A A AA A B B B BB d. Show how your predict ive prser prses the string. (Top of stck is on the left.) Stck Input Remrk S$ $ AS$ $ S AS AS$ $ mtch S$ $ A S$ $ mtch BS$ $ S B S BS$ $ mtch S$ $ B S$ $ mtch AS$ $ S AS AS$ $ mtch S$ $ A S$ $ mtch $ $ S 5. Augmented grmmr (0) S S (1) S SA (2) S A (3) A

9 9 LR(0) utomton: I 0 : I 1 : I 2 : I 3 : I 4 : S S S SA S A A S S S S A A S A A S SA SLR(1) prsing tle: Stte Action Goto $ S A 0 s s3 cc 4 2 r2 r2 3 r3 r3 4 r1 r1 Since ech entry in this tle is uniquely defined, this grmmr is SLR(1). However, this grmmr is not LL(1) since S SA A nd SA nd A cn oth derive strings eginning with the terminl.