FIT2014解析留学生、辅导C语言编程、C/C++语言辅导、lex和yacc实现解析器教程讲解

日期：2018-09-29 09:51

Monash University

Faculty of Information Technology

2

nd Semester 2018

FIT2014

Assignment 2

Regular Languages, Context-Free Languages, Lexical analysis, Parsing,

and Turing machines

DUE: 11:55pm, Friday 5 October 2018

In these exercises, you will

? implement lexical analysers using lex (Problem 3);

? implement parsers using lex and yacc (Problems 1, 4);

? practise proof by induction (Problem 2);

? practise using the Pumping Lemma for regular languages (Problem 7);

? build a Turing machine (Problem 5);

? learn more about Turing machines (Problems 5–7).

Solutions to Problem 5 must be implemented in the simulator tuataraMonash (version 2.1 of

Tuatara), which is available on Moodle.

How to manage this assignment

? You should start working on this assignment now, and spread the work over the time until

it is due. Aim to do at least three questions before the mid-semester break. Do as much as

possible before your week 10 prac class. There will not be time during the class itself to do

the assignment from scratch; there will only be time to get some help and clarification.

? Don’t be deterred by the length of this document! Much of it is an extended tutorial to get

you started with lex and yacc (pp. 2–5) and documentation for functions, written in C, that

are provided for you to use (pp. 5–7); some sample outputs also take up a fair bit of space.

Although lex and yacc are new to you, the questions about them only require you to modify

some existing input files for them rather than write your own input files from scratch.

Instructions

Instructions are as for Assignment 1, except that some of the filenames have changed. The file

to download is now asgn2.tar.gz, and unpacking it will create the directory asgn2 within your

FIT2014 directory. You need to construct new lex files, using chain.l as a starting point, for

Problems 1, 3 & 4, and you’ll need to construct a new yacc file from chain.y for Problem 4. Your

submission must include (as well as the appropriate PDF files for the exercises requiring written

solutions):

? a lex file prob1.l which should be obtained by modifying a copy of chain.l

? a lex file prob3.l which should also be obtained by modifying a copy of chain.l

? a lex file prob4.l which should be obtained by modifying a copy of prob3.l

? a yacc file prob4.y which should be obtained by modifying a copy of chain.y

? a Tuatara Turing machine file prob5.tm

1

? PDF files for the exercises requiring written solutions, namely, prob1.pdf, prob2.pdf, prob6.pdf,

and prob7.pdf.

Each of the problem directories under the asgn2 directory contains empty files with the required

filenames. These much each be replaced by the files you write, as described above. Before submission,

check that each of these empty files is, indeed, replaced by your own file.

To submit your work, simply enter the command ‘make’ from within the asgn2 directory, and

then submit the resulting .tar.gz file to Moodle. As last time, make sure that you have tested the

submission mechanism and that you understand the effect of make on your directory tree.

INTRODUCTION: Lex, Yacc and the CHAIN language

In this part of the Assignment, you will use the lexical analyser generator lex or its variant flex,

initially by itself, and then with the parser generator yacc.

Some useful references on Lex and Yacc:

? T. Niemann, Lex & Yacc Tutorial, http://epaperpress.com/lexandyacc/

? Doug Brown, John Levine, and Tony Mason, lex and yacc (2nd edn.), O’Reilly, 2012.

? the lex and yacc manpages

We will illustrate the use of these programs with a language CHAIN based on certain expressions

involving strings. Then you will use lex and yacc on a language CRYPT of expressions based on

cryptographic operations.

CHAIN

The language CHAIN consists of expressions of the following type. An expression consists of a

number of terms, with # between each pair of consecutive terms, where each term is either a string

of lower-case letters or an application of the Reverse function to such a string. Examples of such

expressions include

mala # y # Reverse(mala)

block # drive # cut # pull # hook # sweep # Reverse(sweep)

Reverse(side) # Reverse(direction) # Reverse(gear)

For lexical analysis, we wish to treat every lower-case alphabetical string as a lexeme for the token

STRING, and the word Reverse as a lexeme for the token REVERSE.

Lex

An input file to lex is, by convention, given a name ending in .l. Such a file has three parts:

? definitions,

? rules,

? C code.

These are separated by double-percent, %%. Comments begin with /* and end with */. Any

comments are ignored when lex is run on the file.

You will find an input file, chain.l, among the files for this Assignment. Study its structure

now, identifying the three sections and noticing that various pieces of code have been commented

out. Those pieces of code are not needed yet, but some will be needed later.

We focus mainly on the Rules section, in the middle of the file. It consists of a series of statements

of the form

pattern { action }

2

where the pattern is a regular expression and the action consists of instructions, written in C,

specifying what to do with text that matches the pattern.

1

In our file, each pattern represents a set

of possible lexemes which we wish to identify. These are:

? a string of lower-case letters;

– This is taken to be an instance of the token STRING (i.e., a lexeme for that token).

? the specific string Reverse;

– Such a string is taken to be an instance of the token REVERSE.

? certain specific characters: #, (, );

? white space, being any sequence of spaces and tabs;

? the newline character.

Note that all matching is case-sensitive.

Our action is, in most cases, to print a message saying what token and lexeme have been found.

For white space, we take no action at all. A character that cannot be matched by any pattern yields

an error message.

If you run lex on the file chain.l, then lex generates the C program lex.yy.c.

2 This is the

source code for the lexical analyser. You compile it using a C compiler such as cc.

$ flex chain.l

$ cc lex.yy.c

By default, cc puts the executable program in a file called a.out.

3 This can be executed in

the usual way, by just entering ./a.out at the command line. If you prefer to give the executable

program another name, such as chain-lex, then you can tell this to the compiler using the -o

option: cc lex.yy.c -o chain-lex.

When you run the program, it will initially wait for you to input a line of text to analyse. Do

so, pressing Return at the end of the line. Then the lexical analyser will print, to standard output,

messages showing how it has analysed your input. The printing of these messages is done by the

printf statements from the file chain.l. Note how it skips over white space, and only reports on

the lexemes and tokens.

$ ./a.out

mala # y #Reverse( mala)

Token: STRING; Lexeme: mala

Token and Lexeme: #

Token: STRING; Lexeme: y

Token and Lexeme: #

Token: REVERSE; Lexeme: Reverse

Token and Lexeme: (

Token: STRING; Lexeme: mala

Token and Lexeme: )

Token and Lexeme: <newline>

Try running this program with some input expressions of your own.

1This may seem reminiscent of awk, but note that: the pattern is not delimited by slashes, /. . . /, as in awk; the

action code is in C, whereas in awk the actions are specified in awk’s own language, which has similarities with C but

is not the same; and the action pertains only to the text that matches the pattern, whereas in awk the action pertains

to the entire line in which the matching text is found.

2The C program will have this same name, lex.yy.c, regardless of the name you gave to the lex input file.

3a.out is short for assembler output.

3

Yacc

We now turn to parsing, using yacc.

Consider the following grammar for CHAIN.

S ?→ E

S ?→ ε

E ?→ E#E

E ?→ STRING

E ?→ REVERSE(STRING)

In this grammar, the non-terminals are S and E. Treat STRING and REVERSE as just single

tokens, and hence single terminal symbols in this grammar.

We now generate a parser for this grammar, which will also evaluate the expressions, with #

interpreted as concatenation and Reverse(. . . ) interpreted as reversing a string.

To generate this parser, you need two files, prob1.l (for lex) and chain.y (for yacc):

? Copy chain.l to a new file prob1.l, and then modify prob1.l as follows:

– in the Definitions section, uncomment the statement #include "y.tab.h";

– in the Rules section, in each action:

? uncomment the statements of the form

· yylval.str = ...;

· return TOKENNAME;

· return *yytext;

? Comment out the printf statements. These may still be handy if debugging is

needed, so don’t delete them altogether, but the lexical analyser’s main role now is

to report the tokens and lexemes to the parser, not to the user.

– in the C code section, comment out the function main(), which in this case occupies

about four lines at the end of the file.

? chain.y, the input file for yacc, is provided for you. You don’t need to modify this yet.

An input file for yacc is, by convention, given a name ending in .y, and has three parts, very loosely

analogous to the three parts of a lex file but very different in their details and functionality:

? Declarations,

? Rules,

? Programs.

These are separated by double-percent, %%. Comments begin with /* and end with */.

Peruse the provided file chain.y, identify its main components, and pay particular attention to

the following, since you will need to modify some of them later.

? in the Declarations section:

– lines like

char *reverse(char *);

char *simpleSub(char *, char*);

.

.

.

which are declarations of functions (but they are defined later, in the Programs section);

– declarations of the tokens to be used:

4

%token <str> STRING

%token <str> REVERSE

– declarations of the nonterminal symbols to be used (which don’t need to start with an

upper-case letter):

%type <str> start

%type <str> expr

– nomination of which nonterminal is the Start symbol:

%start start

? in the Rules section, a list of grammar rules in BNF, except that the colon “:” is used instead

of →, and there must be a semicolon at the end of each rule. Rules with a common left-handside

may be written in the usual compact form, by listing their right-hand-sides separated by

vertical bars, and one semicolon at the very end. The terminals may be token names, in which

case they must be declared in the Declarations section and also used in the lex file, or single

characters enclosed in forward-quote symbols. Each rule has an action, enclosed in braces

{. . . }. A rule for a Start symbol may print output, but most other rules will have an action of

the form $$ = . . . . The special variable $$ represents the value to be returned for that rule,

and in effect specifies how that rule is to be interpreted for evaluating the expression. The

variables $1, $2, . . . refer to the values of the first, second, . . . symbols in the right-hand side

of the rule.

? in the Programs section, various functions, written in C, that your parsers will be able to use.

You do not need to modify these functions, and indeed should not try to do so unless you are

an experienced C programmer and know exactly what you are doing! Most of these functions

are not used yet; some will only be used later, in Problem 4.

After constructing the new lex file prob1.l as above, the parser can be generated by:

$ yacc -d chain.y

$ flex prob1.l

$ cc lex.yy.c y.tab.c

The executable program, which is now a parser for CHAIN, is again named a.out by default,

and will replace any other program of that name that happened to be sitting in the same directory.

$ ./a.out

mala # y #Reverse( mala)

malayalam4

Run it with some input expressions of your own.

Problem 1. [7 marks]

(a) Construct prob1.l, as described above, so that it can be used with chain.y to build a

parser for CHAIN.

(b) Show that the grammar for CHAIN given above is ambiguous.

(c) Find an equivalent grammar (i.e., one that generates the same language) that is not

ambiguous.

4Malayalam is the main language of the southern Indian state of Kerala. The word was given as an example of a

palindrome by an FIT2014 student in a lecture last year.

5

Cryptographic expressions

A cryptographic calculator performs simple operations on strings of a kind that are used in

classical cryptosystems. It is also able to combine these operations in a natural way.

Suppose x = x1x2 · · · xn and k = k1k2 · · · kt are two strings, where xi and ki denote the i-th

letters of x and k respectively.

The available operations are:

? sum: this is written x + k in our expressions, though our C function that computes it is

called sum(. . . ). The resulting string has length min{n, t}, and its i-th letter is xi +ki mod 26,

where the letters of the English alphabet correspond to numbers via a = 0, b = 1, . . . , z = 25.

For example,

x = thebushwasalivewithexcitement

k = mrskoalahadabrandnewbabyandthenewsspreadlikewildfire

x + k = fywlisswhsdljmejlglaycjrezhga

? difference: this is written x ? k, and is computed by the C function diff(. . . ). The

resulting string has length min{n, t}, and its i-th letter is xi ? ki mod 26.

? Vigen`ere cypher: this is written Vigenere(x, k), where x is the plaintext and k is the key.

We first concatenate k with itself as many times as necessary in order to make it at least as

long as x. Then we form the sum. The result is a string whose i-th letter is

(xi + k((i?1) mod t)+1) mod 26.

For example, if the plaintext is

inaholeinthegroundtherelivedahobbit

and the key is

bilbo

then Vigenere(inaholeinthegroundtherelivedahobbit, bilbo) returns the cyphertext

jvlicmmtohimrscvvouvfzpmwwmobvpjmjh

? Simple Substitution: this is written SimpleSub(x, k). Again, x is plaintext, and k is the

key, but this time k must be a permutation of the 26-letter English alphabet, represented as

a string in which each letter appears exactly once (and t = 26). Every a in the plaintext is

replaced by the 1st letter k1 of the key; every b in the plaintext is replaced by the 2nd letter,

k2, of the key; and so on. In general, the plaintext letter xi

is replaced by kxi

.

For example, if the plaintext is

thequickbrownfoxjumpsoverthelazydog

and the key is

qwertyuiopasdfghjklzxcvbnm

then SimpleSub(thequickbrownfoxjumpsoverthelazydog,qwertyuiopasdfghjklzxcvbnm) returns

the cyphertext

zitjxoeawkgvfygbpxdhlgctkzitsqmnrgu

6

? Local Transposition: this is written LocTran(x, k). The letters of the plaintext x are not

replaced, as happens in the previous cyphers, but rather rearranged according to a permutation,

which is represented by a string k of w digits, where w = |k|.

5 This permutation string k has

length ≤ 10, and consists of the digits 0, 1, . . . , w ? 1 arranged in some order. For example,

if k has length 3 (so w = 3), then k can be any of the strings 012, 021, 102, 120, 201, 210.

The plaintext x is divided into blocks of w letters each, and the letters within each block are

permuted according to k. If there are extra letters at the end — too few letters to make up

another full block — then these are just copied across, with no change in their positions.

For example, if the plaintext is

thefamilyofdashwood

and the key is

201

then LocTran(thefamilyofdashwood,201) returns the cyphertext

ethmfayildofhasowod

These operations can be combined. Any valid expression can be given as the first argument

to one of the cypher functions, or as any argument of a sum or difference, to give another valid

espression. So you can form expressions like

Vigenere(LocTran(triantiwontigongolope,3201),bunyip) + muldjewangk

Let CRYPT be the language of cryptographic expressions of this type that can be generated by

the following grammar.

S ?→ E

S ?→ ε

E ?→ E + E

E ?→ E ? E

E ?→ (E)

E ?→ SIMPLESUB(E, STRING)

E ?→ VIGENERE(E, STRING)

E ?→ LOCTRAN(E, DIGITS)

E ?→ STRING

In this grammar, the non-terminals are S and E. Treat SIMPLESUB, VIGENERE, LOCTRAN, STRING

and DIGITS as just single tokens. For SIMPLESUB, VIGENERE, and LOCTRAN, we allow any nonempty

prefix of the function name as well as the full name; e.g., S, Si, Sim, . . . , SimpleSu, SimpleSub are

all acceptable lexemes for the token SIMPLESUB.

Problem 2. [7 marks]

For each n ≥ 0, let Vn be the string

VIGENERE( · · · VIGENERE(VIGENERE(

| {z }

n times

STRING ,STRING),STRING) · · · ,STRING)

| {z }

n times

where the Vigen`ere cypher is applied n times.

Prove, by induction on n, that Vn has a derivation, using the above grammar, of length

n + 2.

5w stands for width, the traditional term for the size of a permutation used in local transposition.

7

Problem 3. [7 marks]

Using the file provided for CHAIN as a starting point, construct a lex file, prob3, and use it

to build a lexical analyser for CRYPT.

Sample output:

$ ./a.out

Loc(Sim(Vig(therewasmovementatthestation,banjo),thequickbrownfxjmpsvlazydg),10)

Token: LOCTRAN; Lexeme: Loc

Token and Lexeme: (

Token: SIMPLESUB; Lexeme: Sim

Token and Lexeme: (

Token: VIGENERE; Lexeme: Vig

Token and Lexeme: (

Token: STRING; Lexeme: therewasmovementatthestation

Token and Lexeme: ,

Token: STRING; Lexeme: banjo

Token and Lexeme: )

Token and Lexeme: ,

Token: STRING; Lexeme: thequickbrownfxjmpsvlazydg

Token and Lexeme: )

Token and Lexeme: ,

Token: DIGITS; Lexeme: 10

Token and Lexeme: )

Token and Lexeme: <newline>

Control-D

$ ./a.out

V(twentysix - eleven, eleven)

Token: VIGENERE; Lexeme: V

Token and Lexeme: (

Token: STRING; Lexeme: twentysix

Token and Lexeme: -

Token: STRING; Lexeme: eleven

Token and Lexeme: ,

Token: STRING; Lexeme: eleven

Token and Lexeme: )

Token and Lexeme: <newline>

Problem 4. [7 marks]

Make a copy of prob3.l, call it prob4.l, then modify it so that it can be used with yacc.

Then construct a yacc file prob4.y from chain.y. Then use these lex and yacc files to build

a parser for CRYPT.

Note that you do not have to program any of the cryptographic functions yourself. They

have already been written: see the Programs section of the yacc file. The actions in your

yacc file will need to call these functions, and you can do that by using the function call for

reverse(. . . ) in chain.y as a template.

The core of your task is to write the grammar rules in the Rules section, in yacc format,

with associated actions, using the examples in chain.y as a guide. You also need to do some

8

modifications in the Declarations section to declare all tokens, using %token, and declare all

nonterminal symbols, using %type.

a See page 4.

aYou should still use start as your Start symbol. If you use another name instead, you will need to modify

the %start line too.

Sample output:

$ ./a.out

Loc(Sim(Vig(therewasmovementatthestation,banjo),thequickbrownfxjmpsvlazydg),10)

kltpysiteauzfglhctaesircrktx

Control-D

$ ./a.out

V(twentysix - eleven, eleven)

twenty

Binary local transposition of width 2

Binary local transposition of width 2 is a special case of the Local Transposition cypher described

above (see p. 7), except that it works with the binary alphabet {0,1} and uses width w = 2. Its

input has the form xyk where

? x is a binary string of even length,

x = x1x2 · · · xn, where n is even;

? k is either 01 or 10;

? the character y is used simply to separate x and k.

The cyphertext z = z1z2 · · · zn is given by:

? if k = 01, then z = x, i.e., zi = xi for all i (so cyphertext = plaintext);

? if k = 10, then for each j = 1, . . . , n/2,

z2j?1 = x2j ,

z2j = x2j?1 .

Think of the plaintext as divided into non-overlapping pairs of bits. Then, for each pair, the

bits within it are swapped.

Problem 5. [7 marks]

Build, in Tuatara, a Turing machine to compute binary local transposition of width 2.

Your Turing machine must have q1 as its Start State and q2 as its Accept State.

The output should have no y and/or key k after the cyphertext.

The language of encoded Turing machines

In the next two questions, your Turing machines use input alphabet {a,b} and tape alphabet {a,b,#}.

Tuatara is not used in these questions.

Problem 6 should help prepare for Problem 7.

9

Problem 6. [7 marks]

(a) Write, in table form, a Turing machine that accepts the input string aaaaa and crashes

for every other input string.

(b) On a single line, write down a string over {a,b} that represents your Turing machine from

part (a) in the Code Word Language (CWL). (See Lecture 15 for CWL.)

(c) How many different strings in CWL represent this exact same Turing machine?

? These strings must all represent identical Turing machines, not just Turing machines that

are equivalent in the sense that they always give the same result.

? You may assume that strings in CWL represent each Turing machine table row exactly

once.

(d) Explain how to construct, for each n ∈ N, a Turing machine Mn that accepts the input

string a

n and crashes for every other input string.

Recall from lectures that the language CWL is regular. But CWL contains many strings

that represent Turing machines that are either invalid or useless.

Let GOOD-TM be the language of all strings in CWL that represent Turing machines

that (i) are valid, and (ii) accept at least one string. For example, if you have done part (b)

correctly, the string you constructed there should belong to GOOD-TM.

(e) Give a string in GOOD-TM that has the property that, if at least one ‘a’ is inserted at

the very start of the string, the resulting string is no longer in GOOD-TM.

Problem 7. [8 marks]

Use the Pumping Lemma for Regular Languages to show that GOOD-TM is not regular.

FOR BONUS MARKS: determine, with proof, whether or not GOOD-TM is context-free.

References

? Jane Austen, Sense and Sensibility, Thomas Egerton, London, 1811.

? C. J. Dennis, The Triantiwontigongolope, poem in his book, A Book for Kids, Angus & Robertson,

Sydney, 1921.

? A. B. ‘Banjo’ Patterson, The Man from Snowy River, poem first published in The Bulletin on

26 April 1890.

? J. R. R. Tolkien, The Hobbit, Allen & Unwin, London, 1937.

? Dorothy Wall, Blinky Bill, Angus & Robertson, Sydney, 1933.