polycopié d'informatique fondamentale...

Polycopié d'InformatiqueFondamentale IMA4-SC

� Version 2013 �

Polytech'Lille, Villeneuve d'Ascq

Laure Gonnord

A�n d'améliorer ce poly n'hésitez pas à me

soumettre toute critique, suggestion, remarque

ou correction, dans mon casier ou, électroni-

quement, à l'adresse

[email protected]

A man provided with paper, pencil, and rubber,

and subject to strict discipline, is in e�ect a

universal machine.

Alan Turing, �Intelligent Machinery : A Report

by A. M. Turing� (Summer 1948)

Laure Gonnord Polycopié d'Informatique Fondamentale IMA4SC � S8 � 2013

Table des matières

1 Introduction, automates �nis et langages réguliers 4

2 Automates à pile et grammaires hors contexte 15

3 Automates à compteurs et machines de turing 22

4 Graphes, notions de NP-complétude 31

5 Construction d'un compilateur en Java 48

3/58


Chapitre 1

Introduction, automates �nis et langages

réguliers

Dans ce cours nous abordons le modèle de calcul le plus simple, les automates �nis, ainsi

que la notion de langage régulier.

4/58

Informatique Fondamentale IMA S8Cours 1 - Intro + schedule + finite state machines

Laure Gonnordhttp://laure.gonnord.org/pro/teaching/

[email protected]

Université Lille 1 - Polytech Lille

2013

Course introduction and schedule

Until now

During S5,S6,S7, IMA students have learnt :(programming stuff) C programing and compiling,(conception stuff) algorithm design and encoding,(microelec) circuits and embedded systems hardware,(auto) design flows of industrial processes

I Solved (computation) problems by ad-hoc solutions.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 2 / 38 �Course introduction and schedule

Until now 2/2

But :No general scheme to design algorithms.(manual) evaluation of the cost of our programs.worse no assurance of correctness.even worse it there always a solution ?

I All these problems will be addressed in this course

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 3 / 38 �


Schedule

Finite state machines (regular automata), regularlanguages. Notion of non determinism. Link with circuits.I/O automata, stack automata and grammars. Link to“simple” languages.Counter automata, Turing machines and undecidableproblems. Link to “classical” programs.Graphs and classical problems/algorithms on graphs.Compiler Construction. (+ lab)



I - Regular languages and automata


Finite state machines

1 Finite state machines

2 Regular Languages

3 The notion of non determinism

4 Classical Algorithms

5 Expressivity of regular languages

6 Link to other models

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 6 / 38 �Finite state machines

What for ?

We want to model :the behaviour of systems with behavioural modes.the behaviour of (Boolean) circuits.sets of words.

I A finite representation.



ExampleOpening/Closing door - Source Wikipedia.

1

Opened

E: open door

2

Closed

E: close door

state

transition

transition condition

close_door open_door

entry action



General definition

Finite state machine (FSM) or regular automataStates are labeled, and there are finitely many (s ∈ Q)Initial state(s) (i ∈ I) and accepting (terminating/finishing)states (t ∈ F )Transitions are finitely many.Transitions are labeled with letters (a ∈ A).

I The transition function is δ : Q×A→ Q.

s0 s1 s2a

b

a



Accepted language 1/2

Accepted wordA word w on the alphabet A is accepted by the automaton iffthere exists a finite path from an initial state to an acceptingstate which is labeled by w.

s0 s1 s2a

b

a

I w = aa is accepted/recognised.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 10 / 38 �Finite state machines

Accepted language 2/2

Accepted languageThe accepted language of a given automaton A is the set ofall accepted words and is denoted by L(A).

I Remark : it can be infinite.

s0 s1 s2a

b

a

I L(A) = {abka, k ∈ N}.



Data Structure for implementation

Problem : how to encode an automata ?

s0 s1 s2a

b

a

The transition function can be (for instance) encoded as atransition table :

s0 s1 s2s0 −− a −−s1 −− b a

s2 −− −− −−


Regular Languages


2 Regular Languages






Regular Languages

Goal

Problem : how to describe languages easily in a textual “linear”way ?I use regular expressions.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 14 / 38 �Regular Languages

Regular expressions

Regular expression (recursive def)A regular expression e on the alphabet A is defined byinduction. It can be of any of the following kinds :

the empty word εa letter a ∈ Aa choice between an expression e1 and another expressione2 : e1 + e2

two successive expressions : e1 · e20,1 or more successive occurrences of e1 : e∗1.

Example with A = {a, b, c, d} : e = a · (b+ c · d)∗


Regular Languages

Regular expression vs word

A regular expression “encodes” the form of a word. Forinstance :

i ·m · a · (3 + 4 + 5)

(on the alphabet {i,m, a, 3, 4, 5}) describes all words beginningby the prefix “ima” and finishing by one of the numbers 3, 4 or 5.

I A regular expression describes a language


Regular Languages

Regular language

Regular languageGiven a regular expression e, L(e) denotes the set of words(the language) that are described by the regular expression e.

Example with A = {0, . . . , 9} :e = (1 + 2 + . . .+ 9) · (0 + 1 + 2 + . . .+ 9)∗. What is L(e)?


Regular Languages

Linux world

Some commands use (extended) regular expressions(regexp) :

ls *.pdf lists all pdfs of the current directorygrep ta*.* *.c find all lines in .c files that contains wordsthat begin with t + some a’s.sed ''s ta*.*/toto/g'' file replace all occurrences...

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 18 / 38 �Regular Languages

Relationship between automata and languages - 1/3

First, some experiments.

Given the following automata A, are you able to give a regularexpression e such that L(e) = L(A)?

s0 s1 s2a

b

a

I e =?


Regular Languages


And the converse :

Given the following regular expression e = b∗ · (c+ a)∗, are youable to give an automaton A such that L(A) = L(e)?


Regular Languages


General result - Kleene TheoremThe regular languages are exactly the languages that aredescribed by finite automata.

I What we’ve done before is always possible.


The notion of non determinism


2 Regular Languages





Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 22 / 38 �The notion of non determinism

Goal

Sometimes some info lacks to make a choice between twotransitions :

q0 q1 q2a

c

c

I From state q1, there is a non deterministic choice whilereading c : either go to state q2 or stay in q1.



Definition

Non deterministic FSMA deterministic automaton is A =< A,Q, I, F, δ > with

δ : A×Q→ Q

A non deterministic automata is the same with :

δ : A×Q→ P (Q)

A non deterministic automata with ε-transitions is thesame with

δ : (A ∪ {ε})×Q→ P (Q)



Example

A non deterministic automata with ε-transitions :

p q ra

c

d

ε

I Then L(A) = a · (c∗ · d)∗.


Classical Algorithms


2 Regular Languages





Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 26 / 38 �Classical Algorithms

Find the associated language

Goal : Given A an automaton, find the associated language.

I Exercises !



Construction of automaton from a regular expression

Goal : Given a regular expression, construct a regularautomaton that “recognises” it.

I Exercises !



Determinisation

Goal : transform a non deterministic automaton into adeterministic one.

I Exercises !



Other Algorithms

In the literature you will easily find :algorithms to eliminate ε transitions without determinising(ε closure) ;algorithms to minimise automata (the number of states) ;algorithms to use automata to find words in a text ;algorithms to test language inclusion (if they are regular). . .

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 30 / 38 �Expressivity of regular languages


2 Regular Languages






Expressivity of regular languages

Non regular languages

Important resultThere exists some non-regular languages.

Examples of non-regular languages :{anbn, n ∈ N}.palindromes on a non-singleton alphabet.{ap, p prime}

I There exists a quite systematic way to prove that a givenlanguage is not regular (Pumping Lemma).


Link to other models


2 Regular Languages







Moore and Mealy Machines - 1Moore : I/O machine whose output values are determinedsolely by the current state :

source Wikipedia

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 34 / 38 �Link to other models

Moore and Mealy Machines - 2Mealy : output values are determined both by the current stateand the value of the input.

source Wikipedia



UML Implementation of FSMs

UML variants of FSMs are hierarchical, react to messages andcall functions.

source Wikipedia



Hardware Implementation of FSMs

It requires :a register to store state variablesa block of combinational logic for the state transition(optional) a block of combinatorial logic for the output


Conclusion

Summary

Regular Automata or Finite State Machines are :Acceptors for regular languages. But some languages arenot regular.Algorithmically efficient.Useful to describe (simple) behaviours of systems.Closely linked to circuits.



Chapitre 2

Automates à pile et grammaires hors

contexte

Les automates à pile sont un modèle de calcul un peu plus puissant que les automates �nis.

Nous caractériserons leur pouvoir d'expressivité en terme de grammaire.

15/58

Informatique Fondamentale IMA S8Cours 2 - Stack automata + Grammars


[email protected]


2013

II - Stack automata and Grammars

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 - S8SC � 2 / 22 �Stack automata

1 Stack automata

2 Grammars

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 - S8SC � 3 / 22 �

Stack automata

What for ?

Express more than regular languages !

I Give a way for automata to “count”, to “have a memory”.


Stack automata

Example

(source : Wikipedia)


Stack automata

General definition

Stack/P automataStates (Q), initial state (q0), finite states (F).Two alphabets (one for read : Σ one for stack : Γ)γ0 ∈ Γ the end of stack character.Transitions are finitely many.A stack to write into.Transitions use the stack.

I The transition function is :

Q× (Σ ∪ ε)× Γ→ P(Q× Γ∗)

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 - S8SC � 6 / 22 �Stack automata

How it works ? 1/3

ConfigurationA configuration is a triple (q, w, α) where :

q is the current statew ∈ Σ∗ the part of the input tape which is not yet readα ∈ Γ∗ the current stack word

Two different conditions for accepting words :“accepting state” : the read word leads to a configurationof the form (q, ε, α) where q ∈ F (with any α)“empty stack” : ... (q, ε, γ0) (with any q, but γ0 is the “emptystack” character.


Stack automata

How it works ? 2/3

Given a configuration (q, w, α), the next one is computed by thehelp of the transition function Q× (Σ ∪ ε)× Γ→ P(Q× Σ∗) :

enabled transitions : those of the form (q, a, A)→ (q′, A′)where a is the next letter to read on the input tape (or ε),and A the letter on top of the stack : w = aw′ and α = Aα′.Pick one enabled transition, and compute the nextconfiguration (q′, w′, A′α′).


Stack automata

How it works ? 3/3

(source : wikipedia)

I Try to derive 0011 and 00111 !I The PDA recognises {0n1n|n ∈ N} by accepting state.


Stack automata

Important theoretical results on PDA

Important Stack automata are non deterministic !The two acceptance criteria define the same class oflanguages

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 - S8SC � 10 / 22 �Grammars

1 Stack automata

2 Grammars


Grammars

Goal

Problem : Express languages with the same expressivity asstack automata.I use grammars


Grammars

General grammars

Grammar ruleA grammar rule (production rule) is of the form

w −→ w′

where w and w′ are words.

A grammar is a set of rules.


Grammars

Grammars

GrammarA grammar is composed of :

A finite set N of non terminal symbolsA finite set Σ of terminal symbols (disjoint from N )A finite set of production rules, each rule of the formw → w′ where w is a word on Σ ∪N with at least one letterof N . w′ is a word on Σ ∪N .A start symbol S ∈ N .


Grammars

Example :S → aSb

S → ε

is a grammar with N = .... and ....


Grammars

Associated Language

DerivationG a grammar defines the relation :

x⇒G y iff ∃u, v, p, qx = upv and y = uqv and (p→ q) ∈ P

I A grammar describes a language (the set of words on Σ thatcan be derived from the start symbol).


Grammars

Examples

S → aSb

S → ε

The grammar defines the language {anbn, n ∈ N}

S → aBSc

S → abc

Ba→ aB

Bb→ bb

The grammar defines the language {anbncn, n ∈ N}I Exercises


Grammars

Context-free grammars

Context-free grammarA CF-grammar is a grammar where all production rules are ofthe form N → (Σ ∪N)∗.


Example of CF-grammar

S → S + S|S ∗ S|aThe grammar defines a language of arithmetical expressions.

I Notion of derivation tree.Draw a derivation tree of a*a+a, of S+S !


Grammars

Relationship between stack automata and grammars

General resultThe context-free/algebraic languages are exactly the languagesthat are described by stack automata.

I The proof is not difficult.

I Exercises


Grammars

Some other results/definitions

Regular languages are algebraic languages, but not theconverse.There exists normal forms for algebraic grammarsA grammar can be ambiguous.


Conclusion

Summary

Stack Automata or PushDown Automata are :Acceptors for context-free languages. But some languagesare not context free.Non deterministic.

I http://en.wikipedia.org/wiki/Pushdown_automaton foruseful pointers



Chapitre 3

Automates à compteurs et machines de

turing

Dans ce chapitre nous étudions des modèles plus proches de nos programmes habituels, les

automates à compteurs, et introduisons le modèle couramment utilisé pour formaliser la notion

de calcul, les machines de Turing.

22/58

Informatique Fondamentale IMA S8Cours 3: Counter automata, Turing Machines and decidable

problems


[email protected]


2013

III - Counter automata, Turing Machines and decidable problems

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8SC � 2 / 30 �Counter automata

1 Counter automata

2 Programs

3 Turing Machines

4 Decidability - Complexity

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8SC � 3 / 30 �

Counter automata

What for ?

Express more than context-free languages !

I Give a way for automata to “count more” : include variables.


Counter automata

Example

kinit k1

cos(y) ≤ x → x := x+ 1

x 6= y → y := yx

true → y := y + 2


Counter automata

General definition

Counter AutomataA finite number of counters (variables)A finite number of control pointsTransitions between them that operate on counters.

I The transition function is of the form g → a (g is the guard, ais the action)


How it works ? 1/3

StateA state is (q, σ)

q is the current control pointσ : V ar → V al is a function that assigns a value (a realone or ⊥) to all counters.

I Non deterministic/No notion of acceptance/Notion ofreachability.


Counter automata

How it works ? 2/3

Given a state (q, σ), the next one is computed by the help of thetransition function :

enabled transitions are transition of the current controlpoints where the current valuations of variables satisfy theguard.Pick one enabled transition, and compute the next state :(q′, σ′). The new values for the variables are computedw.r.t. the action.


Counter automata

How it works ? 3/3

Example of an affine automata :

kinit k1 k2true → x := 0; y := 0 x ≤ 1 → x := x+ 1; y := 2y − 4

y + 2x ≤ 2 → x := 2x+ 5

I Compute successive states.

I Exercises.


Counter automata

What kind of systems ?

These automata are used to encode, for example :Simple systems (coffee machine, ...) specificationsEnergy consumption of sensors :

Sleep Idle

Transmit

Receive

35.1 mW

145.8 mW

140.4 mW

140.4 mW

140.4 mW

145.8 mW

140.4 mW

140.4 mW

140.4 mW

140.4 mW

400 µs

332 µs

144 µs

144 µs

100 µs 100 µs


What for ?

Some classical problems :

Automatically finding invariantsReachability analysisFormula proving.(deterministic) Code generation.


Programs

1 Counter automata

2 Programs

3 Turing Machines



Programs

Expressing programs as counter automata

An example :x:=0;y:=0

while (x<=100) do

read(b);

if b then

x:=x+2

else begin

x:=x+1;

y:=y+1;

end;

endif

endwhile

p

pin

x ≤ 100 →

x := x+ 1y := y + 1

x ≤ 100 →

x := x+ 2

(x, y) := (0, 0)

pout

x ≥ 100

I Some approximations are made (arrays, Boolean, . . . )


Programs

Expressing properties of programsSome (safety) properties of programs can be expressed insidethe counter automaton :

p

pin

x ≤ 100 →

x := x+ 1y := y + 1

x ≤ 100 →

x := x+ 2

(x, y) := (0, 0)

pout

x ≥ 100

I Encode the fact that state = pout ∧ y > 100 is “bad”I Some modifications of the automaton can be automaticallydone.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8SC � 14 / 30 �Programs

Expressing properties of programs - 2Automatically deriving bad states :

int j;

char user[USERSZ];

for(j = 0; line[j] != EOS; ++j)

if (!strchr("-", line[j]))

break;

if(j == J && line[j] == ' ') { /* long list */

/* BUG! No bounds check. */

assert(USERSZ>=N-j,"badstate");

r_strcpy (user, line + j);

}

I “badstate” is reachable.Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8SC � 15 / 30 �

Programs

Expressing programs as counter automata

Pros : finding and proving program propertiesCons : how do we define the operations ? the fact thatintegers are encoded on bits ?

I There is a need for a more accurate (not specialised) model !


Turing Machines

1 Counter automata

2 Programs

3 Turing Machines



Turing Machines

Turing Machine, def - 1

Turing Machine, Turing, 1935A tape (infinite in both sides) : the memory. (it has aworking alphabet which contains a blank symbol).A head : read/write symbols on the tape.A finite transition table (or function)

I A PushDown automaton which is more flexible.

image credits : Wikipedia.org

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8SC � 18 / 30 �Turing Machines

Turing Machine, def - 2

Turing Machine elementsStates (Q), initial state (q0), final (accepting) states (F).One alphabet Γ for the tape.b ∈ Γ the blank letter.Transitions are finitely many : read the letter under thehead, and then :

Erase a symbol or write one under the headMove the head (read, write, or stays)The “state” can be modified.

I The transition function is :

Q× Γ× → Q× Γ× {L,R, S}


Turing Machines

An Example

TM that decides if x is odd : (final State : q4)State/char 0 1 B

q1 (q1,0,R) (q1,1,R) (q2,B,L)q2 (q3,0,L) (q4,1,L) —

Play on BBBBBBB11BBBBB and BBBBBBB10BBBBB

I A configuration is a tuple (word on the tape, position of thehead, state).Adapted from : http://www.madchat.fr/coding/algo/algo_epfl.pdf slide 4


Turing Machines

Another Example

Demo of a TM recognising a language : anbncn

Program found here :http://www.cs.columbia.edu/~zeph/software/BJDweck/


Turing Machines

General results 1/2

There exists Turing machines for the following languages :palindromesanbncn (non algebraic language)ai with i prime (non algebraic)an

2, with n > 0

I Turing machines are more powerful than all other models (wehave seen yet)

Decidable LanguagesA language that is recognised by a Turing Machine is said to bedecidable.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8SC � 22 / 30 �Turing Machines

Accepting or computing

A TM can also compute functions

TM that writes 1 if x is odd, 0 else (q6 is final state) :State/char 0 1 B

q1 (q1,0,R) (q1,1,R) (q2,B,L)q2 (q3,B,L) (q4,B,L) —q3 (q3,B,L) (q3,B,L) (q5,1,R)q4 (q4,B,L) (q4,B,L) (q5,0,R)q5 — — (q6,B,R)

Play on BBBBBBB11BBBBB and BBBBBBB10BBBBB


Turing Machines

Another Example

Demo of a TM computing the subtraction.


Turing Machines

General results 2/2

There exists Turing machines for the computation of :x 7→ x+ 1

(x, y) 7→ x+ y

x 7→ x2

all the functions you are able to write on computers

Computable functionsA function (defined for all its input) that is computable by aTuring Machine is said to be TM computable

I A Model of what can be computed with machines. (ChurchThesis)


Decidability - Complexity

1 Counter automata

2 Programs

3 Turing Machines


Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8SC � 26 / 30 �Decidability - Complexity

Decidable - Semi-decidable languages

If L is a language, L is decidable if there exists a TuringMachine (or an algorithm) that outputs for all w :

1 if w ∈ Lelse 0.

Semi-decidable :1 if w ∈ Lelse does not terminate

I equivalent definition for problems.



Complexity

Link with computational complexity :The number of steps in the execution of a TM gives thecomplexity in time,The number of seen squares in the execution of a TMgives the complexity in space.



Examples of undecidable problems

The halting problem for TM or counter automata (for morethat 2 counters)Given a program, does it loops ?Is a given algebraic expression (with log, *, exp, sin, abs)equal to 0 ? (Richardson, 1968)The 10th Hilbert Problem (Diophantine equations)


Conclusion

SummaryTuring Machines :

A model closed to programming languages.Non deterministic.Acceptors for decidable languages. But some languagesare not decidable !(equivalently) Computes solutions to problems. But ...

Important factSome common problems are undecidable !

Counter automata :are a more simpler modelhave the same power of expression as Turing Machines.Reachability is undecidable too. But we can doapproximations.



Chapitre 4

Graphes, notions de NP-complétude

Les graphes o�rent une modélisation intéressante de beaucoup de problèmes d'informatique.

Nous abordons leur représentation machine et présentons rapidement les algorithmes classiques.

Nous pro�tons du modèle pour parler de complexité théorique, de problèmes �di�ciles�, et de

NP-complétude

Figure 4.1 � Pillow talk, http://xkcd.com/69/, sous License Creative Commons

Figure 4.2 � Travelling salesman Problem, http://xkcd.com/399/, sous License CreativeCommons

31/58

http://xkcd.com/69/

http://xkcd.com/399/

Informatique Fondamentale IMA S8Cours 4 : graphs, problems and algorithms on graphs,

(notions of) NP completeness


[email protected]


2013

IV - Graphs, problems and algorithms on graphs, NP-completeness

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 2 / 64 �Some generalities on graphs

1 Some generalities on graphsGeneralities and first definitionsInternal representation

2 Paths finding - search, and applications

3 Algorithms specific to oriented graphs

4 More difficult problems on graphs

5 NP - completeness

6 Docs


Some generalities on graphs Generalities and first definitions

Goal

Model :relationships between objects (“I know him”, “Thisregister is in conflict with with one”, “there is a roadbetween these two towns”)classical problems of paths (“Do I have a friend that knowsa friend ..... who knows the Dalai-Lama ?”, “It is possible togo from this town to this another in less than 600 km ?”,“how to get out of the labyrinth ?”)



Who ?

Euler (1740) formalised the theory and the Konisberg problem :

I IProblem : compute an eulerian path.Source (fr) : http://fr.wikipedia.org/wiki/Th%C3%A9orie_des_graphes



Other classical problems

The travelling salesman problem : TSPTask scheduling (on parallel machines or not !), the PERTmethod.Coloring mapsCovering (telecom) treesMinimal paths ...

I Graphs are everywhere.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 6 / 64 �Some generalities on graphs Generalities and first definitions

General definition

GraphA finite set of vertices (a vertex) : VA finite set of edges. E ⊆ V × V

v1

v2

v3

e1 e3

I A simplier model than the previous ones !



Other definitions

Non Oriented Graph iff∀(v1, v2), (v1, v2) ∈ E ⇒ (v2, v1) ∈ E.Tree (non oriented acyclic connex) - Forest .Directed Acyclic Graph : DAG . Oriented Graph with nocycle.

I trees are dags !

v1

v2

v3

v4


Some generalities on graphs Internal representation

Sources

Picture and Java codes fromhttp://pauillac.inria.fr/~levy/courses/X/IF/a1/a1.html



Adjacency matrix

0

1

2

3 4

0 1 0 0 01 0 1 0 00 1 0 0 00 0 0 0 10 0 0 1 0

I If the graph is oriented, then the matrix is NOT symmetricI Memory complexity : O(V 2)

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 10 / 64 �Some generalities on graphs Internal representation

Adjacency matrix - Java

boolean [ ] [ ] m;

Graph ( i n t n ) {m = new boolean [ n ] [ n ] ;for ( i n t i = 0 ; i < n ; ++ i )

for ( i n t j = 0 ; j < n ; ++ j )m[ x ] [ y ] = f a l s e ;

}

s t a t i c void addEdge ( Graph g , i n t x , i n t y ){ g .m[ x ] [ y ] = t rue ;

g .m[ y ] [ x ] = t rue / / on ly i f non o r ien ted }



Array of lists

0

1

2

3 4

I Memory complexity : O(V + E)



Array of lists - Java

L i s t e [ ] succ ;

Graph ( i n t n ) { succ = new L i s t e [ n ] ; }

s t a t i c void addEdge ( Graph g , i n t x , i n t y ) {g . succ [ x ] = new L i s t e ( y , g . succ [ x ] ) ;/ / i f non o r ien ted also modify g . succ [ y ] !

}}


Paths finding - search, and applications

1 Some generalities on graphs

2 Paths finding - search, and applicationsLooking for all nodesConnexityTopological Sort



5 NP - completeness

6 Docs

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Search Algorithms

! Translation problem ! (en) Search Algorithms - (fr) Parcours deGraphesReminder : search/do algorithm on a binary tree :

1 If the tree is not empty, do f(root).2 recursively apply the function on the left sub-tree3 recursively apply the function on the right sub-tree

I Deep-First Search


Paths finding - search, and applications Looking for all nodes

Deep First Search on Graphs

On graphs, there can be cycles !



Deep First Search on Graphs - Javas t a t i c i n t [ ] num; / / numbering the edges according to the v i s i t i n g order

s t a t i c void v i s i t e r ( Graphe g ) {i n t n = g . succ . leng th ;num = new i n t [ n ] ; numOrdre = −1;for ( i n t x = 0; x < n ; ++x ) num[ x ] = −1; / / −1 = ‘ ‘ not seen ’ ’for ( i n t x = 0; x < n ; ++x )

i f (num[ x ] == −1)dfs ( g , x ) ;

}

s t a t i c void dfs ( Graphe g , i n t x ) {num[ x ] = ++numOrdre ;for ( L i s t e l s = g . succ [ x ] ; l s != n u l l ; l s = l s . su i van t ) {

i n t y = l s . va l ;i f (num[ y ] == −1)

dfs ( g , y ) ;}

}

I Complexity = O(V + E). Tarjan [1972]



Breadth First Search on Graphs

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 18 / 64 �Paths finding - search, and applications Looking for all nodes

Conventions

BLANC = not (yet) processed NOIR = has been processedGRIS = being processed

f i n a l s t a t i c i n t BLANC = 0 , GRIS = 1 , NOIR = 2;s t a t i c i n t [ ] cou leur ;



Breadth First Search on Graphs - Javai n t n = g . succ . leng th ; cou leur = new i n t [ n ] ;FIFO f = new FIFO ( n ) ;for ( i n t x = 0; x < n ; ++x ) cou leur [ x ] = BLANC;for ( i n t x = 0; x < n ; ++x )

i f ( cou leur [ x ] == BLANC) {FIFO . a j o u t e r ( f , x ) ; cou leur [ x ] = GRIS ;b fs ( g , f ) ;

}

s t a t i c void bfs ( Graphe g , FIFO f ) {while ( ! f . v ide ( ) ) {

i n t x = FIFO . supprimer ( f ) ;for ( L i s t e l s = g . succ [ x ] ; l s != n u l l ; l s = l s . su i van t ) {

i n t y = l s . va l ;i f ( cou leur [ y ] == BLANC) {

FIFO . a j o u t e r ( f , y ) ; cou leur [ y ] = GRIS ;}cou leur [ x ] = NOIR ;

} } }

I Complexity ?



Application : Labyrinth

Finds exit , ie a path from d to s ?cou leur [ d ] = GRIS ;i f ( d == s )

return new L i s t e ( d , n u l l ) ;for ( L i s t e l s = g . succ [ d ] ;

l s != n u l l ; l s = l s . su i van t ) {i n t x = l s . va l ;i f (num[ x ] == BLANC) {

L i s t e r = chemin ( g , x , s ) ;i f ( r != n u l l )

return new L i s t e ( d , r ) ;}

}return n u l l ;

}


Paths finding - search, and applications Connexity

Definition - Connex Component

Connex componentA connex component is a maximal set of connected vertices.

I Exercise : Algorithm for computing CCs.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 22 / 64 �Paths finding - search, and applications Connexity

Definition - Connex Graph

Connex graphA graph is connex if it has only one connex component.

I Exercise : Algorithm that tests the connexity of a graph.


Paths finding - search, and applications Topological Sort

Other AlgorithmsThere exists variants of DFS algo to :

Compute Cover Forets (in bold in the picture) :

Detect Cycles.


Algorithms specific to oriented graphs





5 NP - completeness

6 Docs



Topological sorting of a DAG

I Give a list of vertices in an order compatible with the DAGorder : vi < vj ⇒ vj → vi (total order from a partial header).

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 26 / 64 �Algorithms specific to oriented graphs

Topological sorting of a DAG - Example

1

2

3

4

5

6

7

I If we know an entrant node, a dfs gives the result.Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 27 / 64 �


Topological sorting of a DAG - implementation

f i n a l s t a t i c i n t BLANC = 0 , GRIS = 1 , NOIR = 2;

s t a t i c L i s t e aFai re ( Graphe g , i n t x ) {i n t n = g . succ . leng th ;i n t [ ] cou leur = new i n t [ n ] ;for ( i n t x = 0; x < n ; ++x ) cou leur [ x ] = BLANC;return dfs ( g , x , n u l l , cou leur ) ;

}

s t a t i c L i s t e dfs ( Graphe g , i n t x , L i s t e r , i n t [ ] cou leur ) {cou leur [ x ] = GRIS ;for ( L i s t e l s = g . succ [ x ] ; l s != n u l l ; l s = l s . su i van t ) {

i n t y = l s . va l ;i f ( cou leur [ y ] == BLANC)

r = dfs ( g , y , r , cou leur ) ;}return new L i s t e ( x , r ) ;

}

I dfs here returns a list.



Topological sorting of a DAG - Applications

instruction schedulingcell evaluation in spreadsheetsorder of compilation tasks (Makefiles)symbol dependencies in linkerssheduling in project managements (PERT, 1960)

http://en.wikipedia.org/wiki/Topological_sorting



Transitive closure

The graph G+ = (V,E+) of the transitive closure of G = (V,E)is such that E+ is the smallest set verifying :

E ⊆ E+

(x, y) ∈ E+ and (y, z) ∈ E+ ⇔ (x, z) ∈ E+


Algorithm for Transitive closure

If E is the adjacency matrix : E+ = E + E2 + . . . En−1.I An algorithm in O(n4).

I Is it possible to have a better complexity ? Yes (Warshall, inO(n3), see later).



Definition of oriented valuated graph

Valuated graphIt’s a graph with a function d : V × V → N called distance.

I G is represented by an adjacency matrix with integer values,or +∞ if there is no edge.



The shortest path problem

Given G, find a path whose sum of distances is minimal !

more variants on http://en.wikipedia.org/wiki/Shortest_path_problem



Multiple Source Shortest Path with dynamicprogramming -1

Between all the vertices ! Floyd-Warshall algorithmPrinciple : Let dkx,y be the minimal distance between x and ypassing through vertices whose number is < k. Then

d0x,y = dx,y and dkx,y = min{dk−1x,y , dk−1

x,k + dk−1k,y }

I Store all these temporary distances.


Multiple Source Shortest Path with dynamicprogramming -2

s t a t i c Graphe plusCourtsChemins ( Graphe g ) {i n t n = g . d . leng th ;Graphe gplus = copieGraphe ( g ) ;for ( i n t k = 0; k < n ; ++k )

for ( i n t x = 0; x < n ; ++x )for ( i n t y = 0; y < n ; ++y )

gplus . d [ x ] [ y ] = Math . min ( gplus . d [ x ] [ y ] ,gplus . d [ x ] [ k ] + gplus . d [ k ] [ y ] ) ;

return gplus ; }

I Complexity O(n3), space O(n2).



Simple Source Shortest Path

All d are > 0 : Dijkstra, 1959Negative distances : Bellman Ford ' 1960

I Routing algorithms (Bellman uses only local information).


More difficult problems on graphs





5 NP - completeness

6 Docs



Eulerian and Hamiltonian paths

Eulerian/Hamiltonian PathAn Eulerian path : each edge is seen only once.An Hamiltonien path : each node ...

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 38 / 64 �More difficult problems on graphs

Hamiltonian Tour, one solution

Extend a path until not possibleBacktrack one time, and try with another neighbourAnd so on.

I Worst case complexity time : exponential in n



Application 1 : Knight’s Tour

Is it possible to make a Knight’s tour ? « The knight is placed onthe empty board and, moving according to the rules of chess,must visit each square exactly once.»I A variant of the Hamiltonian tour that can be solved inpolynomial time.



The Travelling Salesman Problem

with valuated (non oriented) graph.I Find an hamiltonian cycle of minimum weight



Travelling Salesman Problem - 2

A (non-polynomial) algorithm for this problem :Transform into a linear programming problem with integersSolve it with the simplex !

I This algorithm is exponential in the size of the graph

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 42 / 64 �More difficult problems on graphs

Graph Coloring Problem

I Application to the register allocation in compilers.I The sudoku problem (9-coloring of a 81-vertices graph)



Graph Coloring Problem - 2

Are you able to design a polynomial algorithm ?



Graph Coloring Problem - 3

We do not know any polynomial algorithm for this problem.


NP - completeness





5 NP - completeness

6 Docs

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Complexity

The complexity of an implementation / an algorithm is linked toTuring machines :

The number of steps in the execution of a TM gives thecomplexity in time,The number of seen squares in the execution of a TMgives the complexity in space.

We can replace the TM by “a C implementation”.


NP - completeness

P Problems

PA problem (or a language) belongs to P if there exists adeterministic Turing Machine that gives the result inpolynomial time.

I there is a polynom p such that for all entry x, the Turingmachine stops (with the good result) in less than p(|x|) steps.

Example : L = anbncn, the TM of the previous course checksany aibici in O(n) steps


NP - completeness

P Problems

Integer multiplicationEulerian tour (each edge once)Linear programming (recall that polynomial diophantineequations are undecidible.)Primality test

http://www.cs.uu.nl/groups/AD/compl-diaz.pdf


NP - completeness

NP Problems

NPA problem (or a language) belongs to NP if there exists adeterministic Turing Machine that checks the result inpolynomial time.

Example : there exists a TM that is able to check if a given pathis an hamiltonien tour, in polynomial time.


NP Problems - Examples

Factoring3-SAT (formulae in CNF with 3 litterals on each term)Hamiltonian tour3-colorability

http://www.cs.uu.nl/groups/AD/compl-diaz.pdf


NP - completeness

NP vs P - some facts

If a problem belongs to NP , then there exists anexponential brute-force deterministic algorithm to solve it(easy to prove)P ⊆ NP (trivial)

I P = NP ? P 6= NP ? Open question ! Problem of the millenium !Is finding more difficult that verifying ?


NP - completeness

NP-complete problems - 1

Given a problem :if we find a polynomial algorithm I Pif we find a polynomial time checking algorithm I NP, butwe want to know more :

“it it really hard to solve it ?”“is it worth looking for a better algorithm ?”


NP - completeness


The key notion is the notion of polynomial reduction

Polynomial reduction of problemsGiven two problems P1 and P2, P1 reduces polynomially to P2 ifthere exists an f such that :

f is polynomialx1 solution of P1 iff f(x1) solution of P2.

Example : Hamiltonian circuit is polynomially reductible to TSP.



The most difficult problems in NP. All problems in NPC arecomputationally equivalent in the sense that, if one problemis easy, all the problems in the class are easy. Therefore, if oneproblem in NPC turns out to be in P, then

P=NP


NP - completeness


Reminder : we want to characterise the most difficult problemsin NP.

NP-completeA problem is said to be NP-complete if :

It belongs to NPAll other problems in NP reduce polynomially to it.

In other words, if we solve any NP-complete in polynomial time,we have proved P = NP (and we become rich)


NP - completeness


Problem : how to prove that a problem is NP-complete ?The very first one is hard to proveThen, we use the following result :

If two problems / languages L1 and L2 belong to NP and L1 isNP-complete, and L1 reduces polynomially to L2, then L2 is

NP-complete.

I Pick one NP complete problem and try to reduce it to yourproblem.


NP - completeness

NP-complete problems - The historical example

SATISFIABILITY is NP-complete (Cook, 1971)SAT reduces polynomially to 3SAT3SAT reduces polynomially to Vertex CoverVertex Cover reduces polynomially to Hamiltonian circuit

I “So”Hamiltonian circuit, then TSP are also NP-completeproblems


Some NP-complete (common) problems

Some classical examples :Hamiltonian cycle and TSPGraph Coloring ProblemVertex coverKnapsack (integer)Flow shop problemSudoku

The book : Computers and Intractability : A Guide to the Theoryof NP-Completeness. A short list on the french webpage :http://fr.wikipedia.org/wiki/Liste_de_probl%C3%A8mes_

NP-complets


NP - completeness

Conclusion - 1


NP - completeness

Conclusion - 2

Knowing that a given problem is NP complete is just thebeginning :

handle less general classes of inputscompute less precise solutions


Docs





5 NP - completeness

6 Docs

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 62 / 64 �Docs

More docs on graphs

In french : http://laure.gonnord.org/site-ens/mim/graphes/cours/cours_graphes.pdf or type “cours deThéorie des graphes” in GoogleIn english : an interactive tutorial here : http://primes.utm.edu/cgi-bin/caldwell/tutor/graph/intro.(en) The theorical book “Graph Theory”, R. Diestel(electronic version :http://diestel-graph-theory.com/basic.html)(en) e-book for algorithms : http://code.google.com/p/graph-theory-algorithms-book/


Docs

More docs on complexity

The Book : computers and intractability, a guide to the theoryof NP completeness, by Garey/Johnson



Chapitre 5

Construction d'un compilateur en Java

Ce chapitre plus pratique utilise les bases fondamentales des chapitres précédents pour construire

le �front-end� d'un compilateur en Java.

48/58

Informatique Fondamentale IMA S8Cours 5 : Lexical and Syntactic Analysis and Compiler

front-end construction


[email protected]


2013

V - Lexical and syntactic analysis : the compiler front-end in practise


Refreshing memories

Compiler Front-End I Abstract Syntax Tree (AST)

int y = 12 + 4*x ;

=⇒ [TKINT, TKVAR("y"), TKEQ, TKINT(12), TKPLUS,TKINT(4), TKFOIS, TKVAR("x"), TKPVIRG]=⇒

=

yint +

12

4 x

*

int


Lexical Analysis aka Lexing

1 Lexical Analysis aka LexingIntro

2 Syntactic Analysis aka Parsing

3 Syntactic Analysis and rules

4 Towards a methodology for designing front-ends and simpleevaluators in Java

5 Other technologies for the front-end


Lexical Analysis aka Lexing Intro

What for ?

int y = 12 + 4*x ;

=⇒ [TKINT, TKVAR("y"), TKEQ, TKINT(12), TKPLUS,TKINT(4), TKFOIS, TKVAR("x"), TKPVIRG]

I The Lexing produces from a flow of characters a list oftokens.



Algorithm

What’s behind ?

From a Regular language, produce an automaton (see course1)

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 6 / 38 �Lexical Analysis aka Lexing Intro

Tools : lexical analysers constructors

Lex(C) JFlex(Java), OCamllex, ... are tools that produces anautomaton that recognises a given language and producestokens :(here we focus on JFlex)

input : a set of regular expressions with actions(toto.jflex).output : a .java that contains the associated automaton.



Lexing tool for java : JFLEX

The official webpage : jflex.de (GPL)

A first example : numbers.flex.



.lex format and compilation.flex construction

%%%p u b l i c <<<−−− generates a p u b l i c c lass%class xx <<<−−− which name i s xx . java%standalone <<<−−− standalone use ( no cup )%unicode <<<−−− encoding

%{<<<−− d e c l a r a t i o n o f l o c a l vars

%}%%

<<<−− f l e x ru l es

Compilation with :

jflex toto.flex // produces xx.java

javac xx.java // compiles into xx.class

java xx filetotest // tests if file2text is recognized



Lexing in java - ex2

An example from the flex distribution (standalone.flex)

%%

%p u b l i c <<<−−− generates a p u b l i c c lass%class Subst <<<−−− which name i s Subst . java%standalone <<<−−− standalone use ( no cup )

%unicode <<<−−− encoding

%{S t r i n g name ; <<<−− d e c l a r a t i o n o f l o c a l var

%}

%%

"name " [ a−zA−Z]+ { name = y y t e x t ( ) . subs t r i ng ( 5 ) ; }[Hh ] " e l l o " { System . out . p r i n t ( y y t e x t ( )+ " "+name+" ! " ) ; }

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 10 / 38 �Lexical Analysis aka Lexing Intro

.flex variables and functions

Access to lexing info :yytext() : last recognized stringyylength() : yytext’s length

(and other, see flex documentation on/usr/share/doc/jflex/)

Functions :yylex() lexing standard function (standalone)lexing function with cup : next_token()



Lexing in java - ex 3 - more than regular languagesCounting in JFLEX - counts.flex

[ . . ]%{ /∗ v a r i a b l e d e c l a r a t i o n ∗ /

i n t num_lines ;%}

%i n i t { /∗ code to be embedded i n the c lass cons t ruc to r ∗ /num_lines =0;

%i n i t }

%eof { /∗When EOF occurs , t h i s code i s executed ∗ /System . out . p r i n t l n ( num_lines+" l i g n e s " ) ;

%eof }

NEWLINE=\ r | \ n | \ r \ nSPACE=\ | \ t%%{NEWLINE} { num_lines ++; }{SPACE} { }


Syntactic Analysis aka Parsing

1 Lexical Analysis aka Lexing

2 Syntactic Analysis aka ParsingParsing ?JFlex : producing tokensJCup : construct acceptors





Syntactic Analysis aka Parsing Parsing ?

What’s Parsing ?

[TKINT, TKVAR("y"), TKEQ, TKINT(12), TKPLUS, TKINT(4),TKFOIS, TKVAR("x"), TKPVIRG]=⇒

=

yint +

12

4 x

*

int

or “yes, it belongs to the grammar !”

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 14 / 38 �Syntactic Analysis aka Parsing Parsing ?

From the grammar to the parser

The grammar must be a context-free grammar

S-> aSb

S-> eps

In this grammar :S is the start symbola and b are terminal tokens (produced by the lexingphase)

I This grammar recognizes anbn.


Syntactic Analysis aka Parsing Parsing ?

So Far ...

JFlex has been used to produce acceptors for (' regular)languages.We want more (general) grammars acceptors !I From a context-free grammar, construct an automaton (seecourse 2).

Thus we need a way :to declare terminal symbols (tokens)to produce them from the input file. (JFlex)to write grammars.

I In .flex and .cup files.


Syntactic Analysis aka Parsing JFlex : producing tokens

Terminal symbols/tokens

For instance :numbersidentifiersoperationskeywordsbraces, brackets


Syntactic Analysis aka Parsing JFlex : producing tokens

Returning tokens With Java/JFLex

Tokens as symbol instances

"+ " { return symbol (sym .PLUS ) ; }

(using Symbol from java_cup.runtime.Symbol)

The token may have values :an integer value for numbersa string value for identifiants

Tokens with values

[0−9]+ { return symbol (sym . TKINT , new In tege r ( y y t e x t ( ) ) ) ; }

Tokens must be declared (in cup file), see later.

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 18 / 38 �Syntactic Analysis aka Parsing JCup : construct acceptors

.cup format and compilation

.cup construction

impor t java_cup . runt ime . ∗ ; <<<−−− impor ts

i n i t w i th { : . . . : } ; <<<−−− o p t i o n a l user codet h i s one to be execut ing before pars ing

te rm ina l . . . . : <<<−−− token d e c l a r a t i o nnon te rm ina l . . . ;

precedence . . . ; <<<−−− precedence and a s s o c i a t i v i t y ( opt )

S : : = . . . ; <<<−− grammar ru l es and ( opt ) ac t i ons

The compilation of the cup file produces parser.java andsym.java

java -jar java-cup-11a.jar toto.cup


Syntactic Analysis aka Parsing JCup : construct acceptors

Recognising anbn - with Flex and Cup -1anbn.flex

impor t java_cup . runt ime . ∗ ; / / impor t Symbol c lass etc%%%class Anbn%unicode%l i n e%column%cup%{/∗ create tokens along wi th l i n e , co l . numbers ∗ /p r i v a t e Symbol symbol ( i n t type ) {

return new Symbol ( type , yy l i ne , yycolumn ) ;}

%}%%

/∗ r u l es ∗ /’ ’ a ’ ’ { return symbol (sym .TKA) ; }’ ’ b ’ ’ { return symbol (sym .TKB) ; }

I flex generates Anbn.javaLaure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 20 / 38 �


Recognising anbn - with Flex and Cup -2

anbn.cup

impor t java_cup . runt ime . ∗ ;

parser code { : /∗ to handle syntax e r r o r s ∗ /p u b l i c void r e p o r t _ f a t a l _ e r r o r ( S t r i n g message , Object i n f o )

throws Except ion {r e p o r t _ e r r o r ( message , i n f o ) ;throw new Except ion ( " Syntax Er ro r " ) ;}

: } ;

t e rm ina l TKA,TKB;non te rm ina l S ;

S : : = TKA S TKB | ;

I cup generates two classes : parser.java and sym.java



Recognising anbn - with Flex and Cup -3The main class :

Analyseur.java

impor t java . i o . ∗ ;

p u b l i c c lass Analyseur {s t a t i c p u b l i c void main ( S t r i n g argv [ ] ) {

t r y {parser p = new parser (new Anbn (new Fi leReader ( argv [ 0 ] ) ) ) ;Object r e s u l t = p . parse ( ) ;System . out . p r i n t l n ( " \ n f i l e OK" ) ;

} catch ( Except ion e ) {System . out . p r i n t l n ( " \ n f i l e not found ? " ) ;

}}

}

I Compiled with all .java. I warning, to run, java-cup-11a.jarmust be in the classpath (or used with the -cp option)

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 22 / 38 �Syntactic Analysis aka Parsing JCup : construct acceptors

A Makefile for flex/cupJFLEX=/home/laure/analyseurs/jflex-1.4.3/bin/jflex

CUPJAR=/home/laure/analyseurs/jflex-1.4.3/java-cup-11a.jar

# directory/package containing your sources

CUPFILE=anbn.cup # the cup file containing your grammar

LEXFILE=anbn.flex # the flex file containing your lexical analyzer

CPATH=.:$(CUPJAR)

SOURCES=*.java

all: cup flex

javac -cp $(CPATH) $(SOURCES)

cup: $(CUPFILE)

java -jar $(CUPJAR) $(CUPFILE)

flex: $(LEXFILE)

$(JFLEX) $(LEXFILE)

clean:

rm *.class parser.java sym.java *~ Anbn.java


Syntactic Analysis and rules








So Far ...

JFLex/Cup have been used to produce acceptors forcontext-free languages

=⇒ the abstract syntax tree remains to be constructed (thenused !)



Semantic actions

Semantic actions : code that are performed each time agrammar rule is matched.

Printing as a semantic action in JavaCup

S: : = TKA S TKB { : System . out . p r i n t l n ( " ru le1 " ) ; : }

I We can do more than pretty print !

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 26 / 38 �Syntactic Analysis and rules

Semantic actions for expressions

a part of expr.flex

{ i n t e g e r } { return symbol (sym . TKINT , new In tege r ( y y t e x t ( ) ) ) ; }"+ " { return symbol (sym .TKPLUS) ; }

a part of expr.cup

non te rm ina l I n tege r E ;

S : : = E : e TKSEMICOL { : System . out . p r i n t l n ( e . i n tVa lue ( ) ) ; : };

E : : = TKINT : n{ : RESULT = new In tege r ( n . i n tVa lue ( ) ) ; : }

| E : e1 TKPLUS E: e2{ : RESULT = new In tege r ( e1 . i n tVa lue ( ) + e2 . i n tVa lue ( ) ) ; : }

I we can attach attributes to (non) terminals.



Semantic actions : priorities

We can tell JavaCup how to solve conflicts while parsing :precedences to solve 3 + 4 * 5

associativity to solve the pb of 3 + 4 + 5

nonassoc : 2==0 but not 3==6==10

a part of expr.cup

precedence l e f t TKPLUS;precedence l e f t TKTIMES ; / / ∗ has more precedence than +[ . . . ]

S : : = . . .



Explicit AST, why ?

Why not program our compilers entirely using semanticactions ?

Because manipulating a tree is easier.Because the semantics actions are not really easy to readBecause of the separation of concernshttp:

//en.wikipedia.org/wiki/Separation_of_concerns

I Parse, then evaluate/print/construct another internalrepresentation, . . .



Semantic actions and explicit AST in Java - 1The class ASTExpr.java : The Tree !

p u b l i c c lass ASTExpr {f i n a l s t a t i c i n t INT=0 , ADD=1 , MUL=2 ;i n t tag ;i n t as In t ; / / value used i f tag = INTASTExpr e1 , e2 ; / / used i f ADD or MUL/ / Const ruc torsASTExpr ( i n t i ) { tag = INT ; as In t = i ; }ASTExpr ( ASTExpr e1 , i n t op , ASTExpr e2 ) {

tag = op ; t h i s . e1 = e1 ; t h i s . e2 = e2 ; }/ / eva lua t i on o f an expressioni n t eval ( ) {

switch ( t h i s . tag ) {case ASTExpr . INT : return t h i s . as I n t ;case ASTExpr .ADD: return t h i s . e1 . eva l ( )+ t h i s . e2 . eva l ( ) ;case ASTExpr .MUL: return t h i s . e1 . eva l ( )∗ t h i s . e2 . eva l ( ) ;}

throw new Er ro r ( " i n c o r r e c t tag " ) ;}

}

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 30 / 38 �Syntactic Analysis and rules

Semantic actions and explicit AST in Java - 2

a part of expr.cup

non te rm ina l ASTExpr E ;

S : : = E : e TKSEMICOL { : System . out . p r i n t l n ( e . eva l ( ) ) ; : };

E : : = TKINT : n{ : RESULT = new ASTExpr ( n . i n tVa lue ( ) ) ; : }

| E : e1 TKPLUS E: e2{ : RESULT = new ASTExpr ( e1 , ASTExpr .ADD, e2 ) ; : }

;


Towards a methodology for designing front-ends and simpleevaluators in Java








The running example

vars x,y,z;

y:=13;

z:=80;

x:=y+z;

z:=x*12;

print(x);

I Parse and evaluate expressions in Java !



Questions

What is the grammar ? (and keywords, and end symbols ...)Write the lex and cup files.Construct the intermediate representation in Java ButHow ?

Laure Gonnord (Lille1/Polytech) Informatique Fondamentale IMA S8 2013 S8-SC � 34 / 38 �Towards a methodology for designing front-ends and simple

evaluators in Java

A class hierarchy as intermediate representation

A quick look at the grammar :

program ::= instruction_l

instruction_l ::= instruction instruction_l

instruction ::= declaration | assignment | print

ThenEach non-terminal is a class.Instruction will be an abstract classclass Declaration extends Instruction

the class Program will have an interpret() function.



Last ( ?) problem

We have to store the variables and their (current) values, thecontextI Use a hashmap !

HashMap<String,Integer> currentContext

I Evaluating an expression requires this context !


Other technologies for the front-end







Other technologies for the front-end

Front-end more recent technologies

XML parsers (java, . . . ) : more for data languagesANTLR (multi languages)ROSE (C/C++ frontend) : source to source translator,provides high level functions in C++LLVM, (C/C++) more for code optimisation, still in researchdomain.


polycopié d'informatique fondamentale...

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