CS252: Systems Programming

Ninghui Li

Topic 3: Programming in a FIZ: Simple Functional Programming Language

slide 2

What is FIZ

FIZF is for functional programming

I is for integer (we only use integer data type)

Z is for zero, denoting the simplicity of the language

In Functional Programming, one

defines functions

writes expressions

Syntax: instead of writing f(a, b, c); we write (f a b c)

slide 3

Basic Features of FIZ

non-negative integer evaluates to its value

(inc exp) evaluates to value of exp + 1

(dec exp) evaluates to halt if value of exp is 0

otherwise evaluates to value of exp - 0

(ifz cexp texp fexp)evaluates to value of texp when cexp evaluates

to 0

and to value of fexp when cexp evaluates to none 0

slide 4

Examples

(inc 3) 4

(inc (dec (inc 1))) 2

(ifz (dec 1) 2 3) 2

slide 5

Defining New Functions(define (name arguments) function-body )

Examples:

(define (add x y)

(ifz y x (add (inc x) (dec y))))

(define (sub x y)

(ifz y x (ifz x halt (sub (dec x) (dec y)))))

(name arguments)

E.g., (add 2 3); (add (inc 2) (dec 3))

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Dealing with Errors(halt) stops the program

Whenever the evaluation could lead to non-integer or negative number, call (halt)

Note:

FIZ has no assignment, except in function invocation.

FIZ has no loop, except in recursion

More Examples: Add

(define (add x y)

(ifz y x

(add (inc x) (dec y))))

; if y==0, x+y=x

; otherwise x+y = (x+1) + (y-1)

We assume lazy evaluation, i.e., in

(ifz cond t_exp e_exp)

We evaluate t_exp only when cond is 0, and evaluate e_exp only when cond is not 0

Lazy versus Eager Evaluation

In Eager evaluation

(add 3 0)

becomes

(ifz 0 3 (add (inc 3) (dec 0)))

We need to evaluate the last argument in the above function call, i.e.,

(add (inc 3) (dec 0)),

Which becomes

(add 4 -1)

This then becomes infinite loop

In Lazy evaluation

(add 3 0)

becomes

(ifz 0 3 (add (inc 3) (dec 0)))

We do not evaluate the last

argument yet, and figures out that we do not need it

A similar issue in C: In “If (cond_1 && func(a,b))”, if cond_1 is false, would func(a,b) be called?

More Examples

(define (mul x y)

(ifz y 0

(add x (mul x (dec y))))

(define (div x y)

(ifz y (halt)

(ifz (lt x y) 0

(inc (div (sub x y) y)))))

Concept of Public Key Encryption

In Traditional Cryptography, encryption key = decryption key, and must be kept secret, and key distribution/agreement is a difficult problem to solve

In public key encryption, each party has a pair (K, K-1) of keys: K is the public key, and used for encryption

K-1 is the private key, and used for decryption

Satisfies DK-1[EK[M]] = M

Knowing K, it is computationally expensive to find K-1

The public key K may be made publicly available, e.g., in a publicly available directoryMany can encrypt, only one can decrypt

Public-key systems aka asymmetric crypto systems

RSA Algorithm

Invented in 1978 by Ron Rivest, Adi Shamir and Leonard AdlemanPublished as R L Rivest, A Shamir, L Adleman, "On

Digital Signatures and Public Key Cryptosystems", Communications of the ACM, vol 21 no 2, pp120-126, Feb 1978

Security relies on the difficulty of factoring large composite numbers

Essentially the same algorithm was discovered in 1973 by Clifford Cocks, who works for the British intelligence

RSA Public Key Crypto System

Key generation:1. Select 2 large prime numbers of about the same

size, p and qTypically each p, q has between 512 and 2048 bits

2. Compute n = pq, and (n) = (q-1)(p-1)3. Select e, 1<e< (n), s.t. gcd(e, (n)) = 1

Typically e=3 or e=65537

4. Compute d, 1< d< (n) s.t. ed 1 mod (n)Knowing (n), d easy to compute.

Public key: (e, n)Private key: d

RSA Description (cont.)

EncryptionGiven a message M, 0 < M < n M Zn

{0}use public key (e, n) compute C = Me mod n C Zn {0}

DecryptionGiven a ciphertext C, use private key (d) Compute Cd mod n = (Me mod n)d mod n =

Med mod n = M

Topic 6: Public Key Encrypption and Digital Signatures

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RSA Example

p = 11, q = 7, n = 77, (n) = 60

d = 13, e = 37 (ed = 481; ed mod 60 = 1)

Let M = 15. Then C Me mod n

C 1537 (mod 77) = 71M Cd mod n

M 7113 (mod 77) = 15

Topic 6: Public Key Encrypption and Digital Signatures

15

We show how to do modular exponentiation in FIZ.

(define (mexp x e y)

(ifz e 1

(ifz (rem e 2)

(rem (square (mexp x (div e 2) y)) y)

(rem (mul x (mexp x (dec e) y)) y))))

If e==0, then result is 1

Else if e%2=0, result is (x^(e/2) mod y) ^ 2 mod y

Else result is (x * (x^(e-1) mod y)) mod y

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FIZ is based on Scheme

Scheme is a functional programming language

Uses the list data structure extensively

Program/Data equivalence

Heavy use of recursion

Garbage-collected, heap-allocated

Usually interpreted, but good compilers exist

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History of Function Programming

Lisp was created in 1958 by John McCarthy at MITStands for LISt Processing

Scheme developed in 1975A dialect of Lisp

RacketMLOCamlHaskell

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Application Areas

Artificial Intelligenceexpert systems

planning

Simulation, modeling

Rapid prototyping

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Functional Languages

Imperative Languages Ex. Fortran, Algol, Pascal, Ada

based on von Neumann computer model

Functional LanguagesEx. Scheme, Lisp, ML, Haskell

Based on mathematical model of computation and lambda calculus

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Review

Be able to write simple programs in FIZ.

Know the concept of lazy evaluation versus eager evaluation.

How RSA work is not required.

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Upcoming Attraction

How to write an interpreter/compiler frontend?

Topic 4: Regular expressions & Lexical Analyzer

Topic 5: Context-free grammar & Parser