Transcript CS252-Slides-2015-to..

CS252: Systems Programming
Ninghui Li
Topic 3: Programming in a FIZ: Simple Functional
Programming Language
What is FIZ
FIZ
F 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 2
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 3
Examples
(inc 3)
(inc (dec (inc 1)))
(ifz (dec 1) 2 3)
4
2
2
slide 4
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))
slide 5
Dealing with Errors
(halt) stops the program
Whenever the evaluation could lead to noninteger or negative number, call (halt)
Note:
FIZ has no assignment, except in function
invocation.
FIZ has no loop, except in recursion
slide 6
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
In Lazy evaluation
(add 3 0)
(add 3 0)
becomes
becomes
(ifz 0 3 (add (inc 3) (dec 0)))
(ifz 0 3 (add (inc 3) (dec 0)))
We do not evaluate the last
We need to evaluate the last
argument in the above function argument yet, and figures out
call, i.e.,
that we do not need it
(add (inc 3) (dec 0)),
Which becomes
(add 4 -1)
This then becomes infinite loop
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 directory
Many 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 Adleman
Published as R L Rivest, A Shamir, L Adleman, "On
Digital Signatures and Public Key Cryptosystems",
Communications of the ACM, vol 21 no 2, pp120126, 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 q
Typically 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.)
Encryption
Given a message M, 0 < M < n M  Zn {0}
use public key (e, n)
compute C = Me mod n
C  Zn {0}
Decryption
Given a ciphertext C, use private key (d)
Compute Cd mod n = (Me mod n)d mod n =
Med mod n = M
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) = 71
M  Cd mod n
M  7113 (mod 77) = 15
Topic 6: Public Key Encrypption
and Digital Signatures
14
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
Topic 6: Public Key Encrypption
and Digital Signatures
15
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
16
History of Function
Programming
Lisp was created in 1958 by John McCarthy
at MIT
Stands for LISt Processing
Scheme developed in 1975
A dialect of Lisp
Racket
ML
OCaml
Haskell
17
Application Areas
Artificial Intelligence
expert systems
planning
Simulation, modeling
Rapid prototyping
18
Functional Languages
Imperative Languages
Ex. Fortran, Algol, Pascal, Ada
based on von Neumann computer model
Functional Languages
Ex. Scheme, Lisp, ML, Haskell
Based on mathematical model of computation
and lambda calculus
19
Review
Be able to write simple programs in FIZ.
Know the concept of lazy evaluation versus
eager evaluation.
How RSA work is not required.
20
Upcoming Attraction
How to write an interpreter/compiler
frontend?
Topic 4: Regular expressions & Lexical
Analyzer
Topic 5: Context-free grammar & Parser
21