Principles Of Programming Languages Lecture 2 Today DesignByContract

Principles Of Programming Languages Lecture 2

Today • Design-By-Contract • Iteration vs. Recursion

Design By Contract • Signature: Name of procedure and arguments • Purpose: Short text explaining what the procedure does • Type • Examples of usage • Tests indicating correctness • Preconditions: Impose a certain obligation to be guaranteed on entry by any client module that calls it • Postcondition: Guarantee a certain property on exit

Example Signature: area-of-square(length) Purpose: to compute the area of a rectangle whose side length is ‘length’ Type: [Number -> Number] Example: (area-of-square 5) should produce 25 Pre-conditions: length>= 0 Post-condition: result = length^2 Tests: (area-of-square 2) expected value: 4 (area-of-square -2) expected value: ? ? ? [What do you think? ]

Design-By-Contract in this course • You are required to write this for every procedure that you write in an exercise (In the example shown in class we will often omit it) • Mandatory: signature, purpose, type and tests

Recursive Procedures • How to create a process of unbounded length? • Needed to solve more complicated problems. • No “for” or “while” loop constructs in scheme! • Answer: Recursion!!! • Start with a simple example. 6

Example: Sum of squares S(n) = 02 + 12 + 22 ………. …… (n-1)2 + n 2 S(n-1) Notice that: S(n) = S(n-1) + n 2 S(0) = 0 These two properties completely define the function Wishful thinking: if I could only solve the smaller instance … 7 2 שיעור - מבוא מורחב

An algorithm for computing sum of squares (define sum-squares (lambda (n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n))))) 8 2 שיעור - מבוא מורחב

Evaluating (sum-squares 3) (define (sum-squares n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n)))) (sum-squares 3) (if (= 3 0) 0 (+ (sum-squares (- 3 1)) (square 3))) (+ (sum-squares (- 3 1)) (square 3)) (+ (sum-squares (- 3 1)) (* 3 3)) (+ (sum-squares (- 3 1)) 9) (+ (sum-squares 2) 9) (+ (if (= 2 0) 0 (+ (sum-squares (- 2 1)) (square 2))) 9) … (+ (+ (sum-squares 1) 4) 9) … (+ (+ (+ (sum-squares 0) 1) 4) 9) (+ (+ (+ (if (= 0 0) 0 (+ (sum-squares (- 0 1)) (square 0))) 1) 4) 9) (+ (+ (+ 0 1) 4) 9) … 14 What would have happened if ‘if’ was a function ? 9 2 שיעור - מבוא מורחב

Evaluating (sum-squares 3) with IF as regular form (define (sum-squares n) (if (= n 0) 0 (+ (sum-squares (- n 1)) (square n)))) (sum-squares 3) (if (= 3 0) 0 (+ (sum-squares (- 3 1)) (square 3))) (if #f 0 (+ (sum-squares 2) 9)) (if #f 0 (+ (sum-squares 1) 4)) 9)) (if #f 0 (+ (sum-squares 0) 1)) 4)) 9)) (if #f 0 (+ (if #t 0 (+ (sum-squares -1) 0)) 1)) 4)) 9)). . We evaluate all operands. We always call (sum-squares) again. We get an infinite loop……. . OOPS 10

General form of recursive algorithms • test, base case, recursive case (define sum-sq (lambda (n) (if (= n 0) ; test for base case 0 ; base case (+ (sum-sq (- n 1)) (square n)) ; recursive case ))) • base case: small (non-decomposable) problem • recursive case: larger (decomposable) problem • at least one base case, and at least one recursive case. 11

Another example of a recursive algorithm • even? (define even? (lambda (n) (if (= n 0) #t ; test for base case ; base case (not (even? (- n 1))) ; recursive case ))) ; not is a primitive procedure that negates a boolean 12

Short summary • Design a recursive algorithm by 1. Solving big instances using the solution to smaller instances. 2. Solving directly the base cases. • Recursive algorithms have 1. test 2. recursive case(s) 3. base case(s) 13

Iteration Vs. Recursion • Problem: Recursive processes are “costly’ memory-wise – Need to “remember” pending operations • Enter Iterative processes • Lots of examples to follow

Compute ab (Recursive Approach) • wishful thinking : ab = a * a(b-1) • base case: a 0 = 1 (define exp-1 (lambda (a b) (if (= b 0) 1 (* a (exp-1 a (- b 1)))))) 15

Compute ab (Iterative Approach) • Another approach: ab = a * a*…*a • Which is: b ab = a 2 *a*…*a= a 3 *…*a • Operationally: product * a counter - 1 • Halting condition: counter = 0 16

Compute ab (Iterative Approach) (define exp-iter (lambda (a counter product) (if (= counter 0) product (exp-iter a (- counter 1) (* a product)))) How then, do the two procedures differ? They give rise to different processes – lets use our model to understand how. 17

Recursive Process (define exp-1 (lambda (a b) (if (= b 0) 1 (* a (exp-1 a (- b 1)))))) (exp-1 3 4) (* 3 (exp-1 3 3)) (* 3 (exp-1 3 2))) (* 3 (exp-1 3 1)))) (* 3 (exp-1 3 0))))) (* 3 1)))) (* 3 3))) (* 3 9)) (* 3 27) 81 18 מבוא מורחב

Iterative Process (define exp-iter (lambda (a counter product) (if (= counter 0) product (exp-iter a (- counter 1) (* a product))))) (exp-iter (exp-iter 81 3 3 3 4 3 2 1 0 1) 3) 9) 27) 81) 19 מבוא מורחב

The Difference (exp-1 3 4) (* 3 (exp-1 3 3)) (* 3 (exp-1 3 2))) (* 3 (exp-1 3 1)))) (* 3 (exp-1 3 0))))) (* 3 1)))) (* 3 3))) (* 3 9)) (* 3 27) 81 Growing (exp-2 3 4) (exp-iter 3 4 1) Constant (exp-iter 3 3 3) (exp-iter 3 2 9) (exp-iter 3 1 27) (exp-iter 3 0 81) amount of space 81 20 מבוא מורחב

Why More Space? • Recursive exponentiation: (define exp-1 operation pending (lambda (a b) (if (= b 0) 1 (* a (exp-1 a (- b 1))))) • Iterative exponentiation: (define (exp-iter a counter product) (if (= counter 0) product (exp-iter a (- counter 1) (* a product))))) no pending operations 21

Frame Space in RAM needed for procedure call evaluation.

Tail Position • A property of a procedure: the recursive call is the last evaluation step. • Compilers can identify it automatically and not open another frame – thus making it iterative.

Example: Factorial • wishful thinking : n! = n * (n-1)! • base case: n=1 (define fact (lambda (n) (if (= n 1) 1 (* n (fact (- n 1)))))) Iterative or Recursive? 24

Computing SQRT: A Numeric Algorithm To find an approximation of square root of x, use the following recipe: • Make a guess G • Improve the guess by averaging G and x/G • Keep improving the guess until it is good enough X=2 G=1 X/G = 2 G = ½ (1+ 2) = 1. 5 X/G = 4/3 G = ½ (3/2 + 4/3) = 17/12 = 1. 416666 X/G = 24/17 G = ½ (17/12 + 24/17) = 577/408 = 1. 4142156 25

(define initial-guess 1. 0) (define precision 0. 0001) (define sqrt-iter (lambda (guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x)))) (define good-enough? (lambda (guess x) (< (abs (- (square guess) x)) precision))) (define improve (lambda (guess x) (average guess (/ x guess)))) (define sqrt (lambda (x) (sqrt-iter initial-guess x))) 26

Towers of Hanoi • Three posts, and a set of disks of different sizes. • A disk can be placed only on a larger disk (or on bottom). • At the beginning all the disks are on the left post. The goal is to move the disks one at a time, while preserving these conditions, until the entire stack has moved from the left post to another You are allowed to move only the topmost disk at a step 27

Use our paradigm • Wishful thinking: • Smaller problem: A problem with one disk less • How do we use it ? To move n disks from post A to post C (using B as aux): Move top n-1 disks from post A to post B (using C as aux) Move the largest disk from post A to post C Move n-1 disks from post B to post C (using A as aux) We solve 2 smaller problems ! 28 4 שיעור - מבוא מורחב

Towers of Hanoi (define (move-tower size from to aux) (cond ((= size 1) (one-move from to)) (else (move-tower (- size 1) from aux to) (one-move from to) (move-tower (- size 1) aux to from)))) (define (one-move from to) (display "Move top disk from ") (display from) (display " To ") (display to) (newline)) 29 4 שיעור - מבוא מורחב

Tree Recursion (mt 3 1 2 3) (mt 2 1 3 2) (one-move 1 3) (mt 1 1 2 3) (mt 1 2 3 1) (mt 2 3 2 1) 3 2) (mt 1 1 2 3) (mt 1 3 1(one-move 2) (one-move 1 2) (one-move 2 3) (one-move 3 1) 30 (one-move 1 2) 4 שיעור - מבוא מורחב

Towers of Hanoi -trace (move-tower 3 1 2 3) Move Move top top disk disk from from 1 1 2 1 3 3 1 to to 2 3 3 2 1 2 2 (move-tower 2 1 3 2) (move-tower 2 3 2 1) 11 31 22 333 a 4 שיעור - מבוא מורחב

Time Complexity for towers of Hanoi Denote by T(n) the number of steps that we need to take to solve the case for n disks. T(n) = 2 T(n-1) + 1 T(1) = 1 This solves to: T(n) = 2 n-1= Q (2 n) What does that mean ? 32

Hanoi Towers Say we want to solve the problem for 400 disks. Say it takes a second to move a disk. We need about 2400 seconds. That’s about 2373 years. That’s about 2363 millenniums. Might be longer then the age of the universe …. Infeasible !!!! 33

Let’s buy a fast computer and make it feasible. Our new computer can move giga billion (260 ) disks a second. Absolutely the last word in the field of computing. We need about 2340 seconds. That’s about 2313 years. That’s about 2303 millenniums. Does not help much. Infeasible !!!! 34