(display "ex-1.1 - solutions in comments\n") ; 10 -> 10 ; (+ 5 3 4) -> 12 ; (- 9 1) -> 8 ; (/ 6 2) -> 3 ; (+ (* 2 4) (- 4 6)) -> 6 ; (define a 3) -> a ; (define b (+ a 1)) -> b ; (+ a b (* a b)) -> 19 ; (= a b) -> #f ; (if (and (> b a) (< b (* a b))) b a) ; -> b -> 4 ; (cond ((= a 4) 6) ; ((= b 4) (+ 6 7 a)) ; (else 25)) -> 126 ; (+ 2 (if (> b a) b a)) -> 6 ; (* (cond ((> a b) a) ; ((< a b) b) ; (else -1)) ; (+ a 1)) -> (display "\nex-1.2 - ") (display (/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5))))) (* 3 (- 6 2) (- 2 7)))) (newline) (display "\nex-1.3 - ") (define (sum-squares a b) (+ (* a a) (* b b))) (define (sum-squares-max a b c) (cond ((and (>= a c) (>= b c)) (sum-squares a b)) ((and (>= a b) (>= c b)) (sum-squares a c)) (else (sum-squares b c)))) (display (sum-squares-max 2 -6 1)) (newline) (display "\nex-1.4 - Operator becomes + or - depending on the value of b\n") ; (define (a-plus-abs-b a b) ; ((if (> b 0) + -) a b)) (display "\nex-1.5 - Only normal-order terminates\n") ;(define (p) (p)) ;(define (test x y) ; (if (= x 0) 0 y)) ;(display (if (= 0 0) 0 (p))) ; Will not terminate: ;(display (test 0 (p))) (display "\nexample - Square roots via Newton's Method") (newline) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (define (improve guess x) (average guess (/ x guess))) (define (average a b) (/ (+ a b) 2)) (define (good-enough? guess x) (< (abs (- (square guess) x)) 0.0001)) (define (square a) (* a a)) (define (sqrt x) (sqrt-iter 1.0 x)) (display (sqrt 9)) (newline) (display (sqrt (+ 100 37))) (newline) (display (sqrt (+ (sqrt 2) (sqrt 3)))) (newline) (display (square (sqrt 1000))) (newline) (display "\nex-1.6 - see comment\n") (define (new-if predicate then-clause else-clause) (cond (predicate then-clause) (else else-clause))) ;(define (sqrt-iter guess x) ; (new-if (good-enough? guess x) ; guess ; (sqrt-iter (improve guess x) ; x))) ;(display (sqrt 9)) ; sqrt-iter doesn't terminate because new-if is evaluated in applicative ; order which results in an endless recursion of sqrt-iter -> new-if. (display "\nex-1.7 - see comments for explanation\nwrong behavior:\n") (display (sqrt 9)) (newline) ; Very small numbers don't work because the delta between the initial ; guess and the expected solution is in a smaller dimension than the ; value used in good-enough? (display (sqrt 0.0000001)) (newline) ; For very large numbers, good-enough? will never return true because the ; representation of floating point numbers is not accurate enough for their ; difference to ever fall below the tolerance value. ;(display (sqrt 9732402478147293489)) (newline) (display "better:\n") (define (good-enough2? guess new-guess) (< (/ (abs (- new-guess guess)) new-guess) 0.00000000001)) (define (sqrt-iter-precise guess x) (if (good-enough2? guess (improve guess x)) guess (sqrt-iter-precise (improve guess x) x))) (display (sqrt 9)) (newline) (display (sqrt 0.00000001)) (newline) ; (display (sqrt 9732402478147293489)) ; works with racket and newer MIT Scheme versions (display "\nex-1.8 - cube-root") (newline) (define (improve-cubic y x) (/ (+ (/ x (* y y)) (* 2 y)) 3)) (define (cbrt-iter guess x) (if (good-enough2? guess (improve-cubic guess x)) guess (cbrt-iter (improve-cubic guess x) x))) (define (cbrt x) (cbrt-iter 1.0 x)) (display (cbrt 27)) (newline) (display (cbrt 0.001)) (newline) (newline) (display "ex-1.9 - see comments in code\n") ;(define (+ a b) ; (if (= a 0) ; b ; (inc (+ (dec a) b)))) ; + 3 2 ; (inc (+ 2 2)) ; (inc (inc (+ 1 2))) ; (inc (inc (inc (+ 0 2))) ; (inc (inc (inc 2))) ; -> recursive process ;(define (+ a b) ; (if (= a 0) ; b ; (+ (dec a) (inc b)))) ; (+ 3 2) ; (+ 2 3) ; (+ 1 4) ; (+ 0 5) ; 5 ; -> iterative process (display "\nex-1.10 - Ackermann") (newline) (define (A x y) (cond ((= y 0) 0) ((= x 0) (* 2 y)) ((= y 1) 2) (else (A (- x 1) (A x (- y 1)))))) (display (A 1 10)) (newline) ; (A 1 10) ; (A 0 (A 1 9)) ; (* 2 (A 1 9)) ; (* 2 (A 0 (A 1 8)) ; (* 2 (* 2 (A 1 8)) ; -> 2^10 = 1024 (display (A 2 4)) (newline) ; (A 2 4) ; (A 1 (A 2 3)) ; (2 ^ (A 2 3)) ; (2 ^ (A 1 (A 2 2)) ; (2 ^ 2 ^ (A 2 2) ; -> 2^2^2^2 = 65536 (display (A 3 3)) (newline) ; (A 3 3) ; (A 2 (A 3 2)) ; (2 ^* (A 3 2)) ; (2 ^* (A 2 (A 3 1))) ; (2 ^* 2 ^* 2) ; (2 ^* (2^2)) ; (2^2^2^2) = 65536 (define (f n) (A 0 n)) ; f(n)=2*n (display (f 111)) (newline) (define (g n) (A 1 n)) ; g(n)=2^n (display (g 12)) (newline) (define (h n) (A 2 n)) ; h(n)=2^2^2^... or 2^(h(n-1)) (display (h 4)) (newline) (newline) (display "example - Couting Change") (newline) (define (dec x) (- x 1)) (define (counting-change-iter amount current-coin) (cond ((= amount 0) 1) ((= current-coin 0) 0) ((< amount 0) 0) (else (+ (counting-change-iter (- amount (list-of-coins current-coin)) current-coin) (counting-change-iter amount (dec current-coin))) ))) (define (list-of-coins coin-index) (cond ((= coin-index 1) 1) ((= coin-index 2) 5) ((= coin-index 3) 10) ((= coin-index 4) 25) ((= coin-index 5) 50))) (define (count-change amount) (counting-change-iter amount 5)) (display "(count-change 100) = ") (display (count-change 100)) (newline) ; Try to implement a better version. Worked in Python. See Euler. ;(define (counting-change-iter amount count-coin current-coin) ; (cond ((= current-coin 0) 0) ; ((<= amount 0) 0) ; (else (+ (counting-change-iter (- amount (list-of-coins current-coin)) (+ count-coin 1) current-coin) ; (if (and (= count-coin 0)(= (modulo amount (list-of-coins current-coin)) 0)) 1 0) ; (counting-change-iter amount 0 (- current-coin 1)) ; )))) ;(display "(count-change 100) = ") ;(display (counting-change-iter 100 0 5)) (newline)