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(define (x1_3 x y z) (- (+ (* x x) (* y y) (* z z)) (expt (min x y z) 2)));Version 2 to appease Ezbez :)(define (x1_3 x y z) (define (** x) (* x x)) (+ (** x) (** y) (** z) (- (** (min x y z)))))
(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 1 3) ) ) ) ) (* 3 (- 6 2) (- 2 7)))(define (square-two-bigger-add x y z) (cond ((and (> x z) (> y z) ) (+ (* x x) (* y y))) ((and (> y x) (> z x) ) (+ (* y y) (* z z))) ((and (> x y) (> z y)) (+ (* y y) (* z z))) ))Quote: Original post by EmrldDrgn
1.2:
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Yeilded -(43)/(180). Seems wierd to me. Reason for this post.
(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5))))) (* 3 (- 6 2) (- 2 7)))Quote: 1.3:This is basically the same as my solution, but I broke it down into smaller procedures:
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(define (square x) (* x x))(define (sum-of-squares x y) (+ (square x) (square y)))(define (sum-of-squares-of-largest-two a b c) (cond ((and (> a c) (> b c)) (sum-of-squares a b)) ((and (> a b) (> c b)) (sum-of-squares a c)) (else (sum-of-squares b c))))Quote: 1.4: If B is positive, add a and b. If not, subtract b from a.Yes.
Quote: 1.5: An applicative-order evaluation returns 0, and a normal-order evaluation never ends. Oh, and because it tries to expand p infinitly, whereas AO never evaluates p.Yes.
(define (1+ x) ( + x 1))(define (1- x) (- x 1))(define (identity x) x)(define (product a step b factor) (define (pgo x ret) (if (> x b) ret (pgo (step x) (* ret (factor x))))) (pgo a 1))(define (factorial n) (product 1 1+ n identity))(define (pi n) (* 4.0 (product 2 1+ (1+ n) (lambda (x) (/ (+ x (modulo x 2)) (+ x (modulo (1+ x) 2))))))); using closures to create counters for 1 1 3 3 5 5 ... type streams(define (make-2-counter start) (let ((plus -1)) (lambda () (set! plus (+ plus 1)) (+ start plus (- (modulo plus 2))))))(define (pi n) (let ((top (make-2-counter 2)) (bottom (make-2-counter 3))) (top) (* 4.0 (product 1 1+ n (lambda (x) (/ (top) (bottom)))))))(defun fib (n) (labels ((fibgo (a b n p q s) (cond ((= 0 n) (list a b)) ((oddp n) (fibgo (+ (* p a) (* q b)) (+ (* q a) (* s b)) (- n 1) p q s)) ((evenp n) (fibgo a b (/ n 2) (+ (* p p) (* q q)) (+ (* p q) (* q s)) (+ (* q q) (* s s))))))) (car (fibgo 1 1 n 1 1 0))))
(define (choice y) (cond ((< 0) (-10)) //it highlights the (< 0) (else 10)))> (choice 67). <: expects at least 2 arguments, given 1: 0Quote: Original post by Alpha_ProgDes
I have no idea what I'm doing wrong here.
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(define (choice y) (cond ((< y 0) -10) (else 10)))Quote: Original post by SamLowryQuote: Original post by Alpha_ProgDes
I have no idea what I'm doing wrong here.
*** Source Snippet Removed ***. <: expects at least 2 arguments, given 1: 0
The < operator takes at least two arguments, and don't put the -10 between parentheses. Parentheses are never optional in Scheme, they always mean "a list", and without quotation it will be interpreted as a function call (in this case, calling the 0-arity function '-10').(define (choice y) (cond ((< y 0) -10) (else 10)))
(define we 12)> (define us (+ we 2))> us14> (define we 20)> us14void define (int*& object, int value){ if (object != NULL) { delete object; } object = new int; *object = value;}
