SICP – cap. 2 – abstraction barriers – esercizi – 4

Da qui oggi un esercizio, copiando qui.

Exercise 2.2.  Consider the problem of representing line segments in a plane. Each segment is represented as a pair of points: a starting point and an ending point. Define a constructor `make-segment` and selectors `start-segment` and `end-segment` that define the representation of segments in terms of points. Furthermore, a point can be represented as a pair of numbers: the `x` coordinate and the `y` coordinate. Accordingly, specify a constructor `make-point` and selectors `x-point` and `y-point` that define this representation. Finally, using your selectors and constructors, define a procedure `midpoint-segment` that takes a line segment as argument and returns its midpoint (the point whose coordinates are the average of the coordinates of the endpoints). To try your procedures, you’ll need a way to print points:

``````(define (print-point p)
(newline)
(display "(")
(display (x-point p))
(display ",")
(display (y-point p))
(display ")"))``````

`<mode nostalgia on>` quando ero piccolo cose simili ne ho fatte parecchie, sigh 😊 `<mode nostalgia off>`. OK, vado.

``````;; point
(define (make-point x y)
(cons x y))

(define (x-point point)
(car point))

(define (y-point point)
(cdr point))

;; segment
(define (make-segment p1 p2)
(list p1 p2))

(define (start-segment segment)
(car segment))

(define (end-segment segment)
(car (cdr segment)))

;; midpoint
(define (midpoint segment)
(cons
(/ ( +
(car (start-segment segment))
(car (end-segment segment))) 2)
(/ (+
(cdr (start-segment segment))
(cdr (end-segment segment))) 2)))

;; print-point
(define (print-point p)
(newline)
(display "(")
(display (x-point p))
(display ",")
(display (y-point p))
(display ")"))``````

La soluzione di Bill the Lizard è simile alla mia ma più sintetica, OK, lui è über.
Meno pulita quella di sicp-ex, usa variabili intermedie.
Bravo Spaceman Aki!

Solo io ho usato la procedura `list`, ma mi sembra che sia OK.

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