Programming Praxis

So we don’t have to deal with the astronomical calculations, which are tricky, we assume you can find the sunrise and sunset times yourself. We did that in a previous exercise, but there seems to be a bug there, as (solar 38.6270 90.1994 2016 7 8 7) returns sunrise at 29:45 instead of 5:45, so you might prefer to look at an almanac. We represent time as a pair containing hour and minute, and use this function to calculate the number of minutes between two times:

(define (duration start end)
  (when (< (cdr end) (cdr start))
    (set! end (cons (- (car end) 1) (+ (cdr end) 60))))
  (+ (* (- (car end) (car start)) 60)
     (- (cdr end) (cdr start))))

> (duration '(5 . 45) '(20 . 29))
884

We also need a function to add a duration to a starting time:

(define (add-duration start duration)
  (let ((end (cons (+ (car start) (quotient duration 60))
                   (+ (cdr start) (modulo duration 60)))))
    (if (< (cdr end) 60) end
      (cons (+ (car end) 1) (- (cdr end) 60)))))

> (add-duration '(5 . 45) 884)
(20 . 29)

And now it’s easy to compute the hours:

(define (greek-time sunrise sunset)
  (let ((hour-length (/ (duration sunrise sunset) 12)))
    (map (lambda (dur) (add-duration sunrise dur))
         (map (lambda (hour) (round (* hour hour-length)))
              (range 13)))))

> (greek-time '(5 . 45) '(20 . 29))
((5 . 45) (6 . 59) (8 . 12) (9 . 26) (10 . 40) (11 . 53) (13 . 7)
  (14 . 21) (15 . 34) (16 . 48) (18 . 2) (19 . 15) (20 . 29))

We used range from the Standard Prelude. You can run the program at http://ideone.com/G5PwOK. Mark Dominus provides a somewhat different answer at his blog.

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