Petar Kapriš
1b2b2dfec1
To be noted: the last two exercises, 2.15 and 2.16 were left unfinished, and will be left for another time.
31 lines
1,011 B
Scheme
31 lines
1,011 B
Scheme
#lang sicp
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(define (add-interval x y)
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(make-interval (+ (lower-bound x) (lower-bound y))
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(+ (upper-bound x) (upper-bound y))))
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(define (sub-interval a b)
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(make-interval (- (lower-bound a) (upper-bound b))
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(- (upper-bound a) (lower-bound b))))
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Notice that in add-interval, the lower-bound is low-x + low-y.
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Meanwhile the higher-bound is high-x + high-y = low-x + width-x
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+ low-y + width-y = low-sum + (width-x + width-y).
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So width-sum = width-x + width-y.
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Similar logic applies to sub-interval.
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For mul-interval it's enough to see that
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(mul-interval (make-interval 1 2) (make-interval 3 4)) =
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(3 . 8) -> width = 5
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doesn't have the same width as:
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(mul-interval (make-interval 3 4) (make-interval 3 4)) =
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(9 . 16) -> width = 7
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despite the fact that in both cases, the intervals have widths:
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1 and 1.
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|#
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; This code is only here to pass hexlet's test, which I'm not sure is
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; necessary.
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(define (width x) (/ (- (upper-bound x) (lower-bound x))
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2))
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