2021-01-27 16:08:51 +01:00
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(load "util.scm")
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2021-01-31 17:53:16 +01:00
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(define (require p)
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(if (not p) (amb)))
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(define (an-element-of items)
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(require (not (null? items)))
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(amb (car items) (an-element-of (cdr items))))
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(define (an-integer-starting-from n)
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(amb n (an-integer-starting-from (+ n 1))))
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(display "\nex-4.35 - an-integer-between\n")
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(define (an-integer-between a b)
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(require (<= a b))
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(amb a (an-integer-between (+ a 1) b)))
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(define (a-pythagorean-triple-between low high)
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(let ((i (an-integer-between low high)))
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(let ((j (an-integer-between i high)))
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(let ((k (an-integer-between j high)))
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(require (= (+ (* i i) (* j j)) (* k k)))
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(list i j k)))))
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(display "[done]\n")
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(display "\nex-4.36 - all-pythagorean-triples\n")
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; If we replace an-integer-between with an-integer-starting-from the variables
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; i and j will stay at their initial value 1 while k will increment endlessly.
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; Hence, only triplets of the form (1 1 n) will be generated.
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(define (all-pythagorean-triples)
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(let ((i (an-integer-starting-from 1)))
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(let ((j (an-integer-starting-from i)))
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(let ((k (an-integer-starting-from j)))
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(require (= (+ (* i i) (* j j)) (* k k)))
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(list i j k)))))
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; Write a procedure that actually will accomplish this. (That is, write a
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; procedure for which repeatedly typing try-again would in principle eventually
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; generate all Pythagorean triples.)
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; (display "[done]\n")
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(display "\nex-4.37\n")
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;(display "\nex-4.38\n")
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2021-01-27 16:08:51 +01:00
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