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28 lines
632 B
28 lines
632 B
2 years ago
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Exercise 4.51: Implement a new kind of
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assignment called permanent-set! that is not undone upon failure. For
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example, we can choose two distinct elements from a list and count the number
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of trials required to make a successful choice as follows:
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(define count 0)
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(let ((x (an-element-of '(a b c)))
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(y (an-element-of '(a b c))))
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(permanent-set! count (+ count 1))
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(require (not (eq? x y)))
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(list x y count))
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;;; Starting a new problem
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;;; Amb-Eval value:
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(a b 2)
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;;; Amb-Eval input:
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try-again
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;;; Amb-Eval value:
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(a c 3)
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What values would have been displayed if we had used set! here
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rather than permanent-set!?
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