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41 lines
906 B
41 lines
906 B
(car (cons x y)) |
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(car (lambda (m) (m x y))) |
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((lambda (m) (m x y)) (lambda (p q) p))) |
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(lambda (x y) x))) |
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x |
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(define (cdr z) (z (lambda (x y) y))) |
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2.5 - Given that we know the bases - 2, and 3, we only need to find the exponents. |
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For example, 6 can only be represented one way - 2^1 3^1, meaning our pair of numbers a and b are 1 |
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So, cons is simple: |
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(define (cons x y) (* (expt 2 x) (expt 3 y))) |
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Now we're left solving for a and b |
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log(2^a * 3^b) = alog(2) + blog(3) |
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call our "cons" value x |
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Because integer factorizations are unique we only need to find the biggest power of two and three that |
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go into the number. |
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(define (power-of-x value base) |
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(define (iter current-power) |
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(if (< ((expt base current-power) value) |
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(iter (+ current-power 1)) |
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(current-power)))) |
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Then with this we can write car and cdr as |
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(define (car int) (power-of-x int 2)) |
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(define (cdr int) (power-of-x int 3))
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