sicp-ex-1.18

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The solution presented here is based on the solutions for sicp-ex-1.16 and sicp-ex-1.17.

  
 ;; ex 1.18. Based on exercises 1.16 and 1.17 
  
 ;; Assume double and halve are defined by the language 
 (define (double x) (+ x x)) 
 (define (halve x) (floor (/ x 2))) 
  
 (define (* a b) 
   (define (iter accumulator a b) 
     (cond ((= b 0) accumulator) 
           ((even? b) (iter accumulator (double a) (halve b))) 
           (else (iter (+ accumulator a) a (- b 1))))) 
   (iter 0 a b)) 
  
 ;; Testing 
 (* 2 4) 
 (* 4 0) 
 (* 5 1) 
 (* 7 10) 
  
 ;; Alternate version, which makes more complete use of the  
 ;; Russian Peasant Algorithm in footnote 40.  Uses roughly half 
 ;; the steps of the above 
 (define (double a) (+ a a)) 
 (define (halve a) (/ a 2)) 
  
 (define (mult3 a b) 
   (define (mult-iter accumulator b c) 
     (cond ((= c 0) accumulator) 
           ((even? c) (mult-iter accumulator (double b) (halve c))) 
           (else (mult-iter (+ accumulator b) (double b) (- (halve c) 0.5))))) 
   (mult-iter 0 a b)) 
  
 ;; Testing 
 (mult3 2 4) 
 (mult3 4 0) 
 (mult3 5 1) 
 (mult3 7 10)