Booth’s Algorithm
April 12, 2013
Our code closely tracks the code from Wikipedia:
(define (booth lt? xs)
(define (eq? a b)
(and (not (lt? a b)) (not (lt? b a))))
(let* ((n (len xs)) ; length of cycle
(xv (make-vector (+ n n))) ; elements of cycle, repeated
(fv (make-vector (+ n n) -1)) ; failure function
(k 0)) ; current minimum rotation
(do ((i 0 (+ i 1)) (xs xs (cdr xs))) ((= n i))
(vector-set! xv i (car xs))
(vector-set! xv (+ i n) (car xs)))
(do ((j 1 (+ j 1))) ((= (+ n n) j))
(let ((i (vector-ref fv (- j k 1))))
(while (and (not (= i -1))
(not (eq? (vector-ref xv j)
(vector-ref xv (+ k i 1)))))
(if (lt? (vector-ref xv j) (vector-ref xv (+ k i 1)))
(set! k (- j i 1)))
(set! i (vector-ref fv i)))
(if (and (= i -1)
(not (eq? (vector-ref xv j)
(vector-ref xv (+ k i 1)))))
(begin
(if (lt? (vector-ref xv j) (vector-ref xv (+ k i 1)))
(set! k j))
(vector-set! fv (- j k) -1))
(vector-set! fv (- j k) (+ i 1)))))
(let loop ((i 0) (zs (list)))
(if (= i n) (reverse zs)
(loop (+ i 1) (cons (vector-ref xv (+ i k)) zs))))))
Then the cyclic equality tester checks that the two cyclic lists have the same length, calls booth on each, and checks if the two outputs are the same.
(define (ceq? lt? xs ys)
(and (= (len xs) (len ys))
(equal? (booth lt? xs) (booth lt? ys))))
Here is the same list of examples as the prior exercise:
> (ceq? < (cycle 1 2 3 4) (cycle 1 2 3 4))
#t
> (ceq? < (cycle 1 2 3 4) (cycle 2 3 4 1))
#t
> (ceq? < (cycle 1 2 3 4) (cycle 3 4 1 2))
#t
> (ceq? < (cycle 1 2 3 4) (cycle 4 1 2 3))
#t
> (ceq? < (cycle 1 2 3 4) (cycle 1 2 3 5))
#f
> (ceq? < (cycle 1 1 1 1) (cycle 1 1 1 1))
#t
> (ceq? < (cycle 1 1 1 1) (cycle 1 1 1 2))
#f
> (ceq? < (cycle 1 1 1 1) (cycle 1 1 1))
#f
You can run the program at http://programmingpraxis.codepad.org/kpl1ksXq.
Programming Praxis 