sicp 2.11 没有完全符合要求的答案

mul-interval-2.scm

; (define (mul-interval x y)
;     (let ((p1 (* (lower-bound x) (lower-bound y)))
;           (p2 (* (lower-bound x) (upper-bound y)))
;           (p3 (* (upper-bound x) (lower-bound y)))
;           (p4 (* (upper-bound x) (upper-bound y)))
;          )
;       (make-interval (min p1 p2 p3 p4) (max p1 p2 p3 p4))
;     )
; )

; 分组依据: 端点和零点关系.
(define (mul-interval-2 x y)
    (let ((lower-x (lower-bound x))
          (upper-x (upper-bound x))
          (lower-y (lower-bound y))
          (upper-y (upper-bound y))
         )

         (cond ((and (>= lower-x 0) (> lower-y 0)) 
                (make-interval (* lower-x lower-y) (* upper-x upper-y)))
               ((and (>= lower-x 0) (< upper-y 0)) 
                (make-interval (* upper-x lower-y) (* lower-x upper-y)))
               ((and (>= lower-x 0) (>= upper-y 0)) 
                (make-interval (* upper-x lower-y) (* upper-x upper-y)))
               
               ((and (< upper-x 0) (> lower-y 0)) 
                (make-interval (* lower-x upper-y) (* upper-x lower-y)))
               ((and (< upper-x 0) (< upper-y 0)) 
                (make-interval (* upper-x upper-y) (* lower-x lower-y)))
               ((and (< upper-x 0) (>= upper-y 0)) 
                (make-interval (* lower-x upper-y) (* lower-x lower-y)))

               ((and (>= upper-x 0) (> lower-y 0)) 
                (make-interval (* lower-x upper-y) (* upper-x upper-y)))
               ((and (>= upper-x 0) (< upper-y 0)) 
                (make-interval (* upper-x lower-y) (* lower-x lower-y)))
               ((and (>= upper-x 0) (>= upper-y 0)) 
                ; 如何只 乘两次? 都穿越零点, 应该绕不开4次乘法吧???  -- 从数据角度分析, 不可能绕过.  -- 看遍社区答案, 没有能绕过去的. http://community.schemewiki.org/?sicp-ex-2.11
                (make-interval (min (* lower-x upper-y) (* upper-x lower-y)) (max (* lower-x lower-y) (* upper-x upper-y))))
        )
    )
)

; 辅助函数.
(define (equal-interval? x y)
    (and (= (lower-bound x) (lower-bound y)) (= (upper-bound x) (upper-bound y)))
)

; test case.
(define (mul-interval-test)
    (let ((x (make-interval 1 9)) (y (make-interval 2 8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval 1 9)) (y (make-interval -2 -8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval 1 9)) (y (make-interval -2 8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval -1 9)) (y (make-interval 2 8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval -1 9)) (y (make-interval -2 -8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval -1 9)) (y (make-interval -2 8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval -1 -9)) (y (make-interval 2 8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval -1 -9)) (y (make-interval -2 -8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    (let ((x (make-interval -1 -9)) (y (make-interval -2 8)))
        (if (not (equal-interval? (mul-interval x y) (mul-interval-2 x y)))
            (error "fail!" x y)
        )
    )

    "All Pass!"
)

问题中的 “每种情况所需的乘法都不超过2次”, 无法满足. 社区答案, 大致看了下, 也无法满足.

特殊的场景在于: 两个区间, 都刚好是 跨域 零点的 区间.