# 2.40

## 练习 2.40 请定义过程unique-pairs，给它整数n，它产生出序对(i, j)，其中$1 \le j < i \le n$。请用unique-pairs去简化上面的prime-sum-pairs的定义。

(define (accumulate op initial sequence)
(if (null? sequence)
initial
(op (car sequence)
(accumulate op initial (cdr sequence)))
)
)


(define (flatmap proc seq)
(accumulate append '()  (map proc seq)))

(define (enumerate-interval low high)
(if (> low high)
'()
(cons low (enumerate-interval (+ low 1) high))))

(define (unique-pairs n)
(define interval (enumerate-interval 1 n))
(flatmap
(lambda (x)
(filter
(lambda (lst) (< (car lst) (cadr lst)))
(map
(lambda (y) (list x y))
(remove x interval))))
interval))

(unique-pairs 10)

(define (remainder a b) (cond ((< (- a b) b) (- a b))
(else (remainder (- a b) b))
))

(define (smallest-divisor n)
(find-divisor n 2))

(define (square x) (* x x))
(define (find-divisor n test-divisor)
(cond ((> (square test-divisor) n) n)
((divides? test-divisor n) test-divisor)
(else (find-divisor n (+ test-divisor 1)))))
(define (divides? x y)
(= (remainder y x) 0))
(define (prime? x) (= x (smallest-divisor x)))

(define (make-pair-sum pair)
(list (car pair) (cadr pair) (+ (car pair) (cadr pair))))

(define (prime-sum? pair)
(prime? (+ (car pair) (cadr pair))))

(define (prime-sum-pairs n)
(map
make-pair-sum
(filter
prime-sum?
(unique-pairs n)))
)

(prime-sum-pairs 10)