Ben*_*enz 1 lisp optimization common-lisp
我试图创建一个prime-factors返回数字素数的函数。为此,我创建了is-prime函数,并将prime-factors-helper对主要因素进行递归检查。
(defun is-prime (n &optional (d (- n 1)))
(if (/= n 1) (or (= d 1)
(and (/= (rem n d) 0)
(is-prime n (- d 1)))) ()))
(defun prime-factors-helper (x n)
(if (is-prime x)
(list x)
(if (is-prime n)
(if (AND (= (mod x n) 0) (<= n (/ x 2)))
(cons n (prime-factors-helper (/ x n) n))
(prime-factors-helper x (+ 1 n)))
(prime-factors-helper x (+ 1 n)))))
(defun prime-factors (x)
(prime-factors-helper x 2))
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题
我有一个优化的问题。当我有大量诸如此类的123456789消息时,我会收到此错误消息Stack overflow (stack size 261120)。我相信,因为正确的答案是(3 3 3607 3803),所以我的程序一旦使用两个前一个元素构造了列表,(3 3)查找下一个主要因素将花费很长时间。如何优化代码?
CL-USER 53 > (prime-factors 512)
(2 2 2 2 2 2 2 2 2)
CL-USER 54 > (prime-factors 123456789)
Stack overflow (stack size 261120).
1 (abort) Return to level 0.
2 Return to top loop level 0.
Type :b for backtrace or :c <option number> to proceed.
Type :bug-form "<subject>" for a bug report template or :? for other options
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从https://codereview.stackexchange.com/a/189932/20936复制:
您的代码有几个问题。
is-prime是C / Java风格。使用者使用primep或prime-number-p。zerop比清楚(= 0 ...)。is-prime 是尾递归的,因此,如果您对其进行编译,它将变成一个简单的循环,并且不应出现堆栈问题。
但是,请不要着急。
让我们用trace来看问题:
> (prime-factors 17)
1. Trace: (IS-PRIME '17)
2. Trace: (IS-PRIME '17 '15)
3. Trace: (IS-PRIME '17 '14)
4. Trace: (IS-PRIME '17 '13)
5. Trace: (IS-PRIME '17 '12)
6. Trace: (IS-PRIME '17 '11)
7. Trace: (IS-PRIME '17 '10)
8. Trace: (IS-PRIME '17 '9)
9. Trace: (IS-PRIME '17 '8)
10. Trace: (IS-PRIME '17 '7)
11. Trace: (IS-PRIME '17 '6)
12. Trace: (IS-PRIME '17 '5)
13. Trace: (IS-PRIME '17 '4)
14. Trace: (IS-PRIME '17 '3)
15. Trace: (IS-PRIME '17 '2)
16. Trace: (IS-PRIME '17 '1)
16. Trace: IS-PRIME ==> T
15. Trace: IS-PRIME ==> T
14. Trace: IS-PRIME ==> T
13. Trace: IS-PRIME ==> T
12. Trace: IS-PRIME ==> T
11. Trace: IS-PRIME ==> T
10. Trace: IS-PRIME ==> T
9. Trace: IS-PRIME ==> T
8. Trace: IS-PRIME ==> T
7. Trace: IS-PRIME ==> T
6. Trace: IS-PRIME ==> T
5. Trace: IS-PRIME ==> T
4. Trace: IS-PRIME ==> T
3. Trace: IS-PRIME ==> T
2. Trace: IS-PRIME ==> T
1. Trace: IS-PRIME ==> T
(17)
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当仅(isqrt 17) = 4需要迭代时,可以执行17次迭代。
现在编译is-prime以将递归转换为循环,请参见:
> (prime-factors 12345)
1. Trace: (IS-PRIME '12345)
1. Trace: IS-PRIME ==> NIL
1. Trace: (IS-PRIME '2)
1. Trace: IS-PRIME ==> T
1. Trace: (IS-PRIME '12345)
1. Trace: IS-PRIME ==> NIL
1. Trace: (IS-PRIME '3)
1. Trace: IS-PRIME ==> T
1. Trace: (IS-PRIME '4115)
1. Trace: IS-PRIME ==> NIL
1. Trace: (IS-PRIME '3)
1. Trace: IS-PRIME ==> T
1. Trace: (IS-PRIME '4115)
1. Trace: IS-PRIME ==> NIL
1. Trace: (IS-PRIME '4)
1. Trace: IS-PRIME ==> NIL
1. Trace: (IS-PRIME '4115)
1. Trace: IS-PRIME ==> NIL
1. Trace: (IS-PRIME '5)
1. Trace: IS-PRIME ==> T
1. Trace: (IS-PRIME '823)
1. Trace: IS-PRIME ==> T
(3 5 823)
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您正在多次检查相同数字的素数!
primep 可以找到除数,而不仅仅是检查素数。
(defun compositep (n &optional (d (isqrt n)))
"If n is composite, return a divisor.
Assumes n is not divisible by anything over d."
(and (> n 1)
(> d 1)
(if (zerop (rem n d))
d
(compositep n (- d 1)))))
(defun prime-decomposition (n)
"Return the prime decomposition of n."
(let ((f (compositep n)))
(if f
(nconc (prime-decomposition (/ n f))
(prime-decomposition f))
(list n))))
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需要注意的是一个最终的优化是可能的-
记忆化的
compositep:
(let ((known-composites (make-hash-table)))
(defun compositep (n &optional (d (isqrt n)))
"If n is composite, return a divisor.
Assumes n is not divisible by anything over d."
(multiple-value-bind (value found-p) (gethash n known-composites)
(if found-p
value
(setf (gethash n known-composites)
(and (> n 1)
(> d 1)
(if (zerop (rem n d))
d
(compositep n (- d 1)))))))))
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或更好的是prime-decomposition:
(let ((known-decompositions (make-hash-table)))
(defun prime-decomposition (n)
"Return the prime decomposition of n."
(or (gethash n known-decompositions)
(setf (gethash n known-decompositions)
(let ((f (compositep n)))
(if f
(append (prime-decomposition (/ n f))
(prime-decomposition f))
(list n)))))))
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请注意使用或append代替
nconc。
另一个有趣的优化是将迭代compositep从降序更改
为升序。这会大大加快速度,因为它会更早地终止:
(let ((known-composites (make-hash-table)))
(defun compositep (n)
"If n is composite, return a divisor.
Assumes n is not divisible by anything over d."
(multiple-value-bind (value found-p) (gethash n known-composites)
(if found-p
value
(setf (gethash n known-composites)
(loop for d from 2 to (isqrt n)
when (zerop (rem n d))
return d))))))
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