我最近偶然发现了Ackermann 函数,它使用一种“嵌套递归”来计算一个值。我在 C++ 中实现了我自己的函数,它缓存中间结果以加快计算速度(比较没有缓存的实现)。
问题:
ackermann
最终将耗尽堆栈空间。您应该如何实现一个执行“深度递归”(多次调用自身)而不用完堆栈空间的函数?
我的实现:
#include <iostream>
#include <map>
#include <tuple>
std::map<std::tuple<int, int>, int> cache;
int ackermann(int n, int m)
{
if (cache.count(std::tuple<int, int>(n, m)))
{
return cache.at(std::tuple<int, int>(n, m));
}
if (n == 0)
{
cache.insert(std::pair<std::tuple<int, int>, int>(std::tuple<int, int>(n, m), m + 1));
return m + 1;
}
else
{
if (m == 0)
{
int tmp = ackermann(n - 1, 1);
cache.insert(std::pair<std::tuple<int, int>, int>(std::tuple<int, int>(n-1, 1), tmp));
return tmp;
}
}
int tmp = ackermann(n, m - 1);
cache.insert(std::pair<std::tuple<int, int>, int>(std::tuple<int, int>(n, m - 1), tmp));
int tmp2 = ackermann(n - 1, tmp);
cache.insert(std::pair<std::tuple<int, int>, int>(std::tuple<int, int>(n - 1, tmp), tmp2));
return tmp2;
}
int main()
{
for (int i = 0; i < 7; ++i)
{
for (int j = 0; j < 7; ++j)
{
std::cout << "ackermann of i=" << std::to_string(i) << ", j=" << std::to_string(j) << " is " << std::to_string(ackermann(i, j)) << '\n';
}
}
}
输出:
ackermann of i=0, j=0 is 1
ackermann of i=0, j=1 is 2
ackermann of i=0, j=2 is 3
ackermann of i=0, j=3 is 4
ackermann of i=0, j=4 is 5
ackermann of i=0, j=5 is 6
ackermann of i=0, j=6 is 7
ackermann of i=1, j=0 is 2
ackermann of i=1, j=1 is 3
ackermann of i=1, j=2 is 4
ackermann of i=1, j=3 is 5
ackermann of i=1, j=4 is 6
ackermann of i=1, j=5 is 7
ackermann of i=1, j=6 is 8
ackermann of i=2, j=0 is 3
ackermann of i=2, j=1 is 5
ackermann of i=2, j=2 is 7
ackermann of i=2, j=3 is 9
ackermann of i=2, j=4 is 11
ackermann of i=2, j=5 is 13
ackermann of i=2, j=6 is 15
ackermann of i=3, j=0 is 5
ackermann of i=3, j=1 is 13
ackermann of i=3, j=2 is 29
ackermann of i=3, j=3 is 61
ackermann of i=3, j=4 is 125
ackermann of i=3, j=5 is 253
ackermann of i=3, j=6 is 509
ackermann of i=4, j=0 is 13
ackermann of i=4, j=1 is 65533
Segmentation fault (core dumped)