It depends on how large a value you want to calculate. For n <= 50000
, the following works:
#include <cmath>
/*
*/
round(1.201*pow(n, 0.618));
As it turns out, due to the nature of this sequence, you need almost every single entry in it to compute g[n]. I coded up a solution that uses a map
to save past calculations, purging it of unneeded values. For n == 500000
, the map still had roughly 496000
entries, and since a map has two values where the array would have one, you end up using about twice as much memory.
#include <iostream>
#include <map>
using namespace std;
class Golomb_Generator {
public:
int next() {
if (n == 1)
return cache[n++] = 1;
int firstTerm = n - 1;
int secondTerm = cache[firstTerm];
int thirdTerm = n - cache[secondTerm];
if (n != 3) {
auto itr = cache.upper_bound(secondTerm - 1);
cache.erase(begin(cache), itr);
}
return cache[n++] = 1 + cache[thirdTerm];
}
void printCacheSize() {
cout << cache.size() << endl;
}
private:
int n = 1;
map<int, int> cache;
};
void printGolomb(long long n)
{
Golomb_Generator g{};
for (int i = 0; i < n - 1; ++i)
g.next();
cout << g.next() << endl;
g.printCacheSize();
}
int main()
{
int n = 500000;
printGolomb(n);
return EXIT_SUCCESS;
}
You can guess as much. n - g(g(n - 1))
uses g(n-1)
as an an argument to g
, which is always much, much smaller than n. At the same time, the recurrence also uses n - 1
as an argument, which is close to n
. You can't delete that many entries.
About the best you can do without O(n) memory is recursion combined with the approximation that is accurate for smaller n
, but it will still become slow quickly. Additionally, as the recursive calls stack up, you will likely use more memory than having an appropriately sized array would.
You might be able to do a little better though. The sequence grows very slowly. Applying that fact to g(n - g(g(n - 1)))
, you can convince yourself that this relationship mostly needs stored values nearer to 1
and stored values nearer to n
-- nearN(n - near1(nearN(n - 1)))
. You can have a tremendous swath in between that do not need to be stored, because they would be used in calculations of g(n)
for much, much larger n
than you care about. Below is an example of maintaining the first 10000
values of g
and the last 20000
values of g
. It works at least for n <= 2000000
, and it stops working for sure at n >= 2500000
. For n == 2000000
, it takes about 5 to 10 seconds to compute.
#include <iostream>
#include <unordered_map>
#include <cmath>
#include <map>
#include <vector>
using namespace std;
class Golomb_Generator {
public:
int next() {
return g(n++);
}
private:
int n = 1;
map<int, int> higherValues{};
vector<int> lowerValues{1, 1};
int g(int n) {
if(n == 1)
return 1;
if (n <= 10000) {
lowerValues.push_back(1 + lowerValues[n - lowerValues[lowerValues[n - 1]]]);
return higherValues[n] = lowerValues[n];
}
removeOldestResults();
return higherValues[n] = 1 + higherValues[n - lowerValues[higherValues[n - 1]]];
}
void removeOldestResults() {
while(higherValues.size() >= 20000)
higherValues.erase(higherValues.begin());
}
};
void printGolomb(int n)
{
Golomb_Generator g{};
for (int i = 0; i < n - 1; ++i)
g.next();
cout << g.next() << endl;
}
int main()
{
int n = 2000000;
printGolomb(n);
return EXIT_SUCCESS;
}