Are loops really faster than recursion? - c++

According to my professor loops are faster and more deficient than using recursion yet I came up with this c++ code that calculates the Fibonacci series using both recursion and loops and the results prove that they are very similar. So I maxed the possible input to see if there was a difference in performance and for some reason recursion clocked in better than using a loop. Anyone know why? Thanks in advanced.
Here's the code:
#include "stdafx.h"
#include "iostream"
#include <time.h>
using namespace std;
double F[200000000];
//double F[5];
/*int Fib(int num)
{
if (num == 0)
{
return 0;
}
if (num == 1)
{
return 1;
}
return Fib(num - 1) + Fib(num - 2);
}*/
double FiboNR(int n) // array of size n
{
for (int i = 2; i <= n; i++)
{
F[i] = F[i - 1] + F[i - 2];
}
return (F[n]);
}
double FibMod(int i,int n) // array of size n
{
if (i==n)
{
return F[i];
}
F[i] = F[i - 1] + F[i - 2];
return (F[n]);
}
int _tmain(int argc, _TCHAR* argv[])
{
/*cout << "----------------Recursion--------------"<<endl;
for (int i = 0; i < 36; i=i+5)
{
clock_t tStart = clock();
cout << Fib(i);
printf("Time taken: %.2fs\n", (double)(clock() - tStart) / CLOCKS_PER_SEC);
cout << " : Fib(" << i << ")" << endl;
}*/
cout << "----------------Linear--------------"<<endl;
for (int i = 0; i < 200000000; i = i + 20000000)
//for (int i = 0; i < 50; i = i + 5)
{
clock_t tStart = clock();
F[0] = 0; F[1] = 1;
cout << FiboNR(i);
printf("Time taken: %.2fs\n", (double)(clock() - tStart) / CLOCKS_PER_SEC);
cout << " : Fib(" << i << ")" << endl;
}
cout << "----------------Recursion Modified--------------" << endl;
for (int i = 0; i < 200000000; i = i + 20000000)
//for (int i = 0; i < 50; i = i + 5)
{
clock_t tStart = clock();
F[0] = 0; F[1] = 1;
cout << FibMod(0,i);
printf("Time taken: %.2fs\n", (double)(clock() - tStart) / CLOCKS_PER_SEC);
cout << " : Fib(" << i << ")" << endl;
}
std::cin.ignore();
return 0;
}

You you go by the conventional programming approach loops are faster. But there is category of languages called functional programming languages which does not contain loops. I am a big fan of functional programming and I am an avid Haskell user. Haskell is a type of functional programming language. In this instead of loops you use recursions. To implement fast recursion there is something known as tail recursion. Basically to prevent having a lot of extra info to the system stack, you write the function such a way that all the computations are stored as function parameters so that nothing needs to be stored on the stack other that the function call pointer. So once the final recursive call has been called, instead of unwinding the stack the program just needs to go to the first function call stack entry. Functional programming language compilers have an inbuilt design to deal with this. Now even non functional programming languages are implementing tail recursion.
For example consider finding the recursive solution for finding the factorial of a positive number. The basic implementation in C would be
int fact(int n)
{
if(n == 1 || n== 0)
return 1
return n*fact(n-1);
}
In the above approach, each time the stack is called n is stored in the stack so that it can be multiplied with the result of fact(n-1). This basically happens during stack unwinding. Now check out the following implementation.
int fact(int n,int result)
{
if(n == 1 || n== 0)
return result
return fact(n-1,n*result);
}
In this approach we are passing the computation result in the variable result. So in the end we directly get the answer in the variable result. The only thing you have to do is that in the initial call pass a value of 1 for the result in this case. The stack can be unwound directly to its first entry. Of course I am not sure that C or C++ allows tail recursion detection, but functional programming languages do.

Your "recursion modified" version doesn't have recursion at all.
In fact, the only thing enabling a non-recursive version that fills in exactly one new entry of the array is the for-loop in your main function -- so it is actually a solution using iteration also (props to immibis and BlastFurnace for noticing that).
But your version doesn't even do that correctly. Rather since it is always called with i == 0, it illegally reads F[-1] and F[-2]. You are lucky (?)1 the program didn't crash.
The reason you are getting correct results is that the entire F array is prefilled by the correct version.
Your attempt to calculate Fib(2000....) isn't successful anyway, since you overflow a double. Did you even try running that code?
Here's a version that works correctly (to the precision of double, anyway) and doesn't use a global array (it really is iteration vs recursion and not iteration vs memoization).
#include <cstdio>
#include <ctime>
#include <utility>
double FiboIterative(int n)
{
double a = 0.0, b = 1.0;
if (n <= 0) return a;
for (int i = 2; i <= n; i++)
{
b += a;
a = b - a;
}
return b;
}
std::pair<double,double> FiboRecursive(int n)
{
if (n <= 0) return {};
if (n == 1) return {0, 1};
auto rec = FiboRecursive(n-1);
return {rec.second, rec.first + rec.second};
}
int main(void)
{
const int repetitions = 1000000;
const int n = 100;
volatile double result;
std::puts("----------------Iterative--------------");
std::clock_t tStart = std::clock();
for( int i = 0; i < repetitions; ++i )
result = FiboIterative(n);
std::printf("[%d] = %f\n", n, result);
std::printf("Time taken: %.2f us\n", (std::clock() - tStart) / 1.0 / CLOCKS_PER_SEC);
std::puts("----------------Recursive--------------");
tStart = std::clock();
for( int i = 0; i < repetitions; ++i )
result = FiboRecursive(n).second;
std::printf("[%d] = %f\n", n, result);
std::printf("Time taken: %.2f us\n", (std::clock() - tStart) / 1.0 / CLOCKS_PER_SEC);
return 0;
}
--
1Arguably anything that hides a bug is actually unlucky.

I don't think this is not a good question. But maybe the answer why is somehow interesting.
At first let me say that generally the statement is probably true. But ...
Questions about performance of c++ programs are very localized. It's never possible to give a good general answer. Every example should be profiled an analyzed separately. It involves lots of technicalities. c++ compilers are allowed to modify program practically as they wish as long as they don't produce visible side effects (whatever precisely that means). So as long as your computation gives the same result is fine. This technically allows to transform one version of your program into an equivalent even from recursive version into the loop based and vice versa. So it depends on compiler optimizations and compiler effort.
Also, to compare one version to another you would need to prove that the versions you compare are actually equivalent.
It might also happen that somehow a recursive implementation of algorithm is faster than a loop based one if it's easier to optimize for the compiler. Usually iterative versions are more complex, and generally the simpler the code is, the easier it is for the compiler to optimize because it can make assumptions about invariants, etc.

Related

How can I test two algorithms and determine which is faster?

Whenever working on a specific problem, I may come across different solutions. I'm not sure how to choose the better of the two options. The first idea is to compute the complexity of the two solutions, but sometimes they may share the same complexity, or they may differ but the range of the input is small that the constant factor matters.
The second idea is to benchmark both solutions. However, I'm not sure how to time them using c++. I have found this question:
How to Calculate Execution Time of a Code Snippet in C++ , but I don't know how to properly deal with compiler optimizations or processor inconsistencies.
In short: is the code provided in the question above sufficient for everyday tests? is there some options that I should enable in the compiler before I run the tests? (I'm using Visual C++) How many tests should I do, and how much time difference between the two benchmarks matters?
Here is an example of a code I want to test. Which of these is faster? How can I calculate that myself?
unsigned long long fiborecursion(int rank){
if (rank == 0) return 1;
else if (rank < 0) return 0;
return fiborecursion(rank-1) + fiborecursion(rank-2);
}
double sq5 = sqrt(5);
unsigned long long fiboconstant(int rank){
return pow((1 + sq5) / 2, rank + 1) / sq5 + 0.5;
}
Using the clock from this answer
#include <iostream>
#include <chrono>
class Timer
{
public:
Timer() : beg_(clock_::now()) {}
void reset() { beg_ = clock_::now(); }
double elapsed() const {
return std::chrono::duration_cast<second_>
(clock_::now() - beg_).count(); }
private:
typedef std::chrono::high_resolution_clock clock_;
typedef std::chrono::duration<double, std::ratio<1> > second_;
std::chrono::time_point<clock_> beg_;
};
You can write a program to time both of your functions.
int main() {
const int N = 10000;
Timer tmr;
tmr.reset();
for (int i = 0; i < N; i++) {
auto value = fiborecursion(i%50);
}
double time1 = tmr.elapsed();
tmr.reset();
for (int i = 0; i < N; i++) {
auto value = fiboconstant(i%50);
}
double time2 = tmr.elapsed();
std::cout << "Recursion"
<< "\n\tTotal: " << time1
<< "\n\tAvg: " << time1 / N
<< "\n"
<< "\nConstant"
<< "\n\tTotal: " << time2
<< "\n\tAvg: " << time2 / N
<< "\n";
}
I would try compiling with no compiler optimizations (-O0) and max compiler optimizations (-O3) just to see what the differences are. It is likely that at max optimizations the compiler may eliminate the loops entirely.

optimize c++ query to calculate Nmin

I have run into a problem where i am trying to optimize my query which is created to calculate Nmin values for the increasing values of N and error approximation.
I am not from programming background and have just started to take it up.
This is the calculation which is inefficient as it calculates Nmin even after finding Nmin.
Now to reduce the time i did below changes reduce function call with no improvement:
#include<iostream>
#include<cmath>
#include<time.h>
#include<iomanip>
using namespace std;
double f(int);
int main(void)
{
double err;
double pi = 4.0*atan(1.0);
cout<<fixed<<setprecision(7);
clock_t start = clock();
for (int n=1;;n++)
{
if((f(n)-pi)>= 1e-6)
{
cout<<"n_min is "<< n <<"\t"<<f(n)-pi<<endl;
}
else
{
break;
}
}
clock_t stop = clock();
//double elapsed = (double)(stop - start) * 1000.0 / CLOCKS_PER_SEC; //this one in ms
cout << "time: " << (stop-start)/double(CLOCKS_PER_SEC)*1000 << endl; //this one in s
return 0;
}
double f(int n)
{
double sum=0;
for (int i=1;i<=n;i++)
{
sum += 1/(1+pow((i-0.5)/n,2));
}
return (4.0/n)*sum;
}
Is there any way to reduce the time and make the second query efficient?
Any help would be greatly appreciated.
I do not see any immediate way of optimizing the algorithm itself. You could however reduce the time significantly by not writing to the standard output for every iteration. Also, do not calculate f(n) more than once per iteration.
for (int n=1;;n++)
{
double val = f(n);
double diff = val-pi;
if(diff < 1e-6)
{
cout<<"n_min is "<< n <<"\t"<<diff<<endl;
break;
}
}
Note however that this will yield a higher n_min (increased by 1 compared to the result of your version) since we changed the condition to diff < 1e-6.

Clock function in search algorithm generally returning 0

I'm sure this may seem like a trivial problem, but I'm having issues with the "clock()" function in my program (please note I have checked similar issues but they didn't seem to relate in the same context). My clock outputs are almost always 0, however there seem to be a few 10's as well (which is why I'm baffled). I considered the fact that maybe the function calling is too quickly processed but judging by the sorting algorithms, surely there should be some time taken.
Thank you all in advance for any help! :)
P.S I'm really sorry about the mess regarding correlation of variables between functions (it's a group code I've merged together, and I'm focusing of correct output before beautifying it) :D
#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;
int bruteForce(int array[], const int& sizeofArray);
int maxSubArray (int array[], int arraySize);
int kadane_Algorithm(int array[], int User_size);
int main()
{
int maxBF, maxDC, maxKD;
clock_t t1, t2, t3;
int arraySize1 = 1;
double t1InMSEC, t2InMSEC, t3InMSEC;
while (arraySize1 <= 30000)
{
int* array = new int [arraySize1];
for (int i = 0; i < arraySize1; i++)
{
array[i] = rand()% 100 + (-50);
}
t1 = clock();
maxBF = bruteForce(array, arraySize1);
t1 = clock() - t1;
t1InMSEC = (static_cast <double>(t1))/CLOCKS_PER_SEC * 1000.00;
t2 = clock();
maxDC = maxSubArray(array, arraySize1);
t2 = clock() - t2;
t2InMSEC = (static_cast <double>(t2))/CLOCKS_PER_SEC * 1000.00;
t3 = clock();
maxKD = kadane_Algorithm(array, arraySize1);
t3 = clock() - t3;
t3InMSEC = (static_cast <double>(t3))/CLOCKS_PER_SEC * 1000.00;
cout << arraySize1 << '\t' << t1InMSEC << '\t' << t2InMSEC << '\t' << t3InMSEC << '\t' << endl;
arraySize1 = arraySize1 + 100;
delete [] array;
}
return 0;
}
int bruteForce(int array[], const int& sizeofArray)
{
int maxSumOfArray = 0;
int runningSum = 0;
int subArrayIndex = sizeofArray;
while(subArrayIndex >= 0)
{
runningSum += array[subArrayIndex];
if (runningSum >= maxSumOfArray)
{
maxSumOfArray = runningSum;
}
subArrayIndex--;
}
return maxSumOfArray;
}
int maxSubArray (int array[], int arraySize)
{
int leftSubArray = 0;
int rightSubArray = 0;
int leftSubArraySum = -50;
int rightSubArraySum = -50;
int sum = 0;
if (arraySize == 1) return array[0];
else
{
int midPosition = arraySize/2;
leftSubArray = maxSubArray(array, midPosition);
rightSubArray = maxSubArray(array, (arraySize - midPosition));
for (int j = midPosition; j < arraySize; j++)
{
sum = sum + array[j];
if (sum > rightSubArraySum)
rightSubArraySum = sum;
}
sum = 0;
for (int k = (midPosition - 1); k >= 0; k--)
{
sum = sum + array[k];
if (sum > leftSubArraySum)
leftSubArraySum = sum;
}
}
int maxSubArraySum = 0;
if (leftSubArraySum > rightSubArraySum)
{
maxSubArraySum = leftSubArraySum;
}
else maxSubArraySum = rightSubArraySum;
return max(maxSubArraySum, (leftSubArraySum + rightSubArraySum));
}
int kadane_Algorithm(int array[], int User_size)
{
int maxsofar=0, maxending=0, i;
for (i=0; i < User_size; i++)
{
maxending += array[i];
if (maxending < 0)
{
maxending = 0 ;
}
if (maxsofar < maxending)
{
maxsofar = maxending;
}
}
return maxending;
}
Output is as follows: (just used a snippet for visualization)
29001 0 0 0
29101 0 10 0
29201 0 0 0
29301 0 0 0
29401 0 10 0
29501 0 0 0
29601 0 0 0
29701 0 0 0
29801 0 10 0
29901 0 0 0
Your problem is probably that clock() doesn't have enough resolution: you truly are taking zero time to within the precision allowed. And if not, then it's almost surely that the compiler is optimizing away the computation since it's computing something that isn't being used.
By default, you really should be using the chrono header rather than the antiquated C facilities for timing. In particular, high_resolution_clock tends to be good for measuring relatively quick things should you find yourself really needing that.
Accurately benchmarking things is a nontrivial exercise, and you really should read up on how to do it properly. There are a surprising number of issues involving things like cache or surprising compiler optimizations or variable CPU frequency that many programmers have never even thought about before, and ignoring them can lead to incorrect conclusions.
One particular aspect is that you should generally arrange to time things that actually take some time; e.g. run whatever it is you're testing on thousands of different inputs, so that the duration is, e.g., on the order of a whole second or more. This tends to improve both the precision and the accuracy of your benchmarks.
Thanks for all the help guys! It seems to be some kind of issue with any windows/windows emulation software.
I decided to boot up ubuntu and rather give it a shot there, and voila! I get a perfect output :)
I guess Open Source really is the best way to go :D

Not Finding Times of Prime Generation / Limited Generation

This program is a c++ program that finds primes using the sieve of eratosthenes to calculate primes. It is then supposed to store the time it takes to do this, and reperform the calculation 100 times, storing the times each time. There are two things that I need help with in this program:
Firstly, I can only test numbers up to 480million I would like to get higher than that.
Secondly, when i time the program it only gets the first timing and then prints zeros as the time. This is not correct and I don't know what the problem with the clock is. -Thanks for the help
Here is my code.
#include <iostream>
#include <ctime>
using namespace std;
int main ()
{
int long MAX_NUM = 1000000;
int long MAX_NUM_ARRAY = MAX_NUM+1;
int long sieve_prime = 2;
int time_store = 0;
while (time_store<=100)
{
int long sieve_prime_constant = 0;
int *Num_Array = new int[MAX_NUM_ARRAY];
std::fill_n(Num_Array, MAX_NUM_ARRAY, 3);
Num_Array [0] = 1;
Num_Array [1] = 1;
clock_t time1,time2;
time1 = clock();
while (sieve_prime_constant <= MAX_NUM_ARRAY)
{
if (Num_Array [sieve_prime_constant] == 1)
{
sieve_prime_constant++;
}
else
{
Num_Array [sieve_prime_constant] = 0;
sieve_prime=sieve_prime_constant;
while (sieve_prime<=MAX_NUM_ARRAY - sieve_prime_constant)
{
sieve_prime = sieve_prime + sieve_prime_constant;
Num_Array [sieve_prime] = 1;
}
if (sieve_prime_constant <= MAX_NUM_ARRAY)
{
sieve_prime_constant++;
sieve_prime = sieve_prime_constant;
}
}
}
time2 = clock();
delete[] Num_Array;
cout << "It took " << (float(time2 - time1)/(CLOCKS_PER_SEC)) << " seconds to execute this loop." << endl;
cout << "This loop has already been executed " << time_store << " times." << endl;
float Time_Array[100];
Time_Array[time_store] = (float(time2 - time1)/(CLOCKS_PER_SEC));
time_store++;
}
return 0;
}
I think the problem is that you don't reset the starting prime:
int long sieve_prime = 2;
Currently that is outside your loop. On second thoughts... That's not the problem. Has this code been edited to incorporate the suggestions in Mats Petersson's answer? I just corrected the bad indentation.
Anyway, for the other part of your question, I suggest you use char instead of int for Num_Array. There is no use using int to store a boolean. By using char you should be able to store about 4 times as many values in the same amount of memory (assuming your int is 32-bit, which it probably is).
That means you could handle numbers up to almost 2 billion. Since you are using signed long as your type instead of unsigned long, that is approaching the numeric limits for your calculation anyway.
If you want to use even less memory, you could use std::bitset, but be aware that performance could be significantly impaired.
By the way, you should declare your timing array at the top of main:
float Time_Array[100];
Putting it inside the loop just before it is used is a bit whack.
Oh, and just in case you're interested, here is my own implementation of the sieve which, personally, I find easier to read than yours....
std::vector<char> isPrime( N, 1 );
for( int i = 2; i < N; i++ )
{
if( !isPrime[i] ) continue;
for( int x = i*2; x < N; x+=i ) isPrime[x] = 0;
}
This section of code is supposed to go inside your loop:
int *Num_Array = new int[MAX_NUM_ARRAY];
std::fill_n(Num_Array, MAX_NUM_ARRAY, 3);
Num_Array [0] = 1;
Num_Array [1] = 1;
Edit: and this one needs be in the loop too:
int long sieve_prime_constant = 0;
When I run this on my machine, it takes 0.2s per loop. If I add two zeros to the MAX_NUM_ARRAY, it takes 4.6 seconds per iteration (up to the 20th loop, I got bored waiting longer than 1.5 minute)
Agree with the earlier comments. If you really want to juice things up you don't store an array of all possible values (as int, or char), but only keep the primes. Then you test each subsequent number for divisibility through all primes found so far. Now you are only limited by the number of primes you can store. Of course, that's not really the algorithm you wanted to implement any more... but since it would be using integer division, it's quite fast. Something like this:
int myPrimes[MAX_PRIME];
int pCount, ii, jj;
ii = 3;
myPrimes[0]=2;
for(pCount=1; pCount<MAX_PRIME; pCount++) {
for(jj = 1; jj<pCount; jj++) {
if (ii%myPrimes[jj]==0) {
// not a prime
ii+=2; // never test even numbers...
jj = 1; // start loop again
}
}
myPrimes[pCount]=ii;
}
Not really what you were asking for, but maybe it is useful.

Using pow() for large number

I am trying to solve a problem, a part of which requires me to calculate (2^n)%1000000007 , where n<=10^9. But my following code gives me output "0" even for input like n=99.
Is there anyway other than having a loop which multilplies the output by 2 every time and finding the modulo every time (this is not I am looking for as this will be very slow for large numbers).
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
int main()
{
unsigned long long gaps,total;
while(1)
{
cin>>gaps;
total=(unsigned long long)powf(2,gaps)%1000000007;
cout<<total<<endl;
}
}
You need a "big num" library, it is not clear what platform you are on, but start here:
http://gmplib.org/
this is not I am looking for as this will be very slow for large numbers
Using a bigint library will be considerably slower pretty much any other solution.
Don't take the modulo every pass through the loop: rather, only take it when the output grows bigger than the modulus, as follows:
#include <iostream>
int main() {
int modulus = 1000000007;
int n = 88888888;
long res = 1;
for(long i=0; i < n; ++i) {
res *= 2;
if(res > modulus)
res %= modulus;
}
std::cout << res << std::endl;
}
This is actually pretty quick:
$ time ./t
./t 1.19s user 0.00s system 99% cpu 1.197 total
I should mention that the reason this works is that if a and b are equivalent mod m (that is, a % m = b % m), then this equality holds multiple k of a and b (that is, the foregoing equality implies (a*k)%m = (b*k)%m).
Chris proposed GMP, but if you need just that and want to do things The C++ Way, not The C Way, and without unnecessary complexity, you may just want to check this out - it generates few warnings when compiling, but is quite simple and Just Works™.
You can split your 2^n into chunks of 2^m. You need to find: `
2^m * 2^m * ... 2^(less than m)
Number m should be 31 is for 32-bit CPU. Then your answer is:
chunk1 % k * chunk2 * k ... where k=1000000007
You are still O(N). But then you can utilize the fact that all chunk % k are equal except last one and you can make it O(1)
I wrote this function. It is very inefficient but it works with very large numbers. It uses my self-made algorithm to store big numbers in arrays using a decimal like system.
mpfr2.cpp
#include "mpfr2.h"
void mpfr2::mpfr::setNumber(std::string a) {
for (int i = a.length() - 1, j = 0; i >= 0; ++j, --i) {
_a[j] = a[i] - '0';
}
res_size = a.length();
}
int mpfr2::mpfr::multiply(mpfr& a, mpfr b)
{
mpfr ans = mpfr();
// One by one multiply n with individual digits of res[]
int i = 0;
for (i = 0; i < b.res_size; ++i)
{
for (int j = 0; j < a.res_size; ++j) {
ans._a[i + j] += b._a[i] * a._a[j];
}
}
for (i = 0; i < a.res_size + b.res_size; i++)
{
int tmp = ans._a[i] / 10;
ans._a[i] = ans._a[i] % 10;
ans._a[i + 1] = ans._a[i + 1] + tmp;
}
for (i = a.res_size + b.res_size; i >= 0; i--)
{
if (ans._a[i] > 0) break;
}
ans.res_size = i+1;
a = ans;
return a.res_size;
}
mpfr2::mpfr mpfr2::mpfr::pow(mpfr a, mpfr b) {
mpfr t = a;
std::string bStr = "";
for (int i = b.res_size - 1; i >= 0; --i) {
bStr += std::to_string(b._a[i]);
}
int i = 1;
while (!0) {
if (bStr == std::to_string(i)) break;
a.res_size = multiply(a, t);
// Debugging
std::cout << "\npow() iteration " << i << std::endl;
++i;
}
return a;
}
mpfr2.h
#pragma once
//#infdef MPFR2_H
//#define MPFR2_H
// C standard includes
#include <iostream>
#include <string>
#define MAX 0x7fffffff/32/4 // 2147483647
namespace mpfr2 {
class mpfr
{
public:
int _a[MAX];
int res_size;
void setNumber(std::string);
static int multiply(mpfr&, mpfr);
static mpfr pow(mpfr, mpfr);
};
}
//#endif
main.cpp
#include <iostream>
#include <fstream>
// Local headers
#include "mpfr2.h" // Defines local mpfr algorithm library
// Namespaces
namespace m = mpfr2; // Reduce the typing a bit later...
m::mpfr tetration(m::mpfr, int);
int main() {
// Hardcoded tests
int x = 7;
std::ofstream f("out.txt");
m::mpfr t;
for(int b=1; b<x;b++) {
std::cout << "2^^" << b << std::endl; // Hardcoded message
t.setNumber("2");
m::mpfr res = tetration(t, b);
for (int i = res.res_size - 1; i >= 0; i--) {
std::cout << res._a[i];
f << res._a[i];
}
f << std::endl << std::endl;
std::cout << std::endl << std::endl;
}
char c; std::cin.ignore(); std::cin >> c;
return 0;
}
m::mpfr tetration(m::mpfr a, int b)
{
m::mpfr tmp = a;
if (b <= 0) return m::mpfr();
for (; b > 1; b--) tmp = m::mpfr::pow(a, tmp);
return tmp;
}
I created this for tetration and eventually hyperoperations. When the numbers get really big it can take ages to calculate and a lot of memory. The #define MAX 0x7fffffff/32/4 is the number of decimals one number can have. I might make another algorithm later to combine multiple of these arrays into one number. On my system the max array length is 0x7fffffff aka 2147486347 aka 2^31-1 aka int32_max (which is usually the standard int size) so I had to divide int32_max by 32 to make the creation of this array possible. I also divided it by 4 to reduce memory usage in the multiply() function.
- Jubiman