Partition_Mikael

To download the codes, please visit our GitHub repository.
Please note that testing a too big size of number could exceed acceptable computation time for online IDEs.
So we recommend you to test the codes in your personal computer environment by installing an IDE.

* Thank you to SageMath for offering an online IDE.

Hardy-Ramanujan-Rademacher formula

* Thank you to JDoodle for offering an online IDE.

Partition Function, Σps(m,2) ver.

C

#include <stdio.h> #include <time.h> long first2terms(long N); long from_3rd_term = 0; long A(long n, long k); long B(long n, long k); long C(long n); long D(long n); int main(void) { double N = 120; // Please input a natural number "N" for P(N) at here. clock_t start, end; double running_time; start = clock(); long n = N-6, k, partition; if(N>=0 && N-(int)N == 0){ for(n; n>=0; n-=2){ k=3+((N-6)-n)/2; A(n,k); } partition = first2terms(N) + from_3rd_term; printf("%ld\n",partition); end = clock(); running_time = ((double)(end - start))/CLOCKS_PER_SEC; printf("%s%lf%s\n", "running time: ", running_time, " seconds"); } else{ printf("%s\n", "undefined"); } return 0; } long first2terms(long N){ if(N==0){ return 1; } else{ long secondterm = 0; if(N%2==1){ secondterm += ((N-3)/2)*((N-3)/2)+((N-3)/2); } else{ secondterm += (N/2-1)*(N/2-1); } return N + secondterm; } } long A(long n, long k){ if(n>=k && k>=3){ return C(n) + A(n-k, k) + B(n,k); } else if(n>=3 && n<k){ k=n; return C(n) + A(n-k, k) + B(n,k); } else{ return C(n); } } long B(long n, long k){ while(k>3){ k--; A(n-k,k); } return 0; } long C(long n){ from_3rd_term += D(n); return from_3rd_term; } long D(long n){ long result = 0; if(n%2==1){ result += ((n+1)/2)*((n+1)/2)+((n+1)/2); } else{ result += (n/2+1)*(n/2+1); } return result; }

C++

#include <iostream> #include <chrono> long first2terms(long N); long from_3rd_term = 0; long A(long n, long k); long B(long n, long k); long C(long n); long D(long n); int main(void) { auto N = 120; // Please input a natural number "N" for P(N) at here. auto start = std::chrono::steady_clock::now(); long n = N-6, k, partition; if(N>=0 && N-(int)N == 0){ for(n; n>=0; n-=2){ k=3+((N-6)-n)/2; A(n,k); } partition = first2terms(N) + from_3rd_term; std::cout << partition << '\n'; auto end = std::chrono::steady_clock::now(); auto diff = end - start; std::cout << "running time: " << std::chrono::duration<double>(diff).count() << " seconds" << '\n'; } else{ std::cout << "undefined" << '\n'; } return 0; } long first2terms(long N){ if(N==0){ return 1; } else{ long secondterm = 0; if(N%2==1){ secondterm += ((N-3)/2)*((N-3)/2)+((N-3)/2); } else{ secondterm += (N/2-1)*(N/2-1); } return N + secondterm; } } long A(long n, long k){ if(n>=k && k>=3){ return C(n) + A(n-k, k) + B(n,k); } else if(n>=3 && n<k){ k=n; return C(n) + A(n-k, k) + B(n,k); } else{ return C(n); } } long B(long n, long k){ while(k>3){ k--; A(n-k,k); } return 0; } long C(long n){ from_3rd_term += D(n); return from_3rd_term; } long D(long n){ long result = 0; if(n%2==1){ result += ((n+1)/2)*((n+1)/2)+((n+1)/2); } else{ result += (n/2+1)*(n/2+1); } return result; }

Python

import timeit timer_start = timeit.default_timer() N = 100 # Please input a natural number "N" for P(N) at here. try: def first2terms(N): if N == 0: return 1 else: if N % 2 == 1: secondterm = ((N-3)/2)*((N-3)/2)+((N-3)/2) else: secondterm = (N/2-1)*(N/2-1) return N + secondterm def from_3rd_term(N): n = N-6 result = 0 for n in range(n, -1, -2): k = 3+((N-6)-n)/2 result += A(n, k) return result def A(n, k): if n >= k and k >= 3: return C(n) + A(n-k, k) + B(n, k) elif n >= 3 and n < k: k = n return C(n) + A(n-k, k) + B(n, k) else: return C(n) def B(n, k): send_back = 0 while k > 3: k -= 1 send_back += A(n-k, k) return send_back def C(n): c_result = 0 c_result += D(n) return c_result def D(n): if n % 2 == 1: return ((n+1)/2)*((n+1)/2)+((n+1)/2) else: return (n/2+1)*(n/2+1) partition = int(first2terms(N) + from_3rd_term(N)) timer_stop = timeit.default_timer() running_time = round(timer_stop - timer_start, 6) if __name__ == '__main__': print(partition, "\n" "running time: ", running_time, " seconds") if N >= 0 else print("undefined") except: print("undefined")

JavaScript (Node.js)

let N = 120 // Please input a natural number "N" for P(N) at here. let startTime = performance.now() if (N >= 0 && N - parseInt(N) == 0) { function first2terms(N) { if (N == 0) { return 1; } else { let secondterm; if (N % 2 == 1) { secondterm = ((N - 3) / 2) * ((N - 3) / 2) + (N - 3) / 2; } else { secondterm = (N / 2 - 1) * (N / 2 - 1); } return N + secondterm; } } function from_3rd_term(N) { let n = N - 6; let result = 0; for (n; n >= 0; n -= 2) { let k = 3 + (N - 6 - n) / 2; result += A(n, k); } return result; } function A(n, k) { if (n >= k && k >= 3) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 3 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } function B(n, k) { let send_back = 0; while (k > 3) { k--; send_back += A(n - k, k); } return send_back; } function C(n) { let c_result = 0; c_result += D(n); return c_result; } function D(n) { let result = 0; if (n % 2 == 1) { result += ((n + 1) / 2) * ((n + 1) / 2) + (n + 1) / 2; } else { result += (n / 2 + 1) * (n / 2 + 1); } return result; } let partition = first2terms(N) + from_3rd_term(N); console.log(partition); let endTime = performance.now(); let duration = ((endTime - startTime) / 1000).toFixed(6); console.log(`running time: ${duration} seconds`); } else { console.log("undefined") }

Java

public class partition_mikael { public static void main(String[] args) { long N = 150; // Please input a natural number "N" for P(N) at here. if (N >= 0) { long startTime = System.nanoTime(); long partition = first2terms(N) + expansion.from_3rd_term(N); long endTime = System.nanoTime(); long totalTime = (endTime - startTime) / 1000000000; System.out.println(partition); System.out.println(totalTime + "seconds"); } else { System.out.println("undefined"); } } static long first2terms(long N) { if (N == 0) { return 1; } else { long second_term = 0; if (N % 2 == 1) { second_term += ((N - 3) / 2) * ((N - 3) / 2) + ((N - 3) / 2); } else { second_term += (N / 2 - 1) * (N / 2 - 1); } return N + second_term; } } } class expansion { static long from_3rd_term(long N) { long result = 0; for (long n = N - 6; n >= 0; n -= 2) { long k = 3 + ((N - 6) - n) / 2; result += A(n, k); } return result; } private static long A(long n, long k) { if (n >= k && k >= 3) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 3 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } private static long B(long n, long k) { long send_back = 0; while (k > 3) { k--; send_back += A(n - k, k); } return send_back; } private static long C(long n) { long c_result = 0; c_result += D(n); return c_result; } private static long D(long n) { long result = 0; if (n % 2 == 1) { result += ((n + 1) / 2) * ((n + 1) / 2) + ((n + 1) / 2); } else { result += (n / 2 + 1) * (n / 2 + 1); } return result; } }

C#

using System; namespace Partition { class I { public static void Main() { long N = 120; // Please input a natural number "N" for P(N) at here. if (N >= 0) { var watch = System.Diagnostics.Stopwatch.StartNew(); long partition = I.first2terms(N) + II.from_3rd_term(N); watch.Stop(); var elapsedMs = (watch.ElapsedMilliseconds) / 1000; Console.WriteLine(partition); Console.WriteLine(elapsedMs + "seconds"); } else { Console.WriteLine("undefined"); } } private static long first2terms(long N) { if (N == 0) { return 1; } else { long secondterm = 0; if (N % 2 == 1) { secondterm += ((N - 3) / 2) * ((N - 3) / 2) + ((N - 3) / 2); } else { secondterm += (N / 2 - 1) * (N / 2 - 1); } return N + secondterm; } } } class II { internal static long from_3rd_term(long N) { long result = 0; for (long n = N - 6; n >= 0; n -= 2) { long k = 3 + ((N - 6) - n) / 2; result += A(n, k); } return result; } private static long A(long n, long k) { if (n >= k && k >= 3) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 3 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } private static long B(long n, long k) { long send_back = 0; while (k > 3) { k--; send_back += A(n - k, k); } return send_back; } private static long C(long n) { long c_result = 0; c_result += D(n); return c_result; } private static long D(long n) { long result = 0; if (n % 2 == 1) { result += ((n + 1) / 2) * ((n + 1) / 2) + ((n + 1) / 2); } else { result += (n / 2 + 1) * (n / 2 + 1); } return result; } } }

revised_C

#include <stdio.h> #include <time.h> long first2terms(long N); long from_3rd_term = 0; long expansion(long n, long k); int main(void) { double N = 150; // Please input a natural number "N" for P(N) at here. clock_t start, end; double running_time; start = clock(); long n = N-6, k, partition; if(N>=0 && N-(int)N == 0){ for(n; n>=0; n-=2){ k=3+((N-6)-n)/2; expansion(n,k); } partition = first2terms(N) + from_3rd_term; printf("%ld\n",partition); end = clock(); running_time = ((double)(end - start))/CLOCKS_PER_SEC; printf("%s%lf%s\n", "running time: ", running_time, " seconds"); } else{ printf("%s\n", "undefined"); } return 0; } long first2terms(long N){ if(N==0){ return 1; } else{ long secondterm = 0; if(N%2==1){ secondterm += ((N-3)/2)*((N-3)/2)+((N-3)/2); } else{ secondterm += (N/2-1)*(N/2-1); } return N + secondterm; } } long expansion(long n, long k){ if(n>=k && k>=3){ for(k; k>=3; k--)expansion(n-k,k);} else if(n>=3 && n<k){k=n; for(k; k>=3; k--)expansion(n-k,k);} else{} if(n%2==1){ from_3rd_term += ((n+1)/2)*((n+1)/2)+((n+1)/2); } else{ from_3rd_term += (n/2+1)*(n/2+1); } return from_3rd_term; }

revised_C++

#include <iostream> #include <chrono> long first2terms(long N); long from_3rd_term = 0; long expansion(long n, long k); int main(void) { auto N = 150; // Please input a natural number "N" for P(N) at here. auto start = std::chrono::steady_clock::now(); long n = N-6, k, partition; if(N>=0 && N-(int)N == 0){ for(n; n>=0; n-=2){ k=3+((N-6)-n)/2; expansion(n,k); } partition = first2terms(N) + from_3rd_term; std::cout << partition << '\n'; auto end = std::chrono::steady_clock::now(); auto diff = end - start; std::cout << "running time: " << std::chrono::duration<double>(diff).count() << " seconds" << '\n'; } else{ std::cout << "undefined" << '\n'; } return 0; } long first2terms(long N){ if(N==0){ return 1; } else{ long secondterm = 0; if(N%2==1){ secondterm += ((N-3)/2)*((N-3)/2)+((N-3)/2); } else{ secondterm += (N/2-1)*(N/2-1); } return N + secondterm; } } long expansion(long n, long k){ if(n>=k && k>=3){ for(k; k>=3; k--)expansion(n-k,k);} else if(n>=3 && n<k){k=n; for(k; k>=3; k--)expansion(n-k,k);} else{} if(n%2==1){ from_3rd_term += ((n+1)/2)*((n+1)/2)+((n+1)/2); } else{ from_3rd_term += (n/2+1)*(n/2+1); } return from_3rd_term; }

Partition Function, Σps(m,3) ver.

C++

#include <iostream> #include <chrono> long first2terms(long N); long from_3rd_term = 0; long A(long n, long k); long B(long n, long k); long C(long n); long D(long n); int main(void) { auto N = 150; // Please input a natural number "N" for P(N) at here. auto start = std::chrono::steady_clock::now(); long n = N-6, k, partition; if(N>=0 && N-(int)N == 0){ for(n; n>=0; n-=2){ k=3+((N-6)-n)/2; A(n,k); } partition = first2terms(N) + from_3rd_term; std::cout << partition << '\n'; auto end = std::chrono::steady_clock::now(); auto diff = end - start; std::cout << "running time: " << std::chrono::duration<double>(diff).count() << " seconds" << '\n'; } else{ std::cout << "undefined" << '\n'; } return 0; } long first2terms(long N){ if(N==0){ return 1; } else{ long secondterm = 0; if(N%2==1){ secondterm += ((N-3)/2)*((N-3)/2)+((N-3)/2); } else{ secondterm += (N/2-1)*(N/2-1); } return N + secondterm; } } long A(long n, long k){ if(n>=k && k>=4){ return C(n) + A(n-k, k) + B(n,k); } else if(n>=4 && n<k){ k=n; return C(n) + A(n-k, k) + B(n,k); } else{ return C(n); } } long B(long n, long k){ while(k>4){ k--; A(n-k,k); } return 0; } long C(long n){ from_3rd_term += D(n); return from_3rd_term; } long D(long n){ double result = 0, a; if(n%6==1){ a = (n-1)/6; from_3rd_term += 6*a*a*a+(27*a*a)/2+(19*a)/2+2; } else if(n%6==2){ a = (n-2)/6; from_3rd_term += 6*a*a*a+(33*a*a)/2+(29*a)/2+4; } else if(n%6==3){ a = (n-3)/6; from_3rd_term += 6*a*a*a+(39*a*a)/2+(41*a)/2+7; } else if(n%6==4){ a = (n-4)/6; from_3rd_term += 6*a*a*a+(45*a*a)/2+(55*a)/2+11; } else if(n%6==5){ a = (n-5)/6; from_3rd_term += 6*a*a*a+(51*a*a)/2+(71*a)/2+16; } else { a = n/6; from_3rd_term += 6*a*a*a+(21*a*a)/2+(11*a)/2+1; } return result; }

revised_C++

#include <iostream> #include <chrono> long first2terms(long N); long from_3rd_term = 0; long expansion(long n, long k); int main(void) { auto N = 150; // Please input a natural number "N" for P(N) at here. auto start = std::chrono::steady_clock::now(); long n = N-6, k, partition; if(N>=0 && N-(int)N == 0){ for(n; n>=0; n-=2){ k=3+((N-6)-n)/2; expansion(n,k); } partition = first2terms(N) + from_3rd_term; std::cout << partition << '\n'; auto end = std::chrono::steady_clock::now(); auto diff = end - start; std::cout << "running time: " << std::chrono::duration<double>(diff).count() << " seconds" << '\n'; } else{ std::cout << "undefined" << '\n'; } return 0; } long first2terms(long N){ if(N==0){ return 1; } else{ long secondterm = 0; if(N%2==1){ secondterm += ((N-3)/2)*((N-3)/2)+((N-3)/2); } else{ secondterm += (N/2-1)*(N/2-1); } return N + secondterm; } } long expansion(long n, long k){ if(n>=k && k>=4){ for(k; k>=4; k--)expansion(n-k,k);} else if(n>=4 && n<k){k=n; for(k; k>=4; k--)expansion(n-k,k);} else{} double a; if(n%6==1){ a = (n-1)/6; from_3rd_term += 6*a*a*a+(27*a*a)/2+(19*a)/2+2; } else if(n%6==2){ a = (n-2)/6; from_3rd_term += 6*a*a*a+(33*a*a)/2+(29*a)/2+4; } else if(n%6==3){ a = (n-3)/6; from_3rd_term += 6*a*a*a+(39*a*a)/2+(41*a)/2+7; } else if(n%6==4){ a = (n-4)/6; from_3rd_term += 6*a*a*a+(45*a*a)/2+(55*a)/2+11; } else if(n%6==5){ a = (n-5)/6; from_3rd_term += 6*a*a*a+(51*a*a)/2+(71*a)/2+16; } else { a = n/6; from_3rd_term += 6*a*a*a+(21*a*a)/2+(11*a)/2+1; } return from_3rd_term; }

JavaScript (Node.js)

let N = 150 // Please input a natural number "N" for P(N) at here. let startTime = performance.now(); if (N >= 0 && N - parseInt(N) == 0) { function first2terms(N) { if (N == 0) { return 1; } else { let secondterm; if (N % 2 == 1) { secondterm = ((N - 3) / 2) * ((N - 3) / 2) + (N - 3) / 2; } else { secondterm = (N / 2 - 1) * (N / 2 - 1); } return N + secondterm; } } function from_3rd_term(N) { let n = N - 6; let result = 0; for (n; n >= 0; n -= 2) { let k = 3 + (N - 6 - n) / 2; result += A(n, k); } return result; } function A(n, k) { if (n >= k && k >= 4) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 4 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } function B(n, k) { let send_back = 0; while (k > 4) { k--; send_back += A(n - k, k); } return send_back; } function C(n) { let c_result = 0; c_result += D(n); return c_result; } function D(n) { let result = 0, a; if (n % 6 == 1) { a = (n - 1) / 6; result += 6 * a * a * a + (27 * a * a) / 2 + (19 * a) / 2 + 2; } else if (n % 6 == 2) { a = (n - 2) / 6; result += 6 * a * a * a + (33 * a * a) / 2 + (29 * a) / 2 + 4; } else if (n % 6 == 3) { a = (n - 3) / 6; result += 6 * a * a * a + (39 * a * a) / 2 + (41 * a) / 2 + 7; } else if (n % 6 == 4) { a = (n - 4) / 6; result += 6 * a * a * a + (45 * a * a) / 2 + (55 * a) / 2 + 11; } else if (n % 6 == 5) { a = (n - 5) / 6; result += 6 * a * a * a + (51 * a * a) / 2 + (71 * a) / 2 + 16; } else { a = n / 6; result += 6 * a * a * a + (21 * a * a) / 2 + (11 * a) / 2 + 1; } return result; } let partition = first2terms(N) + from_3rd_term(N); console.log(partition); let endTime = performance.now(); let duration = ((endTime - startTime) / 1000).toFixed(6); console.log(`running time: ${duration} seconds`); } else { console.log("undefined"); }

Partition by Size Function, ps(n,2) ver.

Python

import timeit timer_start = timeit.default_timer() (n,k) = (50,40) # Please input (n,k) for the ps(n,k) function above here. # Note that when k=n, then ps(n,k) = P(n) holds, # where P(n) is the partition function. if(n-int(n) == 0 and k-int(k) == 0 and n >=0 and k >= 0): def partition_by_size(n, k): m=n result = 0 if n>=k and k>0: for n in range(n-1, n-k-1, -1): k=m-n if k == 1: result += 1 else: result += A(n,k) return result elif k>n and n != 0 and k>0: k=n for n in range(n-1, n-k-1, -1): k=m-n if k == 1: result += 1 else: result += A(n,k) return result else: return 1 def A(n, k): if n >= k and k >= 3: return C(n) + A(n-k, k) + B(n, k) elif n >= 3 and n < k: k = n return C(n) + A(n-k, k) + B(n, k) else: return C(n) def B(n, k): send_back = 0 while k > 3: k -= 1 send_back += A(n-k, k) return send_back def C(n): c_result = 0 c_result += D(n) return c_result def D(n): if n % 2 == 1: return (n+1)/2 else: return (n/2)+1 ps_function = int(partition_by_size(n, k)) timer_stop = timeit.default_timer() running_time = round(timer_stop - timer_start, 6) if __name__ == '__main__': print("ps(n,k): ", ps_function, "\n" "running time: ", running_time, " seconds") else: print("undefined")

Java

public class partition_by_size { public static void main(String[] args) { long n = 100, k = 50; /* Please input (n,k) for the ps(n,k) function above here. Note that when k>=n, then ps(n,k) = P(n) holds, where P(n) is the partition function. */ if (n >= 0 && k >= 0) { long startTime = System.nanoTime(); long result = expansion(n, k); long endTime = System.nanoTime(); long totalTime = (endTime - startTime) / 1000000000; System.out.println(result); System.out.println(totalTime + "seconds"); } else { System.out.println("undefined"); } } private static long expansion(long n, long k) { long m = n, lower_limit = n - k, result = 0; if (n >= k && k > 0) { for (n -= 1; n >= lower_limit; n -= 1) { k = m - n; if (k == 1) { result += 1; } else { result += A(n, k); } } return result; } else if (k > n && n != 0 && k > 0) { k = n; lower_limit = n - k; for (n -= 1; n >= lower_limit; n -= 1) { k = m - n; if (k == 1) { result += 1; } else { result += A(n, k); } } return result; } else { return 1; } } private static long A(long n, long k) { if (n >= k && k >= 3) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 3 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } private static long B(long n, long k) { long send_back = 0; while (k > 3) { k--; send_back += A(n - k, k); } return send_back; } private static long C(long n) { long c_result = 0; c_result += D(n); return c_result; } private static long D(long n) { long result = 0; if (n % 2 == 1) { result += (n + 1) / 2; } else { result += (n / 2) + 1; } return result; } }

JavaScript (Node.js)

let [n, k] = [100, 50]; /* Please input [n,k] for the ps(n,k) function above here. Note that when k>=n, then ps(n,k) = P(n) holds, where P(n) is the partition function. */ let startTime = performance.now(); if (n - parseInt(n) == 0 && k - parseInt(k) == 0 && n >= 0 && k >= 0) { function partition_by_size(n, k) { let m = n; let lowerlimit = n - k; let result = 0; if (n >= k && k > 0) { for (n -= 1; n >= lowerlimit; n -= 1) { k = m - n; if (k == 1) {result += 1;} else {result += A(n, k);} } return result; } else if (k > n && n != 0 && k > 0) { k = n, lowerlimit = n - k; for (n -= 1; n >= lowerlimit; n -= 1) { k = m - n; if (k == 1) {result += 1;} else {result += A(n, k);} } return result; } else { return 1; } } function A(n, k) { if (n >= k && k >= 3) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 3 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } function B(n, k) { let send_back = 0; while (k > 3) { k--; send_back += A(n - k, k); } return send_back; } function C(n) { let c_result = 0; c_result += D(n); return c_result; } function D(n) { let result = 0; if (n % 2 == 1) { result += (n + 1) / 2; } else { result += (n / 2) + 1; } return result; } let ps_function = partition_by_size(n, k); console.log(ps_function); let endTime = performance.now(); let duration = ((endTime - startTime) / 1000).toFixed(6); console.log(`running time: ${duration} seconds`); } else { console.log("undefined"); }

revised_C

#include <stdio.h> #include <time.h> long result = 0; long expansionI(long n, long k); long expansionII(long n, long k); int main(void) { double n = 100, k = 50; /* Please input (n,k) for the ps(n,k) function above here. Note that when k>=n, then ps(n,k) = P(n) holds, where P(n) is the partition function. */ clock_t start, end; double running_time; start = clock(); if (n-(int)n == 0 && k-(int)k == 0 && n >= 0 && k >= 0){ expansionI(n,k); printf("%ld\n", result); end = clock(); running_time = ((double)(end - start))/CLOCKS_PER_SEC; printf("%s%lf%s\n", "running time: ", running_time, " seconds"); } else{ printf("%s\n", "undefined"); } return 0; } long expansionI(long n, long k){ if (n >= k && k > 0) { long m=n, limitlength = n-k; for(n-=1; n>=limitlength; n-=1){ k=m-n; if (k == 1) {result += 1;} else {expansionII(n,k);} } } else if (k > n && n != 0 && k != 0) { long m=n, k=n, limitlength = n-k; for(n-=1; n>=limitlength; n-=1){ k=m-n; if (k == 1) {result += 1;} else {expansionII(n,k);} } } else { return result += 1; } return 0; } long expansionII(long n, long k){ if(n>=k && k>=3){ for(k; k>=3; k--)expansionII(n-k,k);} else if(n>=3 && n<k){k=n; for(k; k>=3; k--)expansionII(n-k,k);} else{} if(n%2==1){ result += (n+1)/2; } else{ result += (n/2)+1; } return result; }

revised_C++

#include <iostream> #include <chrono> long result = 0; long expansionI(long n, long k); long expansionII(long n, long k); int main(void) { double n = 100, k = 50; /* Please input (n,k) for the ps(n,k) function above here. Note that when k>=n, then ps(n,k) = P(n) holds, where P(n) is the partition function. */ auto start = std::chrono::steady_clock::now(); if (n-(int)n == 0 && k-(int)k == 0 && n >= 0 && k >= 0){ expansionI(n,k); std::cout << result << '\n'; auto end = std::chrono::steady_clock::now(); auto diff = end - start; std::cout << "running time: " << std::chrono::duration<double>(diff).count() << " seconds" << '\n'; } else{ std::cout << "undefined" << '\n'; } return 0; } long expansionI(long n, long k){ if (n >= k && k > 0) { long m=n, limitlength = n-k; for(n-=1; n>=limitlength; n-=1){ k=m-n; if (k == 1) {result += 1;} else{expansionII(n,k);} } } else if (k > n && n != 0 && k != 0) { long m=n, k=n, limitlength = n-k; for(n-=1; n>=limitlength; n-=1){ k=m-n; if (k == 1) {result += 1;} else{expansionII(n,k);} } } else { return result += 1; } return 0; } long expansionII(long n, long k){ if(n>=k && k>=3){ for(k; k>=3; k--)expansionII(n-k,k);} else if(n>=3 && n<k){k=n; for(k; k>=3; k--)expansionII(n-k,k);} else{} if(n%2==1){ result += (n+1)/2; } else{ result += (n/2)+1; } return result; }

Partition by Size Function, ps(n,3) ver.

Java

public class partition_by_size_m_3 { public static void main(String[] args) { long n = 100, k = 50; /* Please input (n,k) for the ps(n,k) function above here. Note that when k>=n, then ps(n,k) = P(n) holds, where P(n) is the partition function. */ if (n >= 0 && k >= 0) { long startTime = System.nanoTime(); long result = expansion(n, k); long endTime = System.nanoTime(); long totalTime = (endTime - startTime) / 1000000000; System.out.println(result); System.out.println(totalTime + "seconds"); } else { System.out.println("undefined"); } } private static long expansion(long n, long k) { long m = n, lower_limit = n - k, result = 0; if (n >= k && k > 0) { for (n -= 1; n >= lower_limit; n -= 1) { k = m - n; if (k == 1) { result += 1; } else if (k == 2) { if (n % 2 == 1) { result += (n + 1) / 2; } else { result += (n / 2) + 1; } } else { result += A(n, k); } } return result; } else if (k > n && n != 0 && k > 0) { k = n; lower_limit = n - k; for (n -= 1; n >= lower_limit; n -= 1) { k = m - n; if (k == 1) { result += 1; } else if (k == 2) { if (n % 2 == 1) { result += (n + 1) / 2; } else { result += (n / 2) + 1; } } else { result += A(n, k); } } return result; } else { return 1; } } private static long A(long n, long k) { if (n >= k && k >= 4) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 4 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } private static long B(long n, long k) { long send_back = 0; while (k > 4) { k--; send_back += A(n - k, k); } return send_back; } private static long C(long n) { long c_result = 0; c_result += D(n); return c_result; } private static double D(long n) { long result = 0, a; if (n % 6 == 1) { a = (n - 1) / 6; result += 3 * a * a + 4 * a + 1; } else if (n % 6 == 2) { a = (n - 2) / 6; result += 3 * a * a + 5 * a + 2; } else if (n % 6 == 3) { a = (n - 3) / 6; result += 3 * a * a + 6 * a + 3; } else if (n % 6 == 4) { a = (n - 4) / 6; result += 3 * a * a + 7 * a + 4; } else if (n % 6 == 5) { a = (n - 5) / 6; result += 3 * a * a + 8 * a + 5; } else { a = n / 6; result += 3 * a * a + 3 * a + 1; } return result; } }

JavaScript (Node.js)

let [n, k] = [100, 50]; /* Please input [n,k] for the ps(n,k) function above here. Note that when k>=n, then ps(n,k) = P(n), where P(n) is the partition function. */ let startTime = performance.now(); if (n - parseInt(n) == 0 && k - parseInt(k) == 0 && n >= 0 && k >= 0) { function partition_by_size(n, k) { let m = n; let lowerlimit = n - k; let result = 0; if (n >= k && k > 0) { for (n -= 1; n >= lowerlimit; n -= 1) { k = m - n; if (k == 1) {result += 1;} else if (k == 2) { if (n % 2 == 1) { result += (n + 1) / 2; } else { result += (n / 2) + 1; } } else {result += A(n, k);} } return result; } else if (k > n && n != 0 && k > 0) { k = n, lowerlimit = n - k; for (n -= 1; n >= lowerlimit; n -= 1) { k = m - n; if (k == 1) {result += 1;} else if (k == 2) { if (n % 2 == 1) { result += (n + 1) / 2; } else { result += (n / 2) + 1; } } else {result += A(n, k);} } return result; } else { return 1; } } function A(n, k) { if (n >= k && k >= 4) { return C(n) + A(n - k, k) + B(n, k); } else if (n >= 4 && n < k) { k = n; return C(n) + A(n - k, k) + B(n, k); } else { return C(n); } } function B(n, k) { let send_back = 0; while (k > 4) { k--; send_back += A(n - k, k); } return send_back; } function C(n) { let c_result = 0; c_result += D(n); return c_result; } function D(n) { let result = 0, a; if (n % 6 == 1) { a = (n - 1) / 6; result += 3 * a * a + 4 * a + 1; } else if (n % 6 == 2) { a = (n - 2) / 6; result += 3 * a * a + 5 * a + 2; } else if (n % 6 == 3) { a = (n - 3) / 6; result += 3 * a * a + 6 * a + 3; } else if (n % 6 == 4) { a = (n - 4) / 6; result += 3 * a * a + 7 * a + 4; } else if (n % 6 == 5) { a = (n - 5) / 6; result += 3 * a * a + 8 * a + 5; } else { a = n / 6; result += 3 * a * a + 3 * a + 1; } return result; } let ps_function = partition_by_size(n, k); console.log(ps_function); let endTime = performance.now(); let duration = ((endTime - startTime) / 1000).toFixed(6); console.log(`running time: ${duration} seconds`); } else { console.log("undefined"); }

revised_C

#include <stdio.h> #include <time.h> long result = 0; long expansionI(long n, long k); long expansionII(long n, long k); int main(void) { double n = 100, k = 50; /* Please input (n,k) for the ps(n,k) function above here. Note that when k>=n, then ps(n,k) = P(n) holds, where P(n) is the partition function. */ clock_t start, end; double running_time; start = clock(); if (n-(int)n == 0 && k-(int)k == 0 && n >= 0 && k >= 0){ expansionI(n,k); printf("%ld\n", result); end = clock(); running_time = ((double)(end - start))/CLOCKS_PER_SEC; printf("%s%lf%s\n", "running time: ", running_time, " seconds"); } else{ printf("%s\n", "undefined"); } return 0; } long expansionI(long n, long k){ if (n >= k && k > 0) { long m=n, limitlength = n-k; for(n-=1; n>=limitlength; n-=1){ k=m-n; if (k == 1){result += 1;} else if (k == 2){ if (n % 2 == 1) { result += (n+1) / 2; } else { result += (n/2) + 1; } } else{expansionII(n,k);} } } else if (k > n && n != 0 && k != 0) { long m=n, k=n, limitlength = n-k; for(n-=1; n>=limitlength; n-=1){ k=m-n; if (k == 1){result += 1;} else if (k == 2){ if (n % 2 == 1) { result += (n+1) / 2; } else { result += (n/2) + 1; } } else{expansionII(n,k);} } } else { return result += 1; } return 0; } long expansionII(long n, long k){ if(n>=k && k>=4){ for(k; k>=4; k--)expansionII(n-k,k);} else if(n>=4 && n<k){k=n; for(k; k>=4; k--)expansionII(n-k,k);} else{} long a; if (n % 6 == 1) { a = (n - 1) / 6; result += 3*a*a + 4*a + 1; } else if (n % 6 == 2) { a = (n - 2) / 6; result += 3*a*a + 5*a + 2; } else if (n % 6 == 3) { a = (n - 3) / 6; result += 3*a*a + 6*a + 3; } else if (n % 6 == 4) { a = (n - 4) / 6; result += 3*a*a + 7*a + 4; } else if (n % 6 == 5) { a = (n - 5) / 6; result += 3*a*a + 8*a + 5; } else { a = n / 6; result += 3*a*a + 3*a + 1; } return result; }