using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace Pi_Printer { class Program { static void Main(string[] args) { int[] hs = new int[250]; string[] ds = new string[250]; for (int k = 0; k < 25; k++) { double temp = sFinal(10 * k); //see this function for the pure maths references for (int i = 0; i < 10; i++) { if (temp < 0) { temp -= Math.Floor(temp); } //abrogates negative values if (i == 10 * k && temp > 1) { temp -= Math.Floor(temp); } if (temp > 1) { temp -= Math.Floor(temp); hs[10 * k + i] = 0; } //returns zero if greater than 1, and NOT on the first iteration of the sequence temp *= 16; hs[10 * k + i] = Convert.ToInt32(Math.Floor(temp)); temp -= Math.Floor(temp); } } //base 16 "raw output", truncated at 30 chars Console.WriteLine("Base 16:"); Console.Write("3."); for (int i = 0; i < 30; i++) { Console.Write(hexParse(hs[i])); if ((i + 1) % 5 == 0) { Console.Write(" "); } } /*the rest of the code is dedicated to the arduous task of converting a base 16 decimal to a base 10 one*/ //remove the exponent "E" notation, and preformat with a corresponding number of zeros (10 * k + i - 1). this makes tabular addition possible for (int k = 0; k < 25; k++) { for (int i = 0; i < 10; i++) { double d = hs[10 * k + i] / (Math.Pow(16, 10 * k + i + 1)); string num = null; if (d.ToString().Contains("E")) { int exp = Convert.ToInt32(d.ToString().Substring(d.ToString().IndexOf("E") + 1)); //grabs the exponent number from the end of the double num = d.ToString().Substring(0, d.ToString().IndexOf("E")); //clips the E and number off the end, leaving a decimal num = num.Replace(".", ""); //removes the decimal point for (int j = 1; j < -exp; j++) //places the appropriate number of zeros before, making tabular addition possible { num = num.Insert(0, "0"); } d *= Math.Pow(10, -exp); //preformats } else { num = d.ToString(); num = num.TrimStart('0'); num = num.Replace(".", ""); } ds[10 * k + i] = num; } } //post-format remaining entries with zeros to make strings equal length, and make tabular addition possible for (int i = 0; i < 250; i++) { int deltaA = ds[249].ToString().Length - ds[i].ToString().Length; for (int k = 0; k < deltaA; k++) { ds[i] = ds[i].Insert(ds[i].Length, "0"); } } //peform termwise addition int dA = ds[249].ToString().Length; int[] termSums = new int[dA]; //since dA = 316, this one has much more than 250 elements. due to base 16 extending faster than base 10 for (int k = dA - 1; k >= 0; k--) { for (int i = 0; i < 249; i++) { termSums[dA - 1 - k] += (int)Char.GetNumericValue(ds[i][k]); } } //divide by 10, add result to next term, and save remainder for current term for (int k = 0; k < dA - 1; k++) { termSums[k + 1] += termSums[k] / 10; termSums[k] %= 10; } Array.Reverse(termSums); //reverse the order Console.WriteLine(); Console.WriteLine(); //outputs a processed base 10 format Console.WriteLine("Base 10:"); Console.Write("3."); for (int k = 0; k < 21; k++) { Console.Write(termSums[k]); if ((k + 1) % 5 == 0) { Console.Write(" "); } } Console.WriteLine(); Console.WriteLine(); Console.WriteLine("By exploiting the unlimited nature of strings this program calculates pi to 21 places which is 6 more than can be obtained using double values alone"); //Console.Read(); //waits for the user to exit the program } public static string hexParse(int h) { switch (h) { case 10: return "A"; case 11: return "B"; case 12: return "C"; case 13: return "D"; case 14: return "E"; case 15: return "F"; default: return h.ToString(); //if 9 or less } } //the so-called Bailey–Borwein–Plouffe formula for calculating arbitrary digits around 'd' of pi in base 16 public static double sFinal(int d) { return (4 * S(d, 1) - 2 * S(d, 4) - S(d, 5) - S(d, 6)); } public static double S(int d, int j) { double sum = 0; for (int k = 0; k <= d; k++) { double T = Math.Pow(16, d - k); double q = 8 * k + j; sum += (T % q) / q; } for (int k = d + 1; k < d + 40; k++) { double T = Math.Pow(16, d - k); double q = 8 * k + j; sum += T / q; } return sum; } } }

Pi Printer 2