Added solution to problem 23

Benchmark.ts

@@ -32,7 +32,7 @@ enum BenchmarkOptions{runSpecific = 1, runAllShort, runAll, exit, size};
 
 
 export class Benchmark{
-	private static tooLong: number[] = [14, 15];
+	private static tooLong: number[] = [14, 15, 23];
 	//The driver function for the benchmark selection
 	public static benchmarkMenu(): void{
 		let selection: BenchmarkOptions = BenchmarkOptions.size;

ProblemSelection.ts

@@ -46,12 +46,13 @@ import { Problem19 } from "./Problems/Problem19";
 import { Problem20 } from "./Problems/Problem20";
 import { Problem21 } from "./Problems/Problem21";
 import { Problem22 } from "./Problems/Problem22";
+import { Problem23 } from "./Problems/Problem23";
 
 
 export class ProblemSelection{
 	//Holds the valid problem numbers
 	public static PROBLEM_NUMBERS: number[] = [  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
-												21, 22];
+												21, 22, 23];
 
 	//Returns the problem corresponding to the given problem number
 	public static getProblem(problemNumber: number): Problem{
@@ -79,6 +80,7 @@ export class ProblemSelection{
 			case 20: problem = new Problem20(); break;
 			case 21: problem = new Problem21(); break;
 			case 22: problem = new Problem22(); break;
+			case 23: problem = new Problem23(); break;
 		}
 		return problem;
 	}

Problems/Problem23.ts

@@ -0,0 +1,142 @@
+//ProjectEulerTS/Problems/Problem23.ts
+//Matthew Ellison
+// Created: 04-08-21
+//Modified: 04-08-21
+//Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers
+//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/typescriptClasses
+/*
+	Copyright (C) 2021  Matthew Ellison
+
+	This program is free software: you can redistribute it and/or modify
+	it under the terms of the GNU Lesser General Public License as published by
+	the Free Software Foundation, either version 3 of the License, or
+	(at your option) any later version.
+
+	This program is distributed in the hope that it will be useful,
+	but WITHOUT ANY WARRANTY; without even the implied warranty of
+	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+	GNU Lesser General Public License for more details.
+
+	You should have received a copy of the GNU Lesser General Public License
+	along with this program.  If not, see <https://www.gnu.org/licenses/>.
+*/
+
+
+import { getDivisors, getSum } from "../../../Typescript/typescriptClasses/Algorithms";
+import { Unsolved } from "../Unsolved";
+import { Problem } from "./Problem";
+
+
+export class Problem23 extends Problem{
+	//Variables
+	//Static variables
+	private static MAX_NUM: number = 28123;	//The largest number to be checked
+	//Instance variables
+	private divisorSums: number[];	//This gives the sum of the divisors at subscripts
+	private sum: number;	//The sum of all the numbers we are looking for
+
+	//Functions
+	//Constructor
+	public constructor(){
+		super(`Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers`);
+		this.divisorSums = [];
+		this.reserveArray();
+		this.sum = 0;
+	}
+	//Operational functions
+	//Reserve the size of the array to speed up insertion
+	private reserveArray(): void{
+		//It is faster to reserve the appropriate amount of ram now
+		//Make sure every element has a 0 in it's location
+		while(this.divisorSums.length <= Problem23.MAX_NUM){
+			this.divisorSums.push(0);
+		}
+	}
+	//Solve the problem
+	public solve(): void{
+		//If the problem has already been solved do nothing and end the function
+		if(this.solved){
+			return;
+		}
+
+		//Start the timer
+		this.timer.start();
+
+		//Get the sum of the divisors of all numbers < MAX_NUM
+		for(let cnt = 1;cnt < Problem23.MAX_NUM;++cnt){
+			let div: number[] = getDivisors(cnt);
+			//Remove the last element, which is the number itself. This gives us the propper divisors
+			if(div.length > 1){
+				div.pop();
+			}
+			this.divisorSums[cnt] = getSum(div);
+		}
+
+		//Get the abundant numbers
+		let abund: number[] = [];
+		for(let cnt = 0;cnt < this.divisorSums.length;++cnt){
+			if(this.divisorSums[cnt] > cnt){
+				abund.push(cnt);
+			}
+		}
+
+		//Check if each number can be the sum of 2 abundant numbers and add to the sum if no
+		for(let cnt = 1;cnt < Problem23.MAX_NUM;++cnt){
+			if(!this.isSum(abund, cnt)){
+				this.sum += cnt;
+			}
+		}
+
+		//Stop the timer
+		this.timer.stop();
+
+		//Throw a flag to show the problem is solved
+		this.solved = true;
+	}
+	//A function that returns true if num can be created by adding two elements from abund and false if it cannot
+	private isSum(abund: number[], num: number): boolean{
+		let sum: number = 0;
+		//Pick a number for the first part of the sum
+		for(let firstNum = 0;firstNum < abund.length;++firstNum){
+			//Pick a number for the second part of the sum
+			for(let secondNum = firstNum;secondNum < abund.length;++secondNum){
+				sum = abund[firstNum] + abund[secondNum];
+				if(sum == num){
+					return true;
+				}
+				else if(sum > num){
+					break;
+				}
+			}
+		}
+		//If you have run through the entire list and did not find a sum then it is false
+		return false;
+	}
+	//Reset the problem so it can be run again
+	public reset(): void{
+		super.reset();
+		this.divisorSums = [];
+		this.reserveArray();
+		this.sum = 0;
+	}
+	//Gets
+	//Returns the result of solving the problem
+	public getResult(): string{
+		if(!this.solved){
+			throw new Unsolved();
+		}
+		return `The answer is ${this.sum}`;
+	}
+	//Returns the sum of the numbers asked for
+	public getSum(): number{
+		if(!this.solved){
+			throw new Unsolved();
+		}
+		return this.sum;
+	}
+}
+
+/* Results:
+The answer is 4179871
+It took an average of 7.798 seconds to run this problem through 100 iterations
+*/