I need help with the following code. It builds and runs, but the ending conditions are whacky and I think there is an issue the moveDIsk function. Please help and thanks in advance!

// Peg.h #pragma once #include #include #include using namespace std; class Peg { private: void loadDisk(int numDisks); vector diskStack; // The stack of disks string pegName; // The name of the peg

public: Peg(string newName, int numDisks = 0); int getSize(); void setName(string newName); void printPeg(); // method to display what disks are on the peg string getName(); // accessor for the peg name int topDisk(); // method to return the value of the top disk // methods to add and remove disks from the pegs void removeDisk(); void addDisk(); int lastDisk();

// destructor ~Peg(); };

#include "Peg.h" #include #include #include #include using namespace std; Peg::Peg(string newName, int numDisks) { cout

}

void Peg::loadDisk(int numDisks) { // Load disks onto the pegs assert(numDisks >= 0); for (int i = numDisks; i > 0; i--) { diskStack.push_back(i); }

} // accessors string Peg::getName() { return pegName; }

int Peg::topDisk() // return the value of the top disk on the peg { assert(!diskStack.empty()); return diskStack.back(); } int Peg::getSize() { return diskStack.size(); }

//Mutators

void Peg::setName(string newName) { pegName = newName; }

void Peg::printPeg()

{ cout

void Peg::addDisk() { diskStack.push_back(this->getSize()); }

void Peg::removeDisk() {

return diskStack.pop_back(); } int Peg::lastDisk() { return diskStack.back();

}

//deconstructor Peg::~Peg() { cout

// File Name: Hanoi.cpp // Author: Tanner Crane // Description: Program that recursively solves the Towers of Hanoi with header file for declartions and .cpp for function definitions

#include #include #include "Peg.h" using namespace std;

// Global consts const int NUM_DISKS(1); // Number of disks to simulate

int hanoi(int numDisks, Peg& start, Peg& goal, Peg& temp); void moveDisk(Peg& from, Peg& to);

int main() { // Initialize variables Peg peg1("Peg1", NUM_DISKS), peg2("Peg2"), peg3("Peg3"); int numMoves(0);

// Introduce the program

cout

// Display the starting condition cout

// Solve Tower of Hanoi - move the disks from peg 1 to peg 3, using peg2 temporarily, getting the number of required moves returned cout

cout

cout

// Hanoi - recursive solution to the tower of hanoi problem int hanoi(int numDisks, Peg& start, Peg& goal, Peg& temp) { int numMoves(0);

// Only if there are disks to be moved if (numDisks > 0) { // First, move n-1 disks to the temporary peg, capturing the number of moves required numMoves = hanoi(numDisks - 1, start, temp, goal);

// Next, move the remaining disk to the goal peg; increment the move count cout

// Finally, move n-1 disks to the goal peg, capturing the number of moves required numMoves += hanoi(numDisks - 1, temp, goal, start); }

return(numMoves); } // Function to move a disk from one peg to another void moveDisk(Peg& from, Peg& to) { // Insure the inputs are valid: "from" has a disk, and that disk isn't smaller than the one on "to"(if there's one there). assert(from.getSize() > 0); // if (to.getSize() > 0) // { // assert(from.lastDisk()

The above are my outputs. As you can probably see, The moves required to move the disks and ending conditions aren't making sense.

- o G C:\Users\Tanner Crane source epos\hanoi Debug\hanoi.exe Welcome to Tanner's Tower of Hanoi simulator...this simulation will be running with 1 disks. Starting condition of the three pegs : Pegi has 1 disks: 1 Peg2 has 0 disks: Peg3 has 0 disks: Moves required to move 1 disks from Pegi to Peg3: Move disk 1 from Pegi to Peg3 Ending condition of the three pegs : 1 Pegi has 2 disks: 11 Peg2 has 0 disks: Pegs has 1 disks: 0 1A stack of 1 disks can be transferred in 1 moves. Thanks for using Tanner's Tower of Hanoi simulator! Press any key to continue ... - 0 x GN C:\Users\Tanner Crane source epos\hanoi Debug\hanoi.exe This is my Peg constructor This is my Peg constructor Welcome to Tanner's Tower of Hanoi simulator...this simulation will be running with 3 disks. Starting condition of the three pegs : Pegi has 3 disks: 321 arpeg2 has 0 disks: Pegs has 0 disks: Moves required to move 3 disks from Pegi to Peg3: Move disk 1 from Pegi to Peg3 Move disk 3 from Pegi to Peg2 Move disk from Pegs to Peg2 Move disk 4 from Pegi to Peg3 Move disk 1 from Peg2 to Peg1 Move disk 2 from Peg2 to Peg3 Move disk 6 from Pegi to Peg3 Ending condition of the three pegs : Pegi has 8 disks: 32134567 Peg2 has 4 disks: 0123 Pegs has 5 disks: 01234 A stack of 3 disks can be transferred in 7 moves. Thanks for using Tanner's Tower of Hanoi simulator! Press any key to continue ... 23 // Introduce the program