Iterative Tower of Hanoi - GeeksforGeeks
Tower of Hanoi is a mathematical puzzle. It consists of three poles and a number of disks of different sizes which can slide onto any poles. The puzzle starts with the disk in a neat stack in ascending order of size in one pole, the smallest at the top thus making a conical shape. The objective of the puzzle is to move all the disks from one pole (say 'source pole') to another pole (say 'destination pole') with the help of third pole (say auxiliary pole).
The puzzle has the following two rules:
1. You can't place a larger disk onto smaller disk
2. Only one disk can be moved at a time
https://hellosmallworld123.wordpress.com/2014/05/10/tower-of-hanoi-using-stacks/
For an even number of disks:
make the legal move between pegs A and B
make the legal move between pegs A and C
make the legal move between pegs B and C
repeat until complete
For an odd number of disks:
make the legal move between pegs A and C
make the legal move between pegs A and B
make the legal move between pegs C and B
repeat until complete
In each case, a total of 2n-1 moves are made.
http://simpledeveloper.com/towers-of-hanoi/
http://www.java2s.com/Tutorial/Java/0100__Class-Definition/TheTowersofHanoi.htm
moveDisks(int n, Tower origin, Tower destination, Tower buffer) {
/* Base case */
if (n <= 0) return;
/* move top n - 1 disks from origin to buffer, using destination as a buffer. */
moveDisks(n - 1, origin, buffer, destination);
/* move top from origin to destination
moveTop(origin, destination);
/* move top n - 1 disks from buffer to destination, using origin as a buffer. */
moveDisks(n - 1, buffer, destination, origin);
}
void main(String[] args) {
int n = 3;
Tower[] towers = new Tower[n];
for (int i= 0; i < 3; i++) {
towers[i] = new Tower(i);
}
for (int i= n - 1; i >= 0; i--) {
towers[0].add(i);
}
towers[0].moveDisks(n, towers[2], towers[l]);
}
class Tower {
private Stack<Integer> disks;
private int index;
public Tower(int i) {
disks = new Stack<Integer>();
index = i;
}
public int index() {
return index;
}
public void add(int d) {
if (!disks.isEmpty() && disks.peek() <= d) {
System.out.println("Error placing disk" + d);
} else {
disks.push(d);
}
}
public void moveTopTo(Tower t) {
int top= disks.pop();
t.add(top);
}
public void moveDisks(int n, Tower destination, Tower buffer) {
if (n > 0) {
moveDisks(n - 1, buffer, destination);
moveTopTo( destination);
buffer.moveDisks(n - 1, destination, this);
}
}
}
Read full article from Iterative Tower of Hanoi - GeeksforGeeks
Tower of Hanoi is a mathematical puzzle. It consists of three poles and a number of disks of different sizes which can slide onto any poles. The puzzle starts with the disk in a neat stack in ascending order of size in one pole, the smallest at the top thus making a conical shape. The objective of the puzzle is to move all the disks from one pole (say 'source pole') to another pole (say 'destination pole') with the help of third pole (say auxiliary pole).
The puzzle has the following two rules:
1. You can't place a larger disk onto smaller disk
2. Only one disk can be moved at a time
For n disks, total 2n – 1 moves are required. T(n)=2T(n-1)+1
http://en.wikipedia.org/wiki/Continuation-passing_style
For an even number of disks:
- make the legal move between pegs A and B
- make the legal move between pegs A and C
- make the legal move between pegs B and C
- repeat until complete
For an odd number of disks:
- make the legal move between pegs A and C
- make the legal move between pegs A and B
- make the legal move between pegs C and B
- repeat until complete
1. Calculate the total number of moves required i.e. "pow(2, n)
- 1" here n is number of disks.
2. If number of disks (i.e. n) is even then interchange destination
pole and auxiliary pole.
3. for i = 1 to total number of moves:
if i%3 == 1:
legal movement of top disk between source pole and
destination pole
if i%3 == 2:
legal movement top disk between source pole and
auxiliary pole
if i%3 == 0:
legal movement top disk between auxiliary pole
and destination pole
// Function to implement legal movement between// two polesvoid moveDisksBetweenTwoPoles(struct Stack *src, struct Stack *dest, char s, char d){ int pole1TopDisk = pop(src); int pole2TopDisk = pop(dest); // When pole 1 is empty if (pole1TopDisk == INT_MIN) { push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When pole2 pole is empty else if (pole2TopDisk == INT_MIN) { push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } // When top disk of pole1 > top disk of pole2 else if (pole1TopDisk > pole2TopDisk) { push(src, pole1TopDisk); push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When top disk of pole1 < top disk of pole2 else { push(dest, pole2TopDisk); push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); }}//Function to show the movement of disksvoid moveDisk(char fromPeg, char toPeg, int disk){ printf("Move the disk %d from \'%c\' to \'%c\'\n", disk, fromPeg, toPeg);}//Function to implement TOH puzzlevoid tohIterative(int num_of_disks, struct Stack *src, struct Stack *aux, struct Stack *dest){ int i, total_num_of_moves; char s = 'S', d = 'D', a = 'A'; //If number of disks is even, then interchange //destination pole and auxiliary pole if (num_of_disks % 2 == 0) { char temp = d; d = a; a = temp; } total_num_of_moves = pow(2, num_of_disks) - 1; //Larger disks will be pushed first for (i = num_of_disks; i >= 1; i--) push(src, i); for (i = 1; i <= total_num_of_moves; i++) { if (i % 3 == 1) moveDisksBetweenTwoPoles(src, dest, s, d); else if (i % 3 == 2) moveDisksBetweenTwoPoles(src, aux, s, a); else if (i % 3 == 0) moveDisksBetweenTwoPoles(aux, dest, a, d); }}https://hellosmallworld123.wordpress.com/2014/05/10/tower-of-hanoi-using-stacks/
For an even number of disks:
make the legal move between pegs A and B
make the legal move between pegs A and C
make the legal move between pegs B and C
repeat until complete
For an odd number of disks:
make the legal move between pegs A and C
make the legal move between pegs A and B
make the legal move between pegs C and B
repeat until complete
In each case, a total of 2n-1 moves are made.
public class TowerOfHanoi { Stack<Integer> A = new Stack<Integer>(); Stack<Integer> B = new Stack<Integer>(); Stack<Integer> C = new Stack<Integer>(); int diskNum; public TowerOfHanoi(int n) { for (int i = n; i >= 1; i--) { A.push(i); } diskNum = n; } public int solve() { int moves = 0; if (diskNum%2 == 1) {// odd number while(C.size() != diskNum) { if (C.isEmpty() || (!A.isEmpty() && A.peek() < C.peek())) { C.push(A.pop()); } else { A.push(C.pop()); } moves++; if (C.size() == diskNum) break; if (B.isEmpty() || (!A.isEmpty() && A.peek() < B.peek())) { B.push(A.pop()); } else { A.push(B.pop()); } moves++; if (B.isEmpty() || (!C.isEmpty() && C.peek() < B.peek())) { B.push(C.pop()); } else { C.push(B.pop()); } moves++; } } else { // even number while(C.size() != diskNum) { if (B.isEmpty() || (!A.isEmpty() && A.peek() < B.peek())) { B.push(A.pop()); } else { A.push(B.pop()); } moves++; if (C.isEmpty() || (!A.isEmpty() && A.peek() < C.peek())) { C.push(A.pop()); } else { A.push(C.pop()); } moves++; if (C.size() == diskNum) break; if (B.isEmpty() || (!C.isEmpty() && C.peek() < B.peek())) { B.push(C.pop()); } else { C.push(B.pop()); } moves++; } } return moves; } }http://simpledeveloper.com/towers-of-hanoi/
If n is 1, move disk 1 from origin to destination. Otherwise:
- Move n – 1 disks (one at a time) from origin to temporary
- Move disk n from origin to destination
- Move n – 1 disks (one at a time) from temporary to destination
http://www.java2s.com/Tutorial/Java/0100__Class-Definition/TheTowersofHanoi.htm
public static void doTowers(int topN, char from, char inter, char to) { if (topN == 1){ System.out.println("Disk 1 from " + from + " to " + to); }else { doTowers(topN - 1, from, to, inter); System.out.println("Disk " + topN + " from " + from + " to " + to); doTowers(topN - 1, inter, from, to); }http://www.ideserve.co.in/learn/tower-of-hanoi-algorithm
| 4 | public void solve(int n, String srcTower, String intermediateTower, String destTower) |
| 5 | { |
| 6 | if (n == 1) |
| 7 | { |
| 8 | System.out.println("Move topmost disk from " + srcTower + " to " + destTower); |
| 9 | |
| 10 | steps += 1; |
| 11 | |
| 12 | return; |
| 13 | } |
| 14 | |
| 15 | |
| 16 | solve(n-1, srcTower, destTower, intermediateTower); |
| 17 | |
| 18 | |
| 19 | solve(1, srcTower, intermediateTower, destTower); |
| 20 | |
| 21 | |
| 22 | solve(n-1, intermediateTower, srcTower, destTower); |
| 23 | } |
moveDisks(int n, Tower origin, Tower destination, Tower buffer) {
/* Base case */
if (n <= 0) return;
/* move top n - 1 disks from origin to buffer, using destination as a buffer. */
moveDisks(n - 1, origin, buffer, destination);
/* move top from origin to destination
moveTop(origin, destination);
/* move top n - 1 disks from buffer to destination, using origin as a buffer. */
moveDisks(n - 1, buffer, destination, origin);
}
void main(String[] args) {
int n = 3;
Tower[] towers = new Tower[n];
for (int i= 0; i < 3; i++) {
towers[i] = new Tower(i);
}
for (int i= n - 1; i >= 0; i--) {
towers[0].add(i);
}
towers[0].moveDisks(n, towers[2], towers[l]);
}
class Tower {
private Stack<Integer> disks;
private int index;
public Tower(int i) {
disks = new Stack<Integer>();
index = i;
}
public int index() {
return index;
}
public void add(int d) {
if (!disks.isEmpty() && disks.peek() <= d) {
System.out.println("Error placing disk" + d);
} else {
disks.push(d);
}
}
public void moveTopTo(Tower t) {
int top= disks.pop();
t.add(top);
}
public void moveDisks(int n, Tower destination, Tower buffer) {
if (n > 0) {
moveDisks(n - 1, buffer, destination);
moveTopTo( destination);
buffer.moveDisks(n - 1, destination, this);
}
}
}
Read full article from Iterative Tower of Hanoi - GeeksforGeeks
