Skip to main content

Towers of Hanoi

Towers of Hanoi is a popular mathematical puzzle invented in 1883 by French mathematician Eduoardo Lucas. It is also a popular example in coding world because it is a typical example where a recursive solution is much easier than iterative solution.

In the puzzle, there are 3 rods and n discs of different sizes. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.

Gif file from wikipedia
Aim of the game is to transfer these  n discs to the 3rd rod, using 2nd rod as temporary location, but keeping in mind that
  • only one disc can be moved at a time.
  • a larger disc can not be placed above a smaller disc.

How do we solve this puzzle?

If some how we move n-1 discs to 2nd rod in correct order, we only have largest disc in 1st rod. And it can be moved to 3rd rod.

So now we need to find a method of moving n-1 discs from 1st rod to 2nd rod in correct order. Again we can use 3rd rod for intermediate storage. Why don't we move n-2 rods from 1st to 3rd rod and then we only have to move n-1th disc from 1st to 2nd rod.

So this is recursion. And we can write our algorithm as
  • recursively move n-1 discs from source rod to extra rod
  • move nth disc from source rod to target rod
  • recursively move n-1 discs from extra rod to target rod
And here is C code for the function.

void move_discs(int numdisks,char from,char to, char middle)
{
if(numdisks>0){
move_discs(numdisks-1,from,middle,to);
printf("Move disk %d from rod %c to rod %c\n",numdisks,from,to);
move_discs(numdisks-1,middle,to,from);
}
}

move_discs is called recursively for n-1 discs for moving from from rod to middle rod. And then nth rod is moved - which is represented by printf function.

next move_discs is called recursively for n-1 discs for moving from middle rod to to rod.

The base case is when n is 0, where recursion is not called.


For 3 discs, the output appears like this

aa@dell:~/dsPrograms$ ./a.out
Enter the number of discs3
Move disk 1 from rod A to rod C
Move disk 2 from rod A to rod B
Move disk 1 from rod C to rod B
Move disk 3 from rod A to rod C
Move disk 1 from rod B to rod A
Move disk 2 from rod B to rod C
Move disk 1 from rod A to rod C

And here we have driver program.


#include<stdio.h>
void move_discs(int numdisks,char from,char to, char middle)
{
if(numdisks>0){
move_discs(numdisks-1,from,middle,to);
printf("Move disk %d from rod %c to rod %c\n",numdisks,from,to);
move_discs(numdisks-1,middle,to,from);
}
}
int main()
{
int n;
printf("Enter the number of discs");
scanf("%d",&n);
move_discs(n,'A','C','B');
return 0;
}


Comments

Popular posts from this blog

Delete a node from doubly linked list

Deletion operation in DLL is simpler when compared to SLL. Because we don't have to go in search of previous node of to-be-deleted node.  Here is how you delete a node Link previous node of node of to-be-deleted to next node. Link next node of node of to-be-deleted to previous node. Free the memory of node of to-be-deleted Simple, isn't it. The code can go like this. prevnode = delnode->prev; nextnode = delnode->next; prevnode->next = nextnode; nextnode->prev = prevnode; free(delnode); And that is it. The node delnode is deleted. But we should always consider boundary conditions. What happens if we are trying to delete the first node or last node? If first node is to be deleted, its previous node is NULL. Hence step 3 should not be used.  And also, once head is deleted, nextnode becomes head . Similarly if last node is to be deleted, nextnode is NULL. Hence step 4 is as strict NO NO. And we should set prevnode to tail. After we put these things together, we have...

Program to delete a node from linked list

How do you remove a node from a linked list? If you have to delete a node, first you need to search the node. Then you should find its previous node. Then you should link the previous node to the next node. If node containing 8 has to be deleted, then n1 should be pointing to n2. Looks quite simple. Isn't it? But there are at least two special cases you have to consider. Of course, when the node is not found. If the node is first node of the list viz head. If the node to be deleted is head node, then if you delete, the list would be lost. You should avoid that and make the second node as the head node. So it becomes mandatory that you return the changed head node from the function.   Now let us have a look at the code. #include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; typedef struct node * NODEPTR; NODEPTR create_node ( int value) { NODEPTR temp = (NODEPTR) malloc( size...

Binary tree deletion

Do not get all scared and worried. It is not rocket science (or as I would like to call it - it is not regex). Remember deleting a node from linked list. When you deleted a node, you did the following 2 steps Free the memory of the node Link the previous node of the deleted node to the next node You will have to do these things in binary tree too. But the difficulty here is do you link the previous node - or parent node in tree terminology, to the left child of deleted node? Or to the right child? You can not link both children, as the parent already may have one child node. So before we solve this, let us categorize the deletion with the help of a diagram. Consider the cases The node to be deleted is a leaf node. That is, it does not have left or right child. In this case, link the parent to NULL in place of deleted node.  If you want to delete 13 - which is a leaf node, you set 14->left to NULL and free memory of node 13. If you want to delete 7 - which is also a leaf node and...