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;
}


Labels:

Comments

Post a Comment

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...
Post a Comment
Read more

Function to sort an array using bubble sort

Quick and dirty way of sorting an array is bubble sort. It is very easy to write and follow. But please keep in mind that it is not at all effecient. #include<iostream> using std::cin; using std::cout; void readArray(int arr[],int sz); void printArray(int arr[],int sz); void sortArray(int arr[],int sz); void swap(int &a,int &b); int main() {    int sz;    cout<<"Size of the array=";    cin>>sz;    int arr[sz];    readArray(arr,sz);     sortArray(arr,sz);   cout<<"Sorted array is ";   printArray(arr,sz); } void readArray(int arr[],int sz) {  for(int i=0;i<sz;i++)    {       cout<<"arr["<<i<<"]=";       cin>>arr[i];   } } void printArray(int arr[],int sz) {  for(int i=0;i<sz;i++)    {       cout<<"arr["<<i<<"]=";    ...
Post a Comment
Read more

Merge two binary search trees

How do you merge two binary search trees? I googled about the solutions. Most solutions told me to convert both trees into linked lists. Merge the lists. Then create a tree from the elements of the list. But why lists? Why can't we store the elements in an array? Because if the data of the tree is larger - not just integer keys, array manipulation becomes difficult. But again, we need not convert both the trees into lists. We can convert one tree into list - a doubly linked list. Then insert the elements of this list into the other tree. I tried this approach. To convert a tree into a sorted doubly linked list Create a doubly linked list. Let the prev and next links of nodes in this list be called left and right respectively. This way we can directly use the binary tree nodes in the list. Use a static variable previousnode  call the function recursively for left child of current node. link current node to the previousnode set next pointer of previousnode to curre...
Post a Comment
Read more