Algorithm for prefix in a tree

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November 11th, 2012, 01:16 PM

srsiddiqui

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Algorithm for prefix in a tree

hi everyone!

I have to write a algorithm for creating a prefix using stack.

A
/ \
B C
\\\
DEF

The prefix code is ABDEFC

Can anyone explain me how to do it with using stack.I know this is simple question but I am not able to write a algorithm for it.I do not how to write an algorithm.

After coming to ABDEF by what condition i use to go back to B and then go to C?

Thanks

November 11th, 2012, 01:59 PM

Lux Perpetua

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Use recursion. In pseudocode:

Code:

string prefix(tree x) {
    string val(x.root.name);
    for (int k = 0; k < x.num_children; ++k) {
        val = val + prefix(x.child(k));
    }
    return val;
}

Edit: Oh, sorry, I gather from your post that you don't want to use recursion, though technically, if you were to implement a recursive solution, it would implicitly be using a stack on the assembly level.

Nonrecursive pseudocode:

Code:

string prefix(tree x) {
    string val;
    stack<tree> the_stack;
    the_stack.push(x);
    while (!the_stack.empty()) {
        tree y(the_stack.top());
        the_stack.pop();
        val = val + y.root.name;
        for (int k = y.num_children - 1; k >= 0; --k) {
            the_stack.push(y.child(k));
        }
    }
    return val;
}

Comments on this post

ManiacDan agrees!

Last edited by Lux Perpetua : November 11th, 2012 at 02:10 PM. Reason: Misread OP

November 11th, 2012, 02:04 PM

srsiddiqui

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Thanks alot! but i have to use stack with loops inside.

Quote:

Originally Posted by Lux Perpetua

Use recursion. In pseudocode:

Code:

string prefix(tree x) {
    string val(x.root.name);
    for (int k = 0; k < x.num_children; ++k) {
        val = val + prefix(x.child(k));
    }
    return val;
}

November 11th, 2012, 02:11 PM

Lux Perpetua

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Yes, sorry, I misread it the first time. I've edited my post.

November 11th, 2012, 02:16 PM

srsiddiqui

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Thanks alot for the post .

It is difficult for me to understand the code.

I was trying to do with link list and stack.

I m really sorry I cant understand and i think it is high level programing.
I am new to computer world

Quote:

Originally Posted by Lux Perpetua

Yes, sorry, I misread it the first time. I've edited my post.

November 11th, 2012, 03:50 PM

srsiddiqui

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I am using a struct_node which has
val;
node_son;
node_brother;

my algo is this but i m stucking at one place

current <- father;
if (current != NULL)
{
while( current ! = NULL)
{
insert the value of curent in the stack
current <-- current.son;
} // here the value ABD will be inserted in the stack

while ( current--> brother ! = NULL)
{
current <-- current.brother
insert the value of current into the stack

} // here the stack will have ABDEF

This is where i m stuck how i put the 'C' in the stack , how to hold the value of C .
}

as the final answer should be ABDEFC

Quote:

Originally Posted by srsiddiqui

November 11th, 2012, 10:11 PM

Lux Perpetua

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Quote:

Originally Posted by srsiddiqui

my algo is this but i m stucking at one place

I hate to be the one to tell you this, but you're not simply stuck at one place. The pseudocode you presented is not even close.

Here is what I gather your tree structure is, from your last post:

Code:

A --------------------- *
|
B ------------- C - *
|               |
D - E - F - *   *
|   |   |
*   *   *

where '-' denotes "brother," '|' denotes "son," and '*' denotes NULL. The procedure to do a preorder traversal of this type of tree is actually even simpler than my previous pseudocode: push A onto an empty stack, and then just repeatedly pop a node X from the stack and push X.brother and X.son, in that order, ignoring the ones that are NULL. If you record the nodes in the order in which you pop them, you'll get the correct sequence.

Rather than give you new pseudocode for this, let's just go through the process and see what gets pushed onto and popped from the stack:

Push A onto an empty stack. (Stack = {A})
Pop A from stack. Push A.son (=B) onto the stack. (Stack = {B})
Pop B from stack. Push B.brother (=C) and B.son (=D) onto stack. (Stack = {C, D})
Pop D from stack. Push D.brother (=E) onto stack. (Stack = {C, E})
Pop E from stack. Push E.brother (=F) onto stack. (Stack = {C, F})
Pop F from stack. (Stack = {C})
Pop C from stack. (Stack = {})
Stack is empty, so exit.

November 12th, 2012, 02:31 AM

srsiddiqui

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Thanks!
It is quite simple.Sorry that you have to tell me this things as i suck in computer.

I did this way, i do not know if it is wright or not.

Code:

Procedure Prefix(input father)

      current <-- father;

     if (current != null)
     {
          while current != null)
          { 
                  push(current.value ) in stack
                  currentparent<-- current
           }
          
           if( current.brother != NULL)
           {  
                 current<--(current.brother)
                 push( current.value) in stack
             }
            if( currentparent.brother!= null)
            {
                   current<-- currentparent.brother
                   push(current.value)
            }
      }

I think this is wrong as i do not POP any value.
but i thought alot , this is working with my tree.

Quote:

Originally Posted by Lux Perpetua

I hate to be the one to tell you this, but you're not simply stuck at one place. The pseudocode you presented is not even close.

Here is what I gather your tree structure is, from your last post:

Code:

A --------------------- *
|
B ------------- C - *
|               |
D - E - F - *   *
|   |   |
*   *   *

Push A onto an empty stack. (Stack = {A})
Pop A from stack. Push A.son (=B) onto the stack. (Stack = {B})
Pop B from stack. Push B.brother (=C) and B.son (=D) onto stack. (Stack = {C, D})
Pop D from stack. Push D.brother (=E) onto stack. (Stack = {C, E})
Pop E from stack. Push E.brother (=F) onto stack. (Stack = {C, F})
Pop F from stack. (Stack = {C})
Pop C from stack. (Stack = {})
Stack is empty, so exit.

November 12th, 2012, 02:33 PM

srsiddiqui

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Can you please write the generalised algo for this question.
with using structures.

Quote:

Originally Posted by Lux Perpetua

Yes, sorry, I misread it the first time. I've edited my post.

#10

November 13th, 2012, 12:33 AM

Lux Perpetua

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Quote:

Originally Posted by srsiddiqui

Thanks!
It is quite simple.Sorry that you have to tell me this things as i suck in computer.

I did this way, i do not know if it is wright or not.

Code:

Procedure Prefix(input father)

      current <-- father;

     if (current != null)
     {
          while current != null)
          { 
                  push(current.value ) in stack
                  currentparent<-- current
           }
          
           if( current.brother != NULL)
           {  
                 current<--(current.brother)
                 push( current.value) in stack
             }
            if( currentparent.brother!= null)
            {
                   current<-- currentparent.brother
                   push(current.value)
            }
      }

I think this is wrong as i do not POP any value.
but i thought alot , this is working with my tree.

Again, not even close. I'm surprised, since I already gave you the answer:

Quote:

Originally Posted by Lux Perpetua

...push A onto an empty stack, and then just repeatedly pop a node X from the stack and push X.brother and X.son, in that order, ignoring the ones that are NULL. If you record the nodes in the order in which you pop them, you'll get the correct sequence.

Quote:

Originally Posted by srsiddiqui

Can you please write the generalised algo for this question.
with using structures.

So far, I have provided:

Recursive pseudocode
Nonrecursive pseudocode, using a stack
A complete plain-language description of the stack-based algorithm, using your data structure
A detailed walk-through of the stack algorithm on your example tree

Based on the pseudocode you last posted, you haven't comprehended even 1% of my attempts to point you in the right direction. If you are this clueless, then I'm sorry, you're simply in over your head, and there's no point in my giving you further help on this topic.

#11

November 14th, 2012, 11:57 PM

H2ea5th7er

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I have to write a algorithm for creating a prefix using stack.
URL
URL
URL
URL
URL

#12

November 15th, 2012, 08:19 AM

bolainmarsh5

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I don't see anything wrong with your approach. If you're looking for a spike solution, perhaps the action taken in case 4 is actually feasible for the first three cases, IE find the common prefix to k and k' and rebuild the node with that in mind. If it happens that the keys were prefixes of one-another, the resulting trie will still be correct, only the implementation did a bit more work than it really had to. but then again, without any code to look at it's hard to say if this works in your case.

Algorithm for prefix in a tree

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