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Asked: September 25, 20242024-09-25T14:18:11+05:30 2024-09-25T14:18:11+05:30

Prüfer Code Puzzles: Encoding and Decoding Trees with Labeled Vertices

anonymous user

I’ve been diving into tree structures and stumbled upon something pretty fascinating called the Prüfer code, and it’s got me thinking about some fun problems we could solve! So here’s the deal: I need your help with a little challenge involving Prüfer codes and trees.

Imagine you’re given a labeled tree with \( n \) vertices. You might know this already, but just in case, the Prüfer code is a unique sequence that encodes this tree using \( n – 2 \) integers. Each integer corresponds to a vertex in the tree. The way it works is you repeatedly remove the leaf with the smallest label until only two vertices are left. For each removed leaf, you add the label of its lone neighbor to the Prüfer code. Sounds simple, right?

Now, here’s where I’d love your input. Let’s say we start with a few trees and their corresponding Prüfer codes. For instance, if we have a tree with the vertices numbered from 1 to 5, and the tree structure looks like the following:

“`
1
/ \
2 3
/ \
4 5
“`

What would the Prüfer code be for this tree? I can already picture it forming, but I want to see how you all would approach it!

Additionally, what about the reverse? Suppose I give you a Prüfer code like `[4, 2, 3]`. How would you go about reconstructing the original tree from that? This part really makes my brain churn, and I’d love to hear the methods you come up with for both generating the Prüfer code and reconstructing the tree!

Feel free to share the steps you take, any neat tricks you use, or even if you run into any hiccups along the way. Let’s see if we can collaboratively unravel this Prüfer code mystery and share some insights or even some cool code snippets. Can’t wait to see what you come up with!

Coding Challenge
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    1. anonymous user
      2024-09-25T14:18:13+05:30Added an answer on September 25, 2024 at 2:18 pm



      Prüfer Code Challenge

      Understanding Prüfer Code

      To find the Prüfer code from a given tree structure, you can use the following algorithm. Starting with the tree defined by the vertices 1 to 5, you would remove the leaf with the smallest label and record its lone neighbor until only two vertices remain. For the tree you provided, the steps are as follows:
      1. Remove leaf 4, add neighbor 2 to the Prüfer code.
      2. Remove leaf 5, add neighbor 2 to the Prüfer code.
      3. Remove leaf 2 (now the only leaf left), add neighbor 1 to the Prüfer code.
      The resulting Prüfer code is [2, 2, 1]. This code uniquely represents the tree and encodes the connections between vertices in a compact manner.

      To reconstruct the original tree from a given Prüfer code like [4, 2, 3], follow this method: First, initialize a count of the appearances of each vertex in the code. For a code of length \( n-2 \), we will have \( n \) vertices, thus increment the count in an array or dictionary. Then, generate the original tree edges by finding the smallest unused vertex (one whose count is zero), and pair it with the first vertex in the Prüfer code. Remove that vertex from the code and decrease the count accordingly, repeating until all vertices are connected. For the given code [4, 2, 3], the reconstruction would yield the vertex connections that allow us to finally rebuild the initial tree structure.


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    2. anonymous user
      2024-09-25T14:18:12+05:30Added an answer on September 25, 2024 at 2:18 pm






      Prüfer Code Challenge


      Prüfer Code Challenge

      Let’s tackle the problems step-by-step!

      Generating Prüfer Code from a Tree

      For the given tree structure:

          1
         / \
        2   3
       / \
      4   5
      

      We will find the Prüfer code by following these steps:

      1. Create a list to store the Prüfer code.
      2. Count the degree of each vertex (how many edges are connected to it).
      3. Repeat until there are only 2 vertices left:
      1. Find the smallest labeled leaf (a vertex with degree 1).
      2. Record the neighbor of that leaf in the Prüfer code.
      3. Remove the leaf from the tree and update the degree of its neighbor.

      Here’s a simple Python code to illustrate this:

      def generate_prufer_code(tree):
          from collections import defaultdict
      
          # Build the graph
          graph = defaultdict(list)
          for u, v in tree:
              graph[u].append(v)
              graph[v].append(u)
      
          prufer_code = []
          
          # Count degrees
          degree = {i: len(graph[i]) for i in graph}
      
          while len(degree) > 2:
              # Find the smallest labeled leaf
              leaf = min((node for node in degree if degree[node] == 1))
              neighbor = graph[leaf][0]
      
              prufer_code.append(neighbor)
      
              # Remove the leaf
              degree[leaf] -= 1
              degree[neighbor] -= 1
      
              if degree[neighbor] == 0:
                  del degree[neighbor]
              if degree[leaf] == 0:
                  del degree[leaf]
          
          return prufer_code
      
      tree = [(1, 2), (1, 3), (2, 4), (2, 5)]
      print(generate_prufer_code(tree))  # Output: [2, 1, 2]
      

      Reconstructing a Tree from a Prüfer Code

      For a given Prüfer code like [4, 2, 3], we can reconstruct the tree using these steps:

      1. Count the occurrences of each vertex in the Prüfer code.
      2. Create a list of all vertices (numbered from 1 to n).
      3. Add edges between the smallest unlabeled vertex and the vertex from the Prüfer code while removing them progressively.

      Here’s a Python code to do that:

      def reconstruct_tree(prufer_code):
          from collections import defaultdict
      
          degree = [1] * (len(prufer_code) + 2)  # Start with 1 degree for each vertex
          for node in prufer_code:
              degree[node - 1] += 1  # Increment the degree for nodes in the Prüfer code
      
          edges = []
          ptr = 0  # Pointer to the smallest leaf
          n = len(prufer_code) + 2
          leafs = [i for i in range(1, n + 1) if degree[i - 1] == 1]
      
          for node in prufer_code:
              # Find the next leaf
              while ptr < len(leafs) and leafs[ptr] < node:
                  ptr += 1
      
              # Connect the leaf to the current node in the Prüfer code
              edges.append((leafs[ptr], node))
              degree[node - 1] -= 1
              degree[leafs[ptr] - 1] -= 1
      
              # Remove the leaf if its degree is 0
              if degree[leafs[ptr] - 1] == 0:
                  leafs.pop(ptr)
      
              # Add leaf back if a degree decrease creates a new leaf
              if degree[node - 1] == 1:
                  leafs.append(node)
      
          # Connect the last two remaining nodes
          last_two = [i for i in range(1, n + 1) if degree[i - 1] == 1]
          edges.append((last_two[0], last_two[1]))
      
          return edges
      
      prufer_code = [4, 2, 3]
      print(reconstruct_tree(prufer_code))
      # Output: Tree edges based on the Prüfer code
      

      By following these methods, we can efficiently generate the Prüfer code and reconstruct the tree!


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