Question

Ternary Huffman. Trimedia Disks Inc. has developed “ternary” hard disks. Each cell on a disk can now store values 0,1, or 2(instead of just 0 or 1). To take advantage of this new technology, provide a modified Huffman algorithm for compressing sequences of characters from an alphabet of size n, where the characters occur with known frequencies f1, f2,...., fn. Your algorithm should encode each character with a variable-length codeword over the values 0,1,2, such that no codeword is a prefix of another codeword and so as to obtain the maximum possible compression. Prove that your algorithm is correct

Short answer

The Huffman method for compressing sequences of letters from such a vocabulary with n elements, where even the words appear with known frequency f1, f2,....fn, to take use of this new technology.

Step-by-step solution

Ternary Huffman encoding algorithm

Any ternary tree is the one where the nodes are odd in number. In other words, each node has either 0 or 3 children.

• The left child is labelled " 0," the middle child is labelled " 1," and the right child is labelled " 2."

• Make a list of all potential symbols and their frequencies before building the codes for the ternary Huffman tree.

• Identify the three symbols with the lowest frequency.

Achieved By replacing them with a single set that includes all three symbols and has a frequency that is the sum of their respective frequencies.

• Keep going until you get a list with only one member.

Encapsulated Proof

Check to see if number the symbols is odd at first. If that's the case, then encapsulate the symbol to create the whole ternary tree.

• Add a symbol with frequency 0 if necessary, to make the number of symbols odd.

• To build the whole optimum tree, just use the binary case with the left child as " 0 ," the middle child as " 1," and the right child as " 2 " with the lowest frequency for each step to form the single node.

The number of nodes is thus reduced by two. If the symbol stays strange, the procedure continues.

Expect the number all elements is odd and that the tree with non-leaf node " u " is the best. It doesn't have three leaves, in other words (child nodes). Then, assuming that there is really no loss in generality, consider all children from node " u" to be leaves, and the " u" node to contain the two children.

• We may improve the code by deleting the children of node " u " resulting in a full ternary tree.

• So, this whole ternary tree is built by connecting the three nodes to a current leaf, and it must have an odd number of nodes (child).

• It is possible to start it from the root node.

Conclusion 

The number of sensor nodes increased accordingly for each phase. As little more than a result, every Ternary Huffman compression technique has been updated and is valid.