handouts/Advanced/Compression/parts/3 huffman.tex

\section{Huffman Codes}


\example{}
Now consider the alphabet $\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{E}\}$. \par
With a na\"ive coding scheme, we can encode a length-$n$ string with $3n$ bits, by mapping...
\begin{itemize}
	\item $\texttt{A}$ to $\texttt{000}$
	\item $\texttt{B}$ to $\texttt{001}$
	\item $\texttt{C}$ to $\texttt{010}$
	\item $\texttt{D}$ to $\texttt{011}$
	\item $\texttt{E}$ to $\texttt{100}$
\end{itemize}
For example, this encodes \texttt{ADEBCE} as \texttt{[000 011 100 001 010 100]}. \par
To encoding strings over $\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{E}\}$ with this scheme, we
need an average of three bits per symbol.

\vspace{2mm}

One could argue that this coding scheme is wasteful: \par
we're not using three of the eight possible three-bit sequences!

\example{}
There is, of course, a better way. \par
Consider the following mapping:

\begin{itemize}
	\item $\texttt{A}$ to $\texttt{00}$
	\item $\texttt{B}$ to $\texttt{01}$
	\item $\texttt{C}$ to $\texttt{10}$
	\item $\texttt{D}$ to $\texttt{110}$
	\item $\texttt{E}$ to $\texttt{111}$
\end{itemize}

\problem{}
\begin{itemize}
	\item Using the above code, encode \texttt{ADEBCE}.
	\item Then, decode \texttt{[110011001111]}.
\end{itemize}

\begin{solution}
	\texttt{ADEBCE} becomes \texttt{[00 110 111 01 10 111]}, \par
	and \texttt{[110 01 10 01 111]} is \texttt{DBCBE}.
\end{solution}

\vfill

\problem{}
How many bits does this code need per symbol, on average?

\begin{solution}
	\begin{equation*}
		\frac{2 + 2 + 2 + 3 + 3}{5} = \frac{12}{5} = 2.4
	\end{equation*}
\end{solution}

\vfill

\problem{}
Consider the code below. How is it different from the one above? \par
Is this a good way to encode five-letter strings?
\begin{itemize}
	\item $\texttt{A}$ to $\texttt{00}$
	\item $\texttt{B}$ to $\texttt{01}$
	\item $\texttt{C}$ to $\texttt{10}$
	\item $\texttt{D}$ to $\texttt{110}$
	\item $\texttt{E}$ to $\texttt{11}$
\end{itemize}

\begin{solution}
	No. The code for \texttt{E} occurs inside the code for \texttt{D},
	and we thus can't decode sequences uniquely. For example, we could
	decode the fragment \texttt{[11001$\cdot\cdot\cdot$]} as \texttt{EA}
	or as \texttt{DB}.
\end{solution}

\vfill
\pagebreak


\remark{}
Huffman codes can be visualized as a tree which we traverse while decoding our sequence. \par
We start at the topmost node, taking the left edge if we see a \texttt{0} and the right edge if we see a \texttt{1}. \par
As an example, consider the code for $\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{E}\}$
on the previous page:


\begin{itemize}
	\item $\texttt{A}$ encodes as $\texttt{00}$
	\item $\texttt{B}$ encodes as $\texttt{01}$
	\item $\texttt{C}$ encodes as $\texttt{10}$
	\item $\texttt{D}$ encodes as $\texttt{110}$
	\item $\texttt{E}$ encodes as $\texttt{111}$
\end{itemize}

Drawing this scheme as a tree, we get the following:


\begin{center}
\begin{tikzpicture}[scale=1.0]
	\begin{scope}[layer = nodes]
		\node[int] (x) at (0, 0) {};
		\node[int] (0) at (-0.75, -1) {};
		\node[int] (1) at (0.75, -1) {};
		\node[end] (00) at (-1.25, -2) {\texttt{A}};
		\node[end] (01) at (-0.25, -2) {\texttt{B}};
		\node[end] (10) at (0.25, -2) {\texttt{C}};
		\node[int] (11) at (1.25, -2) {};
		\node[end] (110) at (0.75, -3) {\texttt{D}};
		\node[end] (111) at (1.75, -3) {\texttt{E}};
	\end{scope}

	\draw[-]
		(x) to node[midway, fill=white, text=gray] {\texttt{0}} (0)
		(x) to node[midway, fill=white, text=gray] {\texttt{1}} (1)
		(0) to node[midway, fill=white, text=gray] {\texttt{0}} (00)
		(0) to node[midway, fill=white, text=gray] {\texttt{1}} (01)
		(1) to node[midway, fill=white, text=gray] {\texttt{0}} (10)
		(1) to node[midway, fill=white, text=gray] {\texttt{1}} (11)
		(11) to node[midway, fill=white, text=gray] {\texttt{0}} (110)
		(11) to node[midway, fill=white, text=gray] {\texttt{1}} (111)
	;
\end{tikzpicture}
\end{center}



\vfill
\pagebreak