= Solution
A realized message $x$ has <self-information> $I(x)=-\log_2p_x$. The average information per source message is its <Shannon entropy>,
$$
\boxed{H(M)=-p_A\log_2p_A-p_B\log_2p_B-p_C\log_2p_C.}
$$
The average reaches \b[$\log_2 3\simeq1.58496$ bits per message] at $p_A=p_B=p_C=1/3$. Individual rare realizations can have larger self-information; this bound concerns the average.
The given strings form a <prefix code> with lengths $\ell_A=1$ and $\ell_B=\ell_C=2$. To attain the entropy bound without redundancy, the resulting long bit stream must carry one bit of information per bit: unbiased bits with no predictable dependence. For an independent source, the <dyadic source matched to a complete prefix code> has
$$
\boxed{p_A=2^{-1}=\frac12,\qquad p_B=p_C=2^{-2}=\frac14.}
$$
These probabilities make the encoded stream equivalent to parsing independent fair bits using the code tree. Its <expected codeword length> and source entropy agree:
$$
\boxed{\overline\ell=H(M)=\frac12\cdot1+\frac14\cdot2+\frac14\cdot2=\frac32\ \text{bits per message}.}
$$
Here “optimal” means attaining the Shannon bound. If it means only an <optimal prefix code> among symbol-by-symbol binary prefix codes, the probabilities are not unique: for three positive-probability symbols, this tree is optimal whenever $p_A$ is a largest probability. <Huffman coding> need not attain entropy exactly for nondyadic probabilities.
Concatenating the specified source-message encodings gives
$$
AABCACBA\ \longmapsto\ 0\,|\,0\,|\,10\,|\,11\,|\,0\,|\,11\,|\,10\,|\,0
=001011011100.
$$
Pack consecutive bits in pairs, rather than allocating a fresh Bell pair to each variable-length source word:
$$
00\,|\,10\,|\,11\,|\,01\,|\,11\,|\,00.
$$
Under the mapping in part 1, transmit
$$
\boxed{\Phi^+,\ \Phi^-,\ \Psi^-,\ \Psi^+,\ \Psi^-,\ \Phi^+.}
$$
The block uses six message-bearing qubit transmissions and six preshared <Bell pairs>. Source-message boundaries are recovered by parsing the decoded prefix bit stream.
Back to article page
