A sufficient set of Huffman codes $.$

$($a$)(4$ pts$)$ How many distinct binary Huffman codes does it take to handle all possible

PMF$$s that have exactly three distinct outcomes with nonzero probability? In

other words, what is the smallest number of Huffman codes so that for any threeoutcome

PMF one could label the branches to produce one of these codes?

$($b$)(4$ pts$)$ How many distinct binary Huffman codes does it take to handle all possible

PMF$$s that have exactly four distinct outcomes with nonzero probability?

$1$

$3$

$1$

$3$

$1$

$9$

$1$

$9$

$1$

$27$

$1$

$27$

$1$

$27$

$.(2)$

$($a$)(5$ pts$)$ Find a binary Huffman code for X$.$ Compute its average length in bits.

$($b$)(5$ pts$)$ Find a ternary Huffman code $($for which there are three symbols instead of

two, three brances in every Huffman step instead of two.$)$ Compute it$$s average

length $($average number of ternary symbols, which are sometimes called trits.

$($c$)(2$ pts$)$ Does the ternary Huffman code achieve the corresponding ternary entropy

limit on compression? Which code is more efficient in this case, the ternary or

the binary? Explain your answer.