I am going through some questions and answers regarding Information Theory and I found this question and its solution. Can some one explain this solution to me.
We would like to encode a sequence of symbols that come from an alphabet with $d+3$ symbols. We want to encode symbols $a_1$,$a_2$,and $a_3$ using code words that are three bits long and symbols $a_4$,$a_5$,…,$a_{d+3}$ with code words that are eight bits long. What is the maximum value of $d$ for which this will be possible, if the code must be uniquely decidable?
And the answer is:
The codeword lengths are possible if and only if they satisfy the Kraft-MacMillan inequality,which in this case is
$\frac{1}{2^3} + \frac{1}{2^3} + \frac{1}{2^3} + \frac{d}{2^8} \le 1$
This is equivalent to
$\frac {d}{256} \le \frac{5}{8}$
which is equivalent to $d \le 160$. The maximum value possible value of $d$ is therefore $160$.
what is the function $d+3$ , and where we used it . How does $\frac{5}{8}$ appear in this equation.
What is the effect of d+3 symbols on all this scenario. If I change it to d+1, what is the effect of it
Firstly $5/8$ is $1-(3/8)$ if you manipulate the given equation, entirely trivial.
Clearly there is a typo in the question and symbols $a_4,\ldots,a_{d+3}$ are to be encoded in to codewords $8$ bits long. You could have figured this out by assuming the equation was correct and using Kraft's inequality to observe that the equation contains the term $d/2^8$ so there must be $d$ symbols that are length 8.