# Tag Info

6

Thats kind of hard for a general code but you can use the Sardinas-Patterson algorithm. The algorithm generates all possible "dangling suffixes" and checks to see if any of them is a codeword. A dangling suffix is the bits that are "left over" when you compare two similar sequences of your code. If you want your code to be decodable, there can be no ...

3

Shannon's Information Theory is a fairly broad multidisciplinary subject primarily grounded in the fields of electrical engineering and mathematics with overlap in areas of computer science, physics, statistical mechanics, statistics and even philosophy. I've not run across any writers using the acronym MTC, and supposed (correctly after checking) that the ...

2

$$\log{\binom n{\gamma n}} = \log{\frac{n!}{(n-\gamma n)!(\gamma n)!}}\\ = \left(n+\frac 12\right)\log n - n - \left(n\gamma +\frac 12\right)\log \gamma n + \gamma n \\- \left(n - n\gamma +\frac 12\right)\log (n - \gamma n) + (n - \gamma n) + O(1) \\= n\log n - (n - n\gamma)\log (n - \gamma n) - (n\gamma )\log \gamma n + \frac 12\log n - \frac 12(\log n + ... 1 Consider first the discrete case. Here the joint "density" f_{X,Y} (which I shall call the probability mass function following the terminology of Wikipedia) is defined by:$$f_{X,Y}(x,y) = P(X=x,\ Y=y) where $P$ is the probability measure. (This is a "density" with respect to summation.) Since we clearly have $\{X=x,\ Y=y\} \subseteq \{Y=y\}$, and since ...

1

If $A$ uses more than one bit, there can be at most three symbols using two bits, and this just happens to be the given encoding; so any other code with at least two bits for $A$ is trivially worse than the given one. Hence we should check if an encoding where $A$ uses only one bit could be better than the given one. Then if $B$ uses two bits, the three ...

1

When you build a huffman tree, you always merge the two symbols that are least likely first. So if you use the frequencies of the symbols in the corpus as estimations of their probabilities you could start by merging a and b into ab since both have probability 1/5. Then you will have three symbols in your alphabet $\{ab,c,d\}$ where c has probability 1/5 ...

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