java,compression,nodes,huffman-coding

You can use indices instead of nodes but you need somewhere nodes and/or vertices.

compression,jpeg,huffman-coding

There are two steps. You have to read the data from a DHT market. The structure is defined in the JPEG standard. That gives you an array of counts--the number of codes of a given length. You have to convert the counts to huffman codes. The JPEG standard explains that...

java,encoding,binaryfiles,huffman-coding

Try this, add some zeros until you have blocks of 8 bits, then byte by byte parse and write int b; while ((line2 = br2.readLine()) != null) { String a = Encode.encode(line2, hTree); while (a.length() % 8 != 0) a += "0"; // lets add some extra bits until we...

compression,huffman-coding,run-length-encoding

If RLE works, Huffman will work all the better. There's a proof that if your file is large enough, Huffman will converge to the maximum possible entropy, thus maximizing compression.

compression,gzip,huffman-coding

infgen will show you the dynamic block headers in some detail. infgen -d will show you them in all their detail. I don't know that that will help with what you are trying to do. It sounds like what you're looking for are preset dictionaries. In zlib you can use...

haskell,tuples,reverse,huffman-coding

As you have said yourself, you need to use the reverse function. Wherever you have your code to form the list of tuples, encapsulate that code in a reverse function using brackets (reverse(code)) or reverse $ <code> and the result will be the reverse. Hope this helps!...

algorithm,binary-tree,decoding,huffman-coding

It is not clear why you want to restore the frequency array. As you said, the tree is all you need to decode. (You don't even need to send the tree — you can just send the number of bits for each symbol, and generate a canonical Huffman code from...

java,performance,parsing,huffman-coding

In the end it was quite simple. I support almost all solutions now. One can test every symbol group (same bit length), use a lookup table (10bit + 10bit + 10bit (just tables of 10bit, symbolscount + 1 is the reference to those talbes)) and generating java (and if needed...

There are at least two problems with your code: one serious, one in terms of naming. In the function struct node join_nodes(struct node no1, struct node no2) no1 and no2 are passed by value. They are temporary objects, for all practical purposes. So the lines aux.right = &no1; aux.left =...

You have this problem because you're not correctly using pointers. Look, you create two TreeNode's on a stack, then you take their addresses, and make them children of new node. But, because they are on stack, they have the same address. So all the left and right children will point...

c#,encoding,binary-tree,huffman-coding

I checked your code, and it looks like it should be working. I also ran the code you have on GitHub, and your program correctly builds the tree. The resulting root after your fillTree operation has the correct children, multiple levels deep (as opposed to only one level, as you...

c#,memory,dictionary,huffman-coding

I would move the cache to another process. Even better, I would use a IIS service with MemoryCache (http://msdn.microsoft.com/en-us/library/system.runtime.caching.memorycache(v=vs.110).aspx) and query the service. I am aware that there will be come overhead, but the throughput should be good enough.

algorithm,compression,huffman-coding

the lower bound on the average number of bits per symbol in compressed file is nothing but the entropy H = -sum(p(x)*log(p(x))) for all symbols x in input. P(x) = freq(x)/(filesize). Using this compressed length(lower bound) = filesize*H. This is the lower bound on compressed size of file. But unfortunately...

c#,recursion,traversal,huffman-coding

As I had a little bit of time left, I worked out an example of a Huffman tree, while playing with C# 6.0. It's not optimized (not even by far!), but it works fine as an example. And it will help you to look where your 'challenge' may arise. As...

haskell,decoding,huffman-coding

Try parsing all codes successively, then repeat after a successful match. Repeat until there's no more input. import Control.Monad data Bit = Zero | One deriving (Show, Eq) type HCodeMap = [(Char, [Bit])] decode :: [Bit] -> HCodeMap -> Maybe String decode bits codes = process bits where -- if...

BitSet.cardinality() returns the number of bits set to true in the BitSet. I think you are looking for BitSet.size(). But keep in mind it will return the number of bits, not bytes. Assuming after Huffman encoding you have approximately half of the bits set to true, that means your BitSet...

Just pick 2. Any 2. It doesn't matter which 2 you pick. The only meaningful metric here is their frequency, and the frequencies are all the same regardless of which 2 you pick. You can't combine all 3 of them. Remember, a Huffman tree is a binary tree because each...

haskell,tree,binary-tree,huffman-coding,algebraic-data-types

You're not constructing the Huffman tree correctly. The process is supposed to go like this: Turn all the source symbols into single-element huffman trees Pair each source symbol up with its frequency into a big list of tree/frequency pairs. If there is just one tree/frequency pair left, that tree is...

algorithm,encoding,huffman-coding

Interesting puzzle. As mentioned by j_random_hacker in a comment, it's possible to do this using a backtracking search. There are a few constraints to valid Huffman encodings of the string that we can use to narrow the search down: No two Huffman codes of length n and m can be...

algorithm,text,binary,compression,huffman-coding

If you have a Huffman tree, then you can make many other Huffman trees that assigns the same length to all symbols but different code words by swapping the left and right child of any node. All those trees are equally good - they compress the data just as much...

When writing binary data, you can not use in >> x or out << x, since those are "text read and write" functions. So, instead of: out<<unsigned char(c); out<<int(n); you will need to use: out.write(&c, sizeof(c)); out.write(reinterpret_cast<char*>(&n), sizeof(n)); To explain in more detail: n = 12345; out << n; will...

c,file,compression,huffman-coding

There are two ways to solve this that I can think of: Put the length of the uncompressed data in front of the compressed data. Then, when you're decompressing, count how many characters you have decompressed and stop after the right number. Put a special end symbol into your huffman...

That is not an "overhead", that't the marker that lets Java figure out what type it needs to create when deserializing the object from that file. Since ObjectInputStream has no idea what you have serialized into a file, and has no way for you to provide a "hint", ObjectOutputStream must...

First, let me describe how a Huffman tree works, then I will explain how extended Huffman encoding works. Some terms, codeword means a sequence of bits in our encoded output, that has been compressed. Terms like a1, a2 or a3 are our input characters, we can think of them as...

You messed up your algorithm. You must add to the noder list when the character is not in the list. The number of items (noder.Count) will always be 0 since you only add a Node in that for-loop which iterates from 0 to noder.Count: for (int j = 0; j...

java,recursion,tree,huffman-coding,preorder

I believe this is the problem: makeTree(root.getLeftChild(), start++, treeString); I'm not sure what your approach is, but it looks like if you see an I, your plan is to go to the left node, and start examining the string starting at the next character. The reason for the infinite recursion...

Write the tree as a series of bits: 0 represents a leaf, 1 represents an internal node. The output for a binary tree (Huffman or otherwise) with N leaf nodes and N-1 internal nodes will be a sequence of 2N-1 bits. (You can actually save two bits, since you know...