This may be a stupid question, but compressing an empty file doesn't make any sense right?. The Huffman encoding algorithm on an empty file wouldn't work because it relies on the fact there have to be at least 2 nodes in the priority queue. If we run the algorithm on an empty file, the only node we would get is the one corresponding to EOF.
Ya, that's right, it doesn't make much sense to run the Huffman encoding on it. Depending on the details of the implementation, it may not crash.
But why would you try to compress an empty file?
You need to somehow encode at the start of the compressed data what symbols correspond to what Huffman codes. It is in that representation that the number of symbols would be indicated. If there is only one symbol, which has to be EOF per your description, then the Huffman coding is implied to be zero bits. If there is only one symbol, then you need zero bits to represent it.
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I am writing a Huffman Coding/Decoding algorithm and I am running into the problem that the storing the Huffman tree is taking up way to much room. Currently, I am converting the tree into a hashMap as such -> hashMap<Character(s),Huffman Code> and then storing that hash map. The issue is that, while the string is compressed great, adding the Huffman Tree data stored in the hash map is adding so much overhead that it's actually ending up bigger than the original. Currently I am just naively writing [data, value] pairs to the file, but I imagine there must be some sort of trickier way to do that. Any ideas?
You do not need the tree in order to encode. All you need is the bit lengths for each symbol and a way to order the symbols. See Canonical Huffman Code.
In fact, all you need is the symbols that are coded ordered by bit length, and within bit length sorted by symbol, and then the number of codes of each length. With just those two things you can encode.
So I'm looking into Huffman coding, and it's a pretty simple algorithm to understand, except I was curious about one thing. Given that "a Huffman tree that omits unused symbols produces the most optimal code lengths", I was curious whether the frequency table of a Huffman tree counts towards the total length of the encoded message? I suppose this question in itself boils down to how the frequency table is stored. Is it part of the encoded message, or is it saved as a separate file?
Yes, unless the two sides agree on a pre-determined code book, the frequency table (or equivalent information sufficient to construct the decoding tree on the receiving end) must be included in the message.
Google Canonical Huffman code for a clever way to cut down on the size of this information.
Lets say I have a massive string of just a single character say x. I need to use huffman encoding.
A huffman encoding is a fully binary tree. So how does one create a huffman code for just a single character when we dont need two leaves at all ?
jbr's answer is fine; this is just a longer version of it.
Huffman is meant to produce a minimal-length sequence of bits that contains all the information in the original sequence of symbols, assuming that the decoder already knows the set of symbols. If there's only one symbol, the input data contains no information except its length.
In Huffman-based data formats, length is usually encoded separately, not as part of the Huffman-encoded bit sequence itself. The decoder of a single-symbol Huffman code therefore has all the information it needs to reconstruct the input without needing to read anything from the Huffman-encoded bit sequence. it is logical, then, that the Huffman encoder's output should be 0 bits long.
If you don't have a length encoded separately, then you must have a symbol to represent End Of Sequence so the decoder knows when to stop reading. Then your Huffman tree will have 2 nodes and you won't run into this special case.
If you only have one symbol, then you only need 1 bit per symbol. So you really don't have to do anything except count the number of bits and translate each into your symbol.
You simply could add an edge case in your code.
For example:
check if there is only one character in your hash table, which returns only the root of the tree without any leafs. In this case, you could add a code for this root node in your encoding function, like 0.
In the encoding function, you should refer to this edge case too.
I want to quantify the saving of space I can get by changing the format of a file.
I have a sparse matrix stocked in a text file (30% sparsity). Columns are separated by tabs.
Following an idea in an SO answer, I will change the format to row_id, col_id for the non zero terms only. I know how much space a float takes, but my question is: how much space does a tab take?
CouchDeveloper in his comment is correct. It's impossible to tell from the data you provide.
In a single byte character set encoding you'd save 1 byte per separator from the current ", ".
In a multibyte encoding it'd depend on the way each of those characters is encoded, you could theoretically even lose space. Say a tab is encoded as 4 bytes, a comma and space as 1 each, you'd end up taking 2 more bytes per separator.
Unless you have many separators and relatively very little data, I'd not worry one way or another, it'd be micro optimisation.
If you do, a binary encoding scheme might be more relevant.
1 byte, but significantly less if you're using compression (based on how common they will be, less than a bit on average). Use compression.
My problem is that I have a 100,000+ different elements and as I understand it Huffman works by assigning the most common element a 0 code, and the next 10, the next 110, 1110, 11110 and so on. My question is, if the code for the nth element is n-bits long then surely once I have passed the 32nd term it is more space efficient to just sent 32-bit data types as they are, such as ints for example? Have I missed something in the methodology?
Many thanks for any help you can offer. My current implementation works by doing
code = (code << 1) + 2;
to generate each new code (which seems to be correct!), but the only way I could encode over 100,000 elements would be to have an int[] in a makeshift new data type, where to access the value we would read from the int array as one continuous long symbol... that's not as space efficient as just transporting a 32-bit int? Or is it more a case of Huffmans use being with its prefix codes, and being able to determine each unique value in a continuous bit stream unambiguously?
Thanks
Your understanding is a bit off - take a look at http://en.wikipedia.org/wiki/Huffman_coding. And you have to pack the encoded bits into machine words in order to get compression - Huffman encoded data can best be thought of as a bit-stream.
You seem to understand the principle of prefix codes.
Could you tell us a little more about these 100,000+ different elements you mention?
The fastest prefix codes -- universal codes -- do, in fact, involve a series of bit sequences that can be pre-generated without regard to the actual symbol frequencies. Compression programs that use these codes, as you mentioned, associate the most-frequent input symbol to the shortest bit sequence, the next-most-frequent input symbol to the next-shorted bit sequence, and so on.
What you describe is one particular kind of prefix code: unary coding.
Another popular variant of the unary coding system assigns elements in order of frequency to the fixed codes
"1", "01", "001", "0001", "00001", "000001", etc.
Some compression programs use another popular prefix code: Elias gamma coding.
The Elias gamma coding assigns elements in order of frequency to the fixed set of codewords
1
010
011
00100
00101
00110
00111
0001000
0001001
0001010
0001011
0001100
0001101
0001110
0001111
000010000
000010001
000010010
...
The 32nd Elias gamma codeword is about 10 bits long, about half as long as the 32nd unary codeword.
The 100,000th Elias gamma codeword will be around 32 bits long.
If you look carefully, you can see that each Elias gamma codeword can be split into 2 parts -- the first part is more or less the unary code you are familiar with. That unary code tells the decoder how many more bits follow afterward in the rest of that particular Elias gamma codeword.
There are many other kinds of prefix codes.
Many people (confusingly) refer to all prefix codes as "Huffman codes".
When compressing some particular data file, some prefix codes do better at compression than others.
How do you decide which one to use?
Which prefix code is the best for some particular data file?
The Huffman algorithm -- if you neglect the overhead of the Huffman frequency table -- chooses exactly the best prefix code for each data file.
There is no singular "the" Huffman code that can be pre-generated without regard to the actual symbol frequencies.
The prefix code choosen by the Huffman algorithm is usually different for different files.
The Huffman algorithm doesn't compress very well when we really do have 100,000+ unique elements --
the overhead of the Huffman frequency table becomes so large that we often can find some other "suboptimal" prefix code that actually gives better net compression.
Or perhaps some entirely different data compression algorithm might work even better in your application.
The "Huffword" implementation seems to work with around 32,000 or so unique elements,
but the overwhelming majority of Huffman code implementations I've seen work with around 257 unique elements (the 256 possible byte values, and the end-of-text indicator).
You might consider somehow storing your data on a disk in some raw "uncompressed" format.
(With 100,000+ unique elements, you will inevitably end up storing many of those elements in 3 or more bytes).
Those 257-value implementations of Huffman compression will be able to compress that file;
they re-interpret the bytes of that file as 256 different symbols.
My question is, if the code for the nth element is n-bits long then
surely once I have passed the 32nd term it is more space efficient to
just sent 32-bit data types as they are, such as ints for example?
Have I missed something in the methodology?
One of the more counter-intuitive features of prefix codes is that some symbols (the rare symbols) are "compressed" into much longer bit sequences. If you actually have 2^8 unique symbols (all possible 8 bit numbers), it is not possible to gain any compression if you force the compressor to use prefix codes limited to 8 bits or less. By allowing the compressor to expand rare values -- to use more than 8 bits to store a rare symbol that we know can be stored in 8 bits -- that frees up the compressor to use less than 8 bits to store the more-frequent symbols.
related:
Maximum number of different numbers, Huffman Compression