问题
I am writing a Huffman encoding/decoding tool and am looking for an efficient way to store the Huffman tree that is created to store inside of the output file.
Currently there are two different versions I am implementing.
- This one reads the entire file into memory character by character and builds a frequency table for the whole document. This would only require outputting the tree once, and thus efficiency is not that big of a concern, other than if the input file is small.
- The other method I am using is to read a chunk of data, about 64 kilobyte in size and run the frequency analysis over that, create a tree and encode it. However, in this case before every chunk I will need to output my frequency tree so that the decoder is able to re-build its tree and properly decode the encoded file. This is where the efficiency does come into place since I want to save as much space as possible.
In my searches so far I have not found a good way of storing the tree in as little space as possible, I am hoping the StackOverflow community can help me find a good solution!
回答1:
Since you already have to implement code to handle a bit-wise layer on top of your byte-organized stream/file, here's my proposal.
Do not store the actual frequencies, they're not needed for decoding. You do, however, need the actual tree.
So for each node, starting at root:
- If leaf-node: Output 1-bit + N-bit character/byte
- If not leaf-node, output 0-bit. Then encode both child nodes (left first then right) the same way
To read, do this:
- Read bit. If 1, then read N-bit character/byte, return new node around it with no children
- If bit was 0, decode left and right child-nodes the same way, and return new node around them with those children, but no value
A leaf-node is basically any node that doesn't have children.
With this approach, you can calculate the exact size of your output before writing it, to figure out if the gains are enough to justify the effort. This assumes you have a dictionary of key/value pairs that contains the frequency of each character, where frequency is the actual number of occurrences.
Pseudo-code for calculation:
Tree-size = 10 * NUMBER_OF_CHARACTERS - 1
Encoded-size = Sum(for each char,freq in table: freq * len(PATH(char)))
The tree-size calculation takes the leaf and non-leaf nodes into account, and there's one less inline node than there are characters.
SIZE_OF_ONE_CHARACTER would be number of bits, and those two would give you the number of bits total that my approach for the tree + the encoded data will occupy.
PATH(c) is a function/table that would yield the bit-path from root down to that character in the tree.
Here's a C#-looking pseudo-code to do it, which assumes one character is just a simple byte.
void EncodeNode(Node node, BitWriter writer)
{
if (node.IsLeafNode)
{
writer.WriteBit(1);
writer.WriteByte(node.Value);
}
else
{
writer.WriteBit(0);
EncodeNode(node.LeftChild, writer);
EncodeNode(node.Right, writer);
}
}
To read it back in:
Node ReadNode(BitReader reader)
{
if (reader.ReadBit() == 1)
{
return new Node(reader.ReadByte(), null, null);
}
else
{
Node leftChild = ReadNode(reader);
Node rightChild = ReadNode(reader);
return new Node(0, leftChild, rightChild);
}
}
An example (simplified, use properties, etc.) Node implementation:
public class Node
{
public Byte Value;
public Node LeftChild;
public Node RightChild;
public Node(Byte value, Node leftChild, Node rightChild)
{
Value = value;
LeftChild = leftChild;
RightChild = rightChild;
}
public Boolean IsLeafNode
{
get
{
return LeftChild == null;
}
}
}
Here's a sample output from a specific example.
Input: AAAAAABCCCCCCDDEEEEE
Frequencies:
- A: 6
- B: 1
- C: 6
- D: 2
- E: 5
Each character is just 8 bits, so the size of the tree will be 10 * 5 - 1 = 49 bits.
The tree could look like this:
20
----------
| 8
| -------
12 | 3
----- | -----
A C E B D
6 6 5 1 2
So the paths to each character is as follows (0 is left, 1 is right):
- A: 00
- B: 110
- C: 01
- D: 111
- E: 10
So to calculate the output size:
- A: 6 occurrences * 2 bits = 12 bits
- B: 1 occurrence * 3 bits = 3 bits
- C: 6 occurrences * 2 bits = 12 bits
- D: 2 occurrences * 3 bits = 6 bits
- E: 5 occurrences * 2 bits = 10 bits
Sum of encoded bytes is 12+3+12+6+10 = 43 bits
Add that to the 49 bits from the tree, and the output will be 92 bits, or 12 bytes. Compare that to the 20 * 8 bytes necessary to store the original 20 characters unencoded, you'll save 8 bytes.
The final output, including the tree to begin with, is as follows. Each character in the stream (A-E) is encoded as 8 bits, whereas 0 and 1 is just a single bit. The space in the stream is just to separate the tree from the encoded data and does not take up any space in the final output.
001A1C01E01B1D 0000000000001100101010101011111111010101010
For the concrete example you have in the comments, AABCDEF, you will get this:
Input: AABCDEF
Frequencies:
- A: 2
- B: 1
- C: 1
- D: 1
- E: 1
- F: 1
Tree:
7
-------------
| 4
| ---------
3 2 2
----- ----- -----
A B C D E F
2 1 1 1 1 1
Paths:
- A: 00
- B: 01
- C: 100
- D: 101
- E: 110
- F: 111
Tree: 001A1B001C1D01E1F = 59 bits
Data: 000001100101110111 = 18 bits
Sum: 59 + 18 = 77 bits = 10 bytes
Since the original was 7 characters of 8 bits = 56, you will have too much overhead of such small pieces of data.
回答2:
If you have enough control over the tree generation, you could make it do a canonical tree (the same way DEFLATE does, for example), which basically means you create rules to resolve any ambiguous situations when building the tree. Then, like DEFLATE, all you actually have to store are the lengths of the codes for each character.
That is, if you had the tree/codes Lasse mentioned above:
- A: 00
- B: 110
- C: 01
- D: 111
- E: 10
Then you could store those as: 2, 3, 2, 3, 2
And that's actually enough information to regenerate the huffman table, assuming you're always using the same character set -- say, ASCII. (Which means you couldn't skip letters -- you'd have to list a code length for each one, even if it's zero.)
If you also put a limitation on the bit lengths (say, 7 bits), you could store each of these numbers using short binary strings. So 2,3,2,3,2 becomes 010 011 010 011 010 -- Which fits in 2 bytes.
If you want to get really crazy, you could do what DEFLATE does, and make another huffman table of the lengths of these codes, and store its code lengths beforehand. Especially since they add extra codes for "insert zero N times in a row" to shorten things further.
The RFC for DEFLATE isn't too bad, if you're already familiar with huffman coding: http://www.ietf.org/rfc/rfc1951.txt
回答3:
branches are 0 leaves are 1. Traverse the tree depth first to get its "shape"
e.g. the shape for this tree
0 - 0 - 1 (A)
| \- 1 (E)
\
0 - 1 (C)
\- 0 - 1 (B)
\- 1 (D)
would be 001101011
Follow that with the bits for the characters in the same depth first order AECBD (when reading you'll know how many characters to expect from the shape of the tree). Then output the codes for the message. You then have a long series of bits that you can divide up into characters for output.
If you are chunking it, you could test that storing the tree for the next chuck is as efficient as just reusing the tree for the previous chunk and have the tree shape being "1" as an indicator to just reuse the tree from the previous chunk.
回答4:
The tree is generally created from a frequency table of the bytes. So store that table, or just the bytes themselves sorted by frequency, and re-create the tree on the fly. This of course assumes that you're building the tree to represent single bytes, not larger blocks.
UPDATE: As pointed out by j_random_hacker in a comment, you actually can't do this: you need the frequency values themselves. They are combined and "bubble" upwards as you build the tree. This page describes the way a tree is built from the frequency table. As a bonus, it also saves this answer from being deleted by mentioning a way to save out the tree:
The easiest way to output the huffman tree itself is to, starting at the root, dump first the left hand side then the right hand side. For each node you output a 0, for each leaf you output a 1 followed by N bits representing the value.
回答5:
A better approach
Tree:
7
-------------
| 4
| ---------
3 2 2
----- ----- -----
A B C D E F
2 1 1 1 1 1 : frequencies
2 2 3 3 3 3 : tree depth (encoding bits)
Now just derive this table:
depth number of codes
----- ---------------
2 2 [A B]
3 4 [C D E F]
You don't need to use the same binary tree, just keep the computed tree depth i.e. the number of encoding bits. So just keep the vector of uncompressed values [A B C D E F] ordered by tree depth, use relative indexes instead to this separate vector. Now recreate the aligned bit patterns for each depth:
depth number of codes
----- ---------------
2 2 [00x 01x]
3 4 [100 101 110 111]
What you immediately see is that only the first bit pattern in each row is significant. You get the following lookup table:
first pattern depth first index
------------- ----- -----------
000 2 0
100 3 2
This LUT has a very small size (even if your Huffman codes can be 32-bit long, it will only contain 32 rows), and in fact the first pattern is always null, you can ignore it completely when performing a binary search of patterns in it (here only 1 pattern will need to be compared to know if the bit depth is 2 or 3 and get the first index at which the associated data is stored in the vector). In our example you'll need to perform a fast binary search of input patterns in a search space of 31 values at most, i.e. a maximum of 5 integer compares. These 31 compare routines can be optimized in 31 codes to avoid all loops and having to manage states when browing the integer binary lookup tree. All this table fits in small fixed length (the LUT just needs 31 rows atmost for Huffman codes not longer than 32 bits, and the 2 other columns above will fill at most 32 rows).
In other words the LUT above requires 31 ints of 32-bit size each, 32 bytes to store the bit depth values: but you can avoid it this by implying the depth column (and the first row for depth 1):
first pattern (depth) first index
------------- ------- -----------
(000) (1) (0)
000 (2) 0
100 (3) 2
000 (4) 6
000 (5) 6
... ... ...
000 (32) 6
So your LUT contains [000, 100, 000(30times)]. To search in it you must find the position where the input bits pattern are between two patterns: it must be lower than the pattern at the next position in this LUT but still higher than or equal to the pattern in the current position (if both positions contain the same pattern, the current row will not match, the input pattern fits below). You'll then divide and conquer, and will use 5 tests at most (the binary search requires a single code with 5 embedded if/then/else nested levels, it has 32 branches, the branch reached indicates directly the bit depth that does not need to be stored; you perform then a single directly indexed lookup to the second table for returning the first index; you derive additively the final index in the vector of decoded values).
Once you get a position in the lookup table (search in the 1st column), you immediately have the number of bits to take from the input and then the start index to the vector. The bit depth you get can be used to derive directly the adjusted index position, by basic bitmasking after substracting the first index.
In summary: never store linked binary trees, and you don't need any loop to perform thelookup which just requires 5 nested ifs comparing patterns at fixed positions in a table of 31 patterns, and a table of 31 ints containing the start offset within the vector of decoded values (in the first branch of the nested if/then/else tests, the start offset to the vector is implied, it is always zero; it is also the most frequent branch that will be taken as it matches the shortest code which is for the most frequent decoded values).
来源：https://stackoverflow.com/questions/759707/efficient-way-of-storing-huffman-tree