我在简单的压缩机下使用纯huffman代码实现了Windows.But --我不太了解如何快速解码压缩文件,我的坏算法是:
枚举代码表中的所有huffman代码,然后将其与压缩的file.It中的位进行比较,结果非常糟糕:解压缩3MB文件需要6个小时。
你能提供一个更有效的算法吗?我该用哈希还是别的什么?
更新:我根据我朋友林的建议用状态表实现了译码器,我认为这种方法应该比6s内的3MB的travesal树更好。
谢谢。
发布于 2010-02-21 14:13:47
优化二叉树方法的一种方法是使用查找表。您可以安排表,以便可以直接查找特定的编码位模式,从而允许任何代码的最大位宽。
由于大多数代码没有使用完全的最大宽度,所以它们包含在表中的多个位置--每个未使用比特的组合都包含一个位置。该表指示要从输入和解码的输出中丢弃多少位。
如果最长的代码太长,所以表是不切实际的,那么折衷的方法是使用一棵较小的固定宽度下标查找树。例如,您可以使用256项表来处理一个字节.如果输入代码超过8位,则表条目指示解码是不完整的,并将您引导到一个处理下一个高达8位的表。较大的表用内存交换速度- 256项可能太小了。
我相信这种通用方法被称为“前缀表”,这就是BobMcGees引用的代码所做的事情。一个可能的区别是,一些压缩算法需要在解压缩过程中更新前缀表--这对于简单的Huffman来说并不是必需的。IIRC,我第一次看到它是在一本关于位图图形文件格式的书中,其中包括GIF,在专利恐慌之前的一段时间。
应该可以很容易地从二叉树模型中预先计算出一个完整的查找表、一个哈希表等价物,或者一个小表树。二叉树仍然是代码工作方式的关键表示(心智模型)--这个查找表只是实现它的一种优化方法。
发布于 2010-02-19 16:30:26
为什么不看一看GZIP源是如何实现的,特别是unpack.c中的Huffman解压缩代码?它做的和你完全一样,只不过它做得更快。
据我所知,它使用查找数组和shift/掩码操作,对整个单词进行操作,以提高运行速度。不过,代码相当密集。
编辑:这是完整的源代码
/* unpack.c -- decompress files in pack format.
* Copyright (C) 1992-1993 Jean-loup Gailly
* This is free software; you can redistribute it and/or modify it under the
* terms of the GNU General Public License, see the file COPYING.
*/
#ifdef RCSID
static char rcsid[] = "$Id: unpack.c,v 1.4 1993/06/11 19:25:36 jloup Exp $";
#endif
#include "tailor.h"
#include "gzip.h"
#include "crypt.h"
#define MIN(a,b) ((a) <= (b) ? (a) : (b))
/* The arguments must not have side effects. */
#define MAX_BITLEN 25
/* Maximum length of Huffman codes. (Minor modifications to the code
* would be needed to support 32 bits codes, but pack never generates
* more than 24 bits anyway.)
*/
#define LITERALS 256
/* Number of literals, excluding the End of Block (EOB) code */
#define MAX_PEEK 12
/* Maximum number of 'peek' bits used to optimize traversal of the
* Huffman tree.
*/
local ulg orig_len; /* original uncompressed length */
local int max_len; /* maximum bit length of Huffman codes */
local uch literal[LITERALS];
/* The literal bytes present in the Huffman tree. The EOB code is not
* represented.
*/
local int lit_base[MAX_BITLEN+1];
/* All literals of a given bit length are contiguous in literal[] and
* have contiguous codes. literal[code+lit_base[len]] is the literal
* for a code of len bits.
*/
local int leaves [MAX_BITLEN+1]; /* Number of leaves for each bit length */
local int parents[MAX_BITLEN+1]; /* Number of parents for each bit length */
local int peek_bits; /* Number of peek bits currently used */
/* local uch prefix_len[1 << MAX_PEEK]; */
#define prefix_len outbuf
/* For each bit pattern b of peek_bits bits, prefix_len[b] is the length
* of the Huffman code starting with a prefix of b (upper bits), or 0
* if all codes of prefix b have more than peek_bits bits. It is not
* necessary to have a huge table (large MAX_PEEK) because most of the
* codes encountered in the input stream are short codes (by construction).
* So for most codes a single lookup will be necessary.
*/
#if (1<<MAX_PEEK) > OUTBUFSIZ
error cannot overlay prefix_len and outbuf
#endif
local ulg bitbuf;
/* Bits are added on the low part of bitbuf and read from the high part. */
local int valid; /* number of valid bits in bitbuf */
/* all bits above the last valid bit are always zero */
/* Set code to the next 'bits' input bits without skipping them. code
* must be the name of a simple variable and bits must not have side effects.
* IN assertions: bits <= 25 (so that we still have room for an extra byte
* when valid is only 24), and mask = (1<<bits)-1.
*/
#define look_bits(code,bits,mask) \
{ \
while (valid < (bits)) bitbuf = (bitbuf<<8) | (ulg)get_byte(), valid += 8; \
code = (bitbuf >> (valid-(bits))) & (mask); \
}
/* Skip the given number of bits (after having peeked at them): */
#define skip_bits(bits) (valid -= (bits))
#define clear_bitbuf() (valid = 0, bitbuf = 0)
/* Local functions */
local void read_tree OF((void));
local void build_tree OF((void));
/* ===========================================================================
* Read the Huffman tree.
*/
local void read_tree()
{
int len; /* bit length */
int base; /* base offset for a sequence of leaves */
int n;
/* Read the original input size, MSB first */
orig_len = 0;
for (n = 1; n <= 4; n++) orig_len = (orig_len << 8) | (ulg)get_byte();
max_len = (int)get_byte(); /* maximum bit length of Huffman codes */
if (max_len > MAX_BITLEN) {
error("invalid compressed data -- Huffman code > 32 bits");
}
/* Get the number of leaves at each bit length */
n = 0;
for (len = 1; len <= max_len; len++) {
leaves[len] = (int)get_byte();
n += leaves[len];
}
if (n > LITERALS) {
error("too many leaves in Huffman tree");
}
Trace((stderr, "orig_len %ld, max_len %d, leaves %d\n",
orig_len, max_len, n));
/* There are at least 2 and at most 256 leaves of length max_len.
* (Pack arbitrarily rejects empty files and files consisting of
* a single byte even repeated.) To fit the last leaf count in a
* byte, it is offset by 2. However, the last literal is the EOB
* code, and is not transmitted explicitly in the tree, so we must
* adjust here by one only.
*/
leaves[max_len]++;
/* Now read the leaves themselves */
base = 0;
for (len = 1; len <= max_len; len++) {
/* Remember where the literals of this length start in literal[] : */
lit_base[len] = base;
/* And read the literals: */
for (n = leaves[len]; n > 0; n--) {
literal[base++] = (uch)get_byte();
}
}
leaves[max_len]++; /* Now include the EOB code in the Huffman tree */
}
/* ===========================================================================
* Build the Huffman tree and the prefix table.
*/
local void build_tree()
{
int nodes = 0; /* number of nodes (parents+leaves) at current bit length */
int len; /* current bit length */
uch *prefixp; /* pointer in prefix_len */
for (len = max_len; len >= 1; len--) {
/* The number of parent nodes at this level is half the total
* number of nodes at parent level:
*/
nodes >>= 1;
parents[len] = nodes;
/* Update lit_base by the appropriate bias to skip the parent nodes
* (which are not represented in the literal array):
*/
lit_base[len] -= nodes;
/* Restore nodes to be parents+leaves: */
nodes += leaves[len];
}
/* Construct the prefix table, from shortest leaves to longest ones.
* The shortest code is all ones, so we start at the end of the table.
*/
peek_bits = MIN(max_len, MAX_PEEK);
prefixp = &prefix_len[1<<peek_bits];
for (len = 1; len <= peek_bits; len++) {
int prefixes = leaves[len] << (peek_bits-len); /* may be 0 */
while (prefixes--) *--prefixp = (uch)len;
}
/* The length of all other codes is unknown: */
while (prefixp > prefix_len) *--prefixp = 0;
}
/* ===========================================================================
* Unpack in to out. This routine does not support the old pack format
* with magic header \037\037.
*
* IN assertions: the buffer inbuf contains already the beginning of
* the compressed data, from offsets inptr to insize-1 included.
* The magic header has already been checked. The output buffer is cleared.
*/
int unpack(in, out)
int in, out; /* input and output file descriptors */
{
int len; /* Bit length of current code */
unsigned eob; /* End Of Block code */
register unsigned peek; /* lookahead bits */
unsigned peek_mask; /* Mask for peek_bits bits */
ifd = in;
ofd = out;
read_tree(); /* Read the Huffman tree */
build_tree(); /* Build the prefix table */
clear_bitbuf(); /* Initialize bit input */
peek_mask = (1<<peek_bits)-1;
/* The eob code is the largest code among all leaves of maximal length: */
eob = leaves[max_len]-1;
Trace((stderr, "eob %d %x\n", max_len, eob));
/* Decode the input data: */
for (;;) {
/* Since eob is the longest code and not shorter than max_len,
* we can peek at max_len bits without having the risk of reading
* beyond the end of file.
*/
look_bits(peek, peek_bits, peek_mask);
len = prefix_len[peek];
if (len > 0) {
peek >>= peek_bits - len; /* discard the extra bits */
} else {
/* Code of more than peek_bits bits, we must traverse the tree */
ulg mask = peek_mask;
len = peek_bits;
do {
len++, mask = (mask<<1)+1;
look_bits(peek, len, mask);
} while (peek < (unsigned)parents[len]);
/* loop as long as peek is a parent node */
}
/* At this point, peek is the next complete code, of len bits */
if (peek == eob && len == max_len) break; /* end of file? */
put_ubyte(literal[peek+lit_base[len]]);
Tracev((stderr,"%02d %04x %c\n", len, peek,
literal[peek+lit_base[len]]));
skip_bits(len);
} /* for (;;) */
flush_window();
Trace((stderr, "bytes_out %ld\n", bytes_out));
if (orig_len != (ulg)bytes_out) {
error("invalid compressed data--length error");
}
return OK;
}发布于 2010-02-10 08:10:54
解压缩赫夫曼代码的典型方法是使用二叉树。您可以在树中插入代码,这样代码中的每个位都表示左(0)或右(1)的分支,叶子中有解码的字节(或任何值)。
然后,解码只是从编码内容中读取比特的情况,遍历每一位的树。当您到达叶子时,发出已解码的值,并继续读取,直到输入耗尽为止。
更新: 此页描述了该技术,并具有精美的图形。
https://stackoverflow.com/questions/2235208
复制相似问题