文件列表:

10.4\10-4.pdf (150273, 2005-07-07)
10.4\Debug\ex10.exe (233539, 2011-03-24)
10.4\Debug\ex10.ilk (242292, 2011-03-24)
10.4\Debug\ex10.pch (3588456, 2011-03-24)
10.4\Debug\ex10.pdb (435200, 2011-03-24)
10.4\Debug\in.dat (0, 2012-12-20)
10.4\Debug\main.obj (22989, 2011-03-24)
10.4\Debug\vc60.idb (140288, 2012-12-20)
10.4\Debug\vc60.pdb (77824, 2011-03-24)
10.4\ex10.dsp (4260, 2005-06-28)
10.4\ex10.dsw (533, 2005-06-28)
10.4\ex10.exe (233521, 2005-07-06)
10.4\ex10.ncb (50176, 2012-12-20)
10.4\ex10.opt (48640, 2012-12-20)
10.4\ex10.plg (242, 2012-12-20)
10.4\in.dat (2916, 2012-12-20)
10.4\lena.bmp (66614, 2005-06-29)
10.4\main.cpp (7421, 2005-07-06)
10.4\result.dat (2871, 2012-12-20)
10.4\xiaobobao.m (4631, 2007-10-21)
3\d4_draw1.m (340, 2012-10-18)
3\d4_draw2.m (341, 2012-10-18)
4.2\check.m (881, 2012-10-25)
5.1\wavedec.m (0, 2012-11-08)
5.1\wavedec_test.m (429, 2012-11-08)
5.4\EZW-code\checkchildren.m (337, 2012-11-15)
5.4\EZW-code\checkdescendants1.m (487, 2012-11-15)
5.4\EZW-code\dominantpass.m (3946, 2012-11-15)
5.4\EZW-code\ezw.m (4445, 2012-11-15)
5.4\EZW-code\mapping.m (144, 2003-10-06)
5.4\EZW-code\subordinatepass.m (345, 2012-11-15)
5.4\EZW-code\Text3.txt (2405, 2011-03-29)
5.5\5-5.doc (32768, 2005-07-07)
5.5\SPIHT\contents.m (1340, 2004-04-26)
5.5\SPIHT\func_DWT.m (1588, 2004-04-26)
5.5\SPIHT\func_InvDWT.m (1750, 2004-04-26)
5.5\SPIHT\func_Myappcoef2.m (1683, 2004-04-26)
5.5\SPIHT\func_MyDescendant.m (1158, 2004-04-26)
5.5\SPIHT\func_Mywavedec2.m (2127, 2004-04-26)
... ...

这里的每个文件均定义了一个函数： 函数dominantpass实现主扫描 函数subordinatepass实现辅扫描 函数mapping生成一个与图像同样大小、存储着扫描序号的矩阵函数checkdescendantsl查询某点的子孙是否均小于阈值 函数checkchildren将某T型点的子孙全部标识出来 主函数为ezw(X，LEVEL)，其中X为待处理的小波系数矩阵，LEVEL为欲编码的次数 具体执行过程： 在MATLAB命令窗口中先输入X矩阵的内容，然后键入ezw(X，LEVEL)，LEVEL为某正整数，则会输出每一次编解码的输出结果，以及原系数矩阵、最终重构的矩阵及两者差值。 注：这里的算法与书上的步骤由两点细微的不同：一是对于没有子孙的叶子节点，若其为不重要系数，则用T表示，而书上用Z表示，所以导致一些主扫描表符号的差异；二是没有对重要系数进行重新排序。这些都不会影响到编解码的过程，仅在此作一注明。 程序执行举例： 在命令窗口中输入 >> X=[63 -34 49 10 7 13 -12 7; % 输入系数矩阵X -31 23 14 -13 3 4 6 -1; 15 14 3 -12 5 -7 3 9; -9 -7 -14 8 4 -2 3 2; -5 9 -1 47 4 6 -2 2; 3 0 -3 2 3 -2 0 4; 2 -3 6 -4 3 6 3 6; 5 11 5 6 0 3 -4 4]; >> ezw(X,5) % 对X进行5次ezw编解码 程序的输出显示了每一次编解码的结果： There are the EZW encoding outputs below threshold = 32 D = pnztpttttztttttttptt S = 1 0 1 0 threshold = 16 D = npttttttttttttttt S = 1 0 0 1 1 0 threshold = 8 D = ppnppnttnnptpttnttttttttptttpttttttttptttttttttttt S = Columns 1 through 10 1 0 0 1 1 1 0 1 1 1 Columns 11 through 20 1 0 1 1 0 1 1 0 0 0 threshold = 4 D = nzptppptpnpttttntttttttppnpppptttttptptttpnp S = Columns 1 through 10 1 1 0 1 1 1 1 1 0 1 Columns 11 through 20 1 0 0 1 0 0 0 0 0 1 Columns 21 through 30 1 1 0 1 1 0 1 0 0 0 Columns 31 through 40 1 0 0 1 0 1 0 1 1 0 Column 41 0 threshold = 2 D = pptnppppttnppnpnnptptpp S = Columns 1 through 10 1 0 1 1 1 1 0 0 1 1 Columns 11 through 20 0 1 0 0 0 1 0 1 1 1 Columns 21 through 30 1 1 0 1 0 1 1 0 1 1 Columns 31 through 40 0 0 1 0 0 0 0 0 0 0 Columns 41 through 50 0 1 1 0 1 1 0 1 1 0 Columns 51 through 59 0 1 1 0 0 0 1 1 1 There are the decoding outputs below Threshold = 32 level = 1 XX = 56 -40 56 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 40 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Threshold = 16 level = 2 XX = 60 -36 52 0 0 0 0 0 -28 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 44 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Threshold = 8 level = 3 XX = 62 -34 50 10 0 14 -14 0 -30 22 14 -14 0 0 0 0 14 14 0 -14 0 0 0 10 -10 0 -14 10 0 0 0 0 0 10 0 46 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 Threshold = 4 level = 4 XX = 63 -35 49 11 7 13 -13 7 -31 23 15 -13 0 5 7 0 15 15 0 -13 5 -7 0 9 -9 -7 -15 9 5 0 0 0 -5 9 0 47 5 7 0 0 0 0 0 0 0 0 0 5 0 0 7 -5 0 7 0 7 5 11 5 7 0 0 -5 5 Threshold = 2 level = 5 XX = 63 -34 49 10 7 13 -12 7 -31 23 14 -13 3 4 6 0 15 14 3 -12 5 -7 3 9 -9 -7 -14 8 4 -2 3 2 -5 9 0 47 4 6 -2 2 3 0 -3 2 3 -2 0 4 2 -3 6 -4 3 6 3 6 5 11 5 6 0 3 -4 4 initialiamge = 63 -34 49 10 7 13 -12 7 -31 23 14 -13 3 4 6 -1 15 14 3 -12 5 -7 3 9 -9 -7 -14 8 4 -2 3 2 -5 9 -1 47 4 6 -2 2 3 0 -3 2 3 -2 0 4 2 -3 6 -4 3 6 3 6 5 11 5 6 0 3 -4 4 reconstructedimage = 63 -34 49 10 7 13 -12 7 -31 23 14 -13 3 4 6 0 15 14 3 -12 5 -7 3 9 -9 -7 -14 8 4 -2 3 2 -5 9 0 47 4 6 -2 2 3 0 -3 2 3 -2 0 4 2 -3 6 -4 3 6 3 6 5 11 5 6 0 3 -4 4 difference = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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