MATLAB数字图像处理练习八

1.哈夫曼编码
已知某信源发出的8个消息，其信源概率分布是不均匀的，分别为｛0.1，0.18，0.4，0.05，0.06，0.1，0.07，0.04｝，请对信源进行哈夫曼编码，并求出三个参数：平均码长、熵及编码效率。分析编码结果和书上不一致的原因。

2.算术编码和解码
假设信源符号为{a,b,c,d}，假设这些符号出现的概率分别为{0.2,0.2,0.4,0.2} ，写出对信源符号“bdadcadc”进行算术编码和解码的过程。

3.车牌号码识别
主要步骤：图像预处理、车牌区域定位、车牌字符分割、车牌字符识别。

4.图像拼接
编写程序，将两幅相邻图像（图1和图2）拼接成一幅全景图像（图3）。
图1:

图2:

图3:

二，解答

题目一

huffman.m文：

function CODE = huffman(p)
%HUFFMAN Builds a variable-length Huffman code for a symbol source.
%   CODE = HUFFMAN(P) returns a Huffman code as binary strings in
%   cell array CODE for input symbol probability vector P. Each word
%   in CODE corresponds to a symbol whose probability is at the
%   corresponding index of P. 
%
%   Based on huffman5 by Sean Danaher, University of Northumbria,
%   Newcastle UK. Available at the MATLAB Central File Exchange:
%   Category General DSP in Signal Processing and Communications. 

%   Copyright 2002-2004 R. C. Gonzalez, R. E. Woods, & S. L. Eddins
%   Digital Image Processing Using MATLAB, Prentice-Hall, 2004
%   $Revision: 1.5 $  $Date: 2003/10/26 18:37:16 $

% Check the input arguments for reasonableness.
error(nargchk(1, 1, nargin));
if (ndims(p) ~= 2) | (min(size(p)) > 1) | ~isreal(p) | ~isnumeric(p)
   error('P must be a real numeric vector.');     
end
   
% Global variable surviving all recursions of function 'makecode'
global CODE
CODE = cell(length(p), 1);  % Init the global cell array
                            
if length(p) > 1            % When more than one symbol ...
   p = p / sum(p);          % Normalize the input probabilities
   s = reduce(p);           % Do Huffman source symbol reductions
   makecode(s, []);         % Recursively generate the code
else  
   CODE = {'1'};            % Else, trivial one symbol case!
end;   

%-------------------------------------------------------------------%
function s = reduce(p);
% Create a Huffman source reduction tree in a MATLAB cell structure
% by performing source symbol reductions until there are only two
% reduced symbols remaining

s = cell(length(p), 1);

% Generate a starting tree with symbol nodes 1, 2, 3, ... to 
% reference the symbol probabilities.
for i = 1:length(p)
   s{i} = i; 
end

while numel(s) > 2
   [p, i] = sort(p);    % Sort the symbol probabilities
   p(2) = p(1) + p(2);  % Merge the 2 lowest probabilities
   p(1) = [];           % and prune the lowest one
   
   s = s(i);            % Reorder tree for new probabilities
   s{2} = {s{1}, s{2}}; % and merge & prune its nodes
   s(1) = [];           % to match the probabilities
end

%-------------------------------------------------------------------%
function makecode(sc, codeword)
% Scan the nodes of a Huffman source reduction tree recursively to
% generate the indicated variable length code words.

% Global variable surviving all recursive calls
global CODE                           

if isa(sc, 'cell')                   % For cell array nodes,
   makecode(sc{1}, [codeword 0]);    % add a 0 if the 1st element
   makecode(sc{2}, [codeword 1]);    % or a 1 if the 2nd
else                                 % For leaf (numeric) nodes,
   CODE{sc} = char('0' + codeword);  % create a char code string
end

调用函数

% function huffmanBIANMA 
 p=[0.1 0.18 0.4 0.05 0.06 0.1 0.07 0.04];
%p=[0.12 0.16 0.3 0.05 0.06 0.2 0.07 0.04];
code=huffman(p)

题目二

算数编码函数：

%算数编码函数
function acode = suanshubianma(symbol,ps,inseq)
high_range=[];
for k=1:length(ps)
    high_range=[high_range sum(ps(1:k))];
end
low_range=[0 high_range(1:length(ps)-1)];
sbidx=zeros(size(inseq));
for i=1:length(inseq)
    sbidx(i)=find(symbol==inseq(i));
end
low=0;
high=1;
for i=1:length(inseq)
    range=high-low;
    high=low+range*high_range(sbidx(i));
    low=low+range*low_range(sbidx(i));
end
acode=low;
end

算术解码函数

%算术解码函数
function symbos=suanshujiema(symbol,ps,codeword,symlen)
high_range=[];
for k=1:length(ps)
    high_range=[high_range sum(ps(1:k))];
end
low_range=[0 high_range(1:length(ps)-1)];
psmin=min(ps);
symbos=[];
for i=1:symlen
    idx=max(find(low_range<=codeword));
    codeword=codeword-low_range(idx);
    if abs(codeword-ps(idx))<0.01*psmin
        idx=idx+1;
        codeword=0;
    end
    symbos=[symbos symbol(idx)];
    codeword=codeword/ps(idx);
    if abs(codeword)<0.01*psmin
        i=symlen+1;
    end
end

主函数

%主函数
clc
clear
symbol=['abcd'];
ps=[0.2 0.2 0.4 0.2];
inseq='bdadc';
format long e
codeword=suanshubianma(symbol,ps,inseq);
outseq=suanshujiema(symbol,ps,codeword,length(inseq));

题目三

由于代码较长，这里只展示图片
有需要，请自取：
链接：https://pan.baidu.com/s/17hQBKmhN_mFwL6P4yH5x5w
提取码：sxqy

题目四

由于代码较长，这里只展示图片
有需要，请自取：
链接：https://pan.baidu.com/s/1XMlyH1fMehi4z1QZB0642g
提取码：sxqy