两个集合，两个函数 If we are to develop efficient error-detecting and error-correcting codes,we will need more sophisticated mathematical tools.Group theory will allow faster methods of encoding and decoding messages.A code is an (n,m)- block code if the information that is to be coded can be divided into blocks of m binary digits,each of which can be encoded into n binary digits.More specifically,an (n,m)-block code consists of an encoding function E:Z→Z and a decoding function D:z5→Z 问题58 其实，这n和m之间大小的差别就是“冗余” 关于m,m的大小，你能说点什么？
两个集合，两个函数 其实，这n和m之间大小的差别就是“冗余
两个集合，两个函数 If we are to develop efficient error-detecting and error-correcting codes,we will need more sophisticated mathematical tools.Group theory will allow faster methods of encoding and decoding messages.A code is an (n,m)- block code if the information that is to be coded can be divided into blocks of m binary digits,each of which can be encoded into n binary digits.More specifically,an (n,m)-block code consists of an encoding function E:Z→Z5 and a decoding function D:z→Z. 问题6 顺便问一下；为什么这里说群论会给我们很大带助？
两个集合，两个函数
间题7： 在上面的假设下，采用最大相似度的解码 方法，你认为怎样的code word集合有利 于发现并纠正错码？ 小提示：解码过程可以理解为从receive word定位到 code word,再从code word解码到明文组，其中最迷惑 解码人的是哪一步？
小提示：解码过程可以理解为从receive word定位到 code word，再从code word解码到明文组，其中最迷惑 解码人的是哪一步？
问题8： 你能从第7个间题的思考中，理解码字 hamming距离的本质，以及随后定义出来 的编码系统的最小距离的用意吗？ Ifx=(1100)andy=(1010) are codewords,then d(x,y)=2.If x is transmitted and a single error occurs,then y can never be received. 如果=(1100)，y=(1011),我们收到了 一个码1110，我们可以得到什么结论？
如果x=(1100), y=(1011),我们收到了 一个码1110,我们可以得到什么结论？
最小码距与纠错能力的关系 Theorem 8.3 Let C be a code with dmin 2n+1.Then C can correct any n or fewer errors.Furthermore,any 2n or fewer errors can be detected in C. PROOF.Suppose that a codeword x is sent and the word y is received with at most n errors.Then d(x,y)<n.If z is any codeword other than x,then 2n+1≤d(x,z≤d(x,y)+d(y,z)≤n+d(y,z). Hence,d(y,z)>n+1 and y will be correctly decoded as x.Now suppose that x is transmitted and y is received and that at least one error has occurred,but not more than 2n errors.Then 1 d(x,y)<2n.Since the minimum distance between codewords is 2n +1,y cannot be a codeword. Consequently,the code can detect between 1 and 2n errors. 口
最小码距与纠错能力的关系