问题2： 要发现收到的报文中的错误，最 straight forward的方法是什么？ Example 2.Even parity, Using even parity,the codes for A,B,and C now become A=010000012, B=010000102, C=110000112 Suppose an A is sent and a transmission error in the sixth bit is caused by noise over the communication channel so that (0100 0101)is received
问题2： 要发现甚至纠正收到的报文中的错误， 最“straight forward的方法是什么？ 无论是发现，甚至在某种假设下发现及纠正错误，总有 冗余 的存在，必须的吗？
冗余 无论是发现，甚至在某种假设下发现及纠正错误，总有 的存在，必须的吗？
问题3： 我们必须考虑物理信道会出错，但又假设 出错“不多”，这是为什么？ We will also assume that a received n-tuple is decoded into a codeword that is closest to it;that is,we assume that the receiver uses maximum-likelihood decoding
问题4.1： 一个编码方案”究竞是什么？ 问题4.2 那么，一个好“编码方案”究竟是 好在哪里？
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.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四→Z5 and a decoding function D:Z→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” If we are to develop efficient error-detecting and error-correcting codes, we will need more sophisticated mathematical tools. 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