计算机问题求解一论题4-7 代数编码 2021年04月19日
计算机问题求解 – 论题4-7 - 代数编码 2021年04月19日
问题1: 为什么易于发现错误,甚至易于 纠正错误的编码方案非常重要? 首先当然是因为编码无处不在, 不仅如此.… 无处不在到什么程度?
首先当然是因为编码无处不在, 不仅如此… 无处不在到什么程度?
p U 0 We will assume that transmission errors are q rare,and,that when they 1 do occur,they occur independently in each bit. Figure 8.2.Binary symmetric channel The probability that no errors occur during the transmission of a binary codeword of length n is p".For example,if p =0.999 and a message consisting of 10,000 bits is sent,then the probability of a perfect transmission is (0.999)10,000≈0.00005
We will assume that transmission errors are rare, and, that when they do occur, they occur independently in each bit
即使传输一个bit出错概率不大. Theorem 8.1 If a binary n-tuple (1,...,n)is transmitted across a binary symmetric channel with probability p that no error will occur in each coor- dinate,then the probability that there are errors in exactly k coordinates is pn-k 假如传输1个bit,出错的概率是千分之五,假如要传500个bits,那么: 不出错的概率是:8.2%; 有一位错的概率是:20.4%;有两位错的概率是;25.7%; 而两位以上错误的概率是:45.7%
即使传输一个bit出错概率不大… 假如传输1个bit, 出错的概率是千分之五,假如要传500个bits,那么: 不出错的概率是:8.2%; 有一位错的概率是:20.4%; 有两位错的概率是;25.7%; 而两位以上错误的概率是:45.7%
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 1.One possible coding scheme would be to send a message several times and to compare the received copies with one another.Suppose that the message to be encoded is a binary n-tuple (x1,22,...,n).The message is encoded into a binary 3n-tuple by simply repeating the message three times: x1,c2,.,xn)→(x1,2,,cn,x1,x2,,xn,x1,2,·,xn)
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