计算机问题求解一论题4-7 代数编码 2019年04月24日
计算机问题求解 – 论题4-7 - 代数编码 2019年04月24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 (1,x2,...,n).The message is encoded into a binary 3n-tuple by simply repeating the message three times: (x1,2,.,cn)→(1，c2,.,xn,x1,x2,.,En,1,c2,..,xn)