计算机问题求解一论题4-5 代数编码 2017年04月10日
计算机问题求解 – 论题4-5 - 代数编码 2017年04月10日
间题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
首先当然是因为编码无处不在，不仅如此…
p 0 We will assume that q transmission errors are rare, and,that when they do occur, q they occur independently in 1 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： 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