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