There are several block error correction codes that make a decision on the basis of minimum distance. That is$,$ given a code consisting of $s$ equally likely codewords of length $n,$ for each received sequence $v,$ the receiver selects the codeword $w$ for which the distance $d(w,v)$ is a minimum. We would like to prove that this scheme is "ideal" in the sense that the receiver always selects the codeword for which the probability of $w$ given $(v|),$ is a maximum. Because all codewords are assumed equally likely, the codeword that maximizes $(v|)$ is the same as the codeword that maximizes $(w|).$

$($a$)$ In order that $w$ be received as $v,$ there must be exactly $d(w,v)$ errors in transmission, and these errors must occur in those bits where $w$ and $v$ disagree. Let $$ be the probability that a given bit is transmitted incorrectly and $n$ be the length of a codeword. Write an expression for $(w|)$ as a function of $,d(w,v),$ and $n.$ Hint: The number of bits in error is $d(w,v)$ and the number of bits not in error is $n-d(w,v).$

$($b$)$ Now compare $(w_1|)$ and $(w_2|)$ for two different codewords $w_1$ and $w_2$ by calculating $|)/(p(w_2|).$

$($c$)$ Assume that and show that $(w_1|)(w_2|)$ if and only if This proves that the codeword $w$ that gives the largest value of $(w|)$ is the word whose distance from $v$ is minimum.