Consider a binary code that consists of only four valid codewords as given below: $$00000, 01011, 10101, 11110$$ Let the minimum Hamming distance of the code be $p$ and the maximum number of erroneous bits that can be corrected by the code be $q$. Then the values of $p$ and $q$ are

Given the set of codewords $C = \{00000, 01011, 10101, 11110\}$:

1. Calculate the Hamming distance $d(c_i, c_j)$ between every pair of distinct codewords:

   • $d(00000, 01011) = 3$

   • $d(00000, 10101) = 3$

   • $d(00000, 11110) = 4$

   • $d(01011, 10101) = 4$

   • $d(01011, 11110) = 3$

   • $d(10101, 11110) = 3$

   The minimum Hamming distance is:

   $$p = d_{\min} = 3$$

2. The relationship between the minimum Hamming distance $d_{\min}$ and the error-correcting capability $q$ is:

   $$d_{\min} \ge 2q + 1$$

   $$3 \ge 2q + 1 \implies 2q \le 2 \implies q = 1$$

Therefore, $p = 3$ and $q = 1$, which corresponds to option (A).