Therefore, there exists a set of symplectic partners of the generators of the stabilizer group of the n − qubit bit flip code such that the minimum distance of the code is n . It is easy to verify that (11) is saturated by the parameters [[ n, 1 , n ; n − 1]]. These [[ n, 1 , n ; n − 1]] codes are not equivalent to any standard quantum stabilizer code, for there are no standard [[2 n − 1 , 1 , n ]] quantum codes. Example 2. By Theorem 2 with n = 5 and which is a parity-check matrix of a classical [5 , 1 , 5] repetition code, a set of generators of the stabilizer of the [[5 , 1 , 1]] 5-qubit bit flip code is Then we add the following set of generators: ˆ H$_{5}$ =    $^{1}$ 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1    $^{,}$ g ′ 1 = ZZIII g ′ 2 = IZZII g ′ 3 = IIZZI g ′ 4 = IIIZZ. ˜ h$_{1}$ = XXIII ˜ h$_{2}$ = IXXII ˜ h$_{3}$ = IIXXI ˜ h$_{4}$ = IIIXX. 19 These generators do not form symplectic pairs. Observe that [ ˜ g$_{i}$, ˜ g$_{j}$ ] = 0, [ ˜ h$_{i}$, ˜ h$_{j}$ ] = 0, and [ ˜ g$_{i}$, ˜ h$_{j}$ ] = 0 except { ˜ g$_{1}$, ˜ h$_{2}$ } = 0, { ˜ g$_{2}$, ˜ h$_{1}$ } = 0, { ˜ g$_{2}$, ˜ h$_{3}$ } = 0, { ˜ g$_{3}$, ˜ h$_{2}$ } = 0, { ˜ g$_{3}$, ˜ h$_{4}$ } = 0,