Motivation. I have a bike lock with 4 dials, and I was wondering whether I can reach any combination by always turning a fixed number $k$, say $k=2$, of the dials, by $1$ position, instead of just turning $1$ dial at the time, which makes for a boring challenge.

Formal version. For any integer $n>2$, let $[n] =\{1,\ldots, n\}$ and let $C_n = ([n], E_n)$ denote the cyclic graph on the vertex set $[n] =\{1,\ldots, n\}$ with $$E_n = \big\{\{k, k+1\}: 1\leq k < n\big\}\cup\big\{0,n\big\}.$$

Let $d$ denote the number of dials, $n$ the number of positions that any dial can take, and let $k\leq d$ be the fixed number of dials we have to turn by $1$ position at every step.

The dial itself can be represented by $[n]^d$. For $x,y\in[n]^d$ we let the differing set $D(x,y)$ to be defined by $\{i\in[d]: x_i\neq y_i\}$ where $x_i$ denotes the $i$th component of $x\in [n]^d$.

So we can define the following bike lock graph $B(n, d, k)$ for positive integers $n,d,k>1$ with $k\leq d$:

1. $V(B(n,d,k)) = [n]^d$,
2. $E(B(n,d, k)) = \big\{\{x,y\} \in [n]^d: |D(x,y)| = k \text{, and for all } i\in D(x,y)\text{ we have } \{x_i, y_i\}\in E_n\big\}.$

Every configuration of the bike lock can be reached with the allowed moves if and only if the corresponding graph $B(n,d,k)$ is connected.

Question. Are there infinitely many integers $n>1$ such given an integer $d>2$, the graph $B(n,d,k)$ is connected for some integer $k$ with $2\leq k\leq d-1$?