I'm trying to think of the right way to tackle this situation and will welcome any suggestion.

I'm working with bit vectors, that is, vectors where each element is either 0 or 1. At the moment, I'm interested in 4 dimensions, thus vectors of the form:


with each $b{_i}$ in (0,1)

In my problem, two vectors are equivalent if one can be shifted to become the other. So for instance:

$(1,1,1,0)$ $\cong$ $(0,1,1,1)$


$(1,1,0,0)$ !$\cong$ $(1,0,1,0)$

My question is: what is a good way to give this an algebraic structure so I can reason about the different congruence classes within this space? For instance, given a vector, how many other vectors are congruent to it, and other questions like that?

  • $\begingroup$ If you know about group actions, you can let $\mathbb Z_4$ act on your set of vectors, where the element $k \in \mathbb{Z}_4$ maps $(b_0,b_1,b_2,b_3)$ to $(b_k, b_{k+1}, b_{k+2}, b_{k+3})$ (where the indexing is taken mod 4). Then your equivalence classes are precisely the orbits of the action, and there are nice things like the orbit-stabilizer theorem which will give you insight. $\endgroup$ – Bungo Nov 11 '15 at 2:19

To expand my comment a bit, this sort of problem is well suited to analysis by means of group actions. In case this notion is unfamiliar, I'll give a brief tutorial and apply it to this problem.

Consider the group $G = \mathbb Z_4$, the set of integers modulo 4, under addition. So the elements of $G$ are simply $0,1,2,3$ with the addition rule that $i+j$ is computed modulo $4$, e.g. $2 + 3 = 1$.

Let $\Omega$ be your set of vectors of the form $(b_1,b_2,b_3,b_4)$ where each $b_i$ is $0$ or $1$. Then $\mathbb Z_4$ acts on $\Omega$ via the rule $k \cdot (b_1,b_2,b_3,b_4) = (b_{k+1},b_{k+2},b_{k+3},b_{k+4})$, where the indexing is computed mod $4$. So for example, $2 \cdot (b_1,b_2,b_3,b_4) = (b_3,b_4,b_1,b_2)$. It's easy to check that this rule satisfies the conditions of a group action.

This action, like any group action, partitions $\Omega$ into a disjoint set of equivalence classes, called orbits. The orbit of the vector $v = (b_1,b_2,b_3,b_4)$ is simply the set of all vectors that can be achieved by applying the group action to $v$, in this case shifting $v$ cyclically.

An element $k\in G$ stabilizes a vector $v$ if $k \cdot v = v$, meaning in this case that $v$ shifted by $k$ is just $v$ itself. For example, if $v = (1,0,1,0)$, then both $k=0$ and $k=2$ stabilize $v$, whereas $k=1$ and $k=3$ do not.

Some elementary facts about orbits and stabilizers: the stabilizer of an element $v$, usually denoted $G_v$, is a subgroup of $G$, and therefore its order $|G_v|$ (the number of elements in $G_v$) must be a divisor of $|G|$, the order of the group. Since in your case, $G = \mathbb Z_4$ has $4$ elements, every stabilizer must have order $1$, $2$, or $4$.

The size of any orbit is equal to $|G|/|G_v|$ for any $v$ in the orbit. This means in your case that each orbit will have size $1$, $2$, or $4$.

Your set of vectors $\Omega$ contains $2^4 = 16$ elements total, and each vector in $\Omega$ lies in exactly one orbit, so $|\Omega| = 16$ must equal the sums of the sizes of the orbits.

It's easy to check that there are exactly two orbits which contain one vector each, namely $(0,0,0,0)$ and $(1,1,1,1)$.

What about orbits of size $2$? The only vectors which can be in such an orbit are those which satisfy $(a,b,c,d) = (c,d,a,b)$, or equivalently, $a=c$ and $b=d$. We can choose $a$ and $b$ freely, then $b$ and $d$ are automatically assigned. This means there are $4$ vectors satisfying this condition. They are $(0,0,0,0)$, $(0,1,0,1)$, $(1,0,1,0)$, and $(1,1,1,1)$. Since $(0,0,0,0)$ and $(1,1,1,1)$ lie in their own size-one orbits, the only orbit of size $2$ is the one consisting of $(0,1,0,1)$ and $(1,0,1,0)$.

So far, we have two orbits of size $1$ and one orbit of size $2$, which accounts for $4$ elements of $\Omega$. All other elements of $\Omega$ must be grouped in orbits of size $4$ (the only other possibility). Since there are $12$ remaining elements to divide among these orbits, this means that there are exactly three orbits of size $4$.

To summarize, your group action induces:

  • Two orbits of size $1$
  • One orbit of size $2$
  • Three orbits of size $4$

Of course, it's a bit overkill to apply the theory of group actions to this example; you could just as easily analyze it by brute force. But group actions allow this sort of reasoning to generalize to a wide range of situations which can be difficult to analyze in other ways.

  • $\begingroup$ Wow, this is perfect. It's going to take me a while to digest, but it sounds like exactly what I was looking for. Thanks for the detailed explanation! $\endgroup$ – Joshua Frank Nov 11 '15 at 14:04

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