Saturated Addition Over Tri-state Pulser Input: What kind of algebraic structure is this?

Question

I'm currently working with a "tri-state pulser" in an engineering context. To excite a response from this device, one provides a voltage sequence consisting of elements from $\{-1,0,1\}$. For example, a valid input looks something like this:

One can write out the shape of this input as a sequence: $(-1,1,0,0,-1,0,1)$. In general if we restrict valid inputs to be of length $L$, then a valid input is a tuple of the form $\{-1,0,1\}^L$.

I've been self-learning some algebra on the side, and was realizing that these valid inputs start to look like a vector space. To illustrate in the case of $L=3$:

• we can sometimes add these inputs nicely element-wise: $(0,1,-1) + (1,0,1) = (1,1,0)$
• we have an additive neutral element $(0,0,0)$
• we can multiply inputs element-wise by scalars from $(-1,1,0)$
• under element-wise addition each element has an "inverse" in the sense that another sequence exists that sends it to the neutral element (but elements are not invertible)

But we have a problem here! This is because we can't specify an input bigger than $1$ in absolute value to the pulser. We can model the pulser input as saturating so that the input $(1,-1) + (1,-1) = (1,-1)$. This implies that we do not have have associativity, as $-1+(1+1) = -1 + 1 = 0$ but $(-1+1)+1 = 0+1 = 1$.

Upon realizing this, I began to wonder if there was some other acceptable way of defining the group operation to both faithfully model the tri-state pulser but also get an actual group structure. Inputs are really a sort of "direct sum" of individual inputs, and so we can consider the problem at the individual input level. Unfortunately, there is only one unique group (up to isomorphism) with three elements, and this is the cyclic group $\mathbb{Z}/3\mathbb{Z}$.

Unfortunately, it makes no physical sense to say that providing a very high input voltage level (ex. $6$) to the pulser should be the same thing as providing zero voltage to the pulser. So the cyclic group doesn't seem to provide a good model.

So we have a structure that superficially looks like a vector space, where:

• "addition" is commutative, but not associative
• "addition" has an additive neutral, and "inverses" exist
• multiplication by a scalar is defined, has identity, and is distributive

What algebraic object is this?

(Parenthetical follow-up: Is group theory really the right tool to try and understand this structure? Any introductory book or course note recommendations on the fields of mathematics that study these weird structures are appreciated).

Answer 1

Let me mention first that $\Bbb F_3 = \Bbb Z / 3\Bbb Z = \{ -1, 0, 1\}$ is actually the field with three elements. The only thing you really need to remember to make this work is $1+1 = -1$, and therefore also $-1 - 1=1$. With this in mind, you could view your $L$-sequences as the vector space $\Bbb F_3^L$, but I am quite sure you really do want $1+1$ to be $1$, not $-1$.

In this case, and if you want addition to work nicely, I think you'll have to extend the space where you work to $\Bbb Z^L$, which is a $\Bbb Z$-module if you want to put a nametag on it.

This way your calculation would be $(1,0,-1)+(1,1,0)=(2,1,-1)$ and to obtain the actual sequence of signals, you apply the sign function to each entry (which you would define as $\operatorname{sgn}(0):=0$ and for $x\ne 0$, $$\operatorname{sgn}(x):=\frac{x}{|x|}.$$

Now this on the other hand may not make you very happy because it does not put an algebraic structure on $\{-1,0,1\}^L$. However, it feels to me like this is what is actually happening.

Let me elaborate: I think you want $1+1+1+(-1)+(-1)=1$, i.e. each negative signal can only cancel a single positive signal. This means you have to work in $\Bbb Z$. Now, you also want to collapse all positive values to $1$ and all negative values to $-1$, which means you partition $\Bbb Z=(-\Bbb N)\cup\{0\}\cup\Bbb N$ and divide by the equivalence relation corresponding to this partition. Now the problem is that this partition does not constitute the cosets of an ideal or anything nearly as nice as that, so dividing by it does not maintain a whole lot of the algebraic structure of $\Bbb Z$.

So I am afraid there are only these three options:

1. Define $1+1=1$ and accept that the operation will not satisfy the axioms that "nice" algebraic structures do.
2. Work over $\Bbb F_3$ and have $1+1=-1$ (likely unacceptable)
3. Bite the bullet and work in $\Bbb Z^L$.