There’s a correspondence between the notions of “combinational” and “sequential” in digital circuit engineering and some structure in state machines (and therefore monoids)  that seems interesting.

In digital logic, a “combinational” circuit like a logic gate has can be associated with a time T so that the output depends only on input signals applied over the last T time units. If the propagation delay is T, then signals applied T+n time units ago are irrelevant to current state.  For example, an AND gate with propagation delay T has output 1 if all inputs have been set HIGH over the last T time units. For a sequential circuit, like a latch, there is no magic time T. Suppose we set the stored value in the  latch at time T1 and then signal it to hold the value. No matter how long the latch is held, we have to go back to absolute time T1 to find out what is in the latch. It’s worth noting however, that it’s often very useful to make a sequential circuit have a reset mode where it behaves like a combinatorial circuit. That is, we want to be able to say: no matter what previous signals have done to a memory device, some sequence of signals will clear it and, for example,  put the number 3 in it. More on this below.

Take a Moore machine M and write State(M,w) for the state reached by following input string “w” from the initial state. I’m limiting us to complete Moore machines where State(M,w) must be defined for every w.  Say M is combinational if there is some T so that if w=uz and length(z)>T then State(M,w)= State(M,z) . So the idea is that there is a T so that the last T inputs determine state for M. Call T the “depth” of M. As an example, suppose M represents a memory location and the inputs are just data values to store. Then we only need the last input, T=1.  If M is an AND gate with two pins and a delay of 2, then we can think of inputs as numbers 0,1,2,3 representing their 2 bit encodings: 0 for 00, 1 for 01, 2 for 10 and 3 for 11.  States can be pairs [x1,x2] representing the last two inputs so that if the state is [a,b] and input “c” is received, the new state is [c,b].  The state is actually a shift register. Suppose that the states are all pairs [x1,x2] where x1 and x2 are in the set {-1,0,1,2,3} with “-1” representing  an unknown signal. Then [3,3]  has output 1, [x1,x2] where both x1 and x2 are greater than -1 and less than 3 outputs 0 and the others have undefined output. The initial state is [-1,-1]. Obviously this machine is combinational with depth T=2. The intuition here is that combinatorial state machines will always “swim” to a fixed state given a fixed sequence of inputs of sufficient length. Obviously, this is a useful property of a device and even for sequential devices we are going to want some single or sequence of “reset” operations that, no matter what the current state, will force a reset to a known state. That is, for our latch, we’d want to force it to an assigned latched value, no matter what, if anything, it had previously latched. So properly designed sequential circuits are going to look like some combination of combinatorial and sequential circuit.

Now consider products of state machines. The method I’ve been using for representing state machines as maps f:InputSequences ->Outputs is convenient for products. If f represents a state machine M, then f(w) = Output(State(M,w)). If f represents a combinational state machine of depth T then f(wuz)= f(uz) whenever length(u)>T because State(M,wu)=State(M,u) so State(M,wuz)=State(M,uz) so the outputs must be the same.  Say F(w) = (g1(w1) …, gn(wn)) is a product under connection function Phi and map w |-> w1 … ,wn if when w=empty then wi = empty for all i and if when w maps to wi then wa maps to wi.Phi(a,i,F(w)) with ‘.’ representing string concatenation. Say that F is a cascade product if and only if  Phi(a,i,(x1 … xn)) depends only on xj for j < i. That is, in a cascade product, information only flows from lower numbered factors to higher numbered ones: there is no feedback in a cascade product. If g1 …,gn are combinatorial AND Phi(a,i,(x1 …,xn)) !=empty for any combination of parameters, and F is a cascade product, then F must be combinatorial. For proof, suppose Ti is the depth of each component gi and suppose w=uv where length(v)> T1. Then w1= u1.v1 where length(v1)> T1 so g1(v1)=g1(w1) and more than that g1(w1.z)=g1(v1.z) for all z.  That means Phi(a,i,F(wa))=Phi(a,F(va)) by the cascade property. And so on – eventually we see that the depth of F is the sum of the depths of the factors! For intuition, first we need enough inputs to steer factor 1 to a known state, then factor 2 and so on.

Given a state machine M with states S, we can associate a map from S to S with every string of inputs w so that F(M,w)(s)=s’ if and only if w leads from s to s’. This set of maps forms a monoid under ordinary function compositions  and since F(M,w)(F(z)(s))= F(M,wz)(s) then F(M,w)+F(M,z)= F(M,wz) where + is the monoid operation and not anything necessarily related to arithmetic. The empty string produces the monoid identity F(M,empty)(s)=s.  Look at the monoid structure of combinatorial state machines and note that unless the monoid is trivial, then  F(M,w)=F(M,empty) only if w=empty – there are no other identities.

The proof is easy. Suppose M is combinatorial  with depth T, the monoid is not trivial so that for some v we have F(M,v) != F(M,empty)and suppose F(M,z)=F(M,empty).  Then if z is not empty let Z consist of a string of concatenated z’s that is longer than T  so Z=zzzzzz …. stretched out to a length greater than T. By the combinatorial property, for any u, F(M,uZ)(s)=F(M,Z)(s) but because F(M,z)=F(M,empty) we also know F(M,uZ)(s)=s so F(M,w)(s)=s for all w and s which implies the monoid is trivial after all. It’s also worth pointing out that the combinatorial monoids seem to be a proper subset of the aperiodic monoids ( http://www.liafa.jussieu.fr/~jep/PDF/HandBook.pdf ). Consider the state machine with states {0,1,2} and inputs {a,b} so that, treating the inputs at maps from state to state, a(0)=1 and b(0)=2 for any non-zero state we have x(s)=s. The monoid is clearly aperiodic since a2=a 3 , a2(0)=1 and a2(s!=0)=s. But the state machine is not combinatorial.

Well, more later, but that’s what I have semi-clearly explicated for now. I’ll leave you with this photograph of Camille Jordan who lived in an unenlightened era that did not make clear demarcations between mathematics and mere engineering.