# CSCI 5854: Lecture 4 - Composition of Machines and Synchronization.

We will present compositions of extended finite state machines.

## Deterministic EFSM (Review)

Recall a deterministic EFSM from the previous lecture. $(Q, X, I, O, q_0, x_0, \delta),$

1. $$Q$$ is a finite set of modes (sometimes confusingly called states).
2. $$X$$ represents the typed internal state variables.
3. $$I$$ represents the input variables.
4. $$O$$ represents the typed output variables.
5. $$q_0 \in Q$$ is the starting mode; and $$x_0 \in D(X)$$ is the initial value of the internal state variables.
6. $$\delta: Q \times D(X) \times D(I) \rightarrow Q \times D(X) \times D(O)$$ is the transition function wherein $$\delta(q, x, i) : (q', x', z)$$ denotes that from the mode $$q$$, internal state value $$x$$, input value $$i$$, we move in a step to $$q' \in Q$$, $$x' \in D(X)$$ and output the value $$z \in D(O)$$.

The system is deterministic (since $$\delta$$ is a function). Thus for every pair $$(q, x)$$, with mode $$q \in Q$$ and $$x \in D(X)$$, and each input $$y \in D(I)$$, we have exactly one next state $(q',x') := \delta(q,x,y)\,.$

The model for a nondeterministic system differs in the following ways

1. It may have multiple initial modes $$Q_0 \subseteq Q$$ rather than a single one.
2. It may have multiple initial start states $$X_0 \subseteq D(X)$$ rather than a single start state $$x_0$$ for the deterministic case.
3. For some mode $$s$$, the outgoing guards on the transitions need not be mutually exclusive (permitting multiple transitions to be taken for some states and inputs) or mutually exhaustive (permitting no transitions to be taken for some states and inputs). The transition relation $$\delta$$ is a relation rather than a function: $\delta: \underset{\text{current state}}{\underbrace{Q \times D(X)}} \times D(I) \times \underset{\text{next state}}{\underbrace{Q \times D(X)}} \times D(O)$

## Compositions of EFSMs

Often it is hard to describe a monolithic system but easier to build it from parts using compositions. We will start with two basic types of compositions that can be used to build machines: parallel and sequential compositions. Note that more complex models can be built by combining many components in parallel and sequential compositions.

First, we illustrate the parallel composition $$M_1 \otimes M_2$$ of two systems $$M_1$$, $$M_2$$ as shown in the figure below:

The composed system has the combined inputs of $$M_1, M_2$$, combined internal states and outputs of the two individual systems.

Next, we illustrate the sequential composition $$M_1 \circ M_2$$ of two systems $$M_1$$, $$M_2$$ below:

The composed system has the input of $$M_1$$ and the output of $$M_2$$, but combines the internal states of $$M_1$$ and $$M_2$$.

For simplicity, let us define two deterministic EFSMS:

$M_1: (Q_1, X_1, I_1, O_1, q_{0,1}, x_{0,1}, \delta_1),\ \ M_2: (Q_2, X_2, I_2, O_2, q_{0,2}, x_{0,2}, \delta_2) \,.$

Parallel Composition
The parallel composition $$M_1 \otimes M_2$$ is has the following components 1. Finite set of modes $$Q_1 \times Q_2$$ as the Cartesian product of the modes of each machine. 2. Internal state variables $$X_1 \cup X_2$$. Note however, that this means that the internal states will in fact range over $$D(X_1) \times D(X_2)$$. 3. Input variables $$I_1 \cup I_2$$ 4. Output variables $$O_1 \cup O_2$$. 5. Initial mode: $$(q_{0,1}, q_{0,2})$$ 6. Initial internal state $$(x_{0,1}, x_{0,2}) \in D(X_1) \times D(X_2)$$. 7. Transition function $$\delta_{12}$$ defined as $\delta_{12}( (q_1,q_2), (x_1, x_2), (y_1, y_2)) = \left( (\hat{q_1}, \hat{q_2}), (\hat{x_1},\hat{x_2}), (z_1, z_2) \right) \,,$ wherein $\delta_1(q_1, x_1, y_1) = (\hat{q_1}, \hat{x_1}, z_1)$ and $\delta_2(q_2, x_2, y_2) = (\hat{q_2}, \hat{x_2}, z_2)\,.$

In other words, the parallel composition behaves as if both machines execute in parallel.

### Example of Parallel Composition

Consider two machines shown in the diagram below:

• The parallel composition $$M_1 \otimes M_2$$ has two modes $$(q_1, r_1)$$ and $$(q_2, r_1)$$ given as the product of the two state variables.

• Its inputs are now $$\{y, y_1\}$$, outputs are $$\{z, z_1\}$$ and internal states are $$\{x, x_1\}$$.

• There are six transitions between them given by taking the product of each transition of $$M_1$$ with that of $$M_2$$. They are given by

$\begin{array}{lll} \hline \text{Source/Dest} & Guard & Update \\ \hline (q_1,r_1) \rightarrow (q_1, r_1) & y \leq 0 \land y_1 > 0 & x := x+1, x_1 := x_1 -1, z := x+1 , z_1 := x_1 +1 \\ \hline (q_1,r_1) \rightarrow (q_1, r_1) & y \leq 0 \land y_1 \leq 0 & x := x+1, x_1 := x_1 +1, z := x+1 , z_1 := x_1 -1 \\ \hline (q_1,r_1) \rightarrow (q_2, r_1) & y > 0 \land y_1 > 0 & x := x, x_1 := x_1 -1, z := x , z_1 := x_1 +1 \\ \hline (q_1,r_1) \rightarrow (q_2, r_1) & y > 0 \land y_1 \leq 0 & x := x, x_1 := x_1 +1, z := x , z_1 := x_1 -1 \\ \hline (q_2,r_1) \rightarrow (q_2, r_1) & y_1 > 0 & x := x+1, x_1 := x_1 -1, z := 1 , z_1 := x_1 +1 \\ \hline (q_2,r_1) \rightarrow (q_2, r_1) & y_1 \leq 0 & x := x+1, x_1 := x_1 +1, z := 1, z_1 := x_1 -1 \\ \hline \end{array}$

Sequential Composition
The sequential composition $$M_1 \otimes M_2$$ is only possible provided the following condition holds: > $$O_1 = I_2$$, the outputs of the first machine must be fed as inputs to the second. If the composition is possible, then the result has the following components
1. Finite set of modes $$Q_1 \times Q_2$$ as the Cartesian product of the modes of each machine.
2. Internal state variables $$X_1 \cup X_2$$. Note however, that this means that the internal states will in fact range over $$D(X_1) \times D(X_2)$$.
3. Input variables $$I_1$$ (note the difference from parallel composition).
4. Output variables $$O_2$$ (note the difference from parallel composition).
5. Initial mode: $$(q_{0,1}, q_{0,2})$$
6. Initial internal state $$(x_{0,1}, x_{0,2}) \in D(X_1) \times D(X_2)$$.
7. Transition function $$\delta_{12}$$ defined as $\delta_{12}( (q_1,q_2), (x_1, x_2), y_1) = \left( (\hat{q_1}, \hat{q_2}), (\hat{x_1},\hat{x_2}), z_2 \right) \,,$ wherein $\delta_1(q_1, x_1, y_1) = (\hat{q_1}, \hat{x_1}, z_1)$ and $\delta_2(q_2, x_2, z_1) = (\hat{q_2}, \hat{x_2}, z_2)\,.$

Note here that machine 1 and 2 make a move in a single cycle. First machine $$M_1$$ computs an output $$z_1$$ for the input $$y_1$$ and current state $$(q_1, x_1)$$. Next, machine $$M_2$$ operates on $$z_1$$ as input and produces $$z_2$$, the final output.

### Example of Sequential Composition

Let us revisit the parallel composition example above and instead sequentially compose the two machines $$M_1$$ and $$M_2$$ as defined above.

The sequential composition's structure is the same as that of the parallel composition.

However, the transitions are now going to be completely different. Note that now, the input is $$y$$ and output is $$z_1$$. We can no longer talk about $$z$$ or $$y_1$$ in the composed machine since that connection is internal to the machine.

$\begin{array}{lll} \hline \text{Source/Dest} & Guard & Update \\ \hline (q_1,r_1) \rightarrow (q_1, r_1) & y \leq 0 \land x+1 > 0 & x := x+1, x_1 := x_1 -1, z_1 := x_1 +1 \\ \hline (q_1,r_1) \rightarrow (q_1, r_1) & y \leq 0 \land x+1 \leq 0 & x := x+1, x_1 := x_1 +1, z_1 := x_1 -1 \\ \hline (q_1,r_1) \rightarrow (q_2, r_1) & y > 0 \land x > 0 & x := x, x_1 := x_1 -1, z_1 := x_1 +1 \\ \hline (q_1,r_1) \rightarrow (q_2, r_1) & y > 0 \land x \leq 0 & x := x, x_1 := x_1 +1, z_1 := x_1 -1 \\ \hline (q_2,r_1) \rightarrow (q_2, r_1) & 1 > 0 & x := x+1, x_1 := x_1 -1, z_1 := x_1 +1 \\ \hline (q_2,r_1) \rightarrow (q_2, r_1) & 1 \leq 0 & x := x+1, x_1 := x_1 +1, z_1 := x_1 -1 \\ \hline \end{array}$

In particular, there is a simple way to derive each transition for the sequential composition from the table for the parallel one. Just apply the equality $$y_1 = z$$ to substitute for $$y_1$$ in terms of $$z$$ and the result of $$z$$ in terms of $$y,x$$ every place. Then we eliminate $$z$$ and $$y_1$$ themselves from the transitions. Note that in particular, the last transition has the guard $$1 \leq 0$$ or $$\text{false}$$. It can be safely deleted since it will never be taken.

## Synchronization Using Events

Often machines executing in parallel will need to synchronize their actions by sending or receiving events between each other. This functionality makes the description of many complex systems quite simple and succinct. At the same time, it can lead to some subtle bugs in these systems.

We will study just one type of synchronization for now: synchronous rendezvous that allows machines to send/receive events on transitions.

1. Edge $$s\rightarrow s_1$$ labeled by an event $$\texttt{event!}$$ that denotes the of an event called 'event', whenever we move from $$s$$ to $$s_1$$.
2. In a parallel machine, we have $$t \rightarrow t_1$$ is labeled by an event $$\texttt{event?}$$ that denotes that to take the edge from $$s \rightarrow s_1$$, the event called 'event' must be received.

There are many possible types of communication mechanisms implemented in actual systems in practice. For example, the rendezvous construct is inspired by the ADA programming language. Likewise, Java has wait/notify statements that also have the flavor of sending and receiving events. Similarly, if you are familiar with the message passing interface (MPI) used in large scale distributed computers, the different nodes can communicate by sending and receiving messages, as well.

In EFSMs, we will have the following constraints: 1. Any time an edge $$s \rightarrow s_1$$ wishes to send an event 'event!', there must be a receiver $$t \rightarrow t_1$$ executed in the parallel machine at the same cycle to receive the message. In other words, we will have blocking sends. 2. Any time an edge $$t \rightarrow t_1$$ has to receive a message 'event?' to execute, there must be an edge in the parallel machine that will send that event in that cycle.

Note that each send must be matched by a receive but the send/receive themselves can exchange places in the events.

### Example with Synchronous Rendezvous

Consider two traffic lights $$S_1$$ and $$S_2$$ that control traffic going NS and EW on a road, respectively.

Using message passing, we can write down two separate implementations of each traffic light that can synchronize with each other to avoid both lights simultaneously green or yellow at the same time. Likewise, we can also avoid the lights being simultaneously red.

For convenience, we do not consider inputs, internal states or outputs yet. There is a self loop on each red/yellow node that is to be taken if the events on the transitions cannot be sent/received.

1. Initially, the lights start in the state $$(G_1, R_2)$$.
2. The only move is $$(Y_1, R_2)$$.
3. Next, the first machine sends 'ok2!' event while the other machine receives 'ok2?' to reach the state $$(R_1, G_2)$$.
4. Next we get to $$(R_1, Y_2)$$
5. Next, the event 'ok1!' is sent by the second machine and received ('ok1?') to get to $$(G_1, R_2)$$. The sequence of actions repeat ad infinitum.

## EFSMs with events

We can now extend EFSMs with events. Let $$E := \{ e_1, \ldots, e_k \}$$ be a collection of events that can be sent or received along edges in an EFSM.

An EFSM extended with events has the following information along each transition:

• A guard predicate over inputs/internal state variables, and update assignments over output and new internal state variables, as before.
• A subset $$E_{s} \subseteq E$$ of events that are to be sent and $$E_r \subseteq E$$ that are to be received whenever the edge is taken.

Once the EFSM is extended with events, consider the parallel compositition of two transitions

In particular, each transition is shown with a guard predicate, updates on the internal states and outputs. But additionally, we have

1. First transition sends events $$E_s$$ and receives events $$E_r$$,

2. Second transition sends events $$\widehat{E_s}$$ and receives $$\widehat{E_r}$$.

The composed transition goes from $$(s_1, t_1)$$ to $$(s_2, t_2)$$ with guard $$\Psi_1 \land \Psi_2$$ and actions that simply combine the assignments to the internal states and outputs. The combined transition sends the events:

• $$(E_s \setminus \widehat{E_r}) \cup (\widehat{E_s} \setminus E_r)$$, I.e, every event sent by each transition that is not already received by the other will remain as an event to be sent in the composed transition.

• $$(E_r \setminus \widehat{E_s}) \cup (\widehat{E_r} \setminus E_s)$$, I.e, every event received by each transition that is not already sent by the other will remain as an event to be received when taking the composed transition.