Interleaving possibilities in Promela

Question

I am new to Promela and I'm having some difficulties understanding how to calculate the interleaving possibilities of a model. Which one of the statements is counted in the calculation (how do I know which one to take)? Does process P have 5, as well as Q?

  #define N 2
  proctype P(){
           int counter = 0
           do
           :: counter<N -> printf("P␣"); counter++
  
          :: else -> break
          od
  }
 
  proctype Q(){
           int counter = 0
           do
           :: counter<N -> printf("Q␣"); counter++
           :: else -> break
           od
  }
 
  init {
      atomic{   run P();
                run Q();}
  }```

1. Run the model in random and interactive mode. How many interleaving possibilities does
the program have and which?
2. How many possibilities are there for arbitrary values of N?

score 0 · Answer 1 · answered Jul 20 '23 at 17:05

To calculate the interleaving possibilities of a model in Promela, you need to consider the non-deterministic choices made by the processes during execution. In this case, the processes P and Q run concurrently, and each process has two choices: either to execute the statement inside the loop (if the counter is less than N) or to break out of the loop.

The interleaving possibilities for the given program with N set to 2 are :
P and Q are running concurrently, and they have their own local counters (counter) starting from 0.
Both P and Q can increment their counters inside the loop while the condition counter<N is true.
If the counter reaches N in a process, it will break out of the loop.

Here are the possible interleavings:

P␣P␣Q␣Q␣ (Both P and Q increment their counters twice before breaking out of the loop.)
P␣Q␣P␣Q␣ (One P increments its counter, then Q increments its counter, and this continues.)

The interleaving possibilities are only two because the processes are running concurrently, and they have a deterministic choice to increment their counters until the condition counter<N is satisfied.

For your second question The number of possibilities for arbitrary values of N can be calculated as follows:

Both P and Q can interleave in any order until their counters reach N.
Since there are two processes (P and Q), each having N choices (increment the counter or break out), the total number of interleaving possibilities is 2^N * 2^N = 4^N.

So, for arbitrary values of N, there are 4^N possible interleaving execution sequences.

Not sure why this answer is downvoted, it would be nice people left a comment when they think something can be improved. /OT What about `PQQP`, `QPPQ`? Do you not consider these interleaving? If not, why? Also, your formula says `4^N`, which for `2` would yield `8` but it doesn't seem your example has `8` interleaving options. Perhaps this is a hint that something is not alright — Patrick Trentin, Jul 26 '23 at 15:26

Interleaving possibilities in Promela

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