Copyright | (c) Levent Erkok |
---|---|

License | BSD3 |

Maintainer | erkokl@gmail.com |

Stability | experimental |

Safe Haskell | None |

Language | Haskell2010 |

Demonstrates use of bounded list utilities, proving a simple mutex algorithm correct up to given bounds.

## Synopsis

- data State
- type SState = SBV State
- idle :: SState
- ready :: SState
- critical :: SState
- mutex :: Int -> SList State -> SList State -> SBool
- validSequence :: Int -> Integer -> SList Integer -> SList State -> SBool
- validTurns :: Int -> SList Integer -> SList State -> SList State -> SBool
- checkMutex :: Int -> IO ()
- notFair :: Int -> IO ()

# Documentation

Each agent can be in one of the three states

## Instances

mutex :: Int -> SList State -> SList State -> SBool Source #

A bounded mutex property holds for two sequences of state transitions, if they are not in their critical section at the same time up to that given bound.

validTurns :: Int -> SList Integer -> SList State -> SList State -> SBool Source #

The mutex algorithm, coded implicity as an assignment to turns. Turns start at `1`

, and at each stage is either
`1`

or `2`

; giving preference to that process. The only condition is that if either process is in its critical
section, then the turn value stays the same. Note that this is sufficient to satisfy safety (i.e., mutual
exclusion), though it does not guarantee liveness.

checkMutex :: Int -> IO () Source #

Check that we have the mutex property so long as `validSequence`

and `validTurns`

holds; i.e.,
so long as both the agents and the arbiter act according to the rules. The check is bounded up-to-the
given concrete bound; so this is an example of a bounded-model-checking style proof. We have:

Test turned off. See: https://github.com/Z3Prover/z3/issues/2956 checkMutex 20 All is good!

notFair :: Int -> IO () Source #

Our algorithm is correct, but it is not fair. It does not guarantee that a process that wants to enter its critical-section will always do so eventually. Demonstrate this by trying to show a bounded trace of length 10, such that the second process is ready but never transitions to critical. We have:

ghci> notFair 10 Fairness is violated at bound: 10 P1: [Idle,Idle,Ready,Critical,Idle,Idle,Ready,Critical,Idle,Idle] P2: [Idle,Ready,Ready,Ready,Ready,Ready,Ready,Ready,Ready,Ready] Ts: [1,2,1,1,1,1,1,1,1,1]

As expected, P2 gets ready but never goes critical since the arbiter keeps picking P1 unfairly. (You might get a different trace depending on what z3 happens to produce!)

Exercise for the reader: Change the `validTurns`

function so that it alternates the turns
from the previous value if neither process is in critical. Show that this makes the `notFair`

function below no longer exhibits the issue. Is this sufficient? Concurrent programming is tricky!