Ferryman Problem

Problem

There is a ferryman, goat, cabbage and a wolf on one side of a river. The ferryman can cross a river with at most one passenger in his boat. There is a behavioural conflict between

1. the goat and the cabbage; and
2. the goat and the wolf;

if they are on the same side of the river bank but the ferryman crosses the river or stays on the other bank.

Can ferryman transport all goods to the other side, without any conflicts occuring?

Solution

This is a planning problem but can be solved using model checking.

We describe a transition system in which the states represent which goods are at which side of the river. Then we ask if the goal state is reachable from the initial state: Is there a path from the initial state such that it has a state along it at which all the goods are on the other side, and during the transitions to that state the goods are never left in an unsafe, conflicting situation?

We model all possible behaviour (including that which results in conflicts) as a NuSMV program. The location of each agent is modelled as a boolean variable: FALSE denotes that the agent is on the initial bank, and TRUE the destination bank. Thus, ferryman = FALSE means that the ferryman is on the initial bank, ferryman = TRUE that he is on the destination bank, and similarly for the variables goat, cabbage and wolf.

The variable carry takes a value indicating whether the goat, cabbage, wolf or nothing is carried by the ferryman. The definition of next(carry) works as follows. It is non-deterministic, but the set from which a value is non-deterministically chosen is determined by the values of ferryman, goat, etc., and always includes FALSE. If ferryman = goat (i.e., they are on the same side) then g is a member of the set from which next(carry) is chosen. The situation for cabbage and wolf is similar. Thus, if ferryman = goat = wolf != cabbage then that set is {g, w, 0}. The next value assigned to ferryman is non-deterministic: he can choose to cross or not to cross the river. But the next values of goat, cabbage and wolf are deterministic, since whether they are carried or not is determined by the ferryman’s choice, represented by the non-deterministic assignment to carry; these values follow the same pattern.

Instructions to run

./nusmv.exe ferryman.smv

Sample Solution Path

*** This is NuSMV 2.6.0 (compiled on Wed Oct 14 15:37:51 2015)
*** Enabled addons are: compass
*** For more information on NuSMV see <http://nusmv.fbk.eu>
*** or email to <nusmv-users@list.fbk.eu>.
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*** Copyright (c) 2010-2014, Fondazione Bruno Kessler

*** This version of NuSMV is linked to the CUDD library version 2.4.1
*** Copyright (c) 1995-2004, Regents of the University of Colorado

*** This version of NuSMV is linked to the MiniSat SAT solver.
*** See http://minisat.se/MiniSat.html
*** Copyright (c) 2003-2006, Niklas Een, Niklas Sorensson
*** Copyright (c) 2007-2010, Niklas Sorensson

-- specification !(((goat = cabbage | goat = wolf) -> goat = ferryman) U (((cabbage & goat) & wolf) & ferryman))  is false
-- as demonstrated by the following execution sequence
Trace Description: LTL Counterexample
Trace Type: Counterexample
  -- Loop starts here
  -> State: 1.1 <-
    ferryman = FALSE
    goat = FALSE
    cabbage = FALSE
    wolf = FALSE
    carry = none
  -> State: 1.2 <-
    ferryman = TRUE
    goat = TRUE
    carry = g
  -> State: 1.3 <-
    ferryman = FALSE
    carry = none
  -> State: 1.4 <-
    ferryman = TRUE
    wolf = TRUE
    carry = w
  -> State: 1.5 <-
    ferryman = FALSE
    goat = FALSE
    carry = g
  -> State: 1.6 <-
    ferryman = TRUE
    cabbage = TRUE
    carry = c
  -> State: 1.7 <-
    ferryman = FALSE
    carry = none
  -> State: 1.8 <-
    ferryman = TRUE
    goat = TRUE
    carry = g
  -> State: 1.9 <-
    ferryman = FALSE
    wolf = FALSE
    carry = w
  -> State: 1.10 <-
    ferryman = TRUE
    carry = none
  -> State: 1.11 <-
    ferryman = FALSE
    cabbage = FALSE
    carry = c
  -> State: 1.12 <-
    ferryman = TRUE
    carry = none
  -> State: 1.13 <-
    ferryman = FALSE
    goat = FALSE
    carry = g
  -> State: 1.14 <-
    ferryman = TRUE
    carry = none
  -> State: 1.15 <-
    ferryman = FALSE

Invoking bounded model checking will produce the shortest possible path to violate the property.

References

Michael Huth and Mark Ryan. 2004. Logic in Computer Science: Modelling and Reasoning about Systems.
Cambridge University Press, USA.