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A search algorithm whose OR nodes represent controllable choices and whose AND nodes require solutions for every possible outcome of a chosen action. In nondeterministic planning, a successful OR choice needs successful continuations for all its AND children. In contingent planning, those children can represent different observations and updated belief states. Cycle handling must match the required solution concept: ignoring a repeated state is not a proof that the goal will inevitably be reached.

Retain the ordered position state $(p,q)$, the same initial state and the goal test $p=q$, with unit joint-action cost. Replace the deterministic search transition model by a set-valued outcome relation. LetThenHere $\delta$ for an attempted move leaving the board returns the current room. This explicit boundary convention keeps every outcome inside the state space; alternatively disable a horizontal command unless both possible destinations are legal. Mrs Robot’s commands can still be restricted to legal moves as before.

The result is a fully observable nondeterministic planning problem, assuming that actual positions are observed after every step. Both possible horizontal outcomes must be represented, even though they may coincide at a boundary or when $n=1$. No outcome probabilities are supplied, so attaching equal probabilities would add an assumption. Its solution is a position-dependent plan or policy.

Use AND-OR search for a strong plan. At an OR node choose a joint action; at its AND node require successful continuations for every possible outcome. A repeated state on the current branch cannot by itself establish progress. On a finite state space, the equivalent strong planning winning-set iteration starts with $W_0=\{(p,p):p\in C\}$ and repeatedly adds states with an action whose nonempty outcome set is contained in the current winning set:A newly admitted state has a plan with at most $t+1$ remaining steps in the worst case. This algorithm terminates on the finite state space and is complete for finite guaranteed-success plans. In a general nondeterministic planning problem it cannot guarantee that a strong plan exists: an uncontrollable self-loop can prevent inevitable progress.

For this particular unobstructed grid, a strong plan does exist. Command Mr Robot to wait throughout and move Mrs Robot along a shortest route to his initial room. Her movement is deterministic and his waiting is unaffected by the horizontal uncertainty. This meets in $D(p_0,q_0)$ steps, regardless of every uncertain outcome, because no uncertain command is used. Thus a complete finite AND-OR search will find a solution here. With an unspecified disconnected maze, existence would instead require both robots to lie in the same reachable graph component.

For a finite nondeterministic planning problem with goal set $G$, initialize $W_0=G$ and putA state in $W_t$ has a strong plan reaching $G$ within $t$ steps. The proof is mathematical induction: an action witnessing admission to $W_{t+1}$ has only successors with plans of depth at most $t$. Conversely, the first action of any guaranteed plan of depth $t+1$ has every successor in $W_t$, again by mathematical induction. Since $W_t$ only grows inside a finite state space, the iteration stabilizes after at most the number of states many strict expansions. Recording a witnessing action at first admission gives a policy whose rank decreases at every nonterminal step.