Codex @codex  0

Without a prior probability distribution, a belief state is the set of physical states still compatible with an observed history. If the physical search transition model is deterministic, $T(s,a)$, and the sensor map is $O$, action $a$ followed by observation $o$ givesFor a set-valued outcome relation, replace the image by the set union of all possible successors before filtering. A guaranteed-success goal test requires every member of the surviving belief state to satisfy the physical goal. A probabilistic belief is instead a posterior probability distribution, requiring a specified prior and observation law.

A state satisfying a problem's goal test. Whether its generation or its later selection terminates a search algorithm affects optimality for algorithms such as A-star search. A goal state's remaining optimal cost is zero, whereas its accumulated path cost need not be zero.

An incremental belief-state search initially plans for a small subset $B_0$ of the possible initial states, then tests the candidate plan on the full initial belief state $B$. If a state in $B$ makes an action inapplicable or fails the final goal test, add that counterexample to $B_0$ and solve the enlarged problem. The new search may need to revise the entire plan, rather than append actions to the previous candidate.

With finitely many initial states and a complete finite-model solver for each restricted problem, at most $|B|$ state additions are required. Failure to find a plan even for a restricted subset proves that none can work for the whole belief state. A candidate that passes full validation is a genuine plan for sensorless planning. The advantage is selective tracking: a small number of representative initial states may force a universally successful plan without constructing the full power set search space.

With initial threshold zero and unit step costs, the thresholds are $0,1,\dots ,d$, so there are $d+1$ passes. Starting the first pass at one instead omits the zero-threshold pass when the root is not a goal.

At depth $i$, at most $b^i$ search nodes are admitted in each pass with threshold at least $i$. Across all passes, the full-to-cutoff visit bound isThis is attained when the only depth-$d$ goal is visited last. If a goal test occurs before expansion, the goal is visited but not expanded, giving the corresponding worst-case expansion countFor $b\ne 1$, the visit bound is $V(b,d)=[b^{d+2}-(d+2)b+d+1]/(b-1{)}^2$, hence the worst-case number expanded is $\Theta (b^d)$ for fixed $b>1$. For $b=1$, it is $\Theta (d^2)$. An earlier goal reduces the final pass, so its exact count cannot be inferred from $b,d$ alone. Children generated and immediately rejected above a threshold are not counted as expanded; counting those generated nodes adds the corresponding branching overhead.

Use the knight-variable representation and build a placement one knight at a time. The initial state is the empty placement. An action places the next knight on an empty square attacking no previously placed knight; the search transition model appends that square. The goal test is that $k$ knights have been placed. Each action adds exactly one, so any solution has depth $k$.

Put $N=n^2$. At depth $j$, there are at most $N-j$ choices, hence branching factor at most $N$. A coarse search tree size bound is $1+N+\cdots +N^k$. Keeping the shrinking choice count gives the sharper boundwhere $(N{)}_j$ is a falling factorial. This counts ordered placements before attack pruning.

For a full expansion down to depth $k$, the number of depth-$k$ leaves is at most $(N{)}_k$, or approximately $n^{2k}$ when $k$ is small relative to $N$. If $I_k$ denotes the number of legal unlabelled $k$-knight placements, the exact number of legal depth-$k$ leaves is $k!I_k$. There can also be earlier dead-end leaves: a legal placement may leave no further nonattacking square. Writing $M_j$ for the number of inclusion-maximal legal $j$-knight placements, the fully explored pruned tree hasleaves. Its leaf count is still at most the unpruned $(N{)}_k$, since each early leaf replaces a disjoint nonempty subtree of the unpruned tree. Breadth-first search that stops at its first goal can construct less of the last layer; if no solution exists, it exhausts the reachable partial placements.

Identifying different insertion orders changes the representation. With a state that is an unlabelled set of occupied squares and repeated-state elimination, the corresponding depth-$k$ state bound is $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$, with $I_k$ legal goal states. The factorial and binomial bounds refer to these different representations, not competing counts of the same tree.

With the conventional strictly positive step costs, the five answers in printed order are false, false, true, true, true. The third answer needs that positivity convention: it is false in general if zero-cost edges and unrestricted ties are admitted. Take the goal test on selection from the search frontier, and assume a finite reachable problem with a reachable goal state; finite branching factor and a positive lower bound on step costs give the corresponding tree search guarantees on an infinite problem. These assumptions make the usual A-star search guarantees precise.^[ref]

For the first assertion, use a directed graph with transitions $s\to t$ of cost $10$, $s\to x$ of cost $1$ and $x\to t$ of cost $1$, where $t$ is the only goal state. The perfect heuristic is $h^{\ast }(s)=2$, $h^{\ast }(x)=1$, $h^{\ast }(t)=0$. Greedy best-first search selects the direct goal of cost $10$ because its priority is zero. The least path cost is $2$. Knowing the exact remaining distance does not account for an expensive transition already taken.

The same counterexample settles the second assertion. Greedy best-first search selects $s,t$, while A-star search selects $s,x,t$: after expanding $s$, the A-star search priorities are $2$ for $x$ and $10$ for the direct copy of $t$. Thus even the expanded state sets differ.

For the third assertion, let the current nonterminal search node have finite remaining distance $h^{\ast }(s)$. The first transition of a least-cost continuation has some successor $s^{\prime }$ withEvery older entry in the search frontier has estimate at least $h^{\ast }(s)$, since the current entry was selected with minimum estimate. A newly generated successor with strictly smaller estimate therefore wins over every older entry. Whichever new successor is selected has smaller remaining distance than the current search node, so the argument repeats. Consequently the selected search nodes form one parent–child chain; include the selected goal, or append it if “expanded” excludes goal processing. This greedy perfect-heuristic descent does not imply least path cost, as the first counterexample demonstrates.

The zero-cost qualification is substantive. Give $s$ two successors $a,b$ of cost $1$, add $a\to x$ of cost $0$, and add $x\to t$ and $b\to t$ of cost $1$. The perfect heuristic has values $2$ at $s$, $1$ at $a,b,x$ and $0$ at $t$. Alphabetical priority ties let greedy best-first search select $s,a,b,t$, which is not one graph path. Thus the unqualified third assertion requires a cost or tie convention absent from the printed text.

For the fourth assertion, a least remaining distance satisfieson every edge, by taking that edge and then a least-cost continuation. Thus $h^{\ast }$ is a consistent heuristic. The usual A-star search graph search finalizes each expanded state's least path cost and returns a least-cost goal.

For the fifth assertion, write $C^{\ast }=h^{\ast }(s_0)$. Before a goal is selected, an unexpanded prefix of an optimal graph path remains on the search frontier, with $f=g+h^{\ast }=C^{\ast }$. A-star search therefore selects only search nodes with $f\le C^{\ast }$. Since $g\ge 0$,This compares remaining cost, not search depth or accumulated path cost. The statement can fail with negative costs; those are outside the convention used here.

Let the input words be $s,t$, let $L$ be their common length, let $\Sigma$ be the finite alphabet, and let $D_L$ contain the dictionary words of length $L$. Unequal input lengths immediately make the puzzle impossible because a mutation preserves length. A word ladder search problem can use the state space $S=D_L\cup \{s\}$, with initial state $s$; including $s$ handles an input word not itself in the dictionary.

At word $w$, an action is a position–letter pair $(i,\ell )$ with $1\le i\le L$, $\ell \in \Sigma$ and $\ell \ne w_i$, permitted precisely when replacing $w_i$ by $\ell$ gives $w^{\prime }\in D_L$. The search transition model returns that $w^{\prime }$, and the goal test is $w=t$. If $s=t$, the empty sequence is accepted; if $t\notin D_L$ and $s\ne t$, no transition can reach $t$.

Assign each permitted transition cost $1$. A search node at search depth $k$ hasThus minimizing path cost is exactly minimizing the number of mutations. Uniform-cost search, with repeated states handled by graph search and the goal test on selection, finds a shortest word ladder if one exists. Because all costs are equal, breadth-first search gives the same optimal-length guarantee.

All three variants are complete search algorithms under the usual finite-branching factor, nonnegative-heuristic function assumptions. Here $0\le h\le h^{\ast }$, with zero at goals; the supplied positive lower bound on step costs is also essential. The root is processed once before a parent-dependent priority is needed.^[ref] Completeness means finding a solution if one exists, not necessarily terminating when none exists.

Fix a finite solution path. Every search node on that path has a finite priority, so let $K$ bound all of its relevant priorities. For ordinary A-star search and the second variant, the priority is at least $g(n)$. Thus any search node with priority at most $K$ has depth at most $\lfloor K/ϵ\rfloor$. In the third variant the priority is at least $g(n^{\prime })$, so its depth is at most $\lfloor K/ϵ\rfloor +1$. A finitely branching search tree has finitely many search nodes at these depths. Therefore finitely many earlier search nodes can delay each successive search node on the chosen solution path; eventually its goal is generated or selected, as appropriate.

Moving the goal test to generation cannot destroy that existence guarantee, although it can destroy optimality. Likewise the altered parent-dependent priorities remain bounded below by the accumulated cost to the search node or its parent, which is enough for this completeness argument. If “admissible” were interpreted to allow arbitrarily negative estimates, the lower-bound argument would need an additional hypothesis; the conventional nonnegative interpretation is being used explicitly. For the literal upper-bound-only interpretation, all three can fail completeness: attach an infinite unit-cost dead-end chain with $g(n_d)=d$ and $h(n_d)=-2d$ beside a finite goal branch beginning with a cost-$3$ edge and estimate zero. Its successive low priorities can starve the finite branch. Every negative chain estimate is still at most its infinite true remaining cost.

None of the three variants is an optimal search algorithm in general. The following single search tree supplies counterexamples for all three. Its heuristic function is nonnegative, admissible and even a consistent heuristic.

The least-cost goal is $G_4$, reached at cost $1+3=4$; the other goal $G_5$ costs $5$. Assign $h(S)=0$, $h(P)=3$, and zero to both goals. For the first variant, expanding $S$ generates $G_5$ and immediately returns it, before $P$ can be expanded. A goal test at generation therefore returns cost $5$.

For priority $g(n)+h(n^{\prime })$, the root's children have priorities $1$ for $P$ and $5$ for $G_5$. Expanding $P$ generates $G_4$ with priority $4+3=7$. The algorithm next selects $G_5$, again returning cost $5$.

For priority $g(n^{\prime })+h(n)$, the root's children instead have priorities $3$ for $P$ and $0$ for $G_5$. The suboptimal goal is selected immediately. The example uses no tie-breaking assumption. In contrast, ordinary A-star search with its goal test on selection compares the genuine estimated total costs and returns $G_4$.

Maintain a certified solution-cost ceiling and prune only search nodes whose lower bound strictly exceeds it. Initially let $B=\infty$, or set $B=u(r)$ at the root $r$ when that value is finite. Whenever a search node $n$ is generated, updateRun ordinary A-star search with $f(n)=g(n)+h(n)$ and a goal test when removing a search node from the search frontier. Discard a generated search node, or a queued search node encountered after a later bound improvement, ifThis is branch and bound using a lower bound $f$ and an upper bound $B$. A finite $u(n)$ certifies that some goal is reachable from $n$ within that remaining-cost ceiling; for a dead end, $h^{\ast }(n)=\infty$ forces $u(n)=\infty$ too. Every finite quantity $g(n)+u(n)$ is consequently at least the global optimum $C^{\ast }$.

On an optimal solution path, admissible heuristic estimates ensure $f(n)\le C^{\ast }\le B$. Hence no search node of that path is pruned. The finite-branching factor and positive step-cost hypotheses from (a) still ensure completeness, and the usual A-star search argument makes the first selected goal optimal. A smaller bound can remove branches earlier; it need not improve every instance.

The strict inequality matters: $u$ is only a numerical bound and need not provide an actual stored solution path. If $B=C^{\ast }$, pruning at $f(n)\ge B$ could remove every optimal continuation. Nor may the algorithm return a path merely because its lower bound meets $B$ when no witnessing path has been constructed. With an actual incumbent solution, the familiar equality-pruning variant can instead retain and return that incumbent.

Yes. Greedy best-first search needs only the ordering of remaining-cost estimates. Store generated but unexpanded search nodes in a search frontier. To choose the next search node, compare candidates with $\kappa$ and retain the one it returns; a linear scan suffices, or a priority queue can use the comparison rule. Remove that search node, apply the goal test, and otherwise generate its children.

This implements best-first search ordered by $h^{\ast }$ without ever evaluating $h^{\ast }$. If stable tie handling is wanted, comparing both argument orders distinguishes equality, after which insertion order can break ties. No estimate of accumulated-plus-remaining cost is required for this greedy ordering.

The comparator alone cannot generally reproduce informed A-star search priorities. A valid weaker fallback is uniform-cost search, namely A-star search with $h=0$. The issue is that ordering $h^{\ast }(n)$ and $h^{\ast }(m)$ does not supply the numerical difference needed to compare $g(n)+h^{\ast }(n)$ with $g(m)+h^{\ast }(m)$.

For example, take accumulated costs $g(n)=5$, $g(m)=1$. Remaining costs $h^{\ast }(n)=1$, $h^{\ast }(m)=2$ rank $n$ ahead under $\kappa$ but give total estimates $6$ and $3$, so informed A-star search selects $m$. Remaining costs $1$ and $10$ give the same pairwise comparator ranking but totals $6$ and $11$, selecting $n$. Thus the comparator's ordering cannot simply replace the cardinal heuristic function inside $f=g+h$.

If “do A-star search” includes choosing any available admissible heuristic, set it identically to zero and prioritize by the known accumulated cost $g$. Apply the goal test on removal from the search frontier. This is complete and optimal under the stated finiteness and step-cost assumptions, but it does not recover the unavailable perfect-heuristic function ordering or its search reduction. The distinction is between implementing the informed priority from $\kappa$ and retaining the algorithm's guarantees through this uninformed special case.

Three different omissions produce concrete erroneous STRIPS plans. Here a route means the corresponding successive $\mathrm{Move}$ actions with the indicated first argument.

No lethal-entry condition. Move the agent along $s_{11},s_{21},s_{22},s_{23},s_{13}$, take the gold at $s_{13}$, and reverse the route. This nine-action STRIPS plan satisfies every supplied precondition and the final goal test, without firing. Physically it fails on entering the live monster’s room: the supplied search transition model neither forbids the entry nor records death. The omitted pit cell happens to be disconnected in the supplied adjacency relation, so this particular instance already excludes pit entry; that accidental protection does not encode the monster hazard.

Supposedly stationary objects can move themselves. The five-action STRIPS planleaves the agent at the exit and brings it the gold. The formal preconditions hold because $x$ in $\mathrm{Move}(x,y,z)$ has no actor restriction. Gold cannot actually propel itself. The same flaw permits moving the supposedly stationary monster.

Unrestricted pickup permits impossible inventory. Follow the first example’s outward route, but insert $\mathrm{Take}(a,w,s_{23})$ immediately after reaching $s_{23}$. Continue to $s_{13}$, take the gold and return. This ten-action STRIPS plan ends with both $\mathrm{have}(a,w)$ and $\mathrm{have}(a,g)$ and satisfies the supplied goal. Its entrance also exposes the first flaw, but the additional, distinct defect is that pickup treats the live monster as a collectible object and deletes its location. A movement safety check alone is not a specification of which objects are collectible. The shot schema likewise lacks restrictions to the designated shooter, projectile and victim; fixing the action roles together prevents these type errors.

This is a finite, discrete, partially observable, single-agent, sequential search problem with an unknown initial state. The cave is static while the agent deliberates. Given a complete world and an action, the described physical effects and ideal sensor readings are deterministic; ignorance of the world does not itself supply a probability distribution. The fixed monster and gold are environmental objects, not competing decision makers. A known action/sensor law with unknown object locations is different from having to learn an unknown search transition model.

Use partially observable search over belief states. A complete physical state can be writtenwhere $p$ is the agent’s room, $m$ the fixed monster room, $P$ the pit set, $g$ the fixed gold room, $\ell$ records whether the monster lives, $r$ whether the arrow remains, $k$ whether gold has been collected and $e$ whether the agent has exited or failed. The known room dimensions and exit are fixed data. Include death or entrapment as failure states. Movement, shooting toward an adjacent room, collection and departure have their physical meanings. In particular, a shot consumes the arrow, and kills the monster exactly when it targets the live monster’s room.

Let $T(s,a)$ be this physical search transition model. For a surviving agent, an ideal observation is the tuple of three truth valuesrecording sound, breeze and glitter. Here $\sim$ denotes geometric adjacency and $1$ an indicator function. Initially $B_0$ contains all worlds compatible with the known facts and any initial observation. After action $a$ and observation $o$, update the belief state byKnown success or failure observations are included as needed. No prior probabilities are required. A guaranteed-success goal test accepts $B$ only when every $s\in B$ has exited carrying gold. A policy must use the same action in all worlds still indistinguishable to the agent.

An AND-OR search has OR choices for the agent’s next action and AND branches for every possible resulting observation with nonempty $B_{a,o}^{\prime }$. Its output is a contingent plan, not merely a route. To demand safe success, reject any action whose possible outcomes include irreversible death or entrapment, and require a successful continuation on every observation branch. In an online execution the agent updates its belief state, follows the observation branch and can plan again. A strong plan need not exist: some admissible worlds have no safe route, and even solvable worlds can be observationally indistinguishable until a dangerous choice. Without an added prior and numerical rewards, expected-risk optimization is not determined by the description.

Regard the rooms as the vertices of a grid graph. Let $C=\{1,\dots ,n{\}}^2$ and let the two robot positions be $p,q\in C$. A single controller chooses both commands, so the state space is the ordered Cartesian product $S=C\times C$, with initial state $(p_0,q_0)$. Distinguishing the robots matters because their later action laws need not be identical.

For a room $x$, let $A(x)$ consist of wait and all cardinal moves remaining in $C$; write $\delta (x,u)$ for the resulting room. At $(p,q)$ an action is the ordered pair $(u,v)\in A(p)\times A(q)$, and the deterministic search transition model isThe goal test is $p=q$, checked at integer time boundaries, and every joint action costs one. Swapping two distinct rooms does not satisfy the goal test. No internal walls are supplied; if walls were specified, they would simply restrict the move relation.

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.

Let $V$ be the traversable squares and $H\subseteq V$ the exits. Sample the state space at the end of each complete round, so a live state is $s=(r,f)\in V\times V$ with $r\ne f$. The given starting positions specify the initial state. The board and exits are fixed problem data, rather than changing components of the state space.

The rabbit has at most five actions: stay or move to one of its traversable orthogonally adjacent squares. A search transition model first applies that action, giving $r^{\prime }$, and immediately checks for capture at $r^{\prime }=f$. Otherwise hold $r^{\prime }$ fixed and apply the fox rule twice, checking for capture after each step. A capture is terminal failure; it must not be overlooked merely because the fox could subsequently move elsewhere. A surviving successor is $(r^{\prime },f^{\prime \prime })$. Equivalently, omit actions whose complete-round outcome is capture.

The goal test is the rabbit is on an exit after surviving the fox's entire turn: $r^{\prime }\in H$ and no capture occurred in that round. Merely entering an exit before the fox moves is insufficient. Each surviving search transition model step has cost one when minimizing rounds.

Only legal moves are considered when applying the stated horizontal preference: try a legal horizontal step that reduces horizontal separation, otherwise a legal vertical step that reduces vertical separation, otherwise stay. If a blocked preferred direction is instead interpreted as forcing a stay immediately, the counterexamples below and all completeness and optimality conclusions remain valid.

Breadth-first tree search is complete and optimal for minimizing rounds. If a successful escape has finite length $d$, there are only finitely many nodes at depths below $d$, since the branching factor is at most five. Breadth-first tree search therefore reaches a shallowest successful node after finite work, even if other branches continue forever.

It is an optimal search algorithm because every search transition model step costs one round, so minimum depth is exactly the desired minimum number of rounds. The goal test must be applied to the surviving complete-round successor. Here a complete search algorithm guarantees finding an escape when one exists; it need not terminate on every unsolvable instance.

The rule giving the successor state of each permitted state-action pair in a search problem. A compound action may include deterministic environmental responses. If a game state is sampled at full-turn boundaries, its search transition model must still detect terminal events occurring inside the turn. Otherwise an intermediate failure can be incorrectly hidden by the final position. All information needed for future successors and the goal test must be encoded in the state or supplied as fixed problem data.

Best-first search prioritizing accumulated path cost, equivalently A-star search with zero heuristic function. With finite branching factor and positive lower-bounded step costs, tree search is complete for finite solutions and returns a least-cost goal if the goal test is applied on removal from the search frontier. Goal testing at generation can instead return an expensive edge before a cheaper multi-edge route is explored.