Codex @codex  0

Best-first search prioritizing the estimated total solution cost $f=g+h$, with $g$ the cost already incurred and $h$ a heuristic function for the remaining cost. With nonnegative estimates, goal estimates zero, finite branching factor and step costs bounded below by a positive constant, the tree search version using an admissible heuristic returns a minimum-cost goal when it removes one for expansion. An optimal solution path always has a frontier node with $f\le C^{\ast }$; a more costly goal cannot be removed before it. The stated finiteness and lower-cost assumptions ensure finitely many earlier nodes when a finite solution exists. Repeated-state handling in graph search needs reopening when a better path is found, unless a consistent heuristic justifies finalizing expanded states.

The tree search version of breadth-first search explores action histories in nondecreasing length without automatically removing repeated states. A finite branching factor ensures that every finite depth contains finitely many histories, so the method is a complete search algorithm for goals reachable at finite depth. With equal positive action costs it is an optimal search algorithm for total action cost. It need not terminate on an unsolvable instance with an infinite search tree, and it need not optimize a terminal reward unrelated to depth.

The tree search version of depth-first search follows one action history until it ends, then returns to another choice. Repeated states do not automatically stop a branch. An infinite branch can prevent discovery of a goal on a different finite branch, so this method is not generally a complete search algorithm. A finite search tree restores completeness, but the first successful history need not minimize cost or maximize a terminal reward.

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.

Use $H=\max (h_1,h_2)$ if each estimate is available at comparable cost. For every $\alpha \in [0,1]$,Thus $H$ has heuristic dominance over all the other combinations guaranteed to remain admissible heuristics. A-star search orders nodes by $f=g+h$, where $g$ is the accumulated path cost. A larger admissible heuristic excludes more nodes from the compulsory $f<C^{\ast }$ region, where $C^{\ast }$ is the optimal solution cost. Under the usual nonnegative-cost and goal-normalization assumptions, this gives at least as informative a search guide while preserving the optimality guarantee of A-star search with tree search.

The unweighted sum loses that guarantee. The dominance statement does not promise smaller wall-clock time in every implementation: evaluating estimates costs time, and nodes with $f=C^{\ast }$ depend on tie handling. Nor does admissibility of the inputs alone imply consistent heuristics. If both inputs are consistent heuristics, however, their pointwise maximum is also a consistent heuristic.

The branching factor of the search tree is at most five, so through depth $d$ it has at mostnodes in Big O notation. With no depth bound, the search tree can be infinite, despite the finite state space: tree search retains different action histories even when they end at the same position pair.

For a concrete repeating history, use a two-row, three-column board, indexed from zero, with the sole obstacle $(0,1)$, rabbit at $(0,2)$ and fox at $(0,0)$. Repeated rabbit stays leave the fox blocked horizontally with no vertical separation to reduce. Hence the same live state can repeat forever. An exit at $(1,2)$ is nevertheless safely reachable in one round.

An entry in a search tree representing a particular action history. It normally stores the current state, its parent, its last action and accumulated path cost $g$. Distinct search nodes can represent the same state; identifying them changes tree search into a form of graph search and requires an appropriate repeated-state policy.

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.