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In a finite reachable search problem with strictly positive step costs and a reachable goal, greedy best-first search using a perfect heuristic selects one parent–child chain until it reaches a goal state. At a selected nonterminal state $s$, an optimal first transition has $h^{\ast }(s^{\prime })=h^{\ast }(s)-c(s,s^{\prime })<h^{\ast }(s)$. Every older search frontier entry has estimate at least $h^{\ast }(s)$, so the next selected entry must be a newly generated child with smaller estimate. Repeating proves the chain property. In an infinite finitely branching problem, a positive uniform lower bound on costs also bounds the chain length by the initial remaining cost divided by that lower bound. Zero-cost ties can break the chain property; the lemma does not establish least accumulated path cost.

Iterative lengthening search repeatedly runs cost-bounded depth-first search, starting with threshold zero and replacing each unsuccessful threshold by the smallest rejected path cost. It uses a depth-first search stack rather than the potentially large search frontier of uniform-cost search.

Suppose an optimal plan has cost $C^{\ast }$ and each pass terminates. At every unsuccessful threshold $L<C^{\ast }$, some first rejected prefix of an optimal plan has path cost at most $C^{\ast }$. Therefore the next threshold is no larger than $C^{\ast }$. Any first returned goal state has cost at most $C^{\ast }$ and hence exactly $C^{\ast }$. Finite branching factor and a positive lower bound on step costs ensure that only finitely many search nodes precede that cost, giving completeness as well as optimality. With arbitrary real step costs, the threshold increments can be much smaller than the smallest step cost: the difference between two path costs need not itself be a step cost.

Iterative deepening depth-first search increases a bound on the number of actions, usually through $0,1,2,\dots$. Iterative lengthening search instead bounds accumulated path cost; after an unsuccessful pass, its next bound is the least path cost among the rejected search nodes. Thus it can distinguish shallow expensive paths from deeper cheap ones.

Both repeatedly perform depth-first search and retain its small active-path storage, trading repeated work for avoiding a large search frontier. With unit step costs their bounds coincide. With unequal positive step costs, iterative deepening depth-first search finds a solution with the fewest actions, while iterative lengthening search, under the termination conditions discussed next, finds one of least total cost. Real-valued path costs can make iterative lengthening search perform many more passes than the depth-bounded method.

Backtracking assigns one variable at a time and abandons a partial assignment as soon as it cannot extend to a solution. With forward checking, placing a knight deletes its occupied and attacked squares from every remaining knight's constraint domain. An empty constraint domain rejects the branch immediately; stronger constraint propagation, such as arc consistency, can detect additional incompatibilities among future choices.

This combines early rejection with depth-first storage. It retains the current assignment and domain changes rather than the entire breadth-first layer of partial placements. A direct knight-variable implementation uses $O(k{n}^2)$ space for explicit domains, plus its assignment and restoration trail, instead of an exponentially large breadth-first search search frontier. Propagation has a computational cost and does not guarantee improvement on every input, but often prevents exploration of vast inconsistent subtrees.

The minimum remaining values heuristic selects an unassigned variable with the fewest currently legal values, exposing the likeliest failure first. The degree heuristic selects a variable constrained to the largest number of other unassigned variables, often breaking minimum remaining values heuristic ties. Degree is measured in the remaining constraint graph, not by counting assigned neighbors. In the symmetric knight-variable encoding the remaining degrees can all tie, so the heuristic need not discriminate there.^[ref]

TODO: the classifications, chosen heuristic functions and least-cost outcomes for G1–G5 cannot be established from the corrupted original Figure 1. The missing edges, costs, priority labels and lower portions of the five search trees are required inputs. The surviving portions and the two heuristic function tables do not determine those answers. No reconstructed graph or invented expansion trace is assumed.

For a readable instance, depth-first search follows the deepest pending branch and breadth-first search expands by search depth; neither generally minimizes unequal edge costs. Uniform-cost search orders by path cost $g$, whereas A-star search orders by $g+h$. A displayed A-star search score would identify the estimate by subtracting the accumulated path cost. Under nonnegative costs and the usual terminating-search assumptions, uniform-cost search returns a least-cost goal state when testing goals on removal from the search frontier; A-star search additionally needs an admissible heuristic with appropriate reopening, or a consistent heuristic for the usual closed-set implementation. These rules explain how the missing information would be used, but cannot identify a particular damaged search 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.

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.

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.