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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.

The first assertion is false. Give the root a first child that is a goal state of cost ten and another child leading to many cheaper non-goal search nodes. Depth-first search can take the first child and stop immediately. A-star search with the admissible heuristic $h=0$ behaves as uniform-cost search and expands the cheaper non-goal search nodes before selecting that goal. Thus an admissible heuristic does not make A-star search expand fewer nodes than every possible depth-first search ordering. The latter need not return an optimal solution.

The second assertion is true under the usual finite-branching factor assumption. Breadth-first search orders by search depth, not path cost. Every finite depth contains finitely many search nodes, so a goal at finite depth is eventually reached even if some steps cost zero. Zero-cost steps can affect cost optimality, but do not obstruct this completeness argument. An infinite branching factor would invalidate the argument irrespective of step costs.

The third assertion is true as a worst-case black-box requirement, with a finite optimum assumed. If outgoing actions are known only by expanding a state, an unexamined reachable branch can conceal a sufficiently negative-cost action followed by a route to a goal. Its total cost can beat the current solution by an arbitrarily large amount. Without additional lower bounds or structural information, finding one cheap solution cannot justify stopping; exhaustive examination can be necessary. This does not assert that every already-described irrelevant state must be visited on every input. Moreover, a reachable negative-cost cycle in a graph that can still reach a goal can make costs unbounded below, in which case no finite-cost optimal solution exists at all.

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.

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.

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.

The sum of step costs along the action history represented by a search node. Positive costs give lower bounds on search depth, and the same physical state can have different path costs when reached along different histories. Uniform-cost search orders by accumulated cost; A-star search adds an estimate of remaining cost.