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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 proposed Hamming distance is both an admissible heuristic and a consistent heuristic. WriteEvery position differing from the target must change at least once along any successful word ladder. One mutation changes only one position, so every solution from $w$ has at least $h(w)$ steps. This proves $h(w)\le h^{\ast }(w)$, including an unreachable target with $h^{\ast }(w)=\infty$. Dictionary restrictions may force extra changes, but cannot remove that lower bound.

Adjacent words differ in one position, so their Hamming distances to $t$ differ by at most one. Since every transition costs one,These are exactly the consistent heuristic conditions. Equivalently, allowing all length-$L$ words gives a relaxation heuristic whose exact distance is the Hamming distance; restoring the dictionary can only increase the minimum required number of steps.

The difficulty is representing dictionary membership and the one-letter relationship in positive conjunctive STRIPS preconditions; a finite dictionary nevertheless admits an exact encoding. Pure STRIPS does not provide string indexing, a “differs in one position” operator, arbitrary disjunctive tests or a procedure that calls a dictionary lookup. Those relationships must be represented by ground facts or compiled into ground actions.

One exact whole-word encoding uses a planning fluent $\mathrm{At}(w)$ and a static predicate $\mathrm{Next}(w,w^{\prime })$ containing exactly the permitted dictionary mutations. For the action schema $\mathrm{Change}(w,w^{\prime })$, useInitialize $\mathrm{At}(s)$ and the static transition facts, and use goal $\mathrm{At}(t)$. Only one $\mathrm{At}$ fact is true initially. Each applicable ground action deletes it and adds exactly its permitted successor, so mathematical induction proves that exactly one current-word fact remains true and that every plan is a legal word ladder. Conversely, every legal word ladder directly supplies a plan. An isolated non-dictionary start is handled as in (c). A shortest plan has the fewest mutations.

This encoding moves the hard lexical relationship into preprocessing. For $N$ dictionary words, checking all ordered pairs takes $O(N^{2}L)$ character comparisons, although indexing can do better; the number of stored transitions is at most $NL(|\Sigma |-1)$ when all states are dictionary words. The STRIPS representation is finite, but its size and preprocessing cost matter.

A position-based representation instead uses planning fluents $\mathrm{Letter}(i,\ell )$. Changing position $i$ must delete its old letter and add its new one; all other positions persist through the STRIPS update rule. Merely requiring each position to contain a letter admits non-dictionary strings. A disjunction over all dictionary words is not a positive conjunctive STRIPS precondition. It can be compiled into separate ground actions, each requiring every letter of a valid source word and changing one position to reach a valid destination. This again enumerates lexical cases and can be large. Arbitrary letter changes followed by an unconstrained “validate” action are unsound: several illegal intermediate strings could otherwise lead to a legal goal. Legality must hold after every mutation, not only at the endpoint.