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Map original states to simpler abstract states, include every projected original transition at no greater cost, and map original goals to abstract goals. An original goal-reaching graph path then projects to an abstract goal-reaching walk, so the exact abstract distance is an admissible heuristic. A pattern database stores such distances for repeated lookup.

If $h_1,h_2\le h^{\ast }$, their pointwise minimum, pointwise maximum and each convex combination $\alpha h_1+(1-\alpha )h_2$ with $0\le \alpha \le 1$ are admissible heuristics. Each inequality follows directly by comparison with $h^{\ast }$. The maximum has heuristic dominance over the minimum and every such convex combination.

A sum need not be an admissible heuristic: two exact estimates of the same cost-one remaining action sum to two. A sufficient condition for a valid additive construction is separate nonnegative action-cost allocations $c_1,c_2$ with $c_1+c_2\le c$, and estimates bounded by the corresponding optimal allocated costs. Every goal-reaching graph path then costs at least $h_1+h_2$ under its original costs. Taking the minimum proves admissibility without counting the same work twice.

If $h_1,h_2$ are consistent heuristics, their maximum, minimum and fixed convex combinations are also consistent heuristics. For the maximum, choose an index attaining it at the source and bound its successor by the successor maximum. For the minimum, choose an index attaining it at the successor and bound the source minimum by that component. For a convex combination, multiply the two consistency inequalities by their nonnegative weights and add them. All these combinations vanish at goals when the inputs do.

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.

A heuristic function is consistent if $h(g)=0$ at every goal andfor every permitted transition $s\stackrel{a}{\to }{s}^{\prime }$ with step cost $c(s,a,s^{\prime })$. Summing these inequalities along a finite route to a goal gives $h(s)\le$ its total cost. Taking the minimum over such routes proves that every consistent heuristic is an admissible heuristic.

For positive-precondition STRIPS with a positive conjunctive goal, remove every delete effect and use the exact shortest relaxed plan cost. Every original plan still succeeds in the relaxed model because facts can only accumulate, so its relaxed optimum is an admissible heuristic. The cost of an arbitrary relaxed plan is an upper bound on the relaxed optimum and need not be an admissible heuristic for the original problem.

A heuristic function $h_2$ dominates $h_1$ if $h_2(s)\ge h_1(s)$ everywhere. When both are admissible heuristics, the larger estimate is at least as informative for A-star search: for a fixed accumulated path cost $g$ and optimal solution cost $C^{\ast }$, $g+h_2<C^{\ast }$ implies $g+h_1<C^{\ast }$. This compares the compulsory search region, not arbitrary tie handling at $C^{\ast }$, evaluation overhead or wall-clock time.

A function estimating the remaining cost from a state $s$ to a goal in a search problem. The estimate is an admissible heuristic when it never exceeds the true minimum remaining cost. A consistent heuristic satisfies an additional local inequality along each permitted transition.

Author-provided chapter: aima.cs.berkeley.edu/4th-ed/pdfs/newchap04.pdf

Printed section 4.1, especially pp. 97–100, discusses greedy best-first search, A-star search, admissible heuristics and the finite-branching factor/positive-cost completeness conditions. The printed chapter identity is used without inferring an edition from its host directory.

Four-page University of Oxford Finals examination, Trinity Term 2015, dated Tuesday 16 June 2015, 9:30 am–11:00 am. The original marks Questions 1 and 3 as corrected. Printed p. 2 concerns Horn clause proof search and admissible heuristics; p. 3 concerns a STRIPS encoding of a multi-stack Tower of Hanoi puzzle; p. 4 concerns constraint satisfaction problems, tree decomposition, and all-different constraints.

Public archive: Lawrence Attachment.zip, member 2015-is.pdf.

Held original: pdf/computer-science/finals/intelligent-systems/2015.pdf in the Oxford examinations repository. Editable transcription: tex/computer-science/finals/intelligent-systems/2015.tex.

A-star search is best-first search with priorityHere $u$ is a search node, $s(u)$ is its state, $g(u)$ is its accumulated path cost, and $h^{\ast }(s)$ is the minimum remaining cost to any goal state, interpreted as $\infty$ if no goal is reachable. The inequality defines an admissible heuristic. Thus $f(u)$ is a lower bound on the cost of any solution extending that particular action history.

Initialize the search frontier with the root. Repeatedly remove a search node of least $f$, apply the goal test, and, if it is not a goal, insert its applicable-action children with their updated $g$ and $f$. In tree search, different action histories remain distinct search nodes even when they reach the same state; there is no global visited-state set.

For the finite branching factor and unit costs here, an admissible heuristic ensures that any first goal removed is optimal. If a cheaper solution existed, the search frontier would contain an unexpanded prefix of it with $f$ at most that cheaper cost, contradicting selection of the more expensive goal. Nonnegative $h$ also means only finitely many search nodes can have $f$ below any fixed finite cost, so a finite solution will eventually be reached. This does not guarantee termination when no proof exists.

Termination is not guaranteed. TakeBoth members of $T$ are valid Horn clauses with distinct underlying Boolean variables. The only possible continuation alternatesThe empty query is unreachable. With the admissible heuristic $h=0$, ordinary A-star search keeps generating the next search node forever, despite there being only two reachable states.

A modification requiring no additional stored state collection is to impose the known search depth boundThe unit-cost $g$ already stored by A-star search is also its search depth; the bound is a scalar computed from the input, not a closed set. Apply the goal test before refusing expansion at the bound.

To prove completeness of this modification, remove any segment between two equal states in a successful proof. The successor choices after the later occurrence are available at the earlier occurrence because the search transition model depends only on the current query. Repeating this removal gives a proof with no repeated states, hence at most $2^N-1$ actions. In particular, a shortest proof never needs a repeated state, since every removed segment has positive cost. The depth restriction therefore preserves a shortest proof whenever one exists.

The restricted search tree has finite depth and finite branching factor, so it is finite. A-star search must then either remove a goal or exhaust the search frontier and report failure. Its existing search frontier may still be large; the modification saves a separate visited-state collection, not all storage used by tree search.

Yes: $h_1(R)=|R|$ is an admissible heuristic. One application of the resolution inference rule removes just one current negative Boolean literal. It may add other negative Boolean literals, and some of those may already occur, but it cannot remove any other current obligation. ConsequentlyIf $\ell$ actions lead from $R$ to $\varnothing$, applying this inequality at every step gives $0\ge |R|-\ell$. ThusIt is also a consistent heuristic, since $h_1(R)\le 1+h_1(R^{\prime })$ and $h_1(\varnothing )=0$. In particular, merging duplicate obligations cannot invalidate the lower bound: removing several duplicates from the representation never counts as removing several distinct members of a set.

Use the shortest path distances in the directed dependency graph to defineThe additional $1$ counts the final use of a positive unit Horn clause, which corresponds to a fact rather than an edge of $G$.

To justify the new lower bound, consider any complete proof from $R$ and fix $\neg p\in R$. At some step this obligation must be removed using a Horn clause with head $p$. If that Horn clause is the fact $\{p\}$, we have already found a dependency graph path of length zero ending at a fact. Otherwise select one negative body Boolean literal $\neg q$. Immediately after the step it belongs to the query, whether newly inserted or merged with an existing obligation. It must subsequently be removed. Record the edge $p\to q$ and repeat this argument at its later removal.

The selected removal times strictly increase. Since the proof is finite and ends with no obligations, this chain eventually uses a fact. A proof of $\ell$ actions therefore produces a dependency walk of at most $\ell -1$ edges from $p$ to a fact. Removing any cycles from that walk gives $d_G(p)\le \ell -1$. Hence $1+d_G(p)\le \ell$ for every initial obligation. If $d_G(p)=\infty$, no such proof exists. Combining this bound with the admissible heuristic $|R|$ by taking their maximum preserves admissibility.

For a strict improvement, useHere $d_G(p)=2$, so $h_1(R)=1$ while $h_2(R)=3$. The three successive obligations are $\neg p,\neg q,\neg r$, followed by the empty query, so the improved bound is exact.

Do not sum all the values $1+d_G(p)$ without an additional argument: different obligations may share work. For example, with $T=\{\{p,\neg r\},\{q,\neg r\},\{r\}\}$, the query $\{\neg p,\neg q\}$ takes three actions, while that sum would be four.

When the values of several admissible heuristics are already available, choose their maximum:Every $h_i(R)\le h^{\ast }(R)$, so their maximum is still an admissible heuristic. It has heuristic dominance over their minimum and their arithmetic average, which are also admissible heuristics but give no larger lower bound. Thus the maximum supplies the most informative generally valid choice among these alternatives.

Their sum can overestimate by counting the same remaining action repeatedly. If two heuristic functions both give the exact value $1$ at a state one action from the goal, their sum gives $2$. A valid sum requires extra structure such as dividing each action's cost between the component estimates; admissibility of the components alone does not provide that structure.

Larger admissible values reduce the region $g+h<C^{\ast }$ that A-star search must expand before an optimal solution of cost $C^{\ast }$ can be removed. This is a statement about heuristic dominance, not a strict claim about running time under every tie-breaking rule or evaluation cost.

Two general constructions are delete-relaxation heuristics and abstraction heuristics. Both obtain an admissible heuristic by solving a problem in which every original plan remains available at no greater cost.

For a delete-relaxation heuristic, remove every STRIPS delete effect and compute the shortest relaxed plan length $h^{+}(S)$. A relaxed action never makes a true fact false. Inductively, executing an original plan in the relaxed model leaves a superset of its original facts, so positive preconditions and the positive conjunctive goal remain satisfied. ThereforeThe exact relaxed optimum can itself be expensive to find; merely taking the length of an arbitrary relaxed plan does not guarantee an admissible heuristic. One can instead use a proven lower bound on the relaxed optimum.

For this coin encoding, removing delete effects loses the requirement to free an occupied stack and the requirement to destroy and later rebuild support relations. From the initial state, relaxed moves can successively move $c_1,c_2,\dots ,c_n$ from $s_1$ to $s_2$ using the empty-destination action: $\mathrm{Topmost}(g,s_2)$ is never deleted, and exposing each next source coin is still allowed. The original support-chain goal facts are never deleted either. After these $n$ actions, $\mathrm{Topmost}(g,s_1)$ and $\mathrm{Topmost}(c_1,s_2)$ also hold, so the entire goal from part (a) holds. Thus $h^{+}(I)\le n$, showing why this relaxation misses much of the exponential work needed with three stacks.

For an abstraction heuristic, project states onto selected information and project each action onto its effect on that information. Forgetting a selected collection of planning fluents, and their preconditions and goals, gives one general construction for STRIPS. Define the abstract transition relation to include every projected original transition, with cost at most the original cost. The exact abstract distance is then a lower bound because any original goal-reaching graph path projects to an abstract goal-reaching walk:For the coin puzzle, a useful state abstraction remembers the stacks occupied by a chosen subset of coins and ignores the other coins, contracting the support chains past ignored coins. Moving an ignored coin leaves the abstract state unchanged; moving a retained coin is a legal move in the smaller Tower of Hanoi instance. Precomputing exact abstract goal distances gives a pattern database. This can retain stacking and spare-stack restrictions that the delete-relaxation heuristic loses. Larger retained subsets give potentially stronger information but require more storage and preprocessing. Maxima of these admissible heuristics remain safe; unqualified sums can double-count moves.

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.

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.

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.

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.