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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.

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.

A comparison oracle revealing the ordering of remaining optimal costs without returning their magnitudes. It supplies enough information for greedy best-first search. Its ordering alone does not provide the numerical differences needed to compare sums with independently varying accumulated costs. For $g(n)=5$, $g(m)=1$, remaining costs $(1,2)$ and $(1,10)$ have the same pairwise order but give opposite orderings of $g+h^{\ast }$.

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.

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 exact minimum remaining path cost from a state to a goal state, with value $\infty$ when no goal is reachable. For nonnegative step costs, $h^{\ast }(g)=0$ at goals and $h^{\ast }(s)\le c(s,s^{\prime })+h^{\ast }(s^{\prime })$ on every transition. Thus a perfect heuristic is a consistent heuristic and an admissible heuristic. It measures remaining cost; a greedy best-first search decision using only this number can still take an excessively costly incoming edge.