A mobile agent navigating along edges of a simple connected unweighted graph, either finite or countably infinite, has to find an inert target (treasure) hidden in one of the nodes. This task is known as treasure hunt. The agent has no a priori knowledge of the graph, of the location of the treasure, or of the initial distance to it. The cost of a treasure hunt algorithm is the worst-case number of edge traversals performed by the agent until finding the treasure. Awerbuch et al. [3] considered graph exploration and treasure hunt for finite graphs in a restricted model where the agent has a fuel tank that can be replenished only at the starting node s. The size of the tank is B = 2 (1+α) r, for some positive real constant α, where r, called the radius of the graph, is the maximum distance from s to any other node. The tank of size B allows the agent to make at most ⌊ B ⌋ edge traversals between two consecutive visits at node s.

Let e(d) be the number of edges whose at least one endpoint is at distance less than d from s. Awerbuch et al. [3] conjectured that it is impossible to find a treasure hidden in a node at distance at most d at cost nearly linear in e(d). We first design a deterministic treasure hunt algorithm working in the model without any restrictions on the moves of the agent at cost 𝒪(e(d) log d) and then show how to modify this algorithm to work in the model from Awerbuch et al. [3] with the same complexity. Thus, we refute the preceding 20-year-old conjecture. We observe that no treasure hunt algorithm can beat cost Θ (e(d)) for all graphs, and thus our algorithms are also almost optimal.

沿着有限或可数无限的简单连接未加权图的边缘导航的移动代理必须找到隐藏在其中一个节点中的惰性目标（宝藏）。此任务称为寻宝。智能体对图、宝藏的位置或到宝藏的初始距离没有先验知识。寻宝算法的成本是代理在找到宝藏之前执行的最坏情况下的边遍历次数。Awerbuch 等。[3] 考虑在受限模型中对有限图进行图探索和寻宝，其中代理有一个只能在起始节点 s 补充的燃料箱。水箱的大小为B = 2 (1+α) r，对于某个正实常数 α，其中r，称为图的半径，是从s到任何其他节点的最大距离。大小为B的容器允许代理在节点s的两次连续访问之间最多进行 ⌊ B ⌋ 次边遍历。

令e(d)为其至少一个端点与s的距离小于d的边数。Awerbuch 等。[3] 推测不可能找到隐藏在距离最多为d 的节点中的宝藏，成本几乎与e(d)成线性关系。我们首先设计了一个在模型中工作的确定性寻宝算法，对代理人的移动没有​​任何限制，费用为𝒪(e(d) log d )，然后展示如何修改该算法以在 Awerbuch 等人的模型中工作。[3] 具有相同的复杂性。至此，我们推翻了之前20年的猜想。我们观察到没有寻宝算法可以击败成本 Θ ( e(d)) 对于所有图，因此我们的算法也几乎是最优的。