How good is search?

• Completeness: Does it always find a solution if one exists? 是否通解
• Optimality: Is the solution optimal (i.e., lowest cost)?  是否最优
• Time Complexity: How long does it take to find a solution? 时间多久
• Space Complexity: How much memory does it need to perform the search? 内存多大

Time and space complexity are measured in terms of
• b – maximum branching factor of the search tree
• d – depth of the least-cost solution
• m – maximum depth of the state space (may be ∞)

• b – 搜索树的最大分支因子（branching factor）：这表示在搜索树中，从任何一个节点出发，可以有多少个不同的分支。

• d – 最短路径的深度（depth of the least-cost solution）：这表示从初始状态到达目标状态的最短路径所包含的节点数。

• m – 状态空间的最大深度（maximum depth of the state space）：这表示状态空间中可能的最深层数。对于一些问题，状态空间可能是无限的，这意味着m可以是无穷大。

 Uninformed search strategie

Uninformed strategies use only the information available in the problem definition
• Breadth-first search
• Uniform-cost search
• Depth-first search
• Depth-limited search
• Iterative deepening search

Breadth-first search

广度优先 每层遍历  优先遍历扩展深度最小的节点（为了尽快访问所有状态）

• Root node expanded first
• All successors to root node then expanded, then their successors, etc.
• All nodes at a given depth are expanded before the next level is expanded.

Implementation: fringe is a FIFO (queue)
 

Properties of Breadth-first search

• Complete – Yes
• Time – 1 + b + b2 + b3 + · · · + bd = O(bd)
• Space – O(bd) nodes generated
• Optimal only if the path cost is a non-decreasing function of the depth of the node
深度越深的节点成本越高，那么这个算法是有效的，因为它能够找到成本最低的解。

Tree search

Repeated states

failure to detect repeated states can turn a linear problem into an exponential one

在图搜索中，重复状态（Repeated states）是指在搜索过程中，图中的某些状态被多次访问，但可能没有被正确识别为重复。这种情况下，搜索算法可能会在已经访问过的状态上进行不必要的操作，导致搜索效率降低。

如果算法不能检测到重复状态，那么即使是一个线性问题（问题规模与输入规模成线性关系），也可能变得指数级复杂。这是因为搜索空间会随着重复状态的增加而指数级增长，从而导致算法需要处理更多的状态，增加了搜索的复杂性和所需的时间。

Graph Search

There are many data structures we can use to make things more efficient

— hash table for the closed set,
— priority queue to determine what node to expand in uniform cost Search
— keeping track of shortest path to each node

 Search Strategies Continued
• Breadth-first search is optimal when all step costs are equal as it always expands the
shallowest unexpanded nodes 当每一步成本相等 BFS最优

Uniform Cost Search

• We expand the node with the lowest path cost

• For completeness, we need to always have a path cost (else we can loop forever).
• Implementation: fringe is a FIFO ordered by cost, lowest cost first (priority queue)

使用优先队列（通常是一个最小堆）作为搜索框架，优先扩展代价最小的节点。这意味着它会优先扩展那些最有可能导致最优解的节点。UCS保证找到的路径是最优的，即代价最小的路径。

Depth-first Search

• This type of search expands the deepest node in the current frontier.
• When the end is reached, the search backs up to the next shallowest node that still
has unexplored successors.
• Implementation: fringe is a LIFO (stack）

1. 扩展当前前沿中最深的节点：在深度优先搜索中，算法首先扩展当前前沿中深度最深的节点。这意味着它首先探索尽可能深的路径，直到达到边缘节点。

2. 回溯：当达到一个无法继续扩展的节点（通常是最深的边缘节点）时，搜索会回溯到下一个浅层节点，这个节点仍然有未探索的后继节点。这个过程会一直重复，直到找到目标状态或搜索空间耗尽。

深度优先算法 每个分支遍历  优先拓展深度最深的节点

Properties of Depth-first search

• Complete – No: fails in infinite-depth spaces, spaces with loops
Modify to avoid repeated states along path
⇒ complete in finite spaces
• Time – O(bm) terrible if m is much larger than d
but if solutions are dense, may be much faster than breadth-first
• Space – O(bm), i.e., linear space!
10 exab to 156kb
• Optimal – No

Depth-limited search

• This search solves the infinite path problem of DFS
• But incompleteness is introduced if the shallowest goal is beyond the depth limit

DLS通过设定一个深度限制来避免这种情况。一旦达到这个深度限制，搜索就会停止扩展当前路径。但是如果图中的某个目标是超出深度限制的，DLS可能无法找到该目标的解。

在路由问题中，如果已知有20个城市，那么最大深度就是19（因为不能到达自己）。经过观察发现任何城市都可以在深度9内到达，这是状态空间的最长直径。

Iterative deepening Depth-first search

• This acts as a DFS with a gradually increasing limit until a goal is found.
• Combines benefits of DFS and BFS.
• Note that states are generated multiple times. But this is not costly as most nodes
are at the bottom level.
限制在没有找到目标之前会逐渐增加 eg：limit=1时 按照DFS 发现没有找到目标 就将limit增加到2

每次迭代会从初始状态开始 避免重复状态

Properties of iterative deepening search

• Complete – Yes
• Time – (d)b0 + (d − 1)b1 + (d − 2)b2 + (d − 3)b3 + · · · + (!)bd = O(bd)
• Space – O(bd)
• Optimal – Yes, if step cost = 1
Can be modified to explore uniform-cost tree