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Lecture 2 2022.3.1

The definition of a intelligent agent

  1. reflex agents

percept \(x\) → \(f\) → single action \(y\)

  1. planning agents

precept \(x\) → \(f\) → action sequence (\(a_1, \cdots\))

Search and model

model is still a model, an abstract mathematical description

model is not real thing.

"all in simulation"

Search Problems -- what are they

  • state space \(S\)
  • initial state \(s_0 \in S\)
  • action \(A(s)\) in each case
  • Transition model \(Result(s,a)\)
  • actually a successor function
  • goal test \(G(s)\)
  • action cost \(c(s,a,s')\)

Solution: action sequence that reaches a goal status

optimal solution: least cost

State Space

We cannot put the entire world into the model --- the world model ?

We need to abstract away less-relative details.

The number of state: could be really large

State Space Graph

It's a concept that helps us to think

  • Nodes are states(appear only once)
  • Arcs are transitions
  • the goal test is a set of goal nodes

Search Tree

Structure:

  • root node: start state
  • Children: successors
  • Nodes show states, but correspond to plans that achieve those states(path to the root node)

node(path to root node) is matched uniquely to the "plan"

  • add a "explored/frontier/new" label
    • select an frontier node into the explored set
    • then add new frontier label

→ generate a tree

Things that need to consider:

  • Complete
  • Optimal (least cost)
  • Time complexity
  • Space complexity

DFS & BFS

DFS BFS
Strategy expand a deepest node first expand a shallowest node first
Implementation Frontier is a LIFO stack frontier is a FIFO queue
Argument \(b\) is the branch factor, \(m\) is the maximum depth \(d\) is the depth of the shallowest solution
Time complexity \(O(b^m)\) all nodes in the tree \(O(b^d)\)
Space complexity \(O(bm)\) a path to nodes \(O(b^d)\)
Completeness? NO(could be circles) YES
Optimal NO YES

High-level

both are complete & Optimal

combining DFS and BFS

limit the depth when using DFS, then deepen the maximum depth by 1.

Time: \(O(b^d)\) ，Space: \(O(bd)\)

like BFS.

Implementation of the frontier set:

use a priority queue whose key is the cost from node \(n\) to root \(g(n)\)

Arguments: Solution cost \(C^*\) ，minimum arcs cost \(\varepsilon\)

Time: \(O(b^{C^*/\varepsilon})\) , Space: \(O(b^{C^*/\varepsilon})\)

"Which direction is more likely to be right?"

Heuristic: need a function to evaluate how close a state is to a goal(?)

Strategy: expand a node that you believe is the closest(\(h(n)\)) one to the goal state

Implementation:

  • frontier is an ascending order priority queue by \(h(n)\)

Completeness: NO, Optimal: NO

[Hart/Nilsson/Raphael, 1968]

Combining UCS & Greedy Search

Sorted by \(f(n) = g(n) + h(n)\) \(g\) is the cost from root to \(n\)(backward), \(h\) is the estimated cost from \(n\) to the goal (forward)

Notice: we can only stop when we dequeue a goal node

Optimal?

estimations shouldn't have too much impact!

Admissible Heuristics: (可采纳条件)

\[ 0 \leq h(n) \leq h^*(n) ,\;\; \text{where h* is the {\color\red{true}} cost to a {\color\red{nearest}} goal } \]

We can prove that the optimal goal A with exit "frontier" node before suboptimal node B, with an admissible Heuristics.

a short proof

all ancestor of A (\(n\)) will be expanded before B

  1. \(f(n) \leq f(A)\), using that \(f(n) = g(n) + h(n) \leq g(A) = f(A)\)
  2. \(f(A) < f(B)\), using that \(f(A) = g(A) < g(B) = f(B)\)
  3. \(n\) expands before B, using that \(f(n) \leq f(A) < f(B)\)

Time Effiency?

Theorem

A* will explore all states \(s\) that satisfying \(g(s) \leq g(s_\text{goal}) - h(s)\)

So, larger \(h(s)\) implies better performance

Note

The Search Area of an A* Search is like a oval, while UCS is a perfect circle

How to create Heuristics?

Relaxed problems

Problem \(P_2\) is a relaxed version of \(P_1\) if \(A_2(s) \supseteq A_1(s)\) for every \(s\)

Theorem

\(h_2^*(s) \leq h_1^*(s)\) for every \(s\), so \(h_2^*(s)\) is admissible for \(P_1\)

use a easy version of question, which is easy to compute, to get a heuristic

Example: 8 Puzzles, use the number of different tiles as a heuristic, use the distance of different tiles as a heuristic

Note

There is a trade, in "how easy version" you choose

Easier version is easier to compute, but has less guiding function

You can also combine these several heuristic functions.

Application?

Computer Games, searching for paths

On Search Graph

More strict condition --- requiring a triangle inequality

Consistency: \(h(n) \leq c(n,a,n') + h(n')\)