ARTIFICIAL INTELLIGENCE/Student Engineering Notebook
Page 02·notes
UNINFORMED & INFORMED STATE SPACE SEARCH—~30 MIN READ

State Space Search, UCS & A* Heuristic Admissibility

“f(n) = g(n) + h(n), proving A* optimality on trees vs graphs, and dominance of heuristics.”

1.The A* Search Algorithm & The f(n) Evaluation Function

Uninformed search algorithms suffer from exponential state expansion because they blindly explore all directions. Informed (heuristic) search guides exploration using domain-specific knowledge.

The A Algorithm: A selects for expansion the frontier node that minimizes the estimated total solution cost:

$f(n) = g(n) + h(n)$

Where:

• g(n) is the exact path cost incurred from the start state s_0 to node n.

• h(n) is the heuristic estimate of the remaining path cost from node n to the nearest goal state.

• f(n) represents the estimated total cost of the cheapest solution path passing through node n.

2.Mathematical Proof of A* Tree Search Optimality under Admissibility

Definition (Admissibility): A heuristic function h(n) is admissible if it never overestimates the true cost to reach the goal: 0 ≤ h(n) ≤ h^*(n) for all nodes n, and h(G) = 0 for any goal state G.

Theorem: A* Tree Search is guaranteed to return an optimal solution if h(n) is admissible.

Proof by Contradiction:

1. Let G_2 be a sub-optimal goal state generated during search, with path cost g(G_2) > C^, where C^ is the optimal path cost.

2. Since G_2 is a goal, h(G_2) = 0, so f(G_2) = g(G_2) + h(G_2) = g(G_2) > C^*.

3. Now consider an optimal goal path. At any point before finding G_2, there must exist at least one unexpanded node n on the frontier that lies on this optimal path.

4. For this node n on the optimal path:

$f(n) = g(n) + h(n) ≤ g(n) + h^(n) = C^$

(Since h(n) ≤ h^(n) by admissibility, and g(n) + h^(n) = C^* by definition of an optimal path).

5. Therefore, f(n) ≤ C^* < f(G_2).

6. Since A* always selects the node with the lowest f-value, node n will always be selected for expansion before G_2 could ever be selected for expansion.

7. This holds for every node along the optimal path, culminating in the optimal goal G^* being selected before G_2. Contradiction! Hence, sub-optimal goal G_2 can never be selected.

3.Consistency (Monotonicity) for A* Graph Search

In Graph Search, an explored set (closed list) prevents re-visiting already expanded states. If a heuristic is admissible but inconsistent, the first path found to a state n might NOT be the cheapest. If n is discarded on subsequent encounters, A* can return a sub-optimal solution.

Definition (Consistency / Monotonicity):

A heuristic h(n) is consistent if, for every node n and every successor n' generated by action a:

$h(n) ≤ c(n, a, n') + h(n')$

This is the form of the classical Triangle Inequality: the estimated cost from n to goal cannot exceed the step cost to n' plus the estimated cost from n' to goal.

Consequences of Consistency:

1. Non-decreasing f-costs: f(n') = g(n') + h(n') = g(n) + c(n, a, n') + h(n') ≥ g(n) + h(n) = f(n).

2. First Expansion Optimality: Whenever A* selects a state for expansion in graph search, the path to that state is guaranteed to be optimal. No closed-list state ever needs to be re-opened!

3. Consistency strictly implies Admissibility (by induction along the path to goal, where h(G) = 0).

notice this:— Admissible but Inconsistent Heuristic Example

Consider S -> A (cost 1), A -> G (cost 3), S -> G (cost 5). Let h(S) = 4, h(A) = 1, h(G) = 0. • Admissibility: h(S)=4 <= 4 (optimal is S->A->G cost 4); h(A)=1 <= 3. Admissible! • Consistency check on edge S->A: h(S) <= c(S, A) + h(A) => 4 <= 1 + 1 = 2 (FALSE!). Inconsistent! Triangle inequality is violated.

WORKED EXAM ANALYSIS · STEP-BY-STEP DERIVATION2 Marks (High Frequency in GATE DA)

4. GATE DA Worked Example: A* Node Expansion Trace

⚠️ Pitfall Warning:

Nodes are selected for expansion strictly based on lowest f(n) = g(n) + h(n), not purely h(n) or g(n) alone!

Problem: Trace A* Graph Search from Start S to Goal G. Given transitions with step costs and heuristics h:

Heuristic values: h(S) = 5, h(A) = 3, h(B) = 4, h(C) = 2, h(G) = 0.

Directed Edges with costs:

S -> A (cost 2), S -> B (cost 3)

A -> C (cost 2), B -> C (cost 1)

C -> G (cost 3).

Step 1: Initialize at Start S:

g(S) = 0, h(S) = 5 => f(S) = 0 + 5 = 5.

Frontier: [ S(f=5, g=0) ]. Closed: { }.

Pop S for expansion.

Step 2: Generate successors of S:

Node A: g(A) = 0 + 2 = 2; f(A) = 2 + h(A) = 2 + 3 = 5.

Node B: g(B) = 0 + 3 = 3; f(B) = 3 + h(B) = 3 + 4 = 7.

Frontier: [ A(f=5, g=2), B(f=7, g=3) ]. Closed: { S }.

Lowest f is A(f=5) => Pop A for expansion.

Step 3: Generate successors of A:

Node C: g(C) = g(A) + 2 = 2 + 2 = 4; f(C) = 4 + h(C) = 4 + 2 = 6.

Frontier: [ C(f=6, g=4), B(f=7, g=3) ]. Closed: { S, A }.

Lowest f is C(f=6) => Pop C for expansion.

Step 4: Generate successors of C:

Node G: g(G) = g(C) + 3 = 4 + 3 = 7; f(G) = 7 + h(G) = 7 + 0 = 7.

Frontier: [ G(f=7, g=7), B(f=7, g=3) ]. Closed: { S, A, C }.

Lowest f is tie between G(f=7) and B(f=7).

If tie broken in favor of goal G: Pop G. Goal test succeeds!

If tie broken in favor of B: Pop B. B generates C with g(C) = 3 + 1 = 4 (same cost, C already closed). Then pop G.

Conclusion: Optimal Path is S -> A -> C -> G with total optimal cost C* = 7.

SOURCE TRANSPARENCY & CITATIONS

Primary Academic References & Exam Sources

Every concept in Notes From a B.Tech Brain is synthesized from authoritative academic texts, seminal peer-reviewed research, verified exam archives, and renowned university lectures.

Curated University Lecture Series
Artificial Intelligence: A Modern Approach (4th Edition) ·
LECTURE SERIES
Principles of Artificial Intelligence ·
LECTURE SERIES