reward and once in it we never leave. All of the actions were chosen to take us toward S4., Formulas and forms for Infirmary. Amity Business School
mohamed-khalil
mohamed-khalil

reward and once in it we never leave. All of the actions were chosen to take us toward S4., Formulas and forms for Infirmary. Amity Business School

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We get this answer by noticing that is very close to 1, so we are most interested in the long-term reward. The best spot to be in for the long haul is S4, because it has the highest reward and once in it we never le...
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Homework 3

Informed search

CompSci 171: Intro AI

4.1 A* search: From Lugoj to Bucharest

f(n) = g(n) + h(n) A* search is guided by evaluation function f(n)

100

A* search: From Lugoj to Bucharest

Lugoj

Timisoara Mehadia

f = 111 + 329 = 440 f = 70 + 241 = 311

A* search: From Lugoj to Bucharest

Lugoj

Timisoara Mehadia

f = 111 + 329 = 440 f = 70 + 241 = 311

Dobreta

f = (70 + 75) + 242 = 387

A* search: From Lugoj to Bucharest

Lugoj

Timisoara Mehadia

f = 111 + 329 = 440 f = 70 + 241 = 311

Dobreta

f = (70 + 75) + 242 = 387

Craiova

f = (145 + 120) + 160 = 425

A* search: From Lugoj to Bucharest

Lugoj

Timisoara Mehadia

f = 111 + 329 = 440 f = 70 + 241 = 311

Dobreta

f = (70 + 75) + 242 = 387

Craiova

f = (145 + 120) + 160 = 425

Rimnicu Pitesti

f = 503f = 604

A* search: From Lugoj to Bucharest

Lugoj

Timisoara Mehadia

f = 111 + 329 = 440 f = 70 + 241 = 311

Dobreta

f = (70 + 75) + 242 = 387

Craiova

f = (145 + 120) + 160 = 425

Rimnicu Pitesti

f = 503f = 604

Arad

f = (111 + 118) + 366 = 595

A* search: From Lugoj to Bucharest

Lugoj

Timisoara Mehadia

f = 111 + 329 = 440 f = 70 + 241 = 311

Dobreta

f = (70 + 75) + 242 = 387

Craiova

f = (145 + 120) + 160 = 425

Rimnicu Pitesti

f = 503f = 604

Dobreta

f = (111 + 118) + 366 = 595

Rimnicu Bucharest

f = 504f = 693

f = 70+75+120+138+101 = 504

4.2 Heuristic path algorithm

f(n) = (2 – w)g(n) + wh(n) For what value of w is this algorithm guaranteed to be

optimal?

g(n): a path cost to n from a start state h(n): a heuristic estimate of cost from n to a goal state

4.2 Heuristic path algorithm If h(n) is admissible, the algorithm is guaranteed to be optimal

which behaves exactly like A* search with a heuristic

To be optimal, we require

f n =2−w [g n w 2−w

hn ]

f n =g n  w 2−w

hn

w 2−w

1

w1

4.2 Heuristic path algorithm

For w = 0: f(n) = 2g(n) -> Uniform-cost search

For w = 1: f(n) = g(n) + h(n) -> A* search

For w = 2: f(n) = 2h(n) -> Greedy best search

4.3

(a) Breadth-first search is a special case of uniform-cost search

When all step costs are equal (and let’s assume equal to 1), g(n) is just a multiple of depth n. Thus, breadth-first search and uniform-cost search would behave the same in this case

f(n) = g(n) = 1*(depth of n)

4.3

(b) BFS, DFS and uniform-cost search are special cases of best-first search

BFS: f(n) = depth(n)

DFS: f(n) = -depth(n)

UCS: f(n) = g(n)

4.3

(c) Uniform-cost search is special case of A* search

A* search: f(n) = g(n) + h(n) Uniform-cost search: f(n) = g(n)

Thus, for h(n) = 0, uniform cost search will produce the same result as A* search

4. Prove that the Manhattan Distance heuristic for 8-puzzle is admissible

Manhattan Distance for points P1(x1,y1), P2(x2,y2) is defined by:

d p1 , p2 =∣x1− x2∣∣y1− y2∣

Heuristic:

•Tiles cannot move along diagonals, so each tile has to move at least d(n) steps to its goal

•Any move can only move one tile at a time

h=∑ n=1

8

d n

5. Eight Queens problem

h – number of pairs of queens that are attacking each other

5. Eight Queens problem

h = 2

However, number of conflicts for Queen at “b3” are:

1 2 2 3 0 2 2 2

5. Eight Queens problem

Therefore, the true cost to reach the goal state h* is 1

Thus, h>h* heuristic is not admissible

5. Eight Queens problem

We can propose another heuristic. For example, we can propose heuristic derived form a relaxed (and trivial) version of 8-Queens problem, that the eight queens must be placed in the board so that no two queens are on the same row.

Thus, h = Σ( # queens that are on the same row – 1) for all conflicting rows

4.11

(a) Local beam search with k=1

– We would randomly generate 1 start state

– At each step we would generate all the successors, and retain the 1 best state

– Equivalent to HILL-CLIMBING

4.11

(b) Local beam search with k=∞

– 1 initial state and no limit of the number of states retained

– We start at initial state and generate all successor states (no limit how many)

– If one of those is a goal, we stop – Otherwise, we generate all successors of those states

(2 steps from the initial state), and continue

– Equivalent to BREADTH-FIRST SEARCH

4.11

(c) Simulated annealing with T = 0 at all times

– If T is very small, the probability of accepting an arbitrary neighbor with lower value is approximately 0

– This means that we choose a successor state randomly and move to that state if it is better than the current state

– Equivalent to FIRST-CHOICE HILL CLIMBING

4.11

(d) Genetic algorithm with population size N = 1

– If selection step necessarily chooses the single population member twice, so the crossover steo does nothing.

– Moreover, if we think of the mutation step as selecting a successor at random, there is no guarantee that the successor is an improvement over the parent

– Equivalent to RANDOM WALK

4-Queens problem

Min-conflict algorithm:

1. Randomly choose a variable from set of problematic variables

2. Reassign its value to the one that results in the fewest conflicts overall

3. Continue until there are no conflicts

Q1

Q2

Q3

Q4

4-Queens problem

3 1 2 1

1 3 1 2

2 1 3 1

1 2 1 3

A B C D

4

3

2

1

Q1

Q2

Q3

Q4

Number of conflicts

All queens are attacked.

Pick Q2 randomly

We can move Q2 to B2 or B4

Randomly, move Q2 to B4

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