Homework Assignment # 3 Solution

Description

Question 1: A Water Jug Problem [60 points]

 

This is a programming question. The solution to the programming problem should be coded in Java, and you are required to use only built-in libraries to complete this homework. Please submit a single zip le named hw3.zip, which should contain a source code le named WaterJug.java with no package statements, and make sure your program is runnable from command line on a department Linux machine. We provide a skeleton WaterJug.java code that you can optionally use, or you can write your own.

 

The goal of this assignment is to become familiar with uninformed search { breadth- rst search (BFS), depth- rst search (DFS) and iterative deepening search (IDDFS). The assignment tests your understanding of AI concepts, and your ability to turn conceptual understanding into a computer program. All concepts needed for this homework have been discussed in class, but there may not be existing pseudo-code for you to directly follow. We ask you to implement your own stack for BFS/DFS/IDDFS as we did in class, rather than writing a recursive program.

 

In this question, you will implement uninformed search for a water jug problem.

 

To illustrate, suppose you are given two containers: a 2-gallon water jug and a 1-gallon jug. Neither of them have any measuring marks on them at all. Initially both jugs are empty. You have 3 types of action available to you: 1) ll a jug up completely full (i.e. f1, f2) 2) empty a jug completely (i.e. e1, e2) 3) pour as much water as possible from one jug into the other (i.e. p12 (pour water from jug1 into jug2), p21 (pour water from jug2 into jug1)). The goal is you must end up with a jug having exactly 1 gallons of water. Formally:

 

State space S = f0; 1; 2g f0; 1g Initial state s₀ = (0; 0)

 

Goal states G = f(0; 1); (1; 0); (1; 1); (2; 1)g Actions A = ff1; f2; e1; e2; p12; p21g

Successor function is given by the table below

 

	f1	f2	e1	e2	p12	p21
(0,0)	(2,0)	(0,1)	–	–	–	–
(1,0)	(2,0)	(1,1)	(0,0)	–	(0,1)	–
(2,0)	–	(2,1)	(0,0)	–	(1,1)	–
(0,1)	(2,1)	–	–	(0,0)	–	(1,0)
(1,1)	(2,1)	–	(0,1)	(1,0)	–	(2,0)
(2,1)	–	–	(0,1)	(2,0)	–	–

 

Cost = 1 for all arcs

 

The search problem:  nd a solution path from a state in s₀ to a state in G

 

There are an in nite number of solutions. If a search algorithm is optimal, a solution to the water jug is a path from the initial state to a goal state with the smallest path cost. Possible solutions of the example above:

 

((0,0), (0,1))

 

((0,0), (2,0), (1,1))

 

((0,0), (2,0), (0,0), (2,0), (1,1))

 

Write a program WaterJug.java with the following command-line format:

 

$java WaterJug FLAG cap_jug1 cap_jug2 curr_jug1 curr_jug2 goal

 

where FLAG is an integer that speci es the output of the program (see below). cap jug1 and cap jug2 specify each jug’s capacity. curr jug1 and curr jug2 specify how much water each jug is initially lled and cannot exceed its capacity. goal speci es how much water you must end up with in either jug1 or jug2, or both jugs. These take values in integer 0-9 for command-line argument. For example, given the earlier example and FLAG=100, the command line would be

 

$java WaterJug 100 2 1 0 0 1

 

(Part a, 5 points) When FLAG=100, print out the successor states of an initial state, in the order they are pushed into the stack (see below). Each successor state should be printed as jug1 and jug2 state digits back-to-back on a single line. For example,

 

$java WaterJug 100 2 1 0 0 1

 

01

 

20

 

$java WaterJug 100 5 7 3 1 2

 

01

 

04

 

30

 

37

 

40

 

51

 

Important: We ask you to implement the following order among successors. If we view a state as a 2-digit integer, then there is a natural order among states. Whenever you push successors into the stack, push them from small 2-digit to large 2-digit. This order will be used throughout this program, so that the output is well-de ned.

 

(Part b, 5 points) When FLAG=200, verify if each of the successor states is a goal node. Recall this is true only when either jug1 or jug2, or both have exactly goal gallon. Print each successor state following by either true or false, separated by a whitespace.

 

$java WaterJug 200 2 1 0 0 1

 

• true

 

• false

 

$java WaterJug 200 5 7 3 1 2

 

01 false

 

 

 

 

CS 540

 

 

• false

 

• false

 

• false

 

• false

 

• false

 

(Part c, 15 points) When FLAG=300, perform a breadth- rst search till a goal state is reached. You will need to implement BFS using OPEN and CLOSED lists to keep track of progress through the state space. On the rst line, print the initial state. During each iteration step, take out a state from the fringe, verify whether it is a goal node. If so, print it, followed by the word “Goal” and terminate. Else, check whether it is already expanded. If yes, discard it. Otherwise, expand the state, add its successors to the fringe, and mark that state as already-expanded. At the end of each iteration step, print the expanded state, OPEN list, and CLOSED list with the following format:

 

s_expanded [o₁; o₂; :::; o_m] [c₁; c₂; :::; c_n]

 

The last line, you will print the word “Path”, followed by a sequence of states from the initial to the goal (all space-separated). For example,

 

$java WaterJug 300 4 3 0 0 2

 

00

 

00 [03,40] [00]

 

03 [40,30,43] [00,03]

 

40 [30,43,13] [00,03,40]

 

30 [43,13,33] [00,03,40,30]

 

43 [13,33] [00,03,40,30,43]

 

13 [33,10] [00,03,40,30,43,13]

 

33 [10,42] [00,03,40,30,43,13,33]

 

10 [42,01] [00,03,40,30,43,13,33,10]

 

42 Goal

 

Path 00 03 30 33 42

 

(Part d, 15 points) When FLAG=400, perform a depth- rst search till a goal state is reached. Again, you will need to implement DFS using OPEN and CLOSED lists to keep track of progress through the state space. Print results like in part c. For example,

 

$java WaterJug 400 4 3 0 0 2

 

00

 

00 [03,40] [00]

 

40 [03,13,43] [00,40]

 

43 [03,13] [00,40,43]

 

13 [03,10] [00,40,43,13]

 

10 [03,01] [00,40,43,13,10]

 

01 [03,41] [00,40,43,13,10,01]

 

41 [03,23] [00,40,43,13,10,01,41]

 

23 Goal

 

Path 00 40 13 10 01 41 23

CS 540

 

 

(Part e, 20 points) When FLAG=5XX, perform a depth-limited depth- rst search with cuto depth XX (i.e. this is one outer-loop of iterative deepening). For example, if FLAG=500, the cuto depth is 0. In DFS, you will push the initial state in the stack, pop it out, do a goal test, but will NOT expand it. If FLAG=501, the cuto depth is one. In DFS, you will expand the initial state (i.e. put its successors into the stack in the order we speci ed in Part a). You will pop each successor out, perform goal-check (and terminate the program if goal-check succeeds). But you will not expand any of these successors.

 

• can be 00 to 99. If depth-limited DFS nds a goal before the cuto , it should stop. Print results similar to part d with a current depth parameter pre xed. For example,

 

$java WaterJug 555 4 3 0 0 2

 

0:00

 

0:00 [] [00]

 

1:00

 

1:00 [03,40] [00]

 

1:40 [03] [00,40]

 

1:03 [] [00,40,03]

 

2:00

 

2:00 [03,40] [00]

 

2:40 [03,13,43] [00,40]

 

2:43 [03,13] [00,40,43]

 

2:13 [03] [00,40,43,13]

 

2:03 [30] [00,40,43,13,03]

 

2:30 [] [00,40,43,13,03,30]

 

3:00

 

3:00 [03,40] [00]

 

3:40 [03,13,43] [00,40]

 

3:43 [03,13] [00,40,43]

 

3:13 [03,10] [00,40,43,13]

 

3:10 [03] [00,40,43,13,10]

 

3:03 [30] [00,40,43,13,10,03]

 

3:30 [33] [00,40,43,13,10,03,30]

 

3:33 [] [00,40,43,13,10,03,30,33]

 

4:00

 

4:00 [03,40] [00]

 

4:40 [03,13,43] [00,40]

 

4:43 [03,13] [00,40,43]

 

4:13 [03,10] [00,40,43,13]

 

4:10 [03,01] [00,40,43,13,10]

 

4:01 [03] [00,40,43,13,10,01]

 

4:03 [30] [00,40,43,13,10,01,03]

 

4:30 [33] [00,40,43,13,10,01,03,30]

 

4:33 [42] [00,40,43,13,10,01,03,30,33]

 

4:42 Goal

 

Path 00 03 30 33 42