Water
Jug Problem:
Problem:
You are given two jugs, a 4gallon one and a
3gallon one.Neither has any measuring mark on it.There is a pump that can be
used to fill the jugs with water.How can you get exactly 2 gallons of water
into the 4gallon jug.
Solution:
The state space for this problem can be
described as the set of ordered pairs of integers (x,y)
Where,
X represents the quantity of water in the 4gallon jug X=
0,1,2,3,4
Y represents the quantity of water in
3gallon jug Y=0,1,2,3
Start
State: (0,0)
Goal
State: (2,0)
Generate production rules for the water jug
problem
Production
Rules:
Rule

State

Process

1

(X,Y  X<4)

(4,Y)
{Fill 4gallon jug}

2

(X,Y Y<3)

(X,3)
{Fill 3gallon jug}

3

(X,Y X>0)

(0,Y)
{Empty 4gallon jug}

4

(X,Y  Y>0)

(X,0)
{Empty 3gallon jug}

5

(X,Y  X+Y>=4 ^ Y>0)

(4,Y(4X))
{Pour water from 3gallon jug into 4gallon jug
until 4gallon jug is full}

6

(X,Y  X+Y>=3 ^X>0)

(X(3Y),3)
{Pour water from 4gallon jug into 3gallon jug
until 3gallon jug is full}

7

(X,Y  X+Y<=4 ^Y>0)

(X+Y,0)
{Pour all water from 3gallon jug into 4gallon jug}

8

(X,Y  X+Y <=3^ X>0)

(0,X+Y)
{Pour all water from 4gallon jug into 3gallon jug}

9

(0,2)

(2,0)
{Pour 2 gallon water from 3 gallon jug into 4 gallon
jug}

Initialization:
Start State: (0,0)
Apply Rule 2:
(X,Y  Y<3)
>

(X,3)
{Fill 3gallon jug}

Now the state is (X,3)
Iteration
1:
Current State: (X,3)
Apply Rule 7:
(X,Y  X+Y<=4 ^Y>0)

(X+Y,0)
{Pour all water from 3gallon jug into 4gallon jug}

Now the state is (3,0)
Iteration
2:
Current State : (3,0)
Apply Rule 2:
(X,Y  Y<3)
>

(3,3)
{Fill 3gallon jug}

Now the state is (3,3)
Iteration
3:
Current State:(3,3)
Apply Rule 5:
(X,Y  X+Y>=4 ^ Y>0)

(4,Y(4X))
{Pour water from 3gallon jug into 4gallon jug
until 4gallon jug is full}

Now the state is (4,2)
Iteration
4:
Current State : (4,2)
Apply Rule 3:
(X,Y  X>0)

(0,Y)
{Empty 4gallon jug}

Now state is (0,2)
Iteration
5:
Current State : (0,2)
Apply Rule 9:
(0,2)

(2,0)
{Pour 2 gallon water from 3 gallon jug into 4 gallon
jug}

Now the state is (2,0)
Goal
Achieved.
State
Space Tree:
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Can u do it for 5 gallon jug to 2 gallon jug with exactly one gallon of water where 2 gallon jug is empty
ReplyDelete(0,0) //initial state
Delete(0,2) //rule 2
(2,0) //rule 7
(2,2) //rule 2
(4,0) //rule 7
(4,2) //rule 2
(5,1) //rule 5
(0,1) //rule 3
(1,0) //rule 7 and goal state
Thank You For Sharing this information.
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ReplyDeleteSimple Water Jug Problem
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nice
ReplyDeletekids will do (2,0) and (2,3) ..... send me solutions of (2,1) and (2,2)
ReplyDeleteI made a java program to do this...
ReplyDeleteimport java.util.*;
public class WaterJugProblem
{
static List SolSet=new ArrayList<>();
public static void main(String args[])
{
System.out.println("WATER JUG PROBLEM SOLVER\n");
Scanner sc=new Scanner(System.in);
System.out.println("Enter capcity of the 2 jugs\n");
int x=sc.nextInt();
int y=sc.nextInt();
System.out.println("Enter target capcity of the 2 jugs\n");
int tx=sc.nextInt();
int ty=sc.nextInt();
sc.close();
FindSolution(0,0,x,y,tx,ty,"",0);
Comparator BestSolution = (p,q)> p.solCountq.solCount;
Collections.sort(SolSet,BestSolution);
System.out.println("Total "+SolSet.size()+" possible solutions was found for the problem\n");
if(SolSet.size()!=0)
{
System.out.println("Best Solution is");
System.out.println(SolSet.get(0).sol);
}
System.out.println("Solutions that involve more than 10 steps are ignored to avoid complexity\n");
System.out.println("Designed By Arpan");
}
static void FindSolution(int x,int y,int xCapacity,int yCapacity,int xTarget,int yTarget,String sol,int count)
{
if(x==xTarget && y==yTarget) // If Goal state is found
{
SolSet.add(new Solution(sol,count));
return;
}
int newX=x,newY=y;
String newSol=sol;
if(count>10) // Too many recursion , Ignore this recursion
return;
//Production Rules
if(x0)
{
newSol=sol;
newX=0;
newY=y;
newSol=newSol+"("+newX+","+newY+")";
newSol=newSol+"Empty first jug\n";
FindSolution(newX,newY,xCapacity,yCapacity,xTarget,yTarget,newSol,count+1);
}
if(y>0)
{
newSol=sol;
newX=x;
newY=0;
newSol=newSol+"("+newX+","+newY+")";
newSol=newSol+"Empty Second jug\n";
FindSolution(newX,newY,xCapacity,yCapacity,xTarget,yTarget,newSol,count+1);
}
if((x+y)>xCapacity && y>0)
{
newSol=sol;
newX=xCapacity;
newY=y(xCapacityx);
newSol=newSol+"("+newX+","+newY+")";
newSol=newSol+"Pour water from the second jug into first jug until first jug is full\n";
FindSolution(newX,newY,xCapacity,yCapacity,xTarget,yTarget,newSol,count+1);
}
if((x+y)>yCapacity && x>0)
{
newSol=sol;
newX=x(yCapacityy);
newY=yCapacity;
newSol=newSol+"("+newX+","+newY+")";
newSol=newSol+"Pour water from the first jug into second jug until second jug is full\n";
FindSolution(newX,newY,xCapacity,yCapacity,xTarget,yTarget,newSol,count+1);
}
if((x+y)<=xCapacity && y>0)
{
newSol=sol;
newX=x+y;
newY=0;
newSol=newSol+"("+newX+","+newY+")";
newSol=newSol+"Pour all water from second jug into first jug\n";
FindSolution(newX,newY,xCapacity,yCapacity,xTarget,yTarget,newSol,count+1);
}
if((x+y)0)
{
newSol=sol;
newX=0;
newY=y+x;
newSol=newSol+"("+newX+","+newY+")";
newSol=newSol+"Pour all water from first jug into second jug\n";
FindSolution(newX,newY,xCapacity,yCapacity,xTarget,yTarget,newSol,count+1);
}
}
}
class Solution
{
String sol;
int solCount;
Solution(String sol,int solCount)
{
this.sol=sol;
this.solCount=solCount;
}
}
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ReplyDeletethanks.....
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ReplyDeleteIsn't rule 9 same as rule 7
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