Water Jug Problem

This article discusses the water jug problem and provides a solution using graph modeling and algorithms. It explains how to find 3 gallons of water in a 5 gallon jug using a 3 gallon jug. The article includes step-by-step instructions and references for further reading.

Running head: WATER JUG PROBLEM
WATER JUG PROBLRM
Name of the Student
Name of the University
Author Note

WATER JUG PROBLEM 1
WATER JUG PROBLEM:
Two jugs are given, one is a 5 gallon jug and another one is a 3 gallon jug. The
challenge here is to find 3 gallons of water in a 5 gallon jug.
1) What type of graph would you use to model the problem input and how would you
construct this graph?
For the solution the BFS (Breadth-first search) technique can be used by making a
tree to solve the water jug problem. The graph is a directed graph and it is weighted too and
the graph is vertex labelled.
2. Give pseudo code for an algorithm to compute the amount of water that takes the most
moves to reach?
Pseudo code:
Class Jug {
value as integer
capacity as integer
public:
jug is integer
{
initial value is zero
initial capacity is n
}
void Fill() //fill
{
store capacity to value
}

2WATER JUG PROBLEM
empty function
{
empty value
}
value is full
{
value is greater than or equal to capacity
}
value is empty
{
empty
}
Transfer the contents hold by B to A till A shows full
{
store new value to old value
store value with b's value to value
if value is greater than capacity
omit old value from new value
}
get new value
{
return to value
}
};

3WATER JUG PROBLEM
gcd as integer
{
condition
return value of p
condition
return value
else
condition unsatisfied
}
{
C cannot hold water more than capacity //overflow
return false
}
//if k is multiple i and j hcf then
{
return true;
}
//if k is multiple i and j hcf then
Unable to reach this state with the jugs that are given
//false
}
jug and value
{
//condition
{
C full
{
fill D
store value
}

4WATER JUG PROBLEM
C is full
{
C is empty
store value
}
` Store D to C
store value
}
}
int main()
{
int i, j, result;
Enter the capacity of C
Enter the capacity of D
do{
Enter the water that is required in C
result;
}
condition
Jug C(i), D(j);
print result
solved
cout<<"Done!"<<endl; //done
#endif
//end of program
return 0;

5WATER JUG PROBLEM
}
Here for getting the solution the following steps can be followed. The steps are:
1. Fill 5-gal jug (x, y) → (5, y)
x < 5
2. Fill 3-gal jug (x,y) → (x,3)
y < 3
3. Empty 5-gal jug (x,y) → (0,y)
x > 0
4. Empty 3-gal jug (x,y) → (x,0)
y > 0
5. Pour water from 3 (x,y) → (5, y - (5 - x))
to fill 5-gal jug 0 < x+y ≥ 5 and y > 0
6. Pour water from 5 (x,y) → (x - (3-y), 3)
to fill 3-gal-jug 0 < x+y ≥ 3 and x > 0
7. Pour all water from 3-gal (x,y) → (x+y, 0)
into 5-gal 0 < x+y ≤ 5 and y ≥ 0
8. Pour all water from 5 (x,y) → (0, x+y)
into 3-gal jug 0 < x+y ≤ 3 and x ≥ 0
The solution will be as follows:

6WATER JUG PROBLEM
Gals in 5-gal jug Gals in 5-gal jug Rule Applied
0 0
0 3 1. Fill 3
3 0 7. Pour 3 into 5 to fill
3 3 1. Fill 3
5 1 7. Pour 3 into 5 to fill
0 1 3. Empty 5
1 0 7. Pour 3 into 5 to fill
1 3 1. Fill 3
4 0 7. Pour 3 into 5 to fill
Input:

7WATER JUG PROBLEM
Output:
References:

8WATER JUG PROBLEM
Man, Y. K. (2015, March). Solving the general two water jugs problem via an algorithmic
approach. In Proceedings of the International MultiConference of Engineers and
Computer Scientists (Vol. 1, pp. 18-20).
Saxena, D., Malik, N. K., & Singh, V. R. (2015). A cognitive approach to solve water jugs
problem. International Journal of Computer Applications, 124(17).