- In the water jug problem in Artificial Intelligence, provided two jugs:
- one having the capacity to hold 3 gallons of water and the other has the capacity to hold 4 gallons of water.
- There is no other measuring equipment available and the jugs also do not have any kind of marking on them. So, the agent’s task here is to fill the 4-gallon jug with 2 gallons of water by using only these two jugs and no other material. Initially, both our jugs are empty.
Here, let x denote the 4-gallon jug and y denote the 3-gallon jug.
| S.No. | Initial State | Condition | Final state | Description of action taken |
|---|---|---|---|---|
| 1. | (x,y) | If x<4 | (4,y) | Fill the 4 gallon jug completely |
| 2. | (x,y) | if y<3 | (x,3) | Fill the 3 gallon jug completely |
| 3. | (x,y) | If x>0 | (x-d,y) | Pour some part from the 4 gallon jug |
| 4. | (x,y) | If y>0 | (x,y-d) | Pour some part from the 3 gallon jug |
| 5. | (x,y) | If x>0 | (0,y) | Empty the 4 gallon jug |
| 6. | (x,y) | If y>0 | (x,0) | Empty the 3 gallon jug |
| 7. | (x,y) | If (x+y)<7 | (4, y-[4-x]) | Pour some water from the 3 gallon jug to fill the four gallon jug |
| 8. | (x,y) | If (x+y)<7 | (x-[3-y],y) | Pour some water from the 4 gallon jug to fill the 3 gallon jug. |
| 9. | (x,y) | If (x+y)<4 | (x+y,0) | Pour all water from 3 gallon jug to the 4 gallon jug |
| 10. | (x,y) | if (x+y)<3 | (0, x+y) | Pour all water from the 4 gallon jug to the 3 gallon jug |
| S.No. | 4 gallon jug contents | 3 gallon jug contents | Rule followed |
|---|---|---|---|
| 1. | 0 gallon | 0 gallon | Initial state |
| 2. | 0 gallon | 3 gallons | Rule no.2 |
| 3. | 3 gallons | 0 gallon | Rule no. 9 |
| 4. | 3 gallons | 3 gallons | Rule no. 2 |
| 5. | 4 gallons | 2 gallons | Rule no. 7 |
| 6. | 0 gallon | 2 gallons | Rule no. 5 |
| 7. | 2 gallons | 0 gallon | Rule no. 9 |
On reaching the 7th attempt, we reach a state which is our goal state. Therefore, at this state, our problem is solved.