2 JUG Problem Easy Python Code – The Two Water Jug Puzzle
A systematic approach to solving two jugs of water riddle
Measure 2 liters of water with two empty jugs of 4 liters and 3-liter capacity and a tank full of water. Pour as many times as you want.
How many ways can you do it by repeating no move?
The two jugs problem is an interesting little riddle that has a number of approaches to solve. Go on. Explore.
List of Operations you can PERFORM.
- 1.Fill Jug A Completely
- 2.Fill Jug B Completely
- 3.Empty Jug A Completely
- 4.Empty Jug B Completely
- 5.Pour From Jug A till Jug B filled Completely or A becomes empty
- 6.Pour From Jug B till Jug A filled Completely or B becomes empty
- 7.Pour all From Jug B to Jug A
- 8.Pour all From Jug A to Jug B
PYTHON Code of 2 Water Jug Problem
a=int(input("Enter Jug A Capacity: "));
b=int(input("Enter Jug B Capacity: "));
ai=int(input("Initially Water in Jug A: "));
bi=int(input("Initially Water in Jug B: "));
af=int(input("Final State of Jug A: "));
bf=int(input("Final State of Jug B: "));
print("List Of Operations You can Do:\n");
print("1.Fill Jug A Completely\n");
print("2.Fill Jug B Completely\n");
print("3.Empty Jug A Completely\n");
print("4.Empty Jug B Completely\n");
print("5.Pour From Jug A till Jug B filled Completely or A becomes empty\n");
print("6.Pour From Jug B till Jug A filled Completely or B becomes empty\n");
print("7.Pour all From Jug B to Jug A\n");
print("8.Pour all From Jug A to Jug B\n");
while ((ai!=af or bi!=bf)):
op=int(input("Enter the Operation: "));
if(op==1):
ai=a;
elif(op==2):
bi=b;
elif(op==3):
ai=0;
elif(op==4):
bi=0;
elif(op==5):
if(b-bi>ai):
bi=ai+bi;
ai=0;
else:
ai=ai-(b-bi);
bi=b;
elif(op==6):
if(a-ai>bi):
ai=ai+bi;
bi=0;
else:
bi=bi-(a-ai);
ai=a;
elif(op==7):
ai=ai+bi;
bi=0;
elif(op==8):
bi=bi+ai;
ai=0;
print(ai,bi);Credit – Mani Garg (Github ID) https://github.com/manipta
Checkout: 3 Jug Water Puzzle in Python
Leave in the comments if you have anything in mind.