HOW TO SOLVE AI PROBLEMS AND THINK LIKE COMPUTER?

Date : 08/02/2020

Many AI problems are hard to solve because they are difficult to be characterized. Not only the problem but a path to the solution is also hard to be characterized. While considering chess game ,
If we go and write every chess move, we may not be able to complete it in our lifetime.

Solving AI Problems

The first problem is called 3-4 gallon water jug problem as described in the book by Elaine Rich. We
begin with two empty jugs of the capacity of 3 and 4 gallons respectively. We have an infinite supply of water available and we want the 4-gallon jug to be filled with exactly 2 gallons of water.
Another problem is called 8-5-3 milk jug problem which begins with 8-gallon milk jug completely full while other two are empty. We want exactly 4 gallons of milk in the 8-gallon jug.

In either case, there are no other measuring devices available to us.

One solution for the first problem is

0-0, 0-4, 3-1, 0-1, 1-0, 1-4, 3-2.

(Where n-m indicates quantity of water in 3 and 4-gallon jugs respectively)

One solution to the second problem is

8-0-0, 3-5-0, 3-2-3, 6-2-0, 6-0-2, 1-5-2, 1-4-3, 4-4-0.

Where n-m-l are respective volumes of milk in 8, 5 and 3-gallon jugs.

Applying rules to solve the problem

Now let us see the sequence of rules applied for the solution.

Consider 8-5-3 as (X-Y-Z)

1. 8-0-0 -> 3-5-0 ( X+Y =8 > 5)

2. 3-5-0 -> 3-2-3 ( Y + Z = 5 > 3)

3. 3-2-3 -> 6-2-0 ( X +Z =6)

4. 6-2-0 -> 6-0-2 ( Y+Z = 2 < 3)

5. 6-0-2 -> 1-5-2 ( X + Y = 6 >5)

6. 1-5-2 -> 1-4-3 ( Y + Z = 7 > 3)

7. 1-4-3 -> 4-4-0 ( X + Z =4)

4-4-0 is a final state.

We have an initial state called (8,0,0), we apply a typical rule, record the new status, apply another rule and again look at the status and continue until we reach a state which is qualified as a final state.



code for above question is given :-

FIRST CODE OF AI

#include<iostream>
using namespace std;

int main()
{
int limX = 8, limY = 5, limZ = 3;
int x = 8, y = 0, z = 0;
int a,b,c;
for(int i = 0; i < 2; i++)
{
// Minus water from x
if((x - (limY - y)) >= 0)
{
a = x - (x - (limY - y));
y += a; // Add Required water
x -= a;
}
else
{
y += x;
x = 0;
}

if((y - (limZ - z)) >= 0)
{
b = y - (y - (limZ - z));
z += b;
y -= b;
}
else
{
z += y;
y = 0;
}

if((z - (limX - x)) >= 0)
{
c = z - (z - (limX - x));
x += c;
z -= c;
}
else
{
x += z;
z = 0;
}
if(i>0)
{
break;
}
else
{

z = y;
y = 0;
}

}

cout<<x<<y<<z;
}