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Measure one litre using two vessels and infinite water supplys in C++
In this problem, we are given two vessels with capacities x and y and a supply of infinite water. Our task is to create a program that will be able to calculate exactly 1 liter in one vessel. Given the condition that x and y are co-primes. Co-primes also is known as relatively prime, mutually prime are numbers two numbers that have 1 as their only common divisor. So, this implies that their gcd(greatest common divisor) is 1.
Here, let’s suppose we have two vessels V1 with capacity x and V2 with capacity y. To measure 1 liter using these two vessels we will fill first vessels from water supply and then pour it to the second one. Do this process continues until the vessel V1 contains 1 litre of water.
Let’s take an example to understand the problem,
Input −
V1 = 5, V2 = 8 V1 = 5 ; V2 = 0 -> pour water from V1 to V2 and refill it. V1 = 5 ; V2 = 5 -> pour water from V1 to V2 and refill it. V1 = 2 ; V2 = 0 -> pour water from V1 to V2. Now, V2 is filled, empty it. V1 = 5 ; V2 = 2 -> pour water from V1 to V2 and refill it. V1 = 5 ; V2 = 7 -> pour water from V1 to V2 and refill it. V1 = 4 ; V2 = 0 -> pour water from V1 to V2. Now, V2 is filled, empty it. V1 = 1 ; V2 = 0 -> pour water from V1 to V2 and refill it. Here, V1 measures 1 litre of water.
Example
Program to illustrate the solution,
#include <iostream>
using namespace std;
int x, y, V1, V2 = 0;
int transferWater(int amt1, int amt2) {
if (amt1 + amt2 < y){
V2 += V1;
return V1;
}
int transferred = y - V2;
V2 = 0;
return transferred;
}
void measure1Litre() {
while(V1 != 1){
if (V1 == 0)
V1 = x;
cout<<"Vessel 1: "<<V1<<" | Vessel 2: "<<V2<<endl;
V1 = V1 - transferWater(V1, V2);
}
cout<<"Vessel 1: "<<V1<<" | Vessel 2: "<<V2<<endl;
}
int main() {
x= 5, y = 8;
measure1Litre();
return 0;
}Output
Vessel 1: 5 | Vessel 2: 0 Vessel 1: 5 | Vessel 2: 5 Vessel 1: 2 | Vessel 2: 0 Vessel 1: 5 | Vessel 2: 2 Vessel 1: 5 | Vessel 2: 7 Vessel 1: 4 | Vessel 2: 0 Vessel 1: 5 | Vessel 2: 4 Vessel 1: 1 | Vessel 2: 0
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