You are on the side of the river. You are given a m liter jug and a n liter jug where 0 < m < n. Both the jugs are initially empty. The jugs don’t have markings to allow measuring smaller quantities. You have to use the jugs to measure d liters of water where d < n. Determine the minimum no of operations to be performed to obtain d liters of water in one of jug.
The operations you can perform are:
- Empty a Jug
- Fill a Jug
- Pour water from one jug to the other until one of the jugs is either empty or full.
There are several ways of solving this problem including BFS and DP. In this article, an arithmetic approach to solving the problem is discussed. The problem can be modeled by means of the Diophantine equation of the form mx + ny = d which is solvable if and only if gcd(m, n) divides d. Also, the solution x,y for which equation is satisfied can be given using the Extended Euclid algorithm for GCD.
For example, if we have a jug J1 of 5 liters (n = 5) and another jug J2 of 3 liters (m = 3) and we have to measure 1 liter of water using them. The associated equation will be 5n + 3m = 1. First of all this problem can be solved since gcd(3,5) = 1 which divides 1 (See this for detailed explanation). Using the Extended Euclid algorithm, we get values of n and m for which the equation is satisfied which are n = 2 and m = -3. These values of n, m also have some meaning like here n = 2 and m = -3 means that we have to fill J1 twice and empty J2 thrice.
Now to find the minimum no of operations to be performed we have to decide which jug should be filled first. Depending upon which jug is chosen to be filled and which to be emptied we have two different solutions and the minimum among them would be our answer.
Solution 1 (Always pour from m liter jug into n liter jug)
- Fill the m litre jug and empty it into n liter jug.
- Whenever the m liter jug becomes empty fill it.
- Whenever the n liter jug becomes full empty it.
- Repeat steps 1,2,3 till either n liter jug or the m liter jug contains d litres of water.
Each of steps 1, 2 and 3 are counted as one operation that we perform. Let us say algorithm 1 achieves the task in C1 no of operations.
Solution 2 (Always pour from n liter jug into m liter jug)
- Fill the n liter jug and empty it into m liter jug.
- Whenever the n liter jug becomes empty fill it.
- Whenever the m liter jug becomes full empty it.
- Repeat steps 1, 2 and 3 till either n liter jug or the m liter jug contains d liters of water.
Let us say solution 2 achieves the task in C2 no of operations.
Now our final solution will be a minimum of C1 and C2.
Now we illustrate how both of the solutions work. Suppose there are a 3 liter jug and a 5 liter jug to measure 4 liters water so m = 3,n = 5 and d = 4. The associated Diophantine equation will be 3m + 5n = 4. We use pair (x, y) to represent amounts of water inside the 3-liter jug and 5-liter jug respectively in each pouring step.
Using Solution 1, successive pouring steps are:
(0,0)->(3,0)->(0,3)->(3,3)->(1,5)->(1,0)->(0,1)->(3,1)->(0,4)
Hence the no of operations you need to perform are 8.
Using Solution 2, successive pouring steps are:
(0,0)->(0,5)->(3,2)->(0,2)->(2,0)->(2,5)->(3,4)
Hence the no of operations you need to perform are 6.
Therefore, we would use solution 2 to measure 4 liters of water in 6 operations or moves.
Based on the explanation here is the implementation.
C++
// C++ program to count minimum number of steps// required to measure d litres water using jugs// of m liters and n liters capacity.#include <bits/stdc++.h>using namespace std;// Utility function to return GCD of 'a'// and 'b'.int gcd(int a, int b){ if (b==0) return a; return gcd(b, a%b);}/* fromCap -- Capacity of jug from which water is poured toCap -- Capacity of jug to which water is poured d -- Amount to be measured */int pour(int fromCap, int toCap, int d){ // Initialize current amount of water // in source and destination jugs int from = fromCap; int to = 0; // Initialize count of steps required int step = 1; // Needed to fill "from" Jug // Break the loop when either of the two // jugs has d litre water while (from != d && to != d) { // Find the maximum amount that can be // poured int temp = min(from, toCap - to); // Pour "temp" liters from "from" to "to" to += temp; from -= temp; // Increment count of steps step++; if (from == d || to == d) break; // If first jug becomes empty, fill it if (from == 0) { from = fromCap; step++; } // If second jug becomes full, empty it if (to == toCap) { to = 0; step++; } } return step;}// Returns count of minimum steps needed to// measure d literint minSteps(int m, int n, int d){ // To make sure that m is smaller than n if (m > n) swap(m, n); // For d > n we cant measure the water // using the jugs if (d > n) return -1; // If gcd of n and m does not divide d // then solution is not possible if ((d % gcd(n,m)) != 0) return -1; // Return minimum two cases: // a) Water of n liter jug is poured into // m liter jug // b) Vice versa of "a" return min(pour(n,m,d), // n to m pour(m,n,d)); // m to n}// Driver code to test aboveint main(){ int n = 3, m = 5, d = 4; printf("Minimum number of steps required is %d", minSteps(m, n, d)); return 0;} |
Java
// Java program to count minimum number of// steps required to measure d litres water// using jugs of m liters and n liters capacity.import java.io.*;class GFG{// Utility function to return GCD of 'a'// and 'b'.public static int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}/* fromCap -- Capacity of jug from which water is poured toCap -- Capacity of jug to which water is poured d -- Amount to be measured */public static int pour(int fromCap, int toCap, int d){ // Initialize current amount of water // in source and destination jugs int from = fromCap; int to = 0; // Initialize count of steps required int step = 1; // Needed to fill "from" Jug // Break the loop when either of the two // jugs has d litre water while (from != d && to != d) { // Find the maximum amount that can be // poured int temp = Math.min(from, toCap - to); // Pour "temp" liters from "from" to "to" to += temp; from -= temp; // Increment count of steps step++; if (from == d || to == d) break; // If first jug becomes empty, fill it if (from == 0) { from = fromCap; step++; } // If second jug becomes full, empty it if (to == toCap) { to = 0; step++; } } return step;}// Returns count of minimum steps needed to// measure d literpublic static int minSteps(int m, int n, int d){ // To make sure that m is smaller than n if (m > n) { int t = m; m = n; n = t; } // For d > n we cant measure the water // using the jugs if (d > n) return -1; // If gcd of n and m does not divide d // then solution is not possible if ((d % gcd(n, m)) != 0) return -1; // Return minimum two cases: // a) Water of n liter jug is poured into // m liter jug // b) Vice versa of "a" return Math.min(pour(n, m, d), // n to m pour(m, n, d)); // m to n}// Driver code public static void main(String[] args){ int n = 3, m = 5, d = 4; System.out.println("Minimum number of " + "steps required is " + minSteps(m, n, d));}}// This code is contributed by RohitOberoi |
Python3
# Python3 implementation of program to count # minimum number of steps required to measure# d litre water using jugs of m liters and n # liters capacity.def gcd(a, b): if b==0: return a return gcd(b, a%b)''' fromCap -- Capacity of jug from which water is poured toCap -- Capacity of jug to which water is poured d -- Amount to be measured '''def Pour(toJugCap, fromJugCap, d): # Initialize current amount of water # in source and destination jugs fromJug = fromJugCap toJug = 0 # Initialize steps required step = 1 while ((fromJug is not d) and (toJug is not d)): # Find the maximum amount that can be # poured temp = min(fromJug, toJugCap-toJug) # Pour 'temp' liter from 'fromJug' to 'toJug' toJug = toJug + temp fromJug = fromJug - temp step = step + 1 if ((fromJug == d) or (toJug == d)): break # If first jug becomes empty, fill it if fromJug == 0: fromJug = fromJugCap step = step + 1 # If second jug becomes full, empty it if toJug == toJugCap: toJug = 0 step = step + 1 return step# Returns count of minimum steps needed to# measure d literdef minSteps(n, m, d): if m> n: temp = m m = n n = temp if (d%(gcd(n,m)) is not 0): return -1 # Return minimum two cases: # a) Water of n liter jug is poured into # m liter jug return(min(Pour(n,m,d), Pour(m,n,d)))# Driver codeif __name__ == '__main__': n = 3 m = 5 d = 4 print('Minimum number of steps required is', minSteps(n, m, d)) # This code is contributed by Sanket Badhe |
C#
// C# program to implement// the above approachusing System;class GFG{// Utility function to return GCD of 'a'// and 'b'.public static int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);} /* fromCap -- Capacity of jug from which water is poured toCap -- Capacity of jug to which water is poured d -- Amount to be measured */public static int pour(int fromCap, int toCap, int d){ // Initialize current amount of water // in source and destination jugs int from = fromCap; int to = 0; // Initialize count of steps required int step = 1; // Needed to fill "from" Jug // Break the loop when either of the two // jugs has d litre water while (from != d && to != d) { // Find the maximum amount that can be // poured int temp = Math.Min(from, toCap - to); // Pour "temp" liters from "from" to "to" to += temp; from -= temp; // Increment count of steps step++; if (from == d || to == d) break; // If first jug becomes empty, fill it if (from == 0) { from = fromCap; step++; } // If second jug becomes full, empty it if (to == toCap) { to = 0; step++; } } return step;} // Returns count of minimum steps needed to// measure d literpublic static int minSteps(int m, int n, int d){ // To make sure that m is smaller than n if (m > n) { int t = m; m = n; n = t; } // For d > n we cant measure the water // using the jugs if (d > n) return -1; // If gcd of n and m does not divide d // then solution is not possible if ((d % gcd(n, m)) != 0) return -1; // Return minimum two cases: // a) Water of n liter jug is poured into // m liter jug // b) Vice versa of "a" return Math.Min(pour(n, m, d), // n to m pour(m, n, d)); // m to n}// Driver Codepublic static void Main(){ int n = 3, m = 5, d = 4; Console.Write("Minimum number of " + "steps required is " + minSteps(m, n, d));}}// This code is contributed by code_hunt. |
Javascript
<script>// Javascript program to count minimum number of// steps required to measure d litres water// using jugs of m liters and n liters capacity. // Utility function to return GCD of 'a' // and 'b'. function gcd(a , b) { if (b == 0) return a; return gcd(b, a % b); } /* fromCap -- Capacity of jug from which water is poured toCap -- Capacity of jug to which water is poured d -- Amount to be measured */ function pour(fromCap , toCap , d) { // Initialize current amount of water // in source and destination jugs var from = fromCap; var to = 0; // Initialize count of steps required var step = 1; // Needed to fill "from" Jug // Break the loop when either of the two // jugs has d litre water while (from != d && to != d) { // Find the maximum amount that can be // poured var temp = Math.min(from, toCap - to); // Pour "temp" liters from "from" to "to" to += temp; from -= temp; // Increment count of steps step++; if (from == d || to == d) break; // If first jug becomes empty, fill it if (from == 0) { from = fromCap; step++; } // If second jug becomes full, empty it if (to == toCap) { to = 0; step++; } } return step; } // Returns count of minimum steps needed to // measure d liter function minSteps(m , n , d) { // To make sure that m is smaller than n if (m > n) { var t = m; m = n; n = t; } // For d > n we cant measure the water // using the jugs if (d > n) return -1; // If gcd of n and m does not divide d // then solution is not possible if ((d % gcd(n, m)) != 0) return -1; // Return minimum two cases: // a) Water of n liter jug is poured into // m liter jug // b) Vice versa of "a" return Math.min(pour(n, m, d), // n to m pour(m, n, d)); // m to n } // Driver code var n = 3, m = 5, d = 4; document.write("Minimum number of " + "steps required is " + minSteps(m, n, d));// This code contributed by Rajput-Ji</script> |
Output:
Minimum number of steps required is 6
Time Complexity: O(N + M)
Auxiliary Space: O(1)
Another detailed Explanation:http://web.mit.edu/neboat/Public/6.042/numbertheory1.pdf
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