You are given an m liter jug and a n liter jug. 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 is less than n.
(X, Y) corresponds to a state where X refers to the amount of water in Jug1 and Y refers to the amount of water in Jug2
Determine the path from e initial state (xi, yi) to the final state (xf, yf), where (xi, yi) is (0, 0) which indicates both Jugs are initially empty and (xf, yf) indicates a state which could be (0, d) or (d, 0).
The operations you can perform are:
- Empty a jug (X, 0)->(0, 0) Empty Jug 1.
- Fill a Jug, (0, 0)->(X, 0) Fill Jug 1
- Pour water from one jug to the other until one of the jugs is either empty or full, (X, Y) -> (X-d, Y+d)
Examples:
Input : 4 3 2
Output : {( 0,0),(0,3),(3,0),(3,3),(4,2),(0,2)}
We have discussed the optimal solution in The Two Water Jug Puzzle. In this post, a BFS based solution is discussed.
Here, we keep exploring all the different valid cases of the states of water in the jug simultaneously until and unless we reach the required target water.
As provided in the problem statement, at any given state we can do either of the following operations:
- Fill a jug
- Empty a jug
- Transfer water from one jug to another until either of them gets completely filled or empty.
Examples:
Input: X = 4, Y = 3, Z = 2
Output: {(0, 0), (0, 3), (3, 0), (3, 3), (4, 2), (0, 2)}
Explanation:
Step 1:- First we will fill the 4 litre jug completely with water.
Step 2:- Then optimal approach would be to empty water from 4-litre jug into 3-litre (leaving 1L water in 4L jug and 3L completely full). Hence we got 1L water.
Step 3:- Now, Empty water from 3L.
Step 4:- Pour the water from 4L jug into 3L jug Now 4L container is completely empty and 1L water in present in 3L litre jug.
Step 5:- Fill the 4L jug with water completely again.
Step 6:- On transferring water from 4L jug to 3L jug, we will get 2L water in 4L jug which was our required quantity.
Input: X = 3, Y = 5, Z = 4
Output: 6
Explanation:
Step 1:- First we will fill the 5-litres jug to its maximum capacity.
Step 2:- Then optimal approach would be to transfer 3-litres from 5-litres jug to 3-litres jugs.
Step 3:- Now, Empty the 3-litres jug.
Step 4:- Transfer 2L from 5L jug to 3-L jug.
Step 5:- Now, Fill 5-litres jug to its maximum capacity.
Step 6:- On Pouring water from 5L jug to 3L jug until it’s full we will get 4L water in 5-litre jug which was our required quantity.
Running of the algorithm:
We start at an initial state in the queue where both the jugs are empty. We then continue to explore all the possible intermediate states derived from the current jug state using the operations provided.
We also, maintain a visited matrix of states so that we avoid revisiting the same state of jugs again and again.
|
Cases |
Jug 1 |
Jug 2 |
Is Valid |
|---|---|---|---|
|
Case 1 |
Fill it |
Empty it |
? |
|
Case 2 |
Empty it |
Fill it |
? |
|
Case 3 |
Fill it |
Fill it |
Redundant case |
|
Case 4 |
Empty it |
Empty it |
Already visited (Initial State) |
|
Case 5 |
Unchanged |
Fill it |
? |
|
Case 6 |
Fill it |
Unchanged |
? |
|
Case 7 |
Unchanged |
Empty |
? |
|
Case 8 |
Empty |
Unchanged |
? |
|
Case 9 |
Transfer water from this |
Transfer water into this |
? |
|
Case 10 |
Transfer water into this |
Transfer water from this |
? |
From the table above, we can observe that the state where both the jugs are filled is redundant as we won’t be able to continue ahead / do anything with this state in any possible way.
So, we proceed, keeping in mind all the valid state cases (as shown in the table above) and we do a BFS on them.
In the BFS, we first skip the states which was already visited or if the amount of water in either of the jugs exceeded the jug quantity.
If we continue further, then we first mark the current state as visited and check if in this state, if we have obtained the target quantity of water in either of the jugs, we can empty the other jug and return the current state’s entire path.
But, if we have not yet found the target quantity, we then derive the intermediate states from the current state of jugs i.e. we derive the valid cases, mentioned in the table above (go through the code once if you have some confusion).
We keep repeating all the above steps until we have found our target or there are no more states left to proceed with.
Implementation:
C++
#include <bits/stdc++.h>using namespace std;typedef pair<int, int> pii;void printpath(map<pii, pii> mp, pii u){ if (u.first == 0 && u.second == 0) { cout << 0 << " " << 0 << endl; return; } printpath(mp, mp[u]); cout << u.first << " " << u.second << endl;}void BFS(int a, int b, int target){ map<pii, int> m; bool isSolvable = false; map<pii, pii> mp; queue<pii> q; q.push(make_pair(0, 0)); while (!q.empty()) { auto u = q.front(); // cout<<u.first<<" "<<u.second<<endl; q.pop(); if (m[u] == 1) continue; if ((u.first > a || u.second > b || u.first < 0 || u.second < 0)) continue; // cout<<u.first<<" "<<u.second<<endl; m[{ u.first, u.second }] = 1; if (u.first == target || u.second == target) { isSolvable = true; printpath(mp, u); if (u.first == target) { if (u.second != 0) cout << u.first << " " << 0 << endl; } else { if (u.first != 0) cout << 0 << " " << u.second << endl; } return; } // completely fill the jug 2 if (m[{ u.first, b }] != 1) { q.push({ u.first, b }); mp[{ u.first, b }] = u; } // completely fill the jug 1 if (m[{ a, u.second }] != 1) { q.push({ a, u.second }); mp[{ a, u.second }] = u; } // transfer jug 1 -> jug 2 int d = b - u.second; if (u.first >= d) { int c = u.first - d; if (m[{ c, b }] != 1) { q.push({ c, b }); mp[{ c, b }] = u; } } else { int c = u.first + u.second; if (m[{ 0, c }] != 1) { q.push({ 0, c }); mp[{ 0, c }] = u; } } // transfer jug 2 -> jug 1 d = a - u.first; if (u.second >= d) { int c = u.second - d; if (m[{ a, c }] != 1) { q.push({ a, c }); mp[{ a, c }] = u; } } else { int c = u.first + u.second; if (m[{ c, 0 }] != 1) { q.push({ c, 0 }); mp[{ c, 0 }] = u; } } // empty the jug 2 if (m[{ u.first, 0 }] != 1) { q.push({ u.first, 0 }); mp[{ u.first, 0 }] = u; } // empty the jug 1 if (m[{ 0, u.second }] != 1) { q.push({ 0, u.second }); mp[{ 0, u.second }] = u; } } if (!isSolvable) cout << "No solution";}int main(){ int Jug1 = 4, Jug2 = 3, target = 2; cout << "Path from initial state " "to solution state ::\n"; BFS(Jug1, Jug2, target); return 0;} |
Java
// Java program for water jug problem// using BFS// Code by: Sparsh_CBSimport java.util.*;class Pair { int j1, j2; List<Pair> path; Pair(int j1, int j2) { this.j1 = j1; this.j2 = j2; path = new ArrayList<>(); } Pair(int j1, int j2, List<Pair> _path) { this.j1 = j1; this.j2 = j2; path = new ArrayList<>(); path.addAll(_path); path.add(new Pair(this.j1, this.j2)); }}public class GFG { public static void main(String[] args) throws java.lang.Exception { int jug1 = 4; int jug2 = 3; int target = 2; getPathIfPossible(jug1, jug2, target); } private static void getPathIfPossible(int jug1, int jug2, int target) { boolean[][] visited = new boolean[jug1 + 1][jug2 + 1]; Queue<Pair> queue = new LinkedList<>(); // Initial State: Both Jugs are empty so, // initialise j1 j2 as 0 and put it in the path list Pair initialState = new Pair(0, 0); initialState.path.add(new Pair(0, 0)); queue.offer(initialState); while (!queue.isEmpty()) { Pair curr = queue.poll(); // Skip already visited states and overflowing // water states if (curr.j1 > jug1 || curr.j2 > jug2 || visited[curr.j1][curr.j2]) continue; // mark current jugs state as visited visited[curr.j1][curr.j2] = true; // Check if current state has already reached // the target amount of water or not if (curr.j1 == target || curr.j2 == target) { if (curr.j1 == target) { // If in our current state, jug1 holds // the required amount of water, then we // empty the jug2 and push it into our // path. curr.path.add(new Pair(curr.j1, 0)); } else { // else, If in our current state, jug2 // holds the required amount of water, // then we empty the jug1 and push it // into our path. curr.path.add(new Pair(0, curr.j2)); } int n = curr.path.size(); System.out.println( "Path of states of jugs followed is :"); for (int i = 0; i < n; i++) System.out.println( curr.path.get(i).j1 + " , " + curr.path.get(i).j2); return; } // If we have not yet found the target, then we // have three cases left I. Fill the jug and // Empty the other II. Fill the jug and let the // other remain untouched III. Empty the jug and // let the other remain untouched // IV. Transfer amounts from one jug to another // Please refer to the table attached above to // understand the cases that we are taking into // consideration // Now, // I. Fill the jug and Empty the other queue.offer(new Pair(jug1, 0, curr.path)); queue.offer(new Pair(0, jug2, curr.path)); // II. Fill the jug and let the other remain // untouched queue.offer(new Pair(jug1, curr.j2, curr.path)); queue.offer(new Pair(curr.j1, jug2, curr.path)); // III. Empty the jug and let the other remain // untouched queue.offer(new Pair(0, curr.j2, curr.path)); queue.offer(new Pair(curr.j1, 0, curr.path)); // IV. Transfer water from one to another until // one jug becomes empty or until one jug // becomes full in this process // Transferring water form jug1 to jug2 int emptyJug = jug2 - curr.j2; int amountTransferred = Math.min(curr.j1, emptyJug); int j2 = curr.j2 + amountTransferred; int j1 = curr.j1 - amountTransferred; queue.offer(new Pair(j1, j2, curr.path)); // Tranferring water form jug2 to jug1 emptyJug = jug1 - curr.j1; amountTransferred = Math.min(curr.j2, emptyJug); j2 = curr.j2 - amountTransferred; j1 = curr.j1 + amountTransferred; queue.offer(new Pair(j1, j2, curr.path)); } System.out.println("Not Possible to obtain target"); }} |
Python3
from collections import dequedef BFS(a, b, target): m = {} isSolvable = False path = [] q = deque() q.append((0, 0)) while (len(q) > 0): u = q.popleft()# If this state is already visited if ((u[0], u[1]) in m): continue if ((u[0] > a or u[1] > b or u[0] < 0 or u[1] < 0)): continue # Filling the vector for constructing # the solution path path.append([u[0], u[1]]) # Marking current state as visited m[(u[0], u[1])] = 1 # If we reach solution state, put ans=1 if (u[0] == target or u[1] == target): isSolvable = True if (u[0] == target): if (u[1] != 0): # Fill final state path.append([u[0], 0]) else: if (u[0] != 0): # Fill final state path.append([0, u[1]]) # Print the solution path sz = len(path) for i in range(sz): print("(", path[i][0], ",", path[i][1], ")") break # If we have not reached final state # then, start developing intermediate # states to reach solution state q.append([u[0], b]) # Fill Jug2 q.append([a, u[1]]) # Fill Jug1 for ap in range(max(a, b) + 1): # Pour amount ap from Jug2 to Jug1 c = u[0] + ap d = u[1] - ap # Check if this state is possible or not if (c == a or (d == 0 and d >= 0)): q.append([c, d]) # Pour amount ap from Jug 1 to Jug2 c = u[0] - ap d = u[1] + ap # Check if this state is possible or not if ((c == 0 and c >= 0) or d == b): q.append([c, d]) # Empty Jug2 q.append([a, 0]) # Empty Jug1 q.append([0, b]) # No, solution exists if ans=0 if (not isSolvable): print("No solution")# Driver codeif __name__ == '__main__': Jug1, Jug2, target = 4, 3, 2 print("Path from initial state " "to solution state ::") BFS(Jug1, Jug2, target)# This code is contributed by mohit kumar 29 |
C#
using System;using System.Collections.Generic;class GFG { static void BFS(int a, int b, int target) { // Map is used to store the states, every // state is hashed to binary value to // indicate either that state is visited // before or not Dictionary<Tuple<int, int>, int> m = new Dictionary<Tuple<int, int>, int>(); bool isSolvable = false; List<Tuple<int, int> > path = new List<Tuple<int, int> >(); // Queue to maintain states List<Tuple<int, int> > q = new List<Tuple<int, int> >(); // Initializing with initial state q.Add(new Tuple<int, int>(0, 0)); while (q.Count > 0) { // Current state Tuple<int, int> u = q[0]; // Pop off used state q.RemoveAt(0); // If this state is already visited if (m.ContainsKey(u) && m[u] == 1) continue; // Doesn't met jug constraints if ((u.Item1 > a || u.Item2 > b || u.Item1 < 0 || u.Item2 < 0)) continue; // Filling the vector for constructing // the solution path path.Add(u); // Marking current state as visited m[u] = 1; // If we reach solution state, put ans=1 if (u.Item1 == target || u.Item2 == target) { isSolvable = true; if (u.Item1 == target) { if (u.Item2 != 0) // Fill final state path.Add(new Tuple<int, int>( u.Item1, 0)); } else { if (u.Item1 != 0) // Fill final state path.Add(new Tuple<int, int>( 0, u.Item2)); } // Print the solution path int sz = path.Count; for (int i = 0; i < sz; i++) Console.WriteLine("(" + path[i].Item1 + ", " + path[i].Item2 + ")"); break; } // If we have not reached final state // then, start developing intermediate // states to reach solution state // Fill Jug2 q.Add(new Tuple<int, int>(u.Item1, b)); // Fill Jug1 q.Add(new Tuple<int, int>(a, u.Item2)); for (int ap = 0; ap <= Math.Max(a, b); ap++) { // Pour amount ap from Jug2 to Jug1 int c = u.Item1 + ap; int d = u.Item2 - ap; // Check if this state is possible or not if (c == a || (d == 0 && d >= 0)) q.Add(new Tuple<int, int>(c, d)); // Pour amount ap from Jug 1 to Jug2 c = u.Item1 - ap; d = u.Item2 + ap; // Check if this state is possible or not if ((c == 0 && c >= 0) || d == b) q.Add(new Tuple<int, int>(c, d)); } // Empty Jug2 q.Add(new Tuple<int, int>(a, 0)); // Empty Jug1 q.Add(new Tuple<int, int>(0, b)); } // No, solution exists if ans=0 if (!isSolvable) Console.WriteLine("No solution"); } // Driver code static void Main() { int Jug1 = 4, Jug2 = 3, target = 2; Console.WriteLine("Path from initial state " + "to solution state ::"); BFS(Jug1, Jug2, target); }}// This code is contributed by divyeshrabadiya07 |
Javascript
<script> function BFS(a,b,target) { // This 2d array is used as a hashmap // to keep track of already visited // values and avoid repetition let m = new Array(1000); for(let i=0;i<1000;i++){ m[i]=new Array(1000); for(let j=0;j<1000;j++) { m[i][j]=-1; } } let isSolvable = false; let path = []; let q = []; // queue to maintain states q.push([0,0]); // Initializing with initial state while (q.length!=0) { let u = q[0]; // current state q.shift(); // pop off used state // doesn't met jug constraints if ((u[0] > a || u[1] > b || u[0] < 0 || u[1] < 0)) continue; // if this state is already visited if (m[u[0]][u[1]] > -1) continue; // filling the vector for constructing // the solution path path.push([u[0],u[1]]); // marking current state as visited m[u[0]][u[1]] = 1; // System.out.println(m.get(new Pair(u.first, u.second))); // if we reach solution state, put ans=1 if (u[0] == target || u[1] == target) { isSolvable = true; if (u[0] == target) { if (u[1] != 0) // fill final state path.push([u[0],0]); } else { if (u[0] != 0) // fill final state path.push([0,u[1]]); } // print the solution path let sz = path.length; for (let i = 0; i < sz; i++) document.write("(" + path[i][0] + ", " + path[i][1] + ")<br>"); break; } // if we have not reached final state // then, start developing intermediate // states to reach solution state q.push([u[0],b]); // fill Jug2 q.push([a,u[1]]); // fill Jug1 for (let ap = 0; ap <= Math.max(a, b); ap++) { // pour amount ap from Jug2 to Jug1 let c = u[0] + ap; let d = u[1] - ap; // check if this state is possible or not if (c == a || (d == 0 && d >= 0)) q.push([c,d]); // Pour amount ap from Jug 1 to Jug2 c = u[0] - ap; d = u[1] + ap; // check if this state is possible or not if ((c == 0 && c >= 0) || d == b) q.push([c,d]); } q.push([a,0]); // Empty Jug2 q.push([0,b]); // Empty Jug1 } // No, solution exists if ans=0 if (!isSolvable) document.write("No solution"); } // Driver code let Jug1 = 4, Jug2 = 3, target = 2; document.write("Path from initial state " + "to solution state ::<br>"); BFS(Jug1, Jug2, target); // This code is contributed by unknown2108</script> |
Output
Path of states of jugs followed is : 0 , 0 0 , 3 3 , 0 3 , 3 4 , 2 0 , 2
Time Complexity: O(n*m).
Space Complexity: O(n*m). Where n and m are the quantity of jug1 and jug2, respectively. This article has been improved by Sparsh Sharma.
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