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Water Jug problem using BFS
  • Difficulty Level : Hard
  • Last Updated : 05 Feb, 2021
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You are given a 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 amount of water in Jug1 and Y refers to amount of water in Jug2 
Determine the path from initial state (xi, yi) to 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: 

  1. Empty a Jug, (X, Y)->(0, Y) Empty Jug 1
  2. Fill a Jug, (0, 0)->(X, 0) Fill Jug 1
  3. 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), (4, 0), (4, 3), 
          (3, 0), (1, 3), (3, 3), (4, 2),
          (0, 2)}

We have discussed one solution in The Two Water Jug Puzzle
In this post a BFS based solution is discussed.



We run breadth first search on the states and these states will be created after applying allowed operations and we also use visited map of pair to keep track of states that should be visited only once in the search. This solution can also be achieved using depth first search.

C++




#include <bits/stdc++.h>
#define pii pair<int, int>
#define mp make_pair
using namespace std;
 
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
    map<pii, int> m;
    bool isSolvable = false;
    vector<pii> path;
 
    queue<pii> q; // queue to maintain states
    q.push({ 0, 0 }); // Initialing with initial state
 
    while (!q.empty()) {
 
        pii u = q.front(); // current state
 
        q.pop(); // pop off used state
 
        // if this state is already visited
        if (m[{ u.first, u.second }] == 1)
            continue;
 
        // doesn't met jug constraints
        if ((u.first > a || u.second > b ||
            u.first < 0 || u.second < 0))
            continue;
 
        // filling the vector for constructing
        // the solution path
        path.push_back({ u.first, u.second });
 
        // marking current state as visited
        m[{ u.first, u.second }] = 1;
 
        // if we reach solution state, put ans=1
        if (u.first == target || u.second == target) {
            isSolvable = true;
            if (u.first == target) {
                if (u.second != 0)
 
                    // fill final state
                    path.push_back({ u.first, 0 });
            }
            else {
                if (u.first != 0)
 
                    // fill final state
                    path.push_back({ 0, u.second });
            }
 
            // print the solution path
            int sz = path.size();
            for (int i = 0; i < sz; i++)
                cout << "(" << path[i].first
                    << ", " << path[i].second << ")\n";
            break;
        }
 
        // if we have not reached final state
        // then, start developing intermediate
        // states to reach solution state
        q.push({ u.first, b }); // fill Jug2
        q.push({ a, u.second }); // fill Jug1
 
        for (int ap = 0; ap <= max(a, b); ap++) {
 
            // pour amount ap from Jug2 to Jug1
            int c = u.first + ap;
            int d = u.second - 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.first - ap;
            d = u.second + 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)
        cout << "No solution";
}
 
// Driver code
int main()
{
    int Jug1 = 4, Jug2 = 3, target = 2;
    cout << "Path from initial state "
            "to solution state ::\n";
    BFS(Jug1, Jug2, target);
    return 0;
}

Python3




from collections import deque
 
def BFS(a, b, 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
    m = {}
    isSolvable = False
    path = []
     
    # Queue to maintain states
    q = deque()
     
    # Initialing with initial state
    q.append((0, 0))
 
    while (len(q) > 0):
         
        # Current state
        u = q.popleft()
 
        #q.pop() #pop off used state
 
        # If this state is already visited
        if ((u[0], u[1]) in m):
            continue
 
        # Doesn't met jug constraints
        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 code
if __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

Output: 

Path from initial state to solution state ::
(0, 0)
(0, 3)
(4, 0)
(4, 3)
(3, 0)
(1, 3)
(3, 3)
(4, 2)
(0, 2) 

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