You are given a m litre jug and a n litre 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 litres 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:

- Empty a Jug, (X, Y)->(0, Y) 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), (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.

#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; }

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)

This article is contributed by **Abhishek rajput**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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