Problem:
Let’s say that we have 3 jugs, namely A, B, and C. Jug C will always start out completely full. Furthermore, we will define a goal state where jug A will contain a gallons, B will contain b gallons, and C will contain c gallons upon completion of the search. You need to use breadth-first search to find the minimum number of moves to go from the initial state to the goal state.
My pseudo code:
bool BFS(int a, int b, int c)
add new State(a, b, c) to Queue
pop from the Queue
while (queue is not empty)
if current_state equals goal_state
print the backtracked solution
return true
if current_state has already been seen
continue
mark current as having been visited
try 6 ways to pour water, pushing new states to queue
return false
My solution in C++:
#include <iostream>
#include <sstream>
#include <string>
#include <regex>
#include <queue>
#include <map>
#include <new>
using namespace std;
// Struct to represent state of water in the jugs.
struct State
{
int a, b, c;
int toPour;
char from;
char to;
State *parent;
State(int _a, int _b, int _c, State *_parent) : a(_a), b(_b), c(_c), toPour(0), parent(_parent) {}
State(int _a, int _b, int _c, int _toPour, char _from, char _to, State *_parent) : a(_a), b(_b), c(_c), toPour(_toPour), from(_from), to(_to), parent(_parent) {}
// String representation of state in tuple form.
void to_string()
{
if (toPour > 0)
{
cout << "Pour " << toPour;
if (toPour == 1)
{
cout << " gallon";
} else
{
cout << " gallons";
}
cout << " from " << from << " to " << to;
}
else
{
cout << "Initial state";
}
cout << ". (" << a << ", " << b << ", " << c << ")" << endl;
}
};
void demonstrate(State state) {
if (state.parent != nullptr)
{
demonstrate(*(state.parent));
}
state.to_string();
return;
}
bool BFS(int a, int b, int c, int capacity[], int goal[])
{
map<pair<pair<int, int>, int>, int> m;
queue<State> q;
q.push(*(new State(a, b, c, nullptr)));
int toPour;
while (!q.empty())
{
if (q.front().a == goal[0] && q.front().b == goal[1] && q.front().c == goal[2])
{
demonstrate(q.front());
return true;
}
if (m[make_pair(make_pair(q.front().a, q.front().b), q.front().c)] == 1)
{
q.pop();
continue;
}
m[make_pair(make_pair(q.front().a, q.front().b), q.front().c)] = 1;
// a -> b
if (q.front().a > 0 && q.front().b < capacity[1])
{
toPour = q.front().a > (capacity[1] - q.front().b) ? (capacity[1] - q.front().b) : q.front().a;
q.push(*(new State(q.front().a - toPour, q.front().b + toPour, q.front().c, toPour, 'A', 'B', &q.front())));
}
// a -> c
if (q.front().a > 0 && q.front().c < capacity[2])
{
toPour = q.front().a > (capacity[2] - q.front().c) ? (capacity[2] - q.front().c) : q.front().a;
q.push(*(new State(q.front().a - toPour, q.front().b, q.front().c + toPour, toPour, 'A', 'C', &q.front())));
}
// b -> c
if (q.front().b > 0 && q.front().c < capacity[2])
{
toPour = q.front().b > (capacity[2] - q.front().c) ? (capacity[2] - q.front().c) : q.front().b;
q.push(*(new State(q.front().a, q.front().b - toPour, q.front().c + toPour, toPour, 'B', 'C', &q.front())));
}
// b -> a
if (q.front().b > 0 && q.front().a < capacity[0])
{
toPour = q.front().b > (capacity[0] - q.front().a) ? (capacity[0] - q.front().a) : q.front().b;
q.push(*(new State(q.front().a + toPour, q.front().b - toPour, q.front().c, toPour, 'B', 'A', &q.front())));
}
// c -> a
if (q.front().c > 0 && q.front().a < capacity[0])
{
toPour = q.front().c > (capacity[0] - q.front().a) ? (capacity[0] - q.front().a) : q.front().c;
q.push(*(new State(q.front().a + toPour, q.front().b, q.front().c - toPour, toPour, 'C', 'A', &q.front())));
}
// c -> b
if (q.front().c > 0 && q.front().b < capacity[1])
{
toPour = q.front().c > (capacity[1] - q.front().b) ? (capacity[1] - q.front().b) : q.front().c;
q.push(*(new State(q.front().a, q.front().b + toPour, q.front().c - toPour, toPour, 'C', 'B', &q.front())));
}
q.pop();
}
return false;
}
int main(int argc, char **argv)
{
if (argc != 7)
{
cout << "Usage: ./waterjugpuzzle <cap A> <cap B> <cap C> <goal A> <goal B> <goal C>" << endl;
exit(0);
}
regex digits("\\d+");
int *capacity = new int[3];
for (int i = 1; i < 4; i++)
{
if (!regex_match(argv[i], digits) || stoi(argv[i]) == 0)
{
cout << "Error: Invalid capacity '" << argv[i] << "' for jug " << (char)('A' + i - 1) << "." << endl;
exit(1);
}
capacity[i - 1] = stoi(argv[i]);
}
int *goal = new int[3];
int totalGoal = 0;
for (int i = 4; i < 7; i++)
{
if (!regex_match(argv[i], digits))
{
cout << "Error: Invalid goal '" << argv[i] << "' for jug " << (char)('A' + i - 4) << "." << endl;
exit(1);
}
goal[i - 4] = stoi(argv[i]);
totalGoal += stoi(argv[i]);
}
for (int i = 0; i < 3; i++)
{
if (goal[i] > capacity[i])
{
cout << "Error: Goal cannot exceed capacity of jug " << (char)('A' + i) << "." << endl;
exit(2);
}
}
if (totalGoal != capacity[2])
{
cout << "Error: Total gallons in goal state must be equal to the capacity of jug C." << endl;
exit(3);
}
if (!BFS(0, 0, capacity[2], capacity, goal))
{
cout << "No solution." << endl;
}
return 0;
}
Usage: ./waterjugpuzzle <capacity A> <capacity B> <capacity C> <goal A> <goal B> <goal C>
Example:
./waterjugpuzzle 1 3 4 0 2 2
Initial state. (0, 0, 4)
Pour 3 gallons from C to B. (0, 3, 1)
Pour 1 gallon from B to A. (1, 2, 1)
Pour 1 gallon from A to C. (0, 2, 2)
Error: This solution work perfectly on MacOS, but on Ubuntu, when I try some inputs like 8 17 20 0 10 10 or 4 7 10 0 5 5, the program return Segmentation fault (core dumped) after it finds the solution and starts printing.
