Standard pathfinding is Good Enough -- your states are your current location + your current inventory. "moving" is either changing rooms or changing inventory. Not covered in this answer, but not too much additional effort, is writing a good heuristic for A* -- it can really speed up the search by preferring to pick up things over moving away from it, preferring to unlock a door near the target over searching for a long way around, etc.
This answer has gotten a lot of upvotes since it came first and has a demo, but for a much more optimized and specialized solution, you should also read the "Doing it backwards is much faster" answer https://gamedev.stackexchange.com/a/150155/2624
Fully operational Javascript proof of concept below. Sorry for the answer as a code dump -- I had actually implement this before I was convinced it was a good answer, but it seems pretty flexible to me.
To start off when thinking about pathfinding, remember that the heirarchy of simple pathfinding algorithms is :
- Breadth First Search is about as simple as you can get.
- Djikstra's Algorithm is like Breadth First Search but with varying "distances" between states
- A* is Djikstras where you have a 'general sense of the right direction' available as a heuristic.
In our case, just encoding a "state" as a "location + inventory" and "distances" as a "movement or item usage" allows us to use Djikstra or A* to solve our problem.
Here is some actual code demonstrating your example level. The first snippet is just for comparison -- jump to the second part if you want to see the final solution. We start off with a Djikstra's implementation that finds the correct path, but we've ignored all the obstacles and keys. (Try it out, You can see it just beelines for the finish, from room 0 -> 2 -> 3->4->6->5)
function Transition(cost, state) { this.cost = cost, this.state = state; }
// given a current room, return a room of next rooms we can go to. it costs
// 1 action to move to another room.
function next(n) {
var moves = []
// simulate moving to a room
var move = room => new Transition(1, room)
if (n == 0) moves.push(move(2))
else if ( n == 1) moves.push(move(2))
else if ( n == 2) moves.push(move(0), move(1), move(3))
else if ( n == 3) moves.push(move(2), move(4), move(6))
else if ( n == 4) moves.push(move(3))
else if ( n == 5) moves.push(move(6))
else if ( n == 6) moves.push(move(5), move(3))
return moves
}
// Standard Djikstra's algorithm. keep a list of visited and unvisited nodes
// and iteratively find the "cheapest" next node to visit.
function calc_Djikstra(cost, goal, history, nextStates, visited) {
if (!nextStates.length) return ['did not find goal', history]
var action = nextStates.pop()
cost += action.cost
var cur = action.state
if (cur == goal) return ['found!', history.concat([cur])]
if (history.length > 15) return ['we got lost', history]
var notVisited = (visit) => {
return visited.filter(v => JSON.stringify(v) == JSON.stringify(visit.state)).length === 0;
};
nextStates = nextStates.concat(next(cur).filter(notVisited))
nextStates.sort()
visited.push(cur)
return calc_Djikstra(cost, goal, history.concat([cur]), nextStates, visited)
}
console.log(calc_Djikstra(0, 5, [], [new Transition(0, 0)], []))
So, how do we add items and keys to this code? Simple! instead of every "state" begin just the room number, it's now a tuple of the room and our inventory state:
// Now, each state is a [room, haskey, hasfeather, killedboss] tuple
function State(room, k, f, b) { this.room = room; this.k = k; this.f = f; this.b = b }
Transitions now change from being a (cost, room) tuple to a (cost, state) tuple, so then can encode both "moving to another room" and "picking up an item"
// move(3) keeps inventory but sets the room to 3
var move = room => new Transition(1, new State(room, cur.k, cur.f, cur.b))
// pickup("k") keeps room number but increments the key count
var pickup = (cost, item) => {
var n = Object.assign({}, cur)
n[item]++;
return new Transition(cost, new State(cur.room, n.k, n.f, n.b));
};
finally, we make some minor type-related changes to the Djikstra function (for example, it still is just matching on a goal room number instead of a full state), and we get our full answer! Note the printed result first goes to room 4 to pick up the key, then goes to room 1 to pick up the feather, then goes to room 6, kills the boss, then goes to room 5)
// Now, each state is a [room, haskey, hasfeather, killedboss] tuple
function State(room, k, f, b) { this.room = room; this.k = k; this.f = f; this.b = b }
function Transition(cost, state, msg) { this.cost = cost, this.state = state; this.msg = msg; }
function next(cur) {
var moves = []
// simulate moving to a room
var n = cur.room
var move = room => new Transition(1, new State(room, cur.k, cur.f, cur.b), "move to " + room)
var pickup = (cost, item) => {
var n = Object.assign({}, cur)
n[item]++;
return new Transition(cost, new State(cur.room, n.k, n.f, n.b), {
"k": "pick up key",
"f": "pick up feather",
"b": "SLAY BOSS!!!!"}[item]);
};
if (n == 0) moves.push(move(2))
else if ( n == 1) { }
else if ( n == 2) moves.push(move(0), move(3))
else if ( n == 3) moves.push(move(2), move(4))
else if ( n == 4) moves.push(move(3))
else if ( n == 5) { }
else if ( n == 6) { }
// if we have a key, then we can move between rooms 1 and 2
if (cur.k && n == 1) moves.push(move(2));
if (cur.k && n == 2) moves.push(move(1));
// if we have a feather, then we can move between rooms 3 and 6
if (cur.f && n == 3) moves.push(move(6));
if (cur.f && n == 6) moves.push(move(3));
// if killed the boss, then we can move between rooms 5 and 6
if (cur.b && n == 5) moves.push(move(6));
if (cur.b && n == 6) moves.push(move(5));
if (n == 4 && !cur.k) moves.push(pickup(0, 'k'))
if (n == 1 && !cur.f) moves.push(pickup(0, 'f'))
if (n == 6 && !cur.b) moves.push(pickup(100, 'b'))
return moves
}
var notVisited = (visitedList) => (visit) => {
return visitedList.filter(v => JSON.stringify(v) == JSON.stringify(visit.state)).length === 0;
};
// Standard Djikstra's algorithm. keep a list of visited and unvisited nodes
// and iteratively find the "cheapest" next node to visit.
function calc_Djikstra(cost, goal, history, nextStates, visited) {
if (!nextStates.length) return ['No path exists', history]
var action = nextStates.pop()
cost += action.cost
var cur = action.state
if (cur.room == goal) return history.concat([action.msg])
if (history.length > 15) return ['we got lost', history]
nextStates = nextStates.concat(next(cur).filter(notVisited(visited)))
nextStates.sort()
visited.push(cur)
return calc_Djikstra(cost, goal, history.concat([action.msg]), nextStates, visited)
o}
console.log(calc_Djikstra(0, 5, [], [new Transition(0, new State(0, 0, 0, 0), 'start')], []))
In theory, this works even with BFS and we didn't need the cost function for Djikstra's, but having the cost allows us to say "picking up a key is effortless, but fighting a boss is really hard, and we'd rather backtrack 100 steps rather than fight the boss, if we had the choice":
if (n == 4 && !cur.k) moves.push(pickup(0, 'k'))
if (n == 1 && !cur.f) moves.push(pickup(0, 'f'))
if (n == 6 && !cur.b) moves.push(pickup(100, 'b'))
