Monte Carlo Tree Search Alternating - artificial-intelligence

Could anybody please clarify how (as I have not found any clear example anywhere) The MCTS algorithm iterates for the second player.
Everything I seem just seems to look like it is playing eg P1 move every time.
I understand the steps for one agent but I never find anything showing code where P2 places its counter, which surely must happen when growing the tree.
Essentially I would expect:
for each iter:
select node Player1
expand Player1
select node Player2
expand player 2
rollout
backpropogate
next iter
Is this right?? Could anybody please spell out some psuedocode showing that? Either iteratively or recursion i don't mind.
Thanks for any help.

The trick is in backpropagation part, where you update "wins" variable from the point of view of player whose move led into this position.
Code for MCTS
Notice under UCT function, specially the comments:
#Backpropagate
while node != None: # backpropagate from the expanded node and work back to the root node
node.Update(state.GetResult(node.playerJustMoved)) # state is terminal. Update node with result from POV of node.playerJustMoved
node = node.parentNode
IF you follow the function call, you would realize visit variable is always updated; wins however, is not.

Related

Does Alpha Beta / minimax require that each node be a full copy of the gameboard?

This is not a language-specific question, but for the sake of conversation, I currently work in C# 7.
Over the years I've successfully implemented the Alpha Beta pruning algorithm (even in PASCAL, 35 years ago :)
Each time, I've created semi-deep-copies (discussed below) of the game state which is recursed for each node. I've often wondered if this is necessary and if perhaps I'm not truly understanding the algorithm.
The interweb is full of requests for help for TicTacToe, which makes me think that this must be a common school assignment question - which kinda clogs searches on this fairly basic topic.
Semi-deep-copies ... it appears to me that each node should know:
the full state of the board
the player whose turn it is
the state of play - ie: { playing, Player1 win, Player2 win, draw }
My question is: does each node need its own copy of the board? ... for example Chess has 8x8 grid ... is there something more subtle to the algorithm, or do these nodes each need their own snap-shot of the board state? Is there some cool way (other than copy and apply-possible-move) that nodes can use to derive their state from their parent?
Perhaps someone can explain or point to a "read this, dummy" post, or just confirm that I need to make these instances, as I've attempted to describe, with each recursive call having its own in-memory copy of the game board.
I realize that over the last few decades, memory has become cheap... but combinatorial-explosion is still the main topic. Cheers.
I am not sure if this answers your question. But could you instead of making a copy each turn do the move, make the recursive call, and then undo the move instead? Something like:
board.make_move(move)
eval = minimax(board, ....)
board.unmake_move(move)

Checking every bit is on some cycle

I'm coding a model where :
Node are represented as bitvectors of 10 length each representing some molecules and edges can take any molecules that was present at source to to a target node.
for example
S_Node : 0b0100000011 // Molecule 0 , 1 , 8 present on node
One_Edge : 0b0000000010 // Molecule 1 is going out on edge
I have to enforce condition that each outgoing Molecule on edge is coming back to the source node on some cycle. Molecule has to come back in a cycle means that during taking path of the cycle it has to be present on evry node and evry edge it takes.
* Parallel edges are allowed.
Molecule 1 takes path S_Node -> Node_1 -> Node_2 ... -> S_Node. So Molecule 1 started from S_Node on an edge and traveled through Node_1 ... and came back to S_Node on a cycle. Hence this molecule satisfies the condition.
Similarly i have to check for each molecule on each edge.
I'm doing in trivial possible way of checking for each nodes what are possible edges going out and then for each edge what are possible bits that are present and enforcing each coming back on some cycle.
for (i = 0; i < N; i++) { // for each Node
for (j = 0; j < E; j++) { // for each Edge going out frm node i
// Lets say we have some way of finding E
if(edgeWeight & (1 << j)) { //All outgoing bits
// Enforcing that each will come back
// On some Cycle
Its easily visible that i have to iterate over all nodes and then all edges going out and then for each bits on those edges, have to write code for enforcing the same. Enforcing itself have to iterate over at least no.Of Nodes #N.
Any better way to efficiently do this ? Any other way to check for same thing in graph theory ? Thanks
You seem to have a directed graph per molecule (per bit) Simply do your trick to check for any non-cycles per molecule.
You can take btillys way of checking for cycles, another option is to look at strongly connected components. You essentially want each subgraph (for a given molecule) to be a graph where each connected component is actually strongly connected. There are some good algorithms to for strongly connected components referred to from the wikipedia article linked to earlier.
The representation of nodes is irrelevant to the problem. You have a directed graph. You wish to verify that for every node and edge, there exists a cycle containing that edge. And you want to be reasonably efficient about it (rather than doing a brute force search for all possible cycles from all edges).
Here is an observation. Suppose that you find a cycle in your graph G. Consider the graph G' which is the same as your original graph EXCEPT that the cycle has been collapsed down to a single node. The answer to your question for G is the same as the answer to your question for G' because any cycle in G leads to a cycle in G' (possibly a self-intersecting one that can be turned into 2 cycles), and any cycle in G' leads to a cycle in G (if you hit the collapsed node, then follow its cycle around until you find the exit point to continue).
So now the question goes from brute force discovery of cycles to collapsing cycles until you have a small graph where the question is easily answered. So for each node, for each edge, you start a path. Your path continues until you have discovered a cycle. Any cycle. (Not necessarily back to the original node!) Collapse that cycle, and keep traveling until you either have to backtrack (in which case your condition is not met) or you manage to loop back to your original node, collapse that cycle, and move on to another edge.
If you implement this, you'll have a polynomial algorithm, but not the best you can do. The problem is that creating new graphs with a cycle collapsed is an expensive operation. But there is a trick that helps. Instead of collapsing the graph every time you find a cycle, try to be lazy about it. Create a new "fake node" for that cycle, and mark each node in that cycle as going to that fake one. Every time you see an edge that goes to a node, do a recursive search through those mappings to the most collapsed node that you've found, and mark everything you saw in that search as directly mapping there.
If you implement the lazy bit well, your overall algorithm should wind up O(E) where E is the number of edges in your graph. You actually can't do better than that given that you have to visit every edge no matter what you do.

Cant understand the game tree for Quarto game

I am going to implement quarto game in which each opponent choose the next step for another opponent but I am having trouble to draw min max node and understand which node is max and which is min? the only thing that came to my mind is this :
But I am pretty sure that something is wrong with that?
can anyone help?
Just to clarify, I'm assuming that at the top node P1 is choosing a piece for P2 to place, at the second node P2 is placing the piece, at the third node P2 is choosing a piece for P1 to place, and then at four P1 is placing that piece and so on.
I can see why you might think that you're doing something wrong, because this is not the way that minimax is conventionally set up, but this seems like a logical way to apply it for quarto. You are correctly assigning min and max turns and such, so I don't see anything inherently wrong with this setup. Keeping track of the choosing vs. placing nodes for the purposes of the evaluation function could potentially get tricky, but I think it should be doable. Have you encountered any obstacles with doing it this way? If not, I'd say to give it a shot. It's an interesting twist on standard minimax.

Rush hour - Iterative Deepening

I have to solve the "rush hour puzzle" by iterative deepening algorithm. I have read a lot of topics here on stackoverflow and also on the internet. I think that I understand the iterative deepening algorithm. Basically you just go deeper into the tree and try to find the solution.
I figured that I need to create a graph or a tree from the puzzle, but I really don't have an idea how. Also, if I would have the tree, then how would I tell if something is a valid move or a final state?
There were answers that the nodes should be possible moves and the edges are between the nodes that can be reached in one move. I can imagine this, but somehow I'm getting trouble in see how this can be useful or better yet how can this solve the problem.
Please help me, I'm not asking for complete solution or code sample, I just need some easy explanation of the problem.
There is a reason you need to use the deepening algorithm. Imagine you name each car A, B, C, D... The root node of your tree is the initial board state. Now, move car A. You go down one node in the tree. Move car A back. You are at the initial state, but you made two moves to get here, so you are two nodes down the tree. Repeat over and over. You will never hit a final state.
The root node of your tree is the initial board state. Given that node, add a child node to it for every possible valid move. So, each child node will be what the initial tree looks like after one move. Now, for each of those child nodes, do the same thing: make a child node where each node is one move off the original child node.
Eventually, you will hit a solution to the puzzle. When that happens, you print the moves from the root node to the solution child node and quit. This algorithm ensures that you find a solution with the least number of moves.

Pacman: how do the eyes find their way back to the monster hole?

I found a lot of references to the AI of the ghosts in Pacman, but none of them mentioned how the eyes find their way back to the central ghost hole after a ghost is eaten by Pacman.
In my implementation I implemented a simple but awful solution. I just hard coded on every corner which direction should be taken.
Are there any better/or the best solution? Maybe a generic one that works with different level designs?
Actually, I'd say your approach is a pretty awesome solution, with almost zero-run time cost compared to any sort of pathfinding.
If you need it to generalise to arbitrary maps, you could use any pathfinding algorithm - breadth-first search is simple to implement, for example - and use that to calculate which directions to encode at each of the corners, before the game is run.
EDIT (11th August 2010): I was just referred to a very detailed page on the Pacman system: The Pac-Man Dossier, and since I have the accepted answer here, I felt I should update it. The article doesn't seem to cover the act of returning to the monster house explicitly but it states that the direct pathfinding in Pac-Man is a case of the following:
continue moving towards the next intersection (although this is essentially a special case of 'when given a choice, choose the direction that doesn't involve reversing your direction, as seen in the next step);
at the intersection, look at the adjacent exit squares, except the one you just came from;
picking one which is nearest the goal. If more than one is equally near the goal, pick the first valid direction in this order: up, left, down, right.
I've solved this problem for generic levels that way: Before the level starts, I do some kind of "flood fill" from the monster hole; every tile of the maze that isn't a wall gets a number that says how far it is away from the hole. So when the eyes are on a tile with a distance of 68, they look which of the neighbouring tiles has a distance of 67; that's the way to go then.
For an alternative to more traditional pathfinding algorithms, you could take a look at the (appropriately-named!) Pac-Man Scent Antiobject pattern.
You could diffuse monster-hole-scent around the maze at startup and have the eyes follow it home.
Once the smell is set up, runtime cost is very low.
Edit: sadly the wikipedia article has been deleted, so WayBack Machine to the rescue...
You should take a look a pathfindings algorithm, like Dijsktra's Algorithm or A* algorithm. This is what your problem is : a graph/path problem.
Any simple solution that works is maintainable, reliable and performs well enough is a good solution. It sounds to me like you have already found a good solution ...
An path-finding solution is likely to be more complicated than your current solution, and hence more likely to require debugging. It will probably also be slower.
IMO, if it ain't broken, don't fix it.
EDIT
IMO, if the maze is fixed then your current solution is good / elegant code. Don't make the mistake of equating "good" or "elegant" with "clever". Simple code can also be "good" and "elegant".
If you have configurable maze levels, then maybe you should just do the pathfinding when you initially configure the mazes. Simplest would be to get the maze designer to do it by hand. I'd only bother automating this if you have a bazillion mazes ... or users can design them.
(Aside: if the routes are configured by hand, the maze designer could make a level more interesting by using suboptimal routes ... )
In the original Pacman the Ghost found the yellow pill eater by his "smell" he would leave a trace on the map, the ghost would wander around randomly until they found the smell, then they would simply follow the smell path which lead them directly to the player. Each time Pacman moved, the "smell values" would get decreased by 1.
Now, a simple way to reverse the whole process would be to have a "pyramid of ghost smell", which has its highest point at the center of the map, then the ghost just move in the direction of this smell.
Assuming you already have the logic required for chasing pacman why not reuse that? Just change the target. Seems like it would be a lot less work than trying to create a whole new routine using the exact same logic.
It's a pathfinding problem. For a popular algorithm, see http://wiki.gamedev.net/index.php/A*.
How about each square having a value of distance to the center? This way for each given square you can get values of immediate neighbor squares in all possible directions. You pick the square with the lowest value and move to that square.
Values would be pre-calculated using any available algorithm.
This was the best source that I could find on how it actually worked.
http://gameai.com/wiki/index.php?title=Pac-Man#Respawn
When the ghosts are killed, their disembodied eyes return to their starting location. This is simply accomplished by setting the ghost's target tile to that location. The navigation uses the same rules.
It actually makes sense. Maybe not the most efficient in the world but a pretty nice way to not have to worry about another state or anything along those lines you are just changing the target.
Side note: I did not realize how awesome those pac-man programmers were they basically made an entire message system in a very small space with very limited memory ... that is amazing.
I think your solution is right for the problem, simpler than that, is to make a new version more "realistic" where ghost eyes can go through walls =)
Here's an analog and pseudocode to ammoQ's flood fill idea.
queue q
enqueue q, ghost_origin
set visited
while q has squares
p <= dequeue q
for each square s adjacent to p
if ( s not in visited ) then
add s to visited
s.returndirection <= direction from s to p
enqueue q, s
end if
next
next
The idea is that it's a breadth-first search, so each time you encounter a new adjacent square s, the best path is through p. It's O(N) I do believe.
I don't know much on how you implemented your game but, you could do the following:
Determine the eyes location relative position to the gate. i.e. Is it left above? Right below?
Then move the eyes opposite one of the two directions (such as make it move left if it is right of the gate, and below the gate) and check if there are and walls preventing you from doing so.
If there are walls preventing you from doing so then make it move opposite the other direction (for example, if the coordinates of the eyes relative to the pin is right north and it was currently moving left but there is a wall in the way make it move south.
Remember to keep checking each time to move to keep checking where the eyes are in relative to the gate and check to see when there is no latitudinal coordinate. i.e. it is only above the gate.
In the case it is only above the gate move down if there is a wall, move either left or right and keep doing this number 1 - 4 until the eyes are in the den.
I've never seen a dead end in Pacman this code will not account for dead ends.
Also, I have included a solution to when the eyes would "wobble" between a wall that spans across the origin in my pseudocode.
Some pseudocode:
x = getRelativeOppositeLatitudinalCoord()
y
origX = x
while(eyesNotInPen())
x = getRelativeOppositeLatitudinalCoordofGate()
y = getRelativeOppositeLongitudinalCoordofGate()
if (getRelativeOppositeLatitudinalCoordofGate() == 0 && move(y) == false/*assume zero is neither left or right of the the gate and false means wall is in the way */)
while (move(y) == false)
move(origX)
x = getRelativeOppositeLatitudinalCoordofGate()
else if (move(x) == false) {
move(y)
endWhile
dtb23's suggestion of just picking a random direction at each corner, and eventually you'll find the monster-hole sounds horribly ineficient.
However you could make use of its inefficient return-to-home algorithm to make the game more fun by introducing more variation in the game difficulty. You'd do this by applying one of the above approaches such as your waypoints or the flood fill, but doing so non-deterministically. So at every corner, you could generate a random number to decide whether to take the optimal way, or a random direction.
As the player progresses levels, you reduce the likelihood that a random direction is taken. This would add another lever on the overall difficulty level in addition to the level speed, ghost speed, pill-eating pause (etc). You've got more time to relax while the ghosts are just harmless eyes, but that time becomes shorter and shorter as you progress.
Short answer, not very well. :) If you alter the Pac-man maze the eyes won't necessarily come back. Some of the hacks floating around have that problem. So it's dependent on having a cooperative maze.
I would propose that the ghost stores the path he has taken from the hole to the Pacman. So as soon as the ghost dies, he can follow this stored path in the reverse direction.
Knowing that pacman paths are non-random (ie, each specific level 0-255, inky, blinky, pinky, and clyde will work the exact same path for that level).
I would take this and then guess there are a few master paths that wraps around the entire
maze as a "return path" that an eyeball object takes pending where it is when pac man ate the ghost.
The ghosts in pacman follow more or less predictable patterns in terms of trying to match on X or Y first until the goal was met. I always assumed that this was exactly the same for eyes finding their way back.
Before the game begins save the nodes (intersections) in the map
When the monster dies take the point (coordinates) and find the
nearest node in your node list
Calculate all the paths beginning from that node to the hole
Take the shortest path by length
Add the length of the space between the point and the nearest node
Draw and move on the path
Enjoy!
My approach is a little memory intensive (from the perspective of Pacman era), but you only need to compute once and it works for any level design (including jumps).
Label Nodes Once
When you first load a level, label all the monster lair nodes 0 (representing the distance from the lair). Proceed outward labelling connected nodes 1, nodes connected to them 2, and so on, until all nodes are labelled. (note: this even works if the lair has multiple entrances)
I'm assuming you already have objects representing each node and connections to their neighbours. Pseudo code might look something like this:
public void fillMap(List<Node> nodes) { // call passing lairNodes
int i = 0;
while(nodes.count > 0) {
// Label with distance from lair
nodes.labelAll(i++);
// Find connected unlabelled nodes
nodes = nodes
.flatMap(n -> n.neighbours)
.filter(!n.isDistanceAssigned());
}
}
Eyes Move to Neighbour with Lowest Distance Label
Once all the nodes are labelled, routing the eyes is trivial... just pick the neighbouring node with the lowest distance label (note: if multiple nodes have equal distance, it doesn't matter which is picked). Pseudo code:
public Node moveEyes(final Node current) {
return current.neighbours.min((n1, n2) -> n1.distance - n2.distance);
}
Fully Labelled Example
For my PacMan game I made a somewhat "shortest multiple path home" algorithm which works for what ever labyrinth I provide it with (within my set of rules). It also works across them tunnels.
When the level is loaded, all the path home data in every crossroad is empty (default) and once the ghosts start to explore the labyrinth, them crossroad path home information keeps getting updated every time they run into a "new" crossroad or from a different path stumble again upon their known crossroad.
The original pac-man didn't use path-finding or fancy AI. It just made gamers believe there is more depth to it than it actually was, but in fact it was random. As stated in Artificial Intelligence for Games/Ian Millington, John Funge.
Not sure if it's true or not, but it makes a lot of sense to me. Honestly, I don't see these behaviors that people are talking about. Red/Blinky for ex is not following the player at all times, as they say. Nobody seems to be consistently following the player, on purpose. The chance that they will follow you looks random to me. And it's just very tempting to see behavior in randomness, especially when the chances of getting chased are very high, with 4 enemies and very limited turning options, in a small space. At least in its initial implementation, the game was extremely simple. Check out the book, it's in one of the first chapters.

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