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Disclaimer: this problem came from my past AI final exam. And I found it very interesting but I couldn't figure it out.
There is the description:
Given a maze you are free to move between adjacent white cells, but black cells are blocked. You can try to move UP, DOWN, LEFT, RIGHT. If you are not blocked in that direction, you successfully move. If you are blocked, you stay in the same place.
a) Find the shortest possible sequence of moves that guarantee you will end up at G regardless of where you started.
Before making any moves, you may be in any position. After each move, your set of possible positions will shift, and may shrink when moving from some possible positions would hit a wall. The set of possible positions will never increase in size, though.
You may assume a directed graph, with a vertex for each set of positions that may arise, and an edge connecting each set to the 4 sets that would follow from moving left, right, up, or down.
Your task, then, is to find the shortest path from the vertex for the set of all positions to the vertex for the singleton set containing only the target position.
My first attempt would by to run A* on this graph using an appropriate metric. The number of possible sets is extremely large, though, so this is not guaranteed to be feasible. I would use my human intelligence to try to pick a metric that works quickly.
If you can choose a metric that gets A* to completion in a reasonable amount of time, and conforms to the rules that A* puts on metrics, then that will prove that the path you found is the shortest one.
Off the top of my head, I would first try the length of the shortest path to the target from the furthest position in the set. Maybe in combination with the total number of non-empty rows and columns.
So this is what I would do, but I am not an AI expert. I am guessing that there is probably something you learned in class that can improve upon this procedure.
I tried to apply IDDFS on this graph by first making it in tree form and the result was this :
At level 1: d,e,p
At level 2: d,b,e,c,e,h,r,p,q
At level 3: d,b,a,e,h,c,a,e,h,q,p,r,f,p,q
At level 4: d,b,a,e,h,p,q,c,a,e,h,q,p,q,r,f,c,GOAL
I am confused about those repeated nodes in the path, can we eliminate them or they will appear in the final path ?
Is this the correct approach of traversing the graph to reach the GOAL ? And how we come to know which node to visit next in graph(e.g as in tree we start from left to right).
And what would be the path if we apply DFS and BFS on same graph ?
Will there be any difference in DFS result and IDDFS ? It seems to be similar
Yes, you can and SHOULD get rid of repeated nodes when you implement DFS, by keeping track of which nodes you've already visited. If you don't, your code won't terminate when it finds a cycle. Don't forget to clear the set of visited nodes with each new level. So leave out the visited nodes from your listing, unless it's important to include nodes that are being considered but not re-visited.
If you write out the expansion for BFS and DFS, you'll see that IDDFS starts out looking like BFS and ends up looking more and more like DFS the more you crank up the level. When level = length-of-longest-path, voila, you get DFS, which is not surprising, since IDDFS is DFS, only with paths cut off at a given number; in that particular case, the number has no effect, because there aren't any paths long enough to be cut off.
The order in a graph is not well defined. You choose one order or another yourself. If you choose a next node at random, you get non-deterministic algorithms. If you choose them, dunno, alphabetically, then you get some determinism. Sometimes the distinction doesn't matter, but determinism is good for debugging your code, etc. Now, when you do this exercise, you do it to see patterns, so it's best to leave out the randomness.
Your question really does look like homework. ;)
So I am implementing A* algorithm in C. Here's the procedure.
I am using Priority Queue [using array] for all the open nodes. Since I'll have duplicate distances, that is more than one node with same distance/Priority, hence while inserting a node in PQ, if the parent of the inserted node has the same priority, I still swap them both, so that my newest entered member remains on the top( or as high as possible),so that I keep following a particular direction. Also, on removing, when I swap the topmost element with the last one, then again, if the swapped last element has the same as one of its children, then it gets swapped to the bottom.(I am not sure if this will affect in any way).
Now the problem is say I have a 100*100 matrix, and I have obstacles from (0,20) to (15,20) of the 2D array , in which I am moving. Now for a starting position (2,2) and ending position (16,20) I get a straight path, i.e. firstly go all the way to right, then go down till 15 then move one right and I am done.
But, if I have starting as (2,2) and last as (12,78) i.e. the points are separated by the obstacles and the path has to go around it, I still go via (16,20) and my path after (16,20) is still straight, but my path upto (16,20) is zig zag, i.e. I go some distance straight down, then some right, then down then right and so on, ultimately reaching (16,20) and going straight after that.
Why this zig zag path for the first half of the distance, what can I do to make sure that my path is straight, as it is, when my destination is (16,20) and not (12,78).
Thanks.
void findPath(array[ROW][COLUMN],sourceX,sourceY,destX,destY) {
PQ pq[SIZE];
int x,y;
insert(pq,sourceX,sourceY);
while(!empty(pq)) {
remove(pq);
if(removedIsDestination)
break; //Path Found
insertAdjacent(pq,x,y,destX,destY);
}
}
void insert(PQ pq[SIZE],element){
++sizeOfPQ;
PQ[sizeOfPQ]==element
int i=sizeOfPQ;
while(i>0){
if(pq[i].priority <= pq[(i-1)/2].priority){
swapWithParent
i=(i-1)/2;
}
else
break;
}
}
You should change your scoring part. Right now you calculate absolute distance. Instead calculate min move distance. If you count each move as one then if you were at (x,y) and going to (dX,dY) that would be
distance moved + (max(x,dX) - min(x,dx) + max(y,dY) - min(y,dY))
A lower value is considered a higher score.
This heuristic is a guess at how many moves it would take if there was nothing in the way.
The nice thing about the heuristic is you can change it to get the results you want, for example if you prefer to move in a straight line as you suggest, then you can make this change:
= distance moved + (max(x,dX) - min(x,dx) + max(y,dY) - min(y,dY))
+ (1 if this is a turn from the last move)
This will cause you to "find" solutions which tend to go in the same direction.
If you want to FORCE as few turns as possible:
= distance moved + (max(x,dX) - min(x,dx) + max(y,dY) - min(y,dY))
+ (1 times the number of turns made)
This is what is nice about A* -- the heuristic will inform the search -- you will still always find a solution, but if there is more than one you can influence where you look first -- this makes it good for simulating AI behavior.
Doubt : How is the first one and second calculating way different from
each other?
The first one puts a lower priority on a move that is a turn. The second one puts a lower priority on a path with more turns. In some cases (eg, the first turn) the value will be the same, but over all the 2nd one will pick paths that have as few turns as possible, where the first one might not.
Also, 1 if this is a turn from the last move , for this,
say i have source at top left and destination at bottom right, now my
path normally would be, left,left,left...down,down,down.... Now, 1 if
this is a turn from the last move, according to this, when I change
from left to down, will I add 1?
Yes
Wont it make the total value more and the priority for down will decrease.
Yes, exactly. You want to not look at choices that have a turn in them first. This will make them lower priority and your algorithm will investigate other options with a higher priority -- exactly what you want.
Or 1 if this is a turn from the last move is when I move to a cell, that is not abutting the cell previously worked upon? Thnks –
No, I don't understand this question -- I don't think it makes sense in this context -- all moves have to abut the previous cell, diagonal moves are not allowed.
Though, I'd really appreciate if you could tell me one instance where the first and second methods will give different answers. If you could. Thanks alot. :)
Not so easy without seeing the details of your algorithm but the following might work:
The red are blocks. The green is what I would expect the first one to do, it locally tries to find the least turn. The blue is the least turn solution. Note, how far the red areas are from each other and the details of how your algorithm influence if this will work. As I have it above -- having an extra turn only costs 1 in the heuristic. SO, if you want to be sure this will work change the heuristic like this:
= distance moved + (max(x,dX) - min(x,dx) + max(y,dY) - min(y,dY))
+ (25 times the number of turns made)
Where 25 is bigger than the distance to get past the 2nd turn in the green path. (Thus after the 2nd turn the blue path will be searched.)
The idea is to move all of the right elements into the left and the left into the right with an empty space in the middle. The elements can either jump over one or two pieces into an empty space.
LLL[ ]RRR
I'm trying to think of a heuristic for this task. Is the heuristic meant to aid in finding a possible solution, or actually return a number of moves as the solution? How would I express such a heuristic?
Sounds like you are a bit confused about what a heuristic is.
A rough definition is "a simplifying assumption" or "a decent guess"
For example, let's say you have to put together a basketball team, and you have fact sheets on people who want to play that list their contact info, birth date, and height. You could hold tryouts where you test each candidate's specific skills; that would require bringing in all the candidates, though, and that could take a long time. You use a heuristic to narrow the search -- only call people who are at least 6'2" tall. This might ignore some great basketball players, but it's a pretty decent guess.
Another example of a heuristic: you are trying to use the smallest number of coins to pay a bill. The heuristic (a simplifying approach) is to pick the coin with the biggest value (which is less than the remaining bill) first, subtract the value from the bill, and repeat. This is not guaranteed to work every time, but it'll get you to the right neighborhood most of the time.
A heuristic for your problem might be "never move Ls to the right, and never move Rs to the left" -- it narrows the "search space" of all possible moves by eliminating some of the possibilities from the outset.
Are you looking for a heuristic or an algorithm? A heuristic may or may not solve a given problem. It is really just intended to point you in the direction that the solution probably lies in. An algorithm really should solve a given problem.
A heuristic is generally a "hint" which usually (but not always) will guide your procedure to the correct direction. Using heuristics speeds up your procedures (your algorithms), again, usually, but not always. It's like an "advice" to the algorithm which is correct more often than not.
I'm not sure what you are looking for, as the description is a little vague. If you want the algorithm, you will need to study what effect a particular move will have to the current situation and a way to step forward for all possible moves each time, in effect traversing a tree of states (ie. states that will evolve if you make a particular sequence of moves).
You can also see that it possibly matters how close the current position is to what you want to achieve (your desired final position).So instead of calculating all the possible paths from your initial state until you find the final state, you can guide your algorithm based on the heuristic "how close is the current state to the desired one" and only traverse a part of the tree.
I've read in one of my AI books that popular algorithms (A-Star, Dijkstra) for path-finding in simulation or games is also used to solve the well-known "15-puzzle".
Can anyone give me some pointers on how I would reduce the 15-puzzle to a graph of nodes and edges so that I could apply one of these algorithms?
If I were to treat each node in the graph as a game state then wouldn't that tree become quite large? Or is that just the way to do it?
A good heuristic for A-Star with the 15 puzzle is the number of squares that are in the wrong location. Because you need at least 1 move per square that is out of place, the number of squares out of place is guaranteed to be less than or equal to the number of moves required to solve the puzzle, making it an appropriate heuristic for A-Star.
A quick Google search turns up a couple papers that cover this in some detail: one on Parallel Combinatorial Search, and one on External-Memory Graph Search
General rule of thumb when it comes to algorithmic problems: someone has likely done it before you, and published their findings.
This is an assignment for the 8-puzzle problem talked about using the A* algorithm in some detail, but also fairly straightforward:
http://www.cs.princeton.edu/courses/archive/spring09/cos226/assignments/8puzzle.html
The graph theoretic way to solve the problem is to imagine every configuration of the board as a vertex of the graph and then use a breath-first search with pruning based on something like the Manhatten Distance of the board to derive a shortest path from the starting configuration to the solution.
One problem with this approach is that for any n x n board where n > 3 the game space becomes so large that it is not clear how you can efficiently mark the visited vertices. In other words there is no obvious way to assess if the current configuration of the board is identical to one that has previously been discovered through traversing some other path. Another problem is that the graph size grows so quickly with n (it's approximately (n^2)!) that it is just not suitable for a brue-force attack as the number of paths becomes computationally infeasible to traverse.
This paper by Ian Parberry A Real-Time Algorithm for the (n^2 − 1) - Puzzle describes a simple greedy algorithm that iteritively arrives at a solution by completing the first row, then the first column, then the second row... It arrives at a solution almost immediately, however the solution is far from optimal; essentially it solves the problem the way a human would without leveraging any computational muscle.
This problem is closely related to that of solving the Rubik's cube. The graph of all game states it too large to solve by brue force, but there is a fairly simple 7 step method that can be used to solve any cube in about 1 ~ 2 minutes by a dextrous human. This path is of course non-optimal. By learning to recognise patterns that define sequences of moves the speed can be brought down to 17 seconds. However, this feat by Jiri is somewhat superhuman!
The method Parberry describes moves only one tile at a time; one imagines that the algorithm could be made better up by employing Jiri's dexterity and moving multiple tiles at one time. This would not, as Parberry proves, reduce the path length from n^3, but it would reduce the coefficient of the leading term.
Remember that A* will search through the problem space proceeding down the most likely path to goal as defined by your heurestic.
Only in the worst case will it end up having to flood fill the entire problem space, this tends to happen when there is no actual solution to your problem.
Just use the game tree. Remember that a tree is a special form of graph.
In your case the leaves of each node will be the game position after you make one of the moves that is available at the current node.
Here you go http://www.heyes-jones.com/astar.html
Also. be mindful that with the A-Star algorithm, at least, you will need to figure out a admissible heuristic to determine whether a possible next step is closer to the finished route than another step.
For my current experience, on how to solve an 8 puzzle.
it is required to create nodes. keep track of each step taken
and get the manhattan distance from each following steps, taking/going to the one with the shortest distance.
update the nodes, and continue until reaches the goal