I have to determine a way for a robot to get out of a maze. The thing is that the layout of the maze is unknown, and the position of the exit is unknown too. The robot also start at an unknown position in the maze.
I found 3 solutions but I have a hard time knowing which one should I use, because in the end it seems that the solutions will purely be random anyway.
I have those 3 solutions :
1) The basic "human" strategy(?), where you put your hand on a wall and go through all the maze if necessary. I also keep a variable "turn counter" to avoid situation where the robot loop.
2) Depth first search
3) Making the robot choose direction randomly
The random one seems the worse, because he could take forever to find the exit (but on the other hand he could be the fastest too...). I'm not sure about the other two though.
Also, is there a way to have some kind of heuristic? Again the lack of information makes me think that it's impossible, but maybe I'm missing something.
Last thing : When the robot find the exit, he will have to go back to his start position using A*. This means that during the first part, where he looks for the exit, he will have draw a map of the maze that he will use for the 2nd part. Maybe this can help too chose the best algorithm for the first part, but yeah I don't see why one would be better.
Could someone help me please? Thanks (Also, sorry for my english).
Problems like this are categorised as real-time search, perhaps the best known example is Learning Real-Time A*, where you combine information about what you've seen before (if you've had to backtrack or know a cheaper way to reach a state), and the actions you can take. As is the case in areas like reinforcement learning, some level of randomness helps balance exploration and exploitation.
Assuming your graph is undirected, time invariant, and the initial and exit node exist in the same component, then choosing a direction at random at each vertex is equivalent to a random walk on a graph.
Regardless of whether the graph is initially known or not, this is a very well understood field of mathematics, equivalent to an absorbing Markov chain, the time to reach the exit state in such cases has a Discrete phase-type distribution - often quite slow, but it's also worth noting that in pathological cases it's possible to design a maze where a random walk will outperform DFS.
#beaker is right in that the first two you suggested should lead the the same result. However, you may be able to improve on the search a little by keeping track of any loops you find. If the Robot finds itself in a spot it has already visited and needs to backtrack once coming to a dead end there may be no need go back so far if there is a shortcut it has found. Also use the segments that have been mapped on the way out and apply Dijkstra's algorithm or A* on it to find the most efficient way back. There may be a faster way back on an unexplored path but this would be the safest way to have a quick result.
Obviously this implementing the checks for loops to prevent unneeded back tracking will make thing more complicated to implement. Though for the return to the start using Dijkstra's algorithm should not be as complex.
If you are feeling ambitious now that found the exit you could use this information and give the robot a sense of direction though in a randomly generated maze that may not help much.
Related
I'm working on an optimization problem and attempting to use simulated annealing as a heuristic. My goal is to optimize placement of k objects given some cost function. Solutions take the form of a set of k ordered pairs representing points in an M*N grid. I'm not sure how to best find a neighboring solution given a current solution. I've considered shifting each point by 1 or 0 units in a random direction. What might be a good approach to finding a neighboring solution given a current set of points?
Since I'm also trying to learn more about SA, what makes a good neighbor-finding algorithm and how close to the current solution should the neighbor be? Also, if randomness is involved, why is choosing a "neighbor" better than generating a random solution?
I would split your question into several smaller:
Also, if randomness is involved, why is choosing a "neighbor" better than generating a random solution?
Usually, you pick multiple points from a neighborhood, and you can explore all of them. For example, you generate 10 points randomly and choose the best one. By doing so you can efficiently explore more possible solutions.
Why is it better than a random guess? Good solutions tend to have a lot in common (e.g. they are close to each other in a search space). So by introducing small incremental changes, you would be able to find a good solution, while random guess could send you to completely different part of a search space and you'll never find an appropriate solution. And because of the curse of dimensionality random jumps are not better than brute force - there will be too many places to jump.
What might be a good approach to finding a neighboring solution given a current set of points?
I regret to tell you, that this question seems to be unsolvable in general. :( It's a mix between art and science. Choosing a right way to explore a search space is too problem specific. Even for solving a placement problem under varying constraints different heuristics may lead to completely different results.
You can try following:
Random shifts by fixed amount of steps (1,2...). That's your approach
Swapping two points
You can memorize bad moves for some time (something similar to tabu search), so you will use only 'good' ones next 100 steps
Use a greedy approach to generate a suboptimal placement, then improve it with methods above.
Try random restarts. At some stage, drop all of your progress so far (except for the best solution so far), raise a temperature and start again from a random initial point. You can do this each 10000 steps or something similar
Fix some points. Put an object at point (x,y) and do not move it at all, try searching for the best possible solution under this constraint.
Prohibit some combinations of objects, e.g. "distance between p1 and p2 must be larger than D".
Mix all steps above in different ways
Try to understand your problem in all tiniest details. You can derive some useful information/constraints/insights from your problem description. Assume that you can't solve placement problem in general, so try to reduce it to a more specific (== simpler, == with smaller search space) problem.
I would say that the last bullet is the most important. Look closely to your problem, consider its practical aspects only. For example, a size of your problems might allow you to enumerate something, or, maybe, some placements are not possible for you and so on and so forth. THere is no way for SA to derive such domain-specific knowledge by itself, so help it!
How to understand that your heuristic is a good one? Only by practical evaluation. Prepare a decent set of tests with obvious/well-known answers and try different approaches. Use well-known benchmarks if there are any of them.
I hope that this is helpful. :)
If I tell you the moves for a game of chess and declare who wins, why can't it be checked in polynomial time if the winner does really win? This would make it an NP problem from my understanding.
First of all: The number of positions you can set up with 32 pieces on a 8x8 field is limited. We need to consider any pawn being converted to any other piece and include any such available position, too. Of course, among all these, there are some positions that cannot be reached following the rules of chess, but this does not matter. The important thing is: we have a limit. Lets name this limit simply MaxPositions.
Now for any given position, let's build up a tree as follows:
The given position is the root.
Add any position (legal chess position or not) as child.
For any of these children, add any position as child again.
Continue this way, until your tree reaches a depth of MaxPositions.
I'm now too tired to think of if we need one additional level of depth or not for the idea (proof?), but heck, just let's add it. The important thing is: the tree constructed like this is limited.
Next step: Of this tree, remove any sub-tree that is not reachable from the root via legal chess moves. Repeat this step for the remaining children, grand-children, ..., until there is no unreachable position left in the whole tree. The number of steps must be limited, as the tree is limited.
Now do a breadth-first search and make any node a leaf if it has been found previously. It must be marked as such(!; draw candidate?). Same for any mate position.
How to find out if there is a forced mate? In any sub tree, if it is your turn, there must be at least one child leading to a forced mate. If it is the opponents move, there must be a grand child for every child that leads to a mate. This applies recursively, of course. However, as the tree is limited, this whole algorithm is limited.
[sensored], this whole algorithm is limited! There is some constant limiting the whole stuff. So: although the limit is incredibly high (and far beyond what up-to-date hardware can handle), it is a limit (please do not ask me to calculate it...). So: our problem actually is O(1)!!!
The same for checkers, go, ...
This applies for the forced mate, so far. What is the best move? First, check if we can find a forced mate. If so, fine, we found the best move. If there are several, select the one with the least moves necessary (still there might be more than one...).
If there is no such forced mate, then we need to measure by some means the 'best' one. Possibly count the number of available successions to mate. Other propositions for measurement? As long as operating on this tree from top to down, we still remain limited. So again, we are O(1).
Now what did we miss? Have a look at the link in your comment again. They are talking about an NxN checkers! The author is varying size of the field!
So have a look back at how we constructed the tree. I think it is obvious that the tree grows exponentially with the size of the field (try to prove it yourself...).
I know very well that this answer is not a prove for that the problem is EXP(TIME). Actually, I admit, it is not really an answer at all. But I think what I illustrated still gives quite a good image/impression of the complexity of the problem. And as long as no one provides a better answer, I dare to claim that this is better than nothing at all...
Addendum, considering your comment:
Let me allow to refer to wikipedia. Actually, it should be suffient to transform the other problem in exponential time, not polynomial as in the link, as applying the transformation + solving the resulting problem still remains exponential. But I'm not sure about the exact definition...
It is sufficient to show this for a problem of which you know already it is EXP complete (transforming any other problem to this one and then to the chess problem again remains exponential, if both transformations are exponential).
Apparently, J.M. Robson found a way to do this for NxN checkers. It must be possible for generalized chess, too, probably simply modifying Robsons algorithm. I do not think it is possible for classical 8x8 chess, though...
O(1) applies for classical chess only, not for generalized chess. But it is the latter one for which we assume not being in NP! Actually, in my answer up to this addendum, there is one prove lacking: The size of the limited tree (if N is fix) does not grow faster than exponentially with growing N (so the answer actually is incomplete!).
And to prove that generalized chess is not in NP, we have to prove that there is no polynomial algorithm to solve the problem on a non-deterministic turing machine. This I leave open again, and my answer remains even less complete...
If I tell you the moves for a game of chess and declare who wins, why
can't it be checked in polynomial time if the winner does really win?
This would make it an NP problem from my understanding.
Because in order to check if the winner(white) does really win, you will have to also evaluate all possible moves that the looser(black) could've made in other to also win. That makes the checking also exponential.
I'm writing a program which solves a 24-puzzle (5x5 grid) using two heuristic. The first uses how many blocks the incorrect place and the second uses the Manhattan distance between the blocks current place and desired place.
I have different functions in the program which use each heuristic with an A* and a greedy search and compares the results (so 4 different parts in total).
I'm curious whether my program is wrong or whether it's a limitation of the puzzle. The puzzle is generated randomly with pieces being moved around a few times and most of the time (~70%) a solution is found with most searches, but sometimes they fail.
I can understand why greedy would fail, as it's not complete, but seeing as A* is complete this leads me to believe that there's an error in my code.
So could someone please tell me whether this is an error in my thinking or a limitation of the puzzle? Sorry if this is badly worded, I'll rephrase if necessary.
Thanks
EDIT:
So I"m fairly sure it's something I'm doing wrong. Here's a step-by-step list of how I'm doing the searches, is anything wrong here?
Create a new list for the fringe, sorted by whichever heuristic is being used
Create a set to store visited nodes
Add the initial state of the puzzle to the fringe
while the fringe isn't empty..
pop the first element from the fringe
if the node has been visited before, skip it
if node is the goal, return it
add the node to our visited set
expand the node and add all descendants back to the fringe
If you mean that sliding puzzle: This is solvable if you exchange two pieces from a working solution - so if you don't find a solution this doesn't tell anything about the correctness of your algorithm.
It's just your seed is flawed.
Edit: If you start with the solution and make (random) legal moves, then a correct algorithm would find a solution (as reversing the order is a solution).
It is not completely clear who invented it, but Sam Loyd popularized the 14-15 puzzle, during the late 19th Century, which is the 4x4 version of your 5x5.
From the Wikipedia article, a parity argument proved that half of the possible configurations are unsolvable. You are probably running into something similar when your search fails.
I'm going to assume your code is correct, and you implemented all the algorithms and heuristics correctly.
This leaves us with the "generated randomly" part of your puzzle initialization. Are you sure you are generating correct states of the puzzle? If you generate an illegal state, obviously there will be no solution.
While the steps you have listed seem a little incomplete, you have listed enough to ensure that your A* will reach a solution if there is one (albeit not optimal as long as you are just simply skipping nodes).
It sounds like either your puzzle generation is flawed or your algorithm isn't implemented correctly. To easily verify your puzzle generation, store the steps used to generate the puzzle, and run it in reverse and check if the result is a solution state before allowing the puzzle to be sent to the search routines. If you ever generate an invalid puzzle, dump the puzzle, and expected steps and see where the problem is. If the puzzle passes and the algorithm fails, you have at least narrowed down where the problem is.
If it turns out to be your algorithm, post a more detailed explanation of the steps you have actually implemented (not just how A* works, we all know that), like for instance when you run the evaluation function, and where you resort the list that acts as your queue. That will make it easier to determine a problem within your implementation.
I read about Gomoku that it can be implemented using Minimax and Alpha-Beta Pruning algorithms. So, i read these algorithms and now understand how the game will be solved. But when i sat to down to code, I am facing problem how to approach it.
As in ,
How to design the prototype functions like getNextMove or Max(Move) ?
How will the next move searched?
Till when should i apply the minimax algorithm.
I know i can find the code online, but i want to do it myself.
Can anyone please point me in the right direction?
the minimax algorithm presented in the textbook is usually applied on simple games, e.g. tic-tac-tou, where the final states are reachable within only several turns between min player and max player. However, for Gomoku, it is impossible to reach all final states. Why do we need to reach final states? We need an evaluation for a move, i.e. whether a move is good or not.
So your first step is to design an evaluation function of a move, which tells you how much value you will properly gain if you do one move. e.g. you have a 3 in a row, the move along that row to make a 4 will be very valuable.
Suppose that you have a very clever evaluation function, then you can find an optimal move each time without any search. So before you do any min-max, alpha-beta, you'd probably design a good evaluation function. A good example is Emacs' gomoku source code, which has a good AI player without using any search.
Next you move to min-max and alpha-beta.
It seems that I didn't answer your questions. Actually I don't need. I assume that you know the details of min-max or even alpha-beta search for tic-tac-tao. By designing the evaluation function, you will have a better understanding of gomoku and you design search algorithm for it just as you can do for tic-tac-tou now.
So I've had at least two professors mention that backtracking makes an algorithm non-deterministic without giving too much explanation into why that is. I think I understand how this happens, but I have trouble putting it into words. Could somebody give me a concise explanation of the reason for this?
It's not so much the case that backtracking makes an algorithm non-deterministic.
Rather, you usually need backtracking to process a non-deterministic algorithm, since (by the definition of non-deterministic) you don't know which path to take at a particular time in your processing, but instead you must try several.
I'll just quote wikipedia:
A nondeterministic programming language is a language which can specify, at certain points in the program (called "choice points"), various alternatives for program flow. Unlike an if-then statement, the method of choice between these alternatives is not directly specified by the programmer; the program must decide at runtime between the alternatives, via some general method applied to all choice points. A programmer specifies a limited number of alternatives, but the program must later choose between them. ("Choose" is, in fact, a typical name for the nondeterministic operator.) A hierarchy of choice points may be formed, with higher-level choices leading to branches that contain lower-level choices within them.
One method of choice is embodied in backtracking systems, in which some alternatives may "fail", causing the program to backtrack and try other alternatives. If all alternatives fail at a particular choice point, then an entire branch fails, and the program will backtrack further, to an older choice point. One complication is that, because any choice is tentative and may be remade, the system must be able to restore old program states by undoing side-effects caused by partially executing a branch that eventually failed.
Out of the Nondeterministic Programming article.
Consider an algorithm for coloring a map of the world. No color can be used on adjacent countries. The algorithm arbitrarily starts at a country and colors it an arbitrary color. So it moves along, coloring countries, changing the color on each step until, "uh oh", two adjacent countries have the same color. Well, now we have to backtrack, and make a new color choice. Now we aren't making a choice as a nondeterministic algorithm would, that's not possible for our deterministic computers. Instead, we are simulating the nondeterministic algorithm with backtracking. A nondeterministic algorithm would have made the right choice for every country.
The running time of backtracking on a deterministic computer is factorial, i.e. it is in O(n!).
Where a non-deterministic computer could instantly guess correctly in each step, a deterministic computer has to try all possible combinations of choices.
Since it is impossible to build a non-deterministic computer, what your professor probably meant is the following:
A provenly hard problem in the complexity class NP (all problems that a non-deterministic computer can solve efficiently by always guessing correctly) cannot be solved more efficiently on real computers than by backtracking.
The above statement is true, if the complexity classes P (all problems that a deterministic computer can solve efficiently) and NP are not the same. This is the famous P vs. NP problem. The Clay Mathematics Institute has offered a $1 Million prize for its solution, but the problem has resisted proof for many years. However, most researchers believe that P is not equal to NP.
A simple way to sum it up would be: Most interesting problems a non-deterministic computer could solve efficiently by always guessing correctly, are so hard that a deterministic computer would probably have to try all possible combinations of choices, i.e. use backtracking.
Thought experiment:
1) Hidden from view there is some distribution of electric charges which you feel a force from and you measure the potential field they create. Tell me exactly the positions of all the charges.
2) Take some charges and arrange them. Tell me exactly the potential field they create.
Only the second question has a unique answer. This is the non-uniqueness of vector fields. This situation may be in analogy with some non-deterministic algorithms you are considering. Further consider in math limits which do not exist because they have different answers depending on which direction you approach a discontinuity from.
I wrote a maze runner that uses backtracking (of course), which I'll use as an example.
You walk through the maze. When you reach a junction, you flip a coin to decide which route to follow. If you chose a dead end, trace back to the junction and take another route. If you tried them all, return to the previous junction.
This algorithm is non-deterministic, non because of the backtracking, but because of the coin flipping.
Now change the algorithm: when you reach a junction, always try the leftmost route you haven't tried yet first. If that leads to a dead end, return to the junction and again try the leftmost route you haven't tried yet.
This algorithm is deterministic. There's no chance involved, it's predictable: you'll always follow the same route in the same maze.
If you allow backtracking you allow infinite looping in your program which makes it non-deterministic since the actual path taken may always include one more loop.
Non-Deterministic Turing Machines (NDTMs) could take multiple branches in a single step. DTMs on the other hand follow a trial-and-error process.
You can think of DTMs as regular computers. In contrast, quantum computers are alike to NDTMs and can solve non-deterministic problems much easier (e.g. see their application in breaking cryptography). So backtracking would actually be a linear process for them.
I like the maze analogy. Lets think of the maze, for simplicity, as a binary tree, in which there is only one path out.
Now you want to try a depth first search to find the correct way out of the maze.
A non deterministic computer would, at every branching point, duplicate/clone itself and run each further calculations in parallel. It is like as if the person in the maze would duplicate/clone himself (like in the movie Prestige) at each branching point and send one copy of himself into the left subbranch of the tree and the other copy of himself into the right subbranch of the tree.
The computers/persons who end up at a dead end they die (terminate without answer).
Only one computer will survive (terminate with an answer), the one who gets out of the maze.
The difference between backtracking and non-determinism is the following.
In the case of backtracking there is only one computer alive at any given moment, he does the traditional maze solving trick, simply marking his path with a chalk and when he gets to a dead end he just simply backtracks to a branching point whose sub branches he did not yet explore completely, just like in a depth first search.
IN CONTRAST :
A non deteministic computer can clone himself at every branching point and check for the way out by running paralell searches in the sub branches.
So the backtracking algorithm simulates/emulates the cloning ability of the non-deterministic computer on a sequential/non-parallel/deterministic computer.