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First - I've looked through similar looking questions but they did not solve my problem, this is no repetition (I hope).
I'm building and programming a robot with an Arduino Nano that is supposed to solve a maze. It gets put somewhere in the maze and then has to find an item. The next time it is supposed to go straight to the item (it does not have to be the shortest way but no dead ends allowed).
It is not necessary to know the whole maze because as long as he has one way to the item it is good. As I said, I don't need the shortest way.
The maze is 2D, I just put black tape on a white table and the robot is supposed to use a line sensor to follow the lines.
There are no other sensors to orientate himself. First I thought of making an 2D array and each field of the maze a field in there. But since it's just a normal line sensor the robot doesn't know if a straight line is one or two fields long and the whole thing does not work.
I also tried DFS or something like that but a similar problem here. The maze is circular and how is the robot supposed to know the Node was already found before and it is the same?
It would be nice if anyone had an idea!
Although orientation is a little bit fuzzy it is possible by using the decisions. A decision has to be reproducable. It could be represented by a class:
public class Decision {
boolean[] directions = new boolean[2]; // 0 = left, 1 = straight, 2 = right
// at least 2 of them should be true or it is no decision
int path; // 0-2 to mark the current path
}
Create a stack of decisions.
If there is only one possible direction at the beginning (back doesn't count and is treated later), then move forward until you meet the first decision.
Push the decision with all possible directions on the stack.
Set path to the first possible direction and move that way.
If you end up with another decision: continue at 3.
If you find the token: abort, you found a reproducible way without dead-ends.
If it is a dead-end: return to the previous decision node (the first one one the way back) and continue with 6.
Pop the decision and try the next possible direction (set the new path and push the decision) and continue at 5.
Unless you have tried all directions, then move back another decision and continue with 6.
If there are no more decisions (the special case mentioned above, we went in the wrong direction at the beginning): move forward until you meet the first decision and continue at 3. This means you need another boolean variable to indicate if you should go backwards right at the beginning.
You have to be careful when coming back from left and want to try straight next you would have to turn left and don't go straight. So there is a little calculation involved.
The algorithm has a problem for loop shaped decisions if you start the wrong way at the beginning. I think this could be escaped by setting an upper boundary, e.g. if you still haven't found the token and met 30 decision nodes (going forward), then you are probably running in circles, so go back to start and now instead of trying the directions in increasing order, try them in decreasing order.
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This is an algorithmic problem and I'm not sure it has a solution. I think it's a specific case of a more generic computer science problem that has no solution but I'd rather not disclose which one to avoid planting biases. It came up from a real life situation in which mobile phones were out of credit and thus, we didn't have long range communications.
Two groups of people, each with 2 people (but it might be true for N people) arranged to meet at the center of a park but at the time of meeting, the park is closed. Now, they'll have to meet somewhere else around the park. Is there an algorithm each and every single individual could follow to converge all in one point?
For example, if each group splits in two and goes around and when they find another person keep on going with that person, they would all converge on the other side of the park. But if the other group does the same, then, they wouldn't be able to take the found members of the other group with them. This is not a possible solution.
I'm not sure if I explained well enough. I can try to draw a diagram.
Deterministic Solution for N > 1, K > 1
For N groups of K people each.
Since the problem is based on people whose mobile phones are out of credit, let's assume that each person in each group has their own phone. If that's not acceptable, then substitute the phone with a credit card, social security, driver's license, or any other item with numerical identification that is guaranteed to be unique.
In each group, each person must remember the highest number among that group, and the person with the highest number (labeled leader) must travel clockwise around the perimeter while the rest of the group stays put.
After the leader of each group meets the next group, they compare their number with the group's previous leader number.
If the leader's number is higher than the group's previous leader's number, then the leader and the group all continue along the perimeter of the park. If the group's previous leader's number is higher, then they all stay put.
Eventually the leader with the highest number will continue around the entire perimeter exactly 1 rotation, collecting the entire group.
Deterministic solution for N > 1, K = 1 (with one reasonable assumption of knowledge ahead-of-time)
In this case, each group only contains one person. Let's assume that the number used is a phone number, because it is then reasonable to also assume that at least one pair of people will know each other's numbers and so one of them will stay put.
For N = 2, this becomes trivially reduced to one person staying put and the other person going around clockwise.
For other cases, the fact that at least two people will initially know each other's numbers will effectively increase the maximum K to at least 2 (because the person or people who stay put will continue to stay put if the person they know has a higher number than the leader who shows up to meet them), but we still have to introduce one more step to the algorithm to make sure it will terminate.
The extra step is that if a leader has continued around the perimeter for exactly one rotation without adding anyone to the group, then the leader must leave their group behind and start over for one more rotation around the perimeter. This means that a leader with no group will continue indefinitely until they find someone else, which is good.
With this extra step, it is easy to see why we have to assume that at least one pair of people need to know each other's phone numbers ahead of time, because then we can guarantee that the person who stays put will eventually accumulate the entire group.
Feel free to leave comments or suggestions to improve the algorithm I've laid out or challenge me if you think I missed an edge case. If not, then I hope you liked my answer.
Update
For fun, I decided to write a visual demo of my solutions to the problem using d3. Feel free to play around with the parameters and restart the simulation with any initial state. Here's the link:
https://jsfiddle.net/patrob10114/c3d478ty/show/
Key
black - leader
white - follower
when clicked
blue - selected person
green - known by selected person
red - unknown by selected person
Note that collaboration occurs at the start of every step, so if two groups just combined in the current step, most people won't know the people from the opposite group until after the next step is invoked.
They should move towards the northernmost point of the park.
I'd send both groups in a random direction. If they went a half circle without meeting the other group, rerandomize the directions. This will make them meet in a few rounds most of the time, however there is an infinitely small chance that they still never meet.
It is not possible with a deterministic algorithm if
• we have to meet at some point on the perimeter,
• we are unable to distinguish points on the perimeter (or the algorithm is not allowed to use such a distinction),
• we are unable to distinguish individuals in the groups (or the algorithm is not allowed to use such a distinction),
• the perimeter is circular (see below for a more general case),
• we all follow the same algorithm, and
• the initial points may be anywhere on the perimeter.
Proof: With a deterministic algorithm we can deduce the final positions from the initial positions, but the groups could start evenly spaced around the perimeter, in which case the problem has rotational symmetry and so the solution will be unchanged by a 1/n rotation, which however has no fixed point on the perimeter.
Status of assumptions
Dropping various assumptions leads, as others have observed to various solutions:
Non-deterministic: As others have observed, various non-deterministic algorithms do provide a solution whose probability of termination tends to certainty as time tends to infinity; I suspect almost any random walk would do. (Many answers)
Points indistinguishable: Agree on a fixed point at which to meet if needed: flyx’s answer.
Individuals indistinguishable: If there is a perfect hash algorithm, choose those with the lowest hash to collect others: Patrick Roberts’s solution.
Same algorithm: Choose one in advance to collect the others (adapting Patrick Roberts’s solution).
Other assumptions can be weakened:
Non-circular perimeter: The condition that the perimeter be circular is rather artificial, but if the perimeter is topologically equivalent to a circle, this equivalence can be used to convert any solution to a solution to the circle problem.
Unrestricted initial points: Even if the initial points cannot be evenly spaced, as long as some points are distinct, a topological equivalence (as for a non-circular perimeter) reduces a solution to a solution to the circular case, showing that no solution can exist.
I think this question really belongs on Computer Science Stack Exchange.
This question heavily depends on what kind of operations do we have and what do you consider your environment looks like. I asked your this questions with no reply, so here is my interpretation:
The park is a 2d space, 2 groups are located randomly, each group has the same right/left (both are facing the park). Both have the same operations are programmed to do absolutely the same things (nothing like I go right, and you go left, because this makes the problem obvious). So the operations are: Go right/left/stop for x units of time. They can also figure out that they passed through their original position (the one in which they started). And they can be programmed in a loop.
If you have an ability to use randomness - everything is simple. You can come up with many solutions. For example: with probability 0.5 each of them decide to that they will either do 3 steps right and wait. Or one step right and wait. If you will do this operation in a loop and they will select different options, then clearly they will meet (one is faster than the other, so he will reach a slower person). If they both select the same operation, than they will make a circle and both reach their starting positions. In this case roll the dice one more time. After N circles the probability that they will meet will be 1 - 0.5^n (which approaches 1 very fast)
Surprisingly, there is a way to do it! But first we have to define our terms and assumptions.
We have N=2 "teams" of K=2 "agents" apiece. Each "agent" is running the same program. They can't tell north from south, but they can tell clockwise from counterclockwise. Agents in the same place can talk to each other; agents in different places can't.
Your suggested partial answer was: "If each group splits in two and goes around and when they find another person keep on going with that person, they would all converge on the other side of the park..." This implies that our agents have some (magic, axiomatic) face-to-face decision protocol, such that if Alice and Bob are on the same team and wake up at the same point on the circle, they can (magically, axiomatically) decide amongst themselves that Alice will head clockwise and Bob will head counterclockwise (as opposed to Alice and Bob always heading in exactly the same direction because by definition they react exactly the same way to the situation they're identically in).
One way to implement this magic decision protocol is to give each agent a personal random number generator. Whenever 2 or more agents are gathered at a certain point, they all roll a million-sided die, and whichever one rolls highest is acknowledged as the leader. So in your partial solution, Alice and Bob could each roll: whoever rolls higher (the "leader") goes clockwise and sends the other agent (the "follower") counterclockwise.
Okay, having solved the "how do our agents make decisions" issue, let's solve the actual puzzle!
Suppose our teams are (Alice and Bob) and (Carl and Dave). Alice and Carl are the initially elected leaders.
Step 1: Each team rolls a million-sided die to generate a random number. The semantics of this number are "The team with the higher number is the Master Team," but of course neither team knows right now who's got the higher number. But Alice and Bob both know that their number is let's say 424202, and Carl and Dave both know that their number is 373287.
Step 2: Each team sends its leader around the circle clockwise, while the follower stays stationary. Each leader stops moving when he gets to where the other team's follower is waiting. So now at one point on the circle we have Alice and Dave, and at the other point we have Carl and Bob.
Step 3: Alice and Dave compare numbers and realize that Alice's team is the Master Team. Likewise, Bob and Carl compare numbers and realize that Bob's team is the Master Team.
Step 4: Alice being the leader of the Master Team, she takes Dave with her clockwise around the circle. Bob and Carl (being a follower and a leader of a non-master team respectively) just stay put. When Alice and Dave reach Bob and Carl, the problem is solved!
Notice that Step 1 requires that both teams roll a million-sided die in isolation; if during Step 3 everyone realizes that there was a tie, they'll just have to backtrack and try again. Therefore this solution is still probabilistic... but you can make its expected time arbitrarily small by just replacing everyone's million-sided dice with trillion-sided, quintillion-sided, bazillion-sided... dice.
The general strategy here is to impose a pecking order on all N×K agents, and then bounce them around the circle until everyone is aware of the pecking order; then the top pecker can just sweep around the circle and pick everyone up.
Imposing a pecking order can be done by using the agents' personal random number generators.
The protocol for K>2 agents per team is identical to the K=2 case: you just glom all the followers together in Step 1. Alice (the leader) goes clockwise while Bobneric (the followers) stay still; and so on.
The protocol for K=1 agents per team is... well, it's impossible, because no matter what you do, you can't deterministically ensure that anyone will ever encounter another agent. You need a way for the agents to ensure, without communicating at all, that they won't all just circle clockwise around the park forever.
One thing that would help with (but not technically solve) the K=1 case would be to consider the relative speeds of the agents. You might be familiar with Floyd's "Tortoise and Hare" algorithm for finding a loop in a linked list. Well, if the agents are allowed to move at non-identical speeds, then you could certainly do a "continuous, multi-hare" version of that algorithm:
Step 1: Each agent rolls a million-sided die to generate a random number S, and starts running clockwise around the park at speed S.
Step 2: Whenever one agent catches up to another, both agents glom together and start running clockwise at a new random speed.
Step 3: Eventually, assuming that nobody picked exactly the same random speeds, everyone will have met up.
This protocol requires that Alice and Carl not roll identical numbers on their million-sided dice even when they are across the park from each other. IMHO, this is a very different assumption from the other protocol's assuming that Alice and Bob could roll different numbers on their million-sided dice when they were in the same place. With K=1, we're never guaranteed that two agents will ever be in the same place.
Anyway, I hope this helps. The solution for N>2 teams is left as an exercise for the reader, but my intuition is that it'll be easy to reduce the N>2 case to the N=2 case.
Each group sends out a scout with the remaining group members remaining stationary. Each group remembers the name of their scout. The scouts circle around clockwise, and whenever he meets a group, they compare names of their scouts:
If scout's name is earlier alphabetically: group follows him.
If scout's name is later: he joins the group and gives up his initial group identity.
By the time the lowest named scout makes it back the his starting location, everyone who hasn't stopped at his initial location should be following him.
There are some solutions here that to me are unsatisfactory since they require the two teams to agree a strategy in advance and all follow the same deterministic or probabilistic rules. If you had the opportunity to agree in advance what rules you're all going to follow, then as flyx points out you could just have agreed a backup meeting point. Restrictions that prevent the advance choice of a particular place or a particular leader are standard in the context of some problems with computer networks but distinctly un-natural for four friends planning to meet up. Therefore I will frame a strategy from the POV of only one team, assuming that there has been no prior discussion of the scenario between the two teams.
Note that it is not possible to be robust in the face of any strategy from the other team. The other team can always force a stalemate simply by adopting some pattern of movement that ensures those two will never meet again.
One of you sets out walking around the park. The other stands still, let us say at position X. This ensures that: (a) you will meet each other periodically at X, let us say every T seconds; and (b) for each member of the other team, no matter how they move around the perimeter of the park they must encounter at least one of your team at least every T seconds.
Now you have communication among all members of both groups, and (given sufficient time and passing-on of messages from one person to another) the problem resolves to the same problem as if your mobile phones were working. Choosing a leader by random number is one way to solve it as others have suggested. Note that there are still two issues: the first is a two-generals problem with communication, and I suppose you might feel that a mobile phone conversation allows for the generation of common knowledge whereas these relayed notes do not. The second is the possibility that the other team refuses to co-operate and you cannot agree a meeting point no matter what.
Notwithstanding the above problems, the question supposes that if they had communication that the groups would be able to agree a meeting-point. You have communication: agree a meeting point!
As for how to agree a meeting point, I think it requires some appeal to reason or good intention on the part of the other team. If they are due to meet again, then they will be very reluctant to take any action that results in them breaking their commitment to their partner. Therefore suggest to them both that after their next meeting, when all commitments can be forgiven, they proceed together to X by the shortest route. Listen to their counter-proposal and try to find some common solution.
To help reach a solution, you could pre-agree with your team-mate some variations you'd be willing to make to your plan, provided that they remain within some restrictions that ensure you will meet your team-mate again. For example, if the stationary team-mate agrees that they could be persuaded to set out clockwise, and the moving team-mate sets out anti-clockwise and agrees that they can be persuaded to do something different but not to cross point X in a clockwise direction, then you're guaranteed to meet again and so you can accept certain suggestions from the other team.
Just as an example, if a team following this strategy meets a team (unwisely) following your strategy, then one of my team will agree to go along with the one of your team they meet, and the other will refuse (since it would require them to make the forbidden movement above). This means that when your team meet together, they'll have one of my team with them for a group of three. The loose member of my team is on a collision course with that group of three provided your team doesn't do anything perverse.
I think forming any group of three is a win, so each member should do anything they can to attend a meeting of the other team, subject to the constraints they agreed to guarantee they'll meet up with their own team member again. A group of 3, once formed, should follow whatever agreement is in place to meet the loose member (and if the team of two contained within that 3 refuses to do this then they're saboteurs, there is no good reason for them to refuse). Within these restrictions, any kind of symmetry-breaking will allow the team following these principles to persuade/follow the other team into a 3-way and then a 4-way meeting.
In general some symmetry-breaking is required, if only because both teams might be following my strategy and therefore both have a stationary member at different points.
Assume the park is a circle. (for the sake of clarity)
Group A
Person A.1
Person A.2
Group B
Person B.1
Person B.2
We (group A) are currently at the bottom of the circle (90 degrees). We agree to go towards 0 degrees in opposite directions. I'm person A.1 and I go clockwise. I send Person A.2. counterclockwise.
In any possible scenario (B splits, B doesn't split, B has the same scheme, B has some elaborate scheme), each group might have conflicting information. So unless Group A has a gun to force Group B into submission, the new groups might make conflicting choices upon meeting.
Say for instance, A.1. meets B.1, and A.2. meets B.2. What do we (A.1 and B.1) do if B has the same scheme? Since the new groups can't know what the other group decides (whether to go with A's scheme, or B's scheme), each group might make different decision.
And we'll end up where we started... (i.e. two people at 0 degrees, and two people at 90 degrees). Let's call this checkpoint "First Iteration".
We might account for this and say that we'll come up with a scheme for the "Second Iteration". But then the same thing happens again. And for the third iteration, fourth iteration, ad infinitum.
Each iteration has a 50% chance of not working out.
Which means that after x iterations, your chances of not meeting up at a common point are at most 1-(0.5^x)
N.B. I thought about a bunch of scenarios, such as Group A agreeing to come back to their initial point, and communicating with each other what Group B plans to do. But no cigar, turns out even with very clever schemes the conflicting information issue always arises.
An interesting problem indeed. I'd like to suggest my version of the solution:
0 Every group picks a leader.
1: Leader and followers go opposite directions
2: They meet other group leaders or followers
3: They keep going the same direction as before, 90 degrees magnitude
4: By this time, all groups have made a half-circle around the perimeter, and invariably have met leaders again, theirs, or others'.
5: All Leaders change the next step direction to that of the followers around,and order them to follow.
6: Units from all groups meet at one point.
Refer to the attached file for an in-depth explanation. You will need MS Office Powerpoint 2007 or newer to view it. In case you don't have one, use pptx. viewer (Powerpoint viewer) as a free alternative.
Animated Solution (.pptx)
EDIT: I made a typo in the first slide. It reads "Yellow and red are selected", while it must be "Blue and red" instead.
Each group will split in two parts, and each part will go around the circle in the opposite direction (clockwise and counterclockwise).
Before they start, they choose some kind of random number (in a range large enough so that there is no possibility for two groups to have the same number... or a Guid in computer science : globally unique identifier). So one unique number per group.
If people of the same group meet first (the two parts meet), they are alone, so probably the other groups (if any) gave up.
If two groups meet : they follow the rule that say the biggest number leads the way. So when they meet they continue in the direction that had people with the biggest number.
At the end, the direction of the biggest number will lead them all to one point.
If they have no computer to choose this number, each group could use the full names of the people of the group merged together.
Edit : sorry I just see that this is very close to Patrick Roberts' solution
Another edit : what if each group has its own deterministic strategy ?
In the solution above, all works well if all the groups have the same strategy. But in a real life problem this is not the case (as they cant communicate).
If one group has a deterministic strategy and the others have none, they can agree to follow the deterministic approach and all is ok.
But if two groups have deterministic approaches (simply for instance, the same as above, but one group uses the biggest number and the other group follows the lowest number).
Is there a solution to that ?
You probably know the problem with the two robots dropped on a line when you need to program them to meet.
Two robots are dropped from an airplane and land on a single line (with discrete positions) using a parachute which is left at the landing point. The robots are both facing north, they are an unknown distance apart, and one has landed directly east of the other.
The robots are now to be programmed such that they meet each other. They can be instructed to move left or right to a neighboring position and to check whether a parachute is present at the current location. If the other robot is met both robots stop there and live happily ever after.
The parachute check might conditionally execute any number of instructions and any block of instructions may be repeated unconditionally. Write down a program that both robots can follow simultaneously and which garuantees that they meet.
You have to create a generic algorithm (a little pleonastic) that applied to both robots guarantees that the robots will meet. They leave their parachute on the spot where they are dropped and they can check if in the current position there is a parachute.
The original statement is here: http://en.wikibooks.org/wiki/Puzzles/Logic_puzzles/Parachuted_Robots
There is also a solution that I don't understand. If someone can make any sense of it, please help me with a little explaning. Any other solution would be much appreciated.
My first thought on this problem would be to program the robot to choose randomly to first go right or left, and then make something like an exponential search: first go 2 positions to right, then 4 to left etc. If in one of this "trips" in right or left the robot finds the second parachute (the one that was used by the other robot), the robot will only search in that direction. Does this make any sense?
Thank you very much!
My program is actually shorter, and works like a charm too:
start: left
skipNext
goto start
next: left
goto next
This works because the second loop is faster than the first loop.
You can test your program here: http://david-peter.de/parachuting-robots/
Your "first thought" solution should work too, but it will take a while longer for the robots to meet than the solution you cited at wikibooks. To recap, the wikibooks solution is:
10 Go right
20 Go left
30 Go right
40 If Not Parachute GOTO 10
50 Go right
60 GOTO 50
In case you don't recognize the syntax, the author is trying to mimic BASIC, where the numbers 10-60 are line numbers, and the GOTOs are code jumps.
Lines 10-40 have both robots moving slowly right. The "right, left, right" steps slow down movement to the right. It could have just as easily been "right, wait". Line 40 checks for the parachute. When both robots landed on the line, exactly one of them was to the left of the other. The left robot will eventually find the other parachute. The right never will. When the left robot finds the right robot's parachute, it enters lines 50-60, where it moves right without the slowdown. Now that the left robot is moving right faster than the right robot, the left will eventually catch up.
Personally I think the algorithm you posed is more fun, since both robots would swing back and forth a lot. In a way it's a similar algorithm, but the slowdown grows linearly with each step.
It seems to me that your algorithm ought to work. The idea in the posted solution is that both robots keep a pattern of going right left right, meaning they are advancing right at a certain pace. But when the robot on the left finds the other's parachute, it starts moving to the right at a faster pace since it does not step once to the left as part of its walking pattern but keeps going right, eventually catching up to the robot on the right.
Both robots move left till the right robot finds the left robot's parachute and starts sprinting towards the left robot. Then they collide.
start: left
skipNext
goto start
goto moveLeftFast
moveLeftFast: left
goto moveLeftFast
I made this:
start: left
skipNext
goto start
fastL: left
left
goto fastL
The idea is simple: we go left one bit. One of the robots (the one on the right) will eventually bump into parachute and then it will skip out of the fist loop, and enter the second which makes him go left twice as fast.
It's a homework. I have to design and lights out game using backtracking description is below.
The game consists of a 5-by-5 grid of lights; when the game starts, a set of these lights (random, or one of a set of stored puzzle patterns) are switched on. Pressing one of the lights will toggle it, and the four lights adjacent to it, on and off. (Diagonal neighbours are not affected.) The game provides a puzzle: given some initial configuration where some lights are on and some are off, the goal is to switch all the lights off, preferably in
as few button presses as possible.
My approach is go from 1 to 25 and check if all the lights are off or not. If not then I will check for 1 to 24 and so on until I reach 1 or found solution. No if there is no solution then I will start from 2 to 24 and follow the above process till I reach 2 or found solution.
But through this I am not getting the result ? for example light at (0,0) (1,1) (2,2) (3,3) (4,4) are ON?
If any one need code I can post it.
Can any one tell me correct approach using backtracking to solve this game ?
Thanks.
There is a standard algorithm for solving this problem that is based on Gaussian elimination over GF(2). The idea is to set up a matrix representing the button presses a column vector representing the lights and then to use standard matrix simplification techniques to determine which buttons to press. It runs in polynomial time and does not require any backtracking.
I have an implementation of this algorithm that includes a mathematical description of how it works available on my personal site. I hope you find it useful!
Edit: If you are forced to use backtracking, you can use the following facts to do so:
Any solution will never push the same button twice, since doing so would cancel out a previous move.
Any solution either pushes the first button or does not.
Given this approach, you could solve this using backtracking using a simple recursive algorithm that keeps track of the current state of the board and which buttons you've already made decisions about:
If you've decided about each button, then return whether the board is solved or not.
Otherwise:
Try pushing the next button and seeing if the board is recursively solvable from there.
If so, return success.
Otherwise, try not pushing the next button and seeing if the board is recursively solvable from there.
If so, return success. If not, return failure.
This will explore a search space of size 225, which is about 32 million. That's big, but not insurmountably big.
Hope this helps!
Backtracking means:
Incrementally build a solution, throwing away impossible solutions.
Here is one approach using the fact that there is locality of inputs and outputs (pressing a button affects square around it).
problem = GIVEN
solutions = [[0],[1]] // array of binary matrix answers (two entries, a zero and a one)
for (problemSize = 1; problemSize <= 5; problemSize++) {
newSolutions = [];
foreach (solutions as oldSolution) {
candidateSolutions = arrayOfNByNMatriciesWithMatrixAtTopLeft(min(5,problemSize+1), oldSolution);
// size of candidateSolutions is 2^((problemSize+1)^2 - problemSize^2)
// except last round candidateSolutions == solutions
foreach (candidateSolutions as candidateSolution) {
candidateProblem = boardFromPressingButtonsInSolution(candidateSolution);
if (compareMatrix(problem, candidateProblem, 0, 0, problemSize, problemSize)==0)
newSolutions[] = candidateSolution;
}
}
solutions = newSolutions;
}
return solutions;
As already suggested, you should first form a set of simultaneous equations.
First thing to note is that a particular light button shall be pressed at most once because it does not make sense to toggle the set of lights twice.
Let Aij = Light ij Toggled { Aij = 0 or 1 }
There shall be 25 such variables.
Now for each of the lights, you can form an equation looking like
summation (Amn) = 0. { Amn = 5 light buttons that toggle the light mn }
So you will have 25 variables and 25 unknowns. You can solve these equations simultaneously.
If you need to solve it using backtracking or recursion you can solve the equations that way. Just assume an initial value of variables, see if they satisfy all the equations. If not, then back track.
The Naive Solution
First, you're going to need a way to represent the state of the board and a stack to store all of the states. At each step, make a copy of the board, changed to the new state. Compare that state to all states of the board you've encountered so far. If you haven't seen it, push that state on top of the stack and move on to the next move. If you have seen it, try the next move. Each level will have to try all possible 64 moves before popping the state from the stack (backtracking). You will want to use recursion to manage the state of the next move to check.
There are at most 264 possible board configurations, meaning you could potentially go on a very long chain of unique states and still run out of memory. (For reference, 1 GB is 230 bytes and you need a minimum of 8 bytes to store the board configuration) This algorithm is not likely to terminate in the lifetime of the known universe.
You need to do something clever to reduce your search space...
Greedy-First Search
You can do better by searching the states that are closest to the solved configuration first. At each step, sort the possible moves in order from most lights off to least lights off. Iterate in that order. This should work reasonably well but is not guaranteed to get the optimal solution.
Not All Lights-Out Puzzles Are Solvable
No matter what algorithm you use, there may not be a solution, meaning you might search forever (or several trillion years, at least) without finding a solution.
You will want to check the board for solvability (which is a much faster algorithm as it turns out) before wasting any time trying to find a solution.
Regardless of the layout being used for the tiles, is there any good way to divvy out the tiles so that you can guarantee the user that, at the beginning of the game, there exists at least one path to completing the puzzle and winning the game?
Obviously, depending on the user's moves, they can cut themselves off from winning. I just want to be able to always tell the user that the puzzle is winnable if they play well.
If you randomly place tiles at the beginning of the game, it's possible that the user could make a few moves and not be able to do any more. The knowledge that a puzzle is at least solvable should make it more fun to play.
Place all the tiles in reverse (ie layout out the board starting in the middle, working out)
To tease the player further, you could do it visibly but at very high speed.
Play the game in reverse.
Randomly lay out pieces pair by pair, in places where you could slide them into the heap. You'll need a way to know where you're allowed to place pieces in order to end up with a heap that matches some preset pattern, but you'd need that anyway.
I know this is an old question, but I came across this when solving the problem myself. None of the answers here are quite perfect, and several of them have complicated caveats or will break on pathological layouts. Here is my solution:
Solve the board (forward, not backward) with unmarked tiles. Remove two free tiles at a time. Push each pair you remove onto a "matched pair" stack. Often, this is all you need to do.
If you run into a dead end (numFreeTiles == 1), just reset your generator :) I have found I usually don't hit dead ends, and have so far have a max retry count of 3 for the 10-or-so layouts I have tried. Once I hit 8 retries, I give up and just randomly assign the rest of the tiles. This allows me to use the same generator for both setting up the board, and the shuffle feature, even if the player screwed up and made a 100% unsolvable state.
Another solution when you hit a dead end is to back out (pop off the stack, replacing tiles on the board) until you can take a different path. Take a different path by making sure you match pairs that will remove the original blocking tile.
Unfortunately, depending on the board, this may loop forever. If you end up removing a pair that resembles a "no outlet" road, where all subsequent "roads" are a dead end, and there are multiple dead ends, your algorithm will never complete. I don't know if it is possible to design a board where this would be the case, but if so, there is still a solution.
To solve that bigger problem, treat each possible board state as a node in a DAG, with each selected pair being an edge on that graph. Do a random traversal, until you find a leaf node at depth 72. Keep track of your traversal history so that you never repeat a descent.
Since dead ends are more rare than first-try solutions in the layouts I have used, what immediately comes to mind is a hybrid solution. First try to solve it with minimal memory (store selected pairs on your stack). Once you've hit the first dead end, degrade to doing full marking/edge generation when visiting each node (lazy evaluation where possible).
I've done very little study of graph theory, though, so maybe there's a better solution to the DAG random traversal/search problem :)
Edit: You actually could use any of my solutions w/ generating the board in reverse, ala the Oct 13th 2008 post. You still have the same caveats, because you can still end up with dead ends. Generating a board in reverse has more complicated rules, though. E.g, you are guaranteed to fail your setup if you don't start at least SOME of your rows w/ the first piece in the middle, such as in a layout w/ 1 long row. Picking a completely random (legal) first move in a forward-solving generator is more likely to lead to a solvable board.
The only thing I've been able to come up with is to place the tiles down in matching pairs as kind of a reverse Mahjong Solitaire game. So, at any point during the tile placement, the board should look like it's in the middle of a real game (ie no tiles floating 3 layers up above other tiles).
If the tiles are place in matching pairs in a reverse game, it should always result in at least one forward path to solve the game.
I'd love to hear other ideas.
I believe the best answer has already been pushed up: creating a set by solving it "in reverse" - i.e. starting with a blank board, then adding a pair somewhere, add another pair in a solvable position, and so on...
If you a prefer "Big Bang" approach (generating the whole set randomly at the beginning), are a very macho developer or just feel masochistic today, you could represent all the pairs you can take out from the given set and how they depend on each other via a directed graph.
From there, you'd only have to get the transitive closure of that set and determine if there's at least one path from at least one of the initial legal pairs that leads to the desired end (no tile pairs left).
Implementing this solution is left as an exercise to the reader :D
Here are rules i used in my implementation.
When buildingheap, for each fret in a pair separately, find a cells (places), which are:
has all cells at lower levels already filled
place for second fret does not block first, considering if first fret already put onboard
both places are "at edges" of already built heap:
EITHER has at least one neighbour at left or right side
OR it is first fret in a row (all cells at right and left are recursively free)
These rules does not guarantee a build will always successful - it sometimes leave last 2 free cells self-blocking, and build should be retried (or at least last few frets)
In practice, "turtle" built in no more then 6 retries.
Most of existed games seems to restrict putting first ("first on row") frets somewhere in a middle. This come up with more convenient configurations, when there are no frets at edges of very long rows, staying up until last player moves. However, "middle" is different for different configurations.
Good luck :)
P.S.
If you've found algo that build solvable heap in one turn - please let me know.
You have 144 tiles in the game, each of the 144 tiles has a block list..
(top tile on stack has an empty block list)
All valid moves require that their "current__vertical_Block_list" be empty.. this can be a 144x144 matrix so 20k of memory plus a LEFT and RIGHT block list, also 20 k each.
Generate a valid move table from (remaning_tiles) AND ((empty CURRENT VERTICAL BLOCK LIST) and ((empty CURRENT LEFT BLOCK LIST) OR (empty CURRENT RIGHT BLOCK LIST)))
Pick 2 random tiles from the valid move table, record them
Update the (current tables Vert, left and right), record the Tiles removed to a stack
Now we have a list of moves that constitute a valid game. Assign matching tile types to each of the 72 moves.
for challenging games, track when each tile becomes available. find sets that have are (early early early late) and (late late late early) since it's blank, you find 1 EE 1 LL and 2 LE blocks.. of the 2 LE block, find an EARLY that blocks ANY other EARLY that (except rightblocking a left side piece)
Once youve got a valid game play around with the ordering.
Solitaire? Just a guess, but I would assume that your computer would need to beat the game(or close to it) to determine this.
Another option might be to have several preset layouts(that allow winning, mixed in with your current level.
To some degree you could try making sure that one of the 4 tiles is no more than X layers below another X.
Most games I see have the shuffle command for when someone gets stuck.
I would try a mix of things and see what works best.