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Finding Admissible Heuristics for Martian Rover using A*

I'm trying to solve a problem about AI. I have a "robot" that should go from a point A to B much quickly and cheap as possible. This Rover can't climb heights higher than 10 units and the cost of his route is influenced by the kind of terrain. I need your help becouse I need to find a admissible heuristic for solving my problem. I already try the Euclidian distance but it's not enough. Can you help me?

I would recommend taking a look at this page for some ideas of heuristics. They will not all be applicable in your situation, since I don't believe you have a grid map? But you can at least have a look.
Furthermore, I'd recommend trying to take into account the various costs that are possible in different kinds of terrain. If, for example, you know that every single possible type of terrain has a minimum cost of 2, you can safely multiply your Euclidean distance by 2. If the most common type of terrain has a cost of 2, but there also are some types of terrain with a lower cost, you lose the guarantee of finding an optimal solution if you start multiplying by 2, but you may still find solutions more quickly in practice.
To me, the question sounds like a homework assignment (correct me if I'm wrong), which makes it difficult to give a single answer. I guess that the point of the homework assignment even is to research a bit and try some different things and see how they work (or don't work), which can improve your understanding of how the algorithm works. So, really, just try some things.

Modifing AStar algorithm to connect gates in logic scheme

I've been working on a logic scheme simulator for a while now. The whole thing is pretty much working but there's an issue that I can't seem to solve.
Connections in a logic scheme should be vertical and horizontal. They should avoid logic gates and have as few turns as possible (avoiding the staircase effect). Connections can also intersect but they can never overlap.
I used AStar algorithm to find the shortest and the nicest path between two logic gates. The heuristics for pathfinding is Manhattan distance while the cost between the two nodes is a dimension of a single square on the canvas.
The whole issue arises between two conditions. "Min turns" and "no overlap". I solved "min turns" issue by punishing the algorithm with double the normal cost when it tries to make a turn. That causes the algorithm to postpone all turns to the latest possible moment which causes my next picture.
My condition of no overlapping is forbidding the second input from connecting to the second free input of the AND gate (note: simulator is made for Android and the number of inputs is variable, that's why inputs are close to each other). A situation like this is bound to happen sooner or later but I would like to make it as late as possible.
I have tried to:
introdue "int turnNumber" to count how many turns have been made so far and punish paths that make too many turns. (algorithm takes too long to complete, sometimes very very long)
calculate Manhattan distance from the start to the end. Divide that number by two and then remove the "double cost" punishment from nodes whose heuristics are near that middle. (for some situations algorithm fell into infinite loop)
Are there any ideas on how to redistribute turns in the middle so as many as possible connections can be made between logic gates while still satisfying the "min turn" condition?
In case you'd like to see the code: https://gist.github.com/linaran/c8c493bb54cfca764aeb
Note: The canvas that I'm working with isn't bounded.
EDIT: method for calculating cost and heuristics are -- "calculateCost" and "manhattan"

1. You wrote that you already tried swap start/end position
but my guts tell me if you compute path from In to Out
then the turn is near Output which is in most cases OK because most Gates has single output.
2. Anyway I would change your turn cost policy a bit:
let P0,P1 be the path endpoints
let Pm=(P0+P1)/2 be the mid point
so you want the turn be as close to the mid point as it can be
so change turn cost tc to be dependent to the distance from Pm
tc(x,y)=const0+const1*|(x,y)-Pm|
that should do the trick (but I didn't test this so handle with prejudice)
it could create some weird patterns so try euclidean and manhatan distances
and chose the one with better results
3. Another approach is
fill the map from both start and end points at once
And stop when they meet
you need to distinguish between costs from start and end point
so either use negative for star and positive values for end point origin
or allocate ranges for the two (range should be larger then map size xs*ys in cells/pixels)
or add some flag value to the cost inside map cell
4. you can mix 1. and 2. together
so compute Pm
find nearest free point to Pm and let it be P2
and solve path P0->P2 and P1->P2
so the turns will be near P2 which is near Pm which is desired
[notes]
the 3th approach is the most robust one and should lead to desired results

Guide on how to solve Rush Hour Puzzle in Java obtaining lowest number of moves using A* . Need help on how to start and steps to follow

Firstly, I have read every thread that I could find on stackoverflow or other internet searching. I did learn about different aspects, but it isn't exactly what I need.
I need to solve a Rush Hour puzzle of size no larger than 8 X 8 tiles.
As I have stated in title I want to use A*, as a heuristic for it I was going to use :
number of cars blocking the red car's ( the one that needs to be taken out ) path should decrease or stay the same.
I have read the BFS solution for Rush hour.
I don't know how to start or better said, what steps to follow.
In case anyone needs any explanation, here is the link to the task :
http://www.cs.princeton.edu/courses/archive/fall04/cos402/assignments/rushhour/index.html
So far from what have I read ( especially from polygenelubricants's answer ) I need to generate a graph of stages including initial one and "succes" one and determine the minimum path from initial to final using A* algorithm ?
Should I create a backtracking function to generate all the possible ( valid ) moves ?
As I have previously stated, I need help on outlining the steps I need to take rather than having issues with the implementation.
Edit : Do I need to generate all the possible moves so I convert them into graph nodes, isn't that time consuming ? I need to solve a 8X8 puzzle in less than 10 seconds

A* is an algorithm for searching graphs. Graphs consist of nodes and edges. So we need to represent your problem as a graph.
We can call each possible state of the puzzle a node. Two nodes have an edge between them if they can be reached from each other using exactly one move.
Now we need a start node and an end node. Which puzzle-states would represent our start- and end-nodes?
Finally, A* requires one more thing: an admissable distance heuristic - a guess at how many moves the puzzle will take to complete. The only restriction for this guess is that it must be less than the actual number of moves, so actually what we're looking for is a minimum-bound. Setting the heuristic to 0 would satisfy this, but if we can come up with a better minimum-bound, the algorithm will run faster. Can you come up with a minimum-bound on the number of moves the puzzle will take to complete?

Find location using only distance and bearing?

Triangulation works by checking your angle to three KNOWN targets.
"I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and it's to my right at 90 degrees." Repeat 2 more times for different targets and angles.
Trilateration works by checking your distance from three KNOWN targets.
"I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and I'm 100 meters away from that." Repeat 2 more times for different targets and ranges.
But both of those methods rely on knowing WHAT you're looking at.
Say you're in a forest and you can't differentiate between trees, but you know where key trees are. These trees have been hand picked as "landmarks."
You have a robot moving through that forest slowly.
Do you know of any ways to determine location based solely off of angle and range, exploiting geometry between landmarks? Note, you will see other trees as well, so you won't know which trees are key trees. Ignore the fact that a target may be occluded. Our pre-algorithm takes care of that.
1) If this exists, what's it called? I can't find anything.
2) What do you think the odds are of having two identical location 'hits?' I imagine it's fairly rare.
3) If there are two identical location 'hits,' how can I determine my exact location after I move the robot next. (I assume the chances of having 2 occurrences of EXACT angles in a row, after I reposition the robot, would be statistically impossible, barring a forest growing in rows like corn). Would I just calculate the position again and hope for the best? Or would I somehow incorporate my previous position estimate into my next guess?
If this exists, I'd like to read about it, and if not, develop it as a side project. I just don't have time to reinvent the wheel right now, nor have the time to implement this from scratch. So if it doesn't exist, I'll have to figure out another way to localize the robot since that's not the aim of this research, if it does, lets hope it's semi-easy.

Great question.
The name of the problem you're investigating is localization, and it, together with mapping, are two of the most important and challenging problems in robotics at the moment. Put simply, localization is the problem of "given some sensor observations how do I know where I am?"
Landmark identification is one of the hidden 'tricks' that underpin so much of the practice of robotics. If it isn't possible to uniquely identify a landmark, you can end up with a high proportion of misinformation, particularly given that real sensors are stochastic (ie/ there will be some uncertainty associate with the result). Your choice of an appropriate localisation method, will almost certainly depend on how well you can uniquely identify a landmark, or associate patterns of landmarks with a map.
The simplest method of self-localization in many cases is Monte Carlo localization. One common way to implement this is by using particle filters. The advantage of this is that they cope well when you don't have great models of motion, sensor capability and need something robust that can deal with unexpected effects (like moving obstacles or landmark obscuration). A particle represents one possible state of the vehicle. Initially particles are uniformly distributed, as the vehicle moves and add more sensor observations are incorporated. Particle states are updated to move away from unlikely states - in the example given, particles would move away from areas where the range / bearings don't match what should be visible from the current position estimate. Given sufficient time and observations particles tend to clump together into areas where there is a high probability of the vehicle being located. Look up the work of Sebastian Thrun, particularly the book "probabilistic robotics".

What you're looking for is Monte Carlo localization (also known as a particle filter). Here's a good resource on the subject.
Or nearly anything from the probabilistic robotics crowd, Dellaert, Thrun, Burgard or Fox. If you're feeling ambitious, you could try to go for a full SLAM solution - a bunch of libraries are posted here.
Or if you're really really ambitious, you could implement from first principles using Factor Graphs.

I assume you want to start by turning on the robot inside the forest. I further assume that the robot can calculate the position of every tree using angle and distance.
Then you can identify the landmarks by iterating through the trees and calculating the distance to all its neighbours. In Matlab you can use pdist to get a list of all (unique) pairwise distances.
Then you can iterate through the trees to identify landmarks. For every tree, compare the distances to all its neighbours to the known distances between landmarks. Whenever you find a candidate landmark, you check its possible landmark neighbours for the correct distance signature. Since you say that you always should be able to see five landmarks at any given time, you will be trying to match 20 distances, so I'd say that the chance of false positives is not too high. If the candidate landmark and its candidate fellow landmarks do not match the complete relative distance pattern, you go check the next tree.
Once you have found all the landmarks, you simply triangulate.
Note that depending on how accurately you can measure angles and distances, you need to be able to see more landmark trees at any given time. My guess is that you need to space landmarks with sufficiently density that you can see at least three at a time if you have high measurement accuracy.

I guess you need only distance to two landmarks and the order of seeing them (i.e. from left to right you see point A and B)

(1) "Robotic mapping" and "perceptual aliasing".
(2) Two identical hits are inevitable. Since the robot can only distinguish between a finite number X of distinguishable tree configurations, even if the configurations are completely random, there is almost certainly at least one location that looks "the same" as some other location even if you encounter far fewer than X/2 different trees. Those are called "birthday paradox collisions". You may be lucky that the particular location you are at is in fact actually unique, but I wouldn't bet my robot on it.
So you:
(a) have a map of a large area with
some, but not all trees on it.
(b) a
robot somewhere in the actual forest
that, without looking at the map, has
looked at the nearby trees and
generated an internal map of a all
the trees in a tiny area and its
relative position to them
(c) To the
robot, every tree looks the same as
every other tree.
You want to find: Where is the robot on the large map?
If only each actual tree had a unique name written on it that the robot could read, and then (some of) those trees and their names were on the map, this would be trivial.
One approach is to attach a (not necessarily unique) "signature" to each tree that describes its position relative to nearby trees.
Then, as you travel along, the robot drives up to a tree and finds a "signature" for that tree, and you find all the trees on the map that "match" that signature.
If only one unique tree on the map matches, then the tree the robot is looking might be that tree on the map (yay, you know where the robot is) -- put down a weighty but tentative dot on the map at the robot's relative position to the matching tree -- the tree the robot is next to is certainly not any of the other trees on the map.
If several of the trees on the map match -- they all have the same non-unique signature -- then you could put some less-weighty tentative dots on the map at the robots position relative to each one of them.
Alas, even if find one or more matches, it is still possible that the tree the robot is looking at is not on the map at all, and the signature of that tree is coincidentally the same as one or more trees on the map, and so the robot could be anywhere on the map.
If none of the trees on the map matches, then the tree the robot is looking at is definitely not on the map. (Perhaps later on, once the robot knows exactly where it is, it should start adding these trees to the map?)
As you drive down the path, you push the dots in your estimated direction and speed of travel.
Then as you inspect other trees, possibly after driving down the path a little further, you eventually have lots of dots on the map, and hopefully one heavy, highly overlapping cluster at the actual position, and hopefully each other dot is an easily-ignored isolated coincidences.
The simplest signature is a list of distances from a particular tree to nearby trees.
A particular tree on the map is "matched" to a particular tree in the forest when, for each and every nearby tree on the map, there is a corresponding nearby tree in the forest at "the same" distance, as far as you can tell with your known distance and angular errors.
(By "nearby", I mean "close enough that the robot should be able to definitely confirm that the tree is actually there", although it's probably simpler to approximate this with something like "My robot can see all trees out to a range of R, so I'm only going to bother even trying to match trees that are within a circle of R*1/3 from my robot, and my list of distances only include trees that are within a circle of R*2/3 from the particular tree I'm trying to match").
If you know your north-south orientation even very roughly, you can create signatures that are "more unique", i.e., have fewer spurious matches on the map and (hopefully) in the real forest.
A "match" for the tree the robot is next to occurs when, for each nearby tree on the map, there is a corresponding tree in the forest at "the same" distance and direction, as far as you can tell with your known distance and angular errors.
Say you see that tree "Fred" on the map has another tree 10 meters in the N to W quadrant from it, but the robot is next to a tree that definitely doesn't have any trees at that distance in the N to W quadrant, but it has a tree 10 meters away to the South.
In that case, then (using a more complex signature) you can definitely tell the robot is not next to Fred, even though the simple signature would give a (false) match.
Another approach:
The "digital paper" solves a similar problem ... Can you plant a few trees in a pattern that is specifically designed to be easily recognized?

Why does A* path finding sometimes go in straight lines and sometimes diagonals? (Java)

I'm in the process of developing a simple 2d grid based sim game, and have fully functional path finding.
I used the answer found in my previous question as my basis for implementing A* path finding. (Pathfinding 2D Java game?).
To show you really what I'm asking, I need to show you this video screen capture that I made.
I was just testing to see how the person would move to a location and back again, and this was the result...
http://www.screenjelly.com/watch/Bd7d7pObyFo
Different choice of path depending on the direction, an unexpected result. Any ideas?

If you're looking for a simple-ish solution, may I suggest a bit of randomization?
What I mean is this: in the cokeandcode code example, there is the nested-for-loops that generate the "successor states" (to use the AI term). I refer to the point where it loops over the 3x3 square around the "current" state, adding new locations on the pile to consider.
A relatively simple fix would (should :)) be isolate that code a bit, and have it, say, generated a linkedlist of nodes before the rest of the processing step. Then Containers.Shuffle (or is it Generics.Shuffle?) that linked list, and continue the processing there. Basically, have a routine say,
"createNaiveNeighbors(node)"
that returns a LinkedList = {(node.x-1,node.y), (node.x, node.y-1)... } (please pardon the pidgin Java, I'm trying (and always failing) to be brief.
Once you build the linked list, however, you should just be able to do a "for (Node n : myNewLinkedList)" instead of the
for (int x=-1;x<2;x++) {
for (int y=-1;y<2;y++) {
And still use the exact same body code!
What this would do, ideally, is sort of "shake up" the order of nodes considered, and create paths closer to the diagonal, but without having to change the heuristic. The paths will still be the most efficient, but usually closer to the diagonal.
The downside is, of course, if you go from A to B multiple times, a different path may be taken. If that is unnacceptable, you may need to consider a more drastic modification.
Hope this helps!
-Agor

Both of the paths are of the same length, so the algorithm is doing its job just fine - it's finding a shortest path. However the A* algorithm doesn't specify WHICH shortest path it will take. Implementations normally take the "first" shortest path. Without seeing yours, it's impossible to know exactly why, but if you want the same results each time you're going to have to add priority rules of some sort (so that you're desired path comes up first in the search).

The reason why is actually pretty simple: the path will always try to have the lowest heuristic possible because it searches in a greedy manner. Going closer to the goal is an optimal path.
If you allowed diagonal movement, this wouldn't happen.

The reason is the path you want the algorithm to go.
I don't know the heuristic your A* uses but in the first case it has to go to the end of the tunnel first and then plans the way from the end of the tunnel to the target.
In the second case the simplest moves to the targets are going down till it hits the wall and then it plans the way from the wall to the target.
Most A* I know work with a line of sight heuristic or a Manhattan Distance in the case of a block world. This heuristics give you the shortest way but in case of obstacles that force to go a way that is different from the line of sight the ways depend on your starting point.
The algorithm will go the line of sight as long as possible.

The most likely answer is that going straight south gets it closest to its goal first; going the opposite way, this is not a choice, so it optimizes the sub-path piecewise with the result that alternating up/across moves are seen as best.
If you want it to go along the diagonal going back, you are going to have to identify some points of interest along the path (for example the mouth of the tunnel) and take those into account in your heuristic. Alternatively, you could take them into account in your algorithm by re-computing any sub-path that passes through a point of interest.
Back in the day they used to do a pre-compiled static analysis of maps and placed pathfinding markers at chokepoints. Depending on what your final target is, that might be a good idea here as well.
If you're really interested in learning what's going on, I'd suggest rendering the steps of the A* search. Given your question, it might be very eye-opening for you.

In each case it's preferring the path that takes it closer to its goal node sooner, which is what A* is designed for.

If I saw right, the sphere is moving first to the right in a straigt line, because it cannot got directly toward the goal (path is blocked).
Then, it goes in a straight line toward the goal. It only looks diagonal.

Does your search look in the 'down' direction first? This might explain the algorithm. Try changing it to look 'up' first and I bet you would see the opposite behavior.

Depending on the implementation of your astar you will see different results with the same heuristic, as many people have mentioned. This is because of ties, when two or more paths tie the way you order your open set will determine the way the final path will look. You will always get the optimal path if you have an admissible heuristic, but the nodes visited will increase with the number of ties you have(relative to a heuristic producing not as many ties).
If you dont think visiting more nodes is a problem i would suggest using the randomization (which is your current accepted answer) suggestion. If you think searching more nodes is a problem and want to optimize i would suggest using some sort of tiebreaker. It seems you are using manhattan distance, if you use euclidian distance when two nodes tie as a tiebreaker you will get more straight paths to the goal and you will visit fewer nodes. This is ofcourse given no traps or block of line of sight to the goal.
To avoid visiting nodes with blocking elements in the line of sight path i would suggest finding a heuristic which takes into account these blocking elements. Ofcourse a new heuristic shouldnt do more work than a normal A star search would do.
I would suggest looking at my question as it might produce some ideas and solutions to this problem.