I want to solve a problem similar to N-Queens one, but:
all chess pieces are available
user inputs how many pieces of what kind are to be placed (e.g. 3 rooks, 4 knights, 1 bishop)
I'm lying on the floor for some time now, but can't come up with how to adjust the backtracking algorithm for this purpose. I will be very grateful for any kind of help.
In principle, the same approach as in the classical N-Queen problem should work:
Find an empty, non-attacked square where you can place your next piece
Search the new position recursively (or if you already placed all your pieces, output the solution)
Take back the last placed piece and repeat (goto step 1) until you have tried all squares where you can place the next piece
The only difference to the classical N-Queens problem is that the different pieces have different attack patterns. And some common optimizations might no longer work. For instance, if you have pawns, it breaks symmetry as they only attack the squares in front of them. (Though you still have one symmetry axis even with pawns.)
I would expect the backtracking algorithm to be more efficient if you start with placing the pieces first that cover the most squares: first queens, then rooks and bishops, then knights and kings, and finally pawns.
Related
I am working on developing a shape descriptor module for crack clustering of binary cracks. Long story short, I have engineered several features. To arrange my feature vector for images that contain several cracks, as the biggest crack normally (not always) defines the crack type, I weigh the features based on the area each crack occupies (the number of pixels) and then, take the average.
Take as an example the below image:
enter image description here
The horizontal one decides that this crack image is a transverse crack and at the same time, I can't ignore the one in the corner.
My question is that what is the best way to deal with this? I feel like my feature vector should be multi-dimensional to be able to cover the multi-crack ones, but I haven't seen a problem like this.
Note 1: refining the images and removing the small ones is not an option cuz in some other types, their presence tells a different story. Example:
enter image description here
Here the little ones define that this crack is a meandering one.
Note 2: From an engineering point of view, the simpler the better.
Note 3: The number of cracks in one image can go up to 18, and at the same time, the majority of my data contain one crack.
I am working on what is likely a unique use case - I want to use Skyfield to do some calculations on a hypothetical star system. I would do this by creating my own ephemeris, and using that instead of the actual one. The problem i am finding is that I cannot find documentation on the API to replace the ephemerides with my own.
Is there documentation? Is skyfield something flexible enough to do what I am trying?
Edit:
To clarify what I am asking, I understand that I will have to do some gravitational modeling (and I am perfectly willing to configure every computer, tablet, cable box and toaster in this house to crunch on those numbers for a few days :), but before I really dive into it, I wanted to know what the data looks like. If it is just a module with a number of named numpy 2d arrays... that makes it rather easy, but I didn't see this documented anywhere.
The JPL-issued ephemerides used by Skyfield, like DE405 and DE406 and DE421, simply provide a big table of numbers for each planet. For example, Neptune’s position might be specified in 7-day increments, where for each 7-day period from the beginning to the end of the ephemeris the table provides a set of polynomial coefficients that can be used to estimate Neptune's position at any moment from the beginning to the end of that 7-day period. The polynomials are designed, if I understand correctly, so that their first and second derivative meshes smoothly with the previous and following 7-day polynomial at the moment where one ends and the next begins.
The JPL generates these huge tables by taking the positions of the planets as we have recorded them over human history, taking the rules by which we think an ideal planet would move given gravitational theory, the drag of the solar wind, the planet's own rotation and dynamics, its satellites, and so forth, and trying to choose a “real path” for the planet that agrees with theory while passing as close to the actual observed positions as best as it can.
This is a big computational problem that, I take it, requires quite a bit of finesse. If you cannot match all of the observations perfectly — which you never can — then you have to decide which ones to prioritize, and which ones are probably not as accurate to begin with.
For a hypothetical system, you are going to have to start from scratch by doing (probably?) a gravitational dynamics simulation. There are, if I understand correctly, several possible approaches that are documented in the various textbooks on the subject. Whichever one you choose should let you generate x,y,z positions for your hypothetical planets, and you would probably instantiate these in Skyfield as ICRS positions if you then wanted to use Skyfield to compute distances, observations, or to draw diagrams.
Though I have not myself used it, I have seen good reviews of:
http://www.amazon.com/Solar-System-Dynamics-Carl-Murray/dp/0521575974
I needed some help with a problem I'd been assigned in class. It's our introduction to for loops. Here is the problem:
Consider the following riddle.
This is all I have so far:
function pile = IslandBananas(numpeople, numbears)
for pilesize=1:10000000
end
I would really appreciate your input. Thank you!
I will help you, but you need to try harder than that. And also, you only need one for loop. First, think about how you would construct this algorithm. Well you know you have to use a for loop so that is a start. So let's think about what is going on in the problem.
1) You have a pile.
2) First night someone takes the pile and divides it into 3 and finds that one is left over, this means mod(pile,3) = 1.
3) But he discards the extra banana. This means (pile-1).
4) He takes a third of it, leaving two-thirds left. This means (2/3)*(pile-1).
5) In the morning they take the pile and divide it into 3 and find again that one is left over, so this means mod((2/3)*(pile-1),3) = 1.
6) But they discard the extra banana. This means (2/3)*(pile-1)-1.
7) Finally, they have to each have at least one banana if it is to be the smallest pile possible. Thus, the smallest pile must be such that (1/3)*((2/3)*(pile-1)-1) = 1.
I have essentially given you the answer, the rest you can write with the formula (1/3)*((2/3)*(pile-1)-1) and a simple if statement to test for the smallest possible integer which is 1. This can be done in four lines inside of your for loop.
Now, expanding this to any number of people and any number of bears requires two simple substitutions in that formula! If your teacher demands it, this can easily be split into two nested for loops.
I'm doing a map-coloring problem with Scheme, and I used minimum remaining values (Select the vertex with the fewest legal colors) and degree heuristics select the vertex that has the largest number of neighbors). If there exists a solution for a certain configuration, will these heuristics ensures that it won't need to backtrack?
Let's do a simple theoretical analysis.
Graph coloring is NP-complete for general graphs (if not asking for a coloring with less than 4 colors). This means there exists no known polynomial time algorithm.
Your heuristic is computable in polynomial time.
Assuming you need no backtracking, then you need to make n steps, each of which requires polynomial time (n is number of vertices). Thus you can solve the problem in polynomial time.
Either you have proven P=NP or your assumption is wrong.
I leave it up to you to decide upon which option in point (4) is more plausible.
In general: no, MRV and your other heuristic will not guarantee a straight walk to the goal. (I imagine they might if your problem has some very specific structure, but don't count on it until you've seen the theorem.)
Heuristics prune the search space, or change the order of the search to make an early termination more likely. This is not the same thing as backtracking.
But it's a related concept.
We prune some spaces because we are confident that the solution does not lie in those branches of the search tree, or change the order because we have some reason to believe that it will be quicker if we look in some subtrees before others.
We also cut ourselves off from backtracking because we are confident that the solution is in the branch of the space we are in now (so that if we don't find it in this subtree, we can declare failure and don't bother).
Both kinds of strategies are ultimately about searching less of the space somehow and getting to the answer (positive or negative) without searching everything.
MRV and the degrees heuristic are about reordering the sub-searches, not about avoiding backtracking. Heuristics can be right and make a short search but that's not the same
thing as eliminating backtracking (e.g. the "cut" operator in Prolog). When you find what you're looking for, you can declare success, and of course that eliminates further backtracking. But real backtracking elimination means making a decision not to backtrack no matter what, before the search completes.
E.g. if you're doing a depth-first search, and you find what you're looking for by dumb luck without backtracking, we cannot say that dumb luck is a fence operation that eliminates backtracking. :)
I'm currently writing an optimization algorithm in MATLAB, at which I completely suck, therefore I could really use your help. I'm really struggling to find a good way of representing a graph (or well more like a tree with several roots) which would look more or less like this:
alt text http://img100.imageshack.us/img100/3232/graphe.png
Basically 11/12/13 are our roots (stage 0), 2x is stage1, 3x stage2 and 4x stage3. As you can see nodes from stageX are only connected to several nodes from stage(X+1) (so they don't have to be connected to all of them).
Important: each node has to hold several values (at least 3-4), one will be it's number and at least two other variables (which will be used to optimize the decisions).
I do have a simple representation using matrices but it's really hard to maintain, so I was wondering is there a good way to do it?
Second question: when I'm done with that representation I need to calculate how good each route (from roots to the end) is (like let's say I need to compare is 11-21-31-41 the best or is 11-21-31-42 better) to do that I will be using the variables that each node holds. But the values will have to be calculated recursively, let's say we start at 11 but to calcultate how good 11-21-31-41 is we first need to go to 41, do some calculations, go to 31, do some calculations, go to 21 do some calculations and then we can calculate 11 using all the previous calculations. Same with 11-21-31-42 (we start with 42 then 31->21->11). I need to check all the possible routes that way. And here's the question, how to do it? Maybe a BFS/DFS? But I'm not quite sure how to store all the results.
Those are some lengthy questions, but I hope I'm not asking you for doing my homework (as I got all the algorithms, it's just that I'm not really good at matlab and my teacher wouldn't let me to do it in java).
Granted, it may not be the most efficient solution, but if you have access to Matlab 2008+, you can define a node class to represent your graph.
The Matlab documentation has a nice example on linked lists, which you can use as a template.
Basically, a node would have a property 'linksTo', which points to the index of the node it links to, and a method to calculate the cost of each of the links (possibly with some additional property that describe each link). Then, all you need is a function that moves down each link, and brings the cost(s) with it when it moves back up.