So basically, I have a warehouse represented by a graph and each node in it contains a certain amount of 3 objects (A, B, C). I have to use Dijkstra to find the shortest path the robot should take in order to take an amount of each item provided as the input and minimize the time.
Also, each time the robot picks an object, the robot's speed goes slower so the time it takes for him to travel a vertice isn't equal to its distance anymore. The given equation is Time = Distance * k where k is a constant associated with the robot (k= 1 + mass carried) and type A objects have a mass of 1kg, B objects of 3kg and C objects of 5kg.
My question is how can I modify or use the Dijkstra's algorithm given that I have to take into account the objects that I have to pick and the decrease of speed.
Thanks in advance!
When calculating the cost to go to a node, have another variable to account for time such that more items result in a larger number meaning more costly. So the total cost to go to a node is the sum of the cost between nodes and the variable accounting for time. The rest of Dijkstra should still work.
Related
folks! Apologies if this is a duplicate question and I've done some research on the topic but don't know if I'm heading in the right direction.
I have converted gridded data of population density to a MongoDB collection using a geometry object defining the population density cell as a five node polygon (the fifth node matching the first) and a float value consisting of the population in that geographic region. Even though the database is huge in size, I can quickly retrieve the "records" of the population regions as they are indexed as a 2D Sphere when it intersects a geo-polygon indicating some type of weather event or other geofence polygon.
The issue comes when I try to add all of the boxes up. It takes an exceedingly long amount of time, especially if the polygon is of a significant geographic area. The population data I have are 1km^2 cells. The adding of the data can take several seconds or, in worse case scenario, minutes!
I had a thought of creating a type of quadtree structure in the database by a lower resolution node set as a separate collection and so on and so on. Then when calculating population, I could start with the lowest res set and work my way down the node "tree" by making several database calls until there are no more matches. While I'd increase my database calls significantly, I'd reduce the sheer number of elements that I would need to add up at the end - which is taking the most computational time.
I could try to create these data using bottom-up neighbor finding whilst adding up the four population values that would make up the next lower-resolution node set. This, of course, will explode the database size and will increase the number of queries to the database for a single population request.
I haven't seen too much of this done with databases. I'd like to have it in a database (could also be PostgreSQL) since it gives me the ability to quickly geo-query by point or area. And, I'm returning the result as an API call so the efficiency of time is of the essence!
Any advice or places to research would be greatly appreciated!!!
I have a problem with my android app, I have x value (whatever it is) and I have data in the database, I want to compare the value of x with all the data in the database at the same time in real time
the app is using sqlite.
I used a loop but when the database is large in this case my app lags in comparing all the data.my code is
public void Check_Distance(Location Current_Location,ArrayList<Location> LocationArrayList1)
{
double Distance;
for(int i=0;i<LocationArrayList1.size();i++)
{
Distance=distanceBetween(Current_Location,LocationArrayList1.get(i));
if(Distance<=0.1*1000){ // if distance is less then 100m give a sound
Notification_Sound();
}
}
}
You can't look at every record in the database at the exact same time. That's called quantum computing, and is currently an active research area where people far smarter than you or I are currently spending millions of dollars to try and create a machine that can do this kind of parallel processing.
That being said, you can make your algorithm more efficient, but that takes some effort to do. Both of the below are based on the idea of eliminating the majority of the locations that are obviously too far away very quickly, and performing more in-depth checks on those that could be in range.
One method is to sort the locations in ascending order in two arrays - one by North/South and the other by East/West. Find all entries within a given distance of the current position in each list, then combine the results to get a list of points within a box of X distance from the location. This box will have a much smaller number of points within it that you can then apply an iterative, circular, distance based approach to.
Another is to create a quadtree. This would subdivide the map area into a set of bounding volumes, where each volume would have a set of points, or additional bounding volumes. You can then place down your search area and find all the quadtree volumes that intersect with your circular search area, greatly minimizing the number of locations you need to do a true distance check on.
This question combines math and programming. I will first describe the general problem and then give an example that is (hopefully) simpler to understand.
General Question: Consider a Markov-chain process of N-states with transition matrix Π. Each state is associated with a value x_n (n in {1,…,n}). Our goal is to find the unconditional average of the first two moments (mean and var) along T-period paths conditional on (i) the path starts in a subset of states, N_0, (ii) it ends in a subset of states, N_T, and (iii) it is not going through a subset of states, N_not, in any of the periods between 1 to T-1. By saying we are interested in the unconditional average of these two moments, I basically mean what would be the average of these two moments in the stationary distribution. To be more concrete, let me illustrate the goal of the exercise in a simple case.
Simple Example: Consider a 3-state Markov-chain process with transition matrix Π, and let the three state be denoted by A, B, and C. Each of these states are associated with some value (x_A, x_B, and x_C), respectively. We are interested in what happens along paths that satisfy the following condition. The path starts at point A, after 3 periods are in either points B or C, and between periods 1 to 3 never go again through point A. Denote this condition by (#). So, for example, a path which we are interested in would be {A,B,B,C} with the associated values {x_A, x_B, x_B, x_C}. We are interested in the average and standard deviation along such paths. In particular, we would like to find the unconditional average of these first two moments in paths that satisfy (#).
Let me now propose a solution based on simulating the process. Since both T and N are quite large, this solution is too slow for my purpose.
Simulation Solution: Starting from some initial point simulate the process for a very long time period, and drop the first τ periods. Extract all paths along the simulation that satisfy condition (#) and compute the mean and std along each of these paths. Finally, simply take the average across these paths.
I’m hoping there is a better and more efficient way to achieve the goal. Since I want the solution to be accurate and the size of T and N the simulation takes a long time.
I would love to hear your thoughts and if you know of efficient methods to achieve this goal. Please let me know if something is not clear and I'll try to clarify it.
Thank you!!!
I think I know how to do this if N_0 consists of one state, let's call that state A.
The long run probability of being in A is pi(A) and can be obtained by solving pi = pi*P, with P the transition matrix.
The other thing you need to calculate is the probability of those transient paths. You probably need to introduce a modified P, where all states i in the set N_not are absorbing (i.e. P[i,i]=1 and P[i,j]=0 for j is not i). Then starting from a vector p(0) which has a 1 in the element corresponding to state A and 0 otherwise, you can keep calculating p(n) = p(n-1)*P to get the probabilities of your transient paths.
Multiply the result of that by pi(A) to get the unconditional probability.
You can probably do something like this as well when N_0 is a set, but I don't know how you should select p(0) in that case.
In the Kademlia paper by Petar Maymounkov and David Mazières, it is said that the XOR distance is a valid non-Euclidian metric with limited explanations as to why each of the properties of a valid metric are necessary or interesting, namely:
d(x,x) = 0
d(x,y) > 0, if x != y
forall x,y : d(x,y) = d(y,x) -- symmetry
d(x,z) <= d(x,y) + d(y,z) -- triangle inequality
Why is it important for a metric to have these properties in general? Why is each of these properties necessary in the context of routing queries in the Kademlia Distributed Hash Table implementation?
In addition, the paper mentions that unidirectionality (for a given x, and a distance l, there exist only a single y for which d(x,y) = l) guarantees that all queries will converge along the same path. Why is that so?
I can only speak for Kademlia, maybe someone else can provide a more general answer. In the meantime...
d(x,x) = 0
d(x,y) > 0, if x != y
These two points together effectively mean that the closest point to x is x itself; every other point is further away. (This may seem intuitive, but other aspects of the XOR metric aren't.)
In the context of Kademlia, this is important since a lookup for node with ID x will yield that node as the closest. It would be awkward if that were not the case, since a search converging towards x might not find node x.
forall x,y : d(x,y) = d(y,x)
The structure of the Kademlia routing table is such that nodes maintain detailed knowledge of the address space closest to them, and exponentially decreasing knowledge of more distant address space. In short, a node tries to keep all the k closest contacts it hears about.
The symmetry is useful since it means that each of these closest contacts will be maintaining detailed knowledge of a similar part of the address space, rather than a remote part.
If we didn't have this property, it might be helpful to think of the search as more like the hands of a clock moving in one direction round a clockface. The node at 1 o'clock (Node1) is close to Node2 at 2 o'clock (30°), but Node2 is far from Node1 (330°). So imagine we're looking for the two closest to 3 o'clock (i.e. Node1 and Node2). If the search reaches Node2, it won't know about Node1 since it's far away. The whole lookup and topology would have to change.
d(x,z) <= d(x,y) + d(y,z)
If this weren't the case, it would be impossible for a node to know which contacts from its routing table to return during a lookup. It would know the k closest to the target, but there would be no guarantee that one of the other more distant contacts wouldn't yield a shorter overall path.
Because of this property and unidirectionality, different searches starting from vastly separated points will tend to converge down the same path.
The unidirectionality means that no two nodes can have the same distance from a given point. If that weren't the case, then the target point could be encircled by a bunch of nodes all the same distance from it. Then various different searches would be free pick any of those to pass through. However, unidirectionality guarantees that exactly one of this bunch will be the closest, and any search which chooses between this group will always select the same one.
I've been bashing my head on this for quite some time: how can the XOR - as in the number of differing bits, a proper Hamming distance - be the basis of a total order?
Well it can't, such a metric on its own is not enough for a comparable relationship, all it can do is dump nodes in circles around a point.
Then I read the paper more closely and noticed that it says "the XOR as an integer value" and it dawned on me: the crux is not the "XOR metric", but the length of the common prefix of the ID (of which XOR is a derivation mechanism.)
Take two nodes with the same Hamming distance from "self" and the length of their prefix common to "self": the one with shortest common prefix is the furthest node.
The paper uses "XOR distance metric" but it really should read "ID prefix length total ordering"
I think this may explain it a wee bit, let me know http://metaquestions.me/2014/08/01/shortest-distance-between-two-points-is-not-always-a-straight-line/
Basically each hop if it were only one bit at a time in a fully populated network (extreme) then would have twice the knowledge of the previous hop. As you converge the knowledge is greater until you get to the closest nodes whose knowledge is ultimate in the network.
ByteLand
Byteland consists of N cities numbered 1..N. There are M roads connecting some pairs of cities. There are two army divisions, A and B, which protect the kingdom. Each city is either protected by army division A or by army division B.
You are the ruler of an enemy kingdom and have devised a plan to destroy Byteland. Your plan is to destroy all the roads in Byteland disrupting all communication. If you attack any road, the armies from both the cities that the road connects comes for its defense. You realize that your attack will fail if there are soldiers from both armies A and B defending any road.
So you decide that before carrying out this plan, you will attack some of the cities and defeat the army located in the city to make your plan possible. However, this is considerably more difficult. You have estimated that defeating the army located in city i will take up ci amount of resources. Your aim now is to decide which cities to attack so that your cost is minimum and no road should be protected from both armies A and B.
----Please tell me if this approach is correct----
We need to sort the cities in terms of resources required to destroy the city. For each city we need to ask the following questions:
1) Did deletion of the previous city NOT result into a state which can destroy Byteland?
2) Does it connect any road?
3) Does it connect any road which is armed by a different city?
If all of these conditions are true, we'll proceed towards destroying the city and record the total cost incurred so far and also determine if destruction of this city will lead to overall destruction of Byteland.
Since the cities are arranged in increasing order of the cost incurred, we can stop wherever we find the desired set of deletions.
You need only care about roads that link two cities with different armies - links between A and B or links between B and A, so let's delete all links from A to A or B to B.
You want to find a set of points such that each link has at least one point on it, which is a minimum weight vertex cover. On an arbitrary graph this would be NP-complete. However, your graph only ever has nodes of type A linked to nodes of type B, or the reverse - it is a bipartite graph with these two types of nodes as the two parties. So you can find a minimum weight vertex cover by using an algorithm for finding minimum weight vertex covers on bipartite graphs. Searching for this, I find e.g. http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-854j-advanced-algorithms-fall-2008/assignments/sol5.pdf
mcdowella,
But the vertices have a cost to them and the minimum vertex cover would not produce the right vertices to remove. Imagine 2 vertices (A army) pointing to the third one (B). First two vertices cost 1 each, where the third one costs 5. A minimum vertex cover would return the third one - but removing the third one costs more than removing both nodes with cost 1 + 1.
We would probably need some modified version of a minimum vertex cover here.