I am trying to solve COURIER problem on spoj. I am able to understand that I have to solve TSP for this with dynamic programming approach but I am unable to exactly understand that whether my approach for handling multiple parcel between same pair of cities is correct or not. My pseudocode will be somewhat as following:
1) Use floyd warshall to find all pair shortest path in O(n^3). Some pair of cities are connected by more than one roads, I can just keep the shortest one for my undirected graph.
2) Add the shortest cost for each request start to end.
3) Create a new directed graph for each of 12 requests and homecity. The node of this new graph will be a merge of each request's source and destination. The edge weight between a->b can be calculated by shortest path between 'a' request's destination to 'b' request's source.I am thinking of duplicating the pairs if I have multiple request between them.
4) Use a TSP DP to solve this new undirected 13 city TSP problem. O(n^2 2^n) would come around 1384448. Not sure whether this will time out for multiple test cases.
Can you please give your inputs as am I complicating the problem with my approach of creating this new directed graph? I am not using the information that there are only 5 such different requests. I know I can coed this up and know but I want to get some suggestions on solution first.
Nice problem.
After doing point 1), you can ignore all cities that are not source or address of delivery.
Therefore you have 10 cities where the traveler currently is and 2^12 possible combinations of tasks that are still to complete.
You can just do DP with two arguments: current city and deliveries to complete, that you can store with bit mask.
EDIT:
As mentioned you have two arguments: p which tracks current position and mask which tracks which visits you have already done.
Mask works as bit mask: http://en.wikipedia.org/wiki/Mask_%28computing%29
You start with mask 0, which in binary is 000000000000. When you do for example 5th requested travel you change mask to: 000000010000 etc.
You start by calling f(p=0, mask=0).
When you are solving f(p, mask) you have two options. You can move to any other city p2. You can make travel p -> p2 if this is one of travels you haven't done. Out of all these options you have to choose the best one.
This problem is quite tricky and I would suggest first solving easier problems using bit-masks to begin with. You can find some here: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=778
I don't know if you need the answer now or not, but here is what I did :
Initially your approach is correct, you have to apply floyd-warshall
to get shortest distance b/w all the pairs. Now it is a classic dp+bitmask
problem. There are only 12 operations, you have to arrange these 12 operations
such that you get the minimum.
This can be done by using bitmask : 000000000000
You have these 12 states => if you pick one, you can't pick it again. And since
(2^12=4096) it will not be difficult to store such a number.
My dp states were very straight forward : 'mask number' and 'parent'
PARENT-> the last operation you have done
dp[mask][par]
Hope this will help
Related
I have a weighted undirected graph. Given two vertices in that graph that have no path between them, I want to create a path between then by adding edges to the graph, increasing the total weight of the graph by as little as possible. Is there a known algorithm for determining which edges to add?
An analogous problem would be if I have a graph of a country's road system, where there are two cities that are inaccessible by road from each-other, and I want to build the shortest set of new roads that will connect them. There may be other cities in between them that are connected to neither, and if they exist I want to take advantage of them.
Here's a little illustration; red and green are the vertices I want to connect, black lines are existing edges, and the blue line represents the path I want to exist.
Is there a known algorithm that gives the edges that are missing from that path?
You could use the A* algorithm with a weight of zero for existing edges and distance (or whatever cost makes sense) for missing edges.
https://en.wikipedia.org/wiki/A*_search_algorithm
Actually Dijkstra with some preprocessing is best friend for you.
I answered problem that can be similar to this one: What is the time complexity if it needs to revisit visited nodes in BFS?
What I see in common - you want to use as much existing roads as possible. Also you sometimes need to break that and build new road. The point is in setting the proper weights for existing ones and for "possibly new ones".
What would be my approach - lets say, that each 1 km of existing road has cost of 1. You can sum all existing roads in your graph and lets say there are 1000km in total. Then I would preprocess whole graph and from each node(city) I would create path to all other non-diretly-connected cities and each of them would cost 1000 + 1000 per kilometer and then run Dijkstra on it.
It will automatically use as much existing roads as possible and create as less new roads as possible.
Also you can play with that settings a bit to achieve what you want.
Imagine there are two cities that are only 100m away from each other. And there is actually path between them from existing roads that takes 20 000km. With the settings I have suggested you will get 20000km path as the best one (which satisfy need of "dont build any new roads if not necessary"). Sometimes you actually want this. Sometimes you do not. In case you do not, you can think about "ok, if I build like a little of extra road and it dramatically lowers the distance, its still better solution". If this is the case, you can think about lower the price of new roads (like removing the initial cost, or per kilometer cost or both of it - it depends on what you take as best output).
I don't think that there's an accepted algorithm. But you could try and do the following. First run Vitterbi Triangulation, then run a depth first search on this fully connected graph. Take the sum of the new links in the path from A -> B. Remove the longest new link, and repeat. Once the path from A -> B cannot be reached, check the previous solutions to see which of the solutions had the smallest sum of new links.
I have been playing with Hamiltonian paths for some time and have found some cool uses of it. One being a sort of puzzle where the goal is to connect all the nodes to form a hamiltonian path.
So, being a novice programmer I am, I created a very basic brute force graph generator that creates graphs having a hamiltonian path. But the problem arises when I try to increase the graph size to 10x10. Needless to say, by brute force does not works there.
I understand Hamiltonian path is an NP-Complete problem, but is there a method to optimize the graph generation process. Any sort of trick that can create 15x15 grids in reasonable time.
Thank you very much.
You're looking for what's called the "Backbite algorithm" - start with any Hamiltonian path, a trivial one will do (zig-zag back and forth across your grid, or have a spiral).
Then loop a random number of times:
Select either end point
There are at least two and at most four neighboring vertices; randomly select one that is not already the Hamiltonian neighbor of the end point you started with
If the second point was not the other endpoint
a. connect the two points you picked - this will produce a graph that looks like a loop with a tail
b. Find the edge that causes the loop (not the edge you just added, rather one of its neighbors), and delete it (this is the 'backbite' step)
If in step 3 the second point was the other endpoint, join the two (you will now have a Hamiltonian cycle), and delete any other random edge
At each step, the new graph is still a Hamiltonian path.
I am aware that Dijkstra's algorithm can find the minimum distance between two nodes (or in case of a metro - stations). My question though concerns finding the minimum number of transfers between two stations. Moreover, out of all the minimum transfer paths I want the one with the shortest time.
Now in order to find a minimum-transfer path I utilize a specialized BFS applied to metro lines, but it does not guarantee that the path found is the shortest among all other minimum-transfer paths.
I was thinking that perhaps modifying Dijkstra's algorithm might help - by heuristically adding weight (time) for each transfer, such that it would deter the algorithm from making transfer to a different line. But in this case I would need to find the transfer weights empirically.
Addition to the question:
I have been recommended to add a "penalty" to each time the algorithm wants to transfer to a different subway line. Here I explain some of my concerns about that.
I have put off this problem for a few days and got back to it today. After looking at the problem again it looks like doing Dijkstra algorithm on stations and figuring out where the transfer occurs is hard, it's not as obvious as one might think.
Here's an example:
If here I have a partial graph (just 4 stations) and their metro lines: A (red), B (red, blue), C (red), D (blue). Let station A be the source.
And the connections are :
---- D(blue) - B (blue, red) - A (red) - C (red) -----
If I follow the Dijkstra algorithm: initially I place A into the queue, then dequeue A in the 1st iteration and look at its neighbors :
B and C, I update their distances according to the weights A-B and A-C. Now even though B connects two lines, at this point I don't know
if I need to make a transfer at B, so I do not add the "penalty" for a transfer.
Let's say that the distance between A-B < A-C, which causes on the next iteration for B to be dequeued. Its neighbor is D and only at this
point I see that the transfer had to be made at B. But B has already been processed (dequeued). S
So I am not sure how this "delay" in determining the need for transfer would affect the integrity of the algorithm.
Any thoughts?
You can make each of your weights a pair: (# of transfers, time). You can add these weights in the obvious way, and compare them in lexicographic order (compare # of transfers first, use time as the tiebreaker).
Of course, as others have mentioned, using K * (# of transfers) + time for some large enough K produces the same effect as long as you know the maximum time apriori and you don't run out of bits in your weight storage.
I'm going to be describing my solution using the A* Algorithm, which I consider to be an extension (and an improvement -- please don't shoot me) of Dijkstra's Algorithm that is easier to intuitively understand. The basics of it goes like this:
Add the starting path to the priority queue, weighted by distance-so-far + minimum distance to goal
Every iteration, take the lowest weighted path and explode it into every path that is one step from it (discarding paths that wrap around themselves) and put it back into the queue. Stop if you find a path that ends in the goal.
Instead of making your weight simply distance-so-far + minimum-distance-to-goal, you could use two weights: Stops and Distance/Time, compared this way:
Basically, to compare:
Compare stops first, and report this comparison if possible (i.e., if they aren't the same)
If stops are equal, compare distance traveled
And sort your queue this way.
If you've ever played Mario Party, think of stops as Stars and distance as Coins. In the middle of the game, a person with two stars and ten coins is going to be above someone with one star and fifty coins.
Doing this guarantees that the first node you take out of your priority queue will be the level that has the least amount of stops possible.
You have the right idea, but you don't really need to find the transfer weights empirically -- you just have to ensure that the weight for a single transfer is greater than the weight for the longest possible travel time. You should be pretty safe if you give a transfer a weight equivalent to, say, a year of travel time.
As Amadan noted in a comment, it's all about creating right graph. I'll just describe it in more details.
Consider two vertexes (stations) to have edge if they are on a single line. With this graph (and weights 1) you will find minimum number of transitions with Dijkstra.
Now, lets assume that maximum travel time is always less 10000 (use your constant). Then, weight of edge AB (A and B are on one line) is a time_to_travel_between(A, B) + 10000.
Running Dijkstra on such graph will guarantee that minimal number of transitions is used and minimum time is reached in the second place.
update on comment
Let's "prove" it. There're two solution: with 2 transfers and 40 minutes travel time and with 3 transfers and 25 minutes travel time. In first case you travel on 3 lines, so path weight will be 3*10000 + 40. In second: 4*10000 + 25. First solution will be chosen.
I had the same problem as you, until now. I was using Dijkstra. The penalties for transfers is a very good idea indeed and I've been using it for a while now. The main problem is that you cannot use it directly in the weight as you first you have to identify the transfer. And I didn't want to modify the algorithm.
So what I'be been doing, is that each time and you find a transfer, delete the node, add it with the penalty weight and rerun the graph.
But this way I found out that Dijkstra wont work. And this is where I tried Floyd-Warshall which au contraire to Dijkstra compares all possible paths through the graph between each pair of vertices.
It helped me with my problem switching to Floyd-Warshall. Hope it helps you as well.
Its easier to code and lot more easier to implement.
Problem: I have a large collection of points. Each of these points has a list with references to other points with the distance between them already calculated and stored. I need to determine the shortest route that begins from an origin and passes through a specific number of points to any destination.
Ex: I'm on vacation and I'm staying in a specific city. I'm making a ONE WAY trip to see ANY four cities and I want to travel the least distance possible. I cannot visit the same city more than once.
Current solution: Right now I'm just iterating through every possibility manually and storing the shortest path. This works but feels inefficient. Also, this problem will eventually be expanded to include searching from multiple origin points to multiple destination points, so I think that might explode the search space.
What is the better way to search for the shortest route?
Answering to the updated post, your solution of checking every possibility is optimal (at least, noone has discovered better algorithms so far). Yes, that's a travelling salesman, whose essense is not touching every city, but touching every city once. If you don't want to search for best solution possible, you may find it useful to use heuristics that work faster, but allow for limited discrepancy from ideal solution.
For future answerers: Floyd-Warshall algorithm and all Floyd-like variations are inapplicable here.
In generally you should to strict bad variants...
I think you should use some variations of Branch_and_bound method
http://en.wikipedia.org/wiki/Branch_and_bound
Either bredth first search as norheim.se said or Dijkstra's algorithm would be my suggestion as well.
This sounds Travelling Salesman-esque? One solution is to use an optimisation technique such as an evolutionary algorithm. Currently you are doing an exhaustive search, which will get very slow very quickly. But I think this is pretty much a travelling salesman problem and it has been tackled for several decades if not centuries, and such there are several possible ways of attack. Google is your friend.
Perhaps this is what the original poster means by "iterating through each possibility manually and storing the shortest path", but I thought I'd like to make explicit what appears to be a baseline solution.
Assume you already have a two-point shortest path algorithm--this has classical solutions for various kinds of graphs. Assume all distances are nonnegative and d(A->B->C) = d(A->B) + d(B->C).
The essentials are that the path starts at S goes through one of intermediate cities "abcd" and ends with E:
e.g. SabcdE, SacbdE, etc...
With only 4 intermediate cities, you enumerate all 24 permutations. For each permutation use your shortest two-point algorithm to compute the path from head to tail, and its total distance.
Then given the start and ending point, there are 12 possibilities attaching to one of abcd and for each two possibilities for the interior. You've computed these distances already, so you add on the distance from S to the head and the tail to E. Choose minimum. So once you've precomputed the intermediate distances for a fixed set of interior cities you need to do 12 two point shortest path problems for any pair of start and end points.
This obviously scales poorly with increasing number of intermediate cities. It's not clear to me that it could do better unless you impose greater restrictions on the graph structure (is this in a physical Euclidenan space? Triangle inequality?).
My thought example: suppose all intermediate distances between cities are O(1). With no restriction on the graph, then the distance from S to any intermediate city might be 1000 except for one being 1. Same for the tail. So you can force the first city to be visited to be anything. Now, go one layer down, take the first city as the "start point". Apply the same argument: you can make the best path go to any of the following cities by manipulating the distances in the graph.
So it seems that the complexity can't be helped without additional assumptions.
This is the very common and real time situation any one can fall in.Google map user interface gives you the path in the same order, you add in the destination list. it doesn't give you the optimal path though their own Google maps API provide the solution.
Google maps API provides the solution for this. In the request to find out the path you have to provide the flag 'optimizeWaypoints: true,'. The request will seem like this.
var request = {
origin: start,
destination: end,
waypoints: waypts,
optimizeWaypoints: true,
travelMode: google.maps.TravelMode.DRIVING
};
and you can see whole code of the utility in the view source as complete utility is developed in javascript and HTML.
I hope it will help.
It appears that the edges of your graph are bidirectional. In this case, the algorithm you're looking for is Dijkstra's algorithm.
I have a undirected graph with about 100 nodes and about 200 edges. One node is labelled 'start', one is 'end', and there's about a dozen labelled 'mustpass'.
I need to find the shortest path through this graph that starts at 'start', ends at 'end', and passes through all of the 'mustpass' nodes (in any order).
( http://3e.org/local/maize-graph.png / http://3e.org/local/maize-graph.dot.txt is the graph in question - it represents a corn maze in Lancaster, PA)
Everyone else comparing this to the Travelling Salesman Problem probably hasn't read your question carefully. In TSP, the objective is to find the shortest cycle that visits all the vertices (a Hamiltonian cycle) -- it corresponds to having every node labelled 'mustpass'.
In your case, given that you have only about a dozen labelled 'mustpass', and given that 12! is rather small (479001600), you can simply try all permutations of only the 'mustpass' nodes, and look at the shortest path from 'start' to 'end' that visits the 'mustpass' nodes in that order -- it will simply be the concatenation of the shortest paths between every two consecutive nodes in that list.
In other words, first find the shortest distance between each pair of vertices (you can use Dijkstra's algorithm or others, but with those small numbers (100 nodes), even the simplest-to-code Floyd-Warshall algorithm will run in time). Then, once you have this in a table, try all permutations of your 'mustpass' nodes, and the rest.
Something like this:
//Precomputation: Find all pairs shortest paths, e.g. using Floyd-Warshall
n = number of nodes
for i=1 to n: for j=1 to n: d[i][j]=INF
for k=1 to n:
for i=1 to n:
for j=1 to n:
d[i][j] = min(d[i][j], d[i][k] + d[k][j])
//That *really* gives the shortest distance between every pair of nodes! :-)
//Now try all permutations
shortest = INF
for each permutation a[1],a[2],...a[k] of the 'mustpass' nodes:
shortest = min(shortest, d['start'][a[1]]+d[a[1]][a[2]]+...+d[a[k]]['end'])
print shortest
(Of course that's not real code, and if you want the actual path you'll have to keep track of which permutation gives the shortest distance, and also what the all-pairs shortest paths are, but you get the idea.)
It will run in at most a few seconds on any reasonable language :)
[If you have n nodes and k 'mustpass' nodes, its running time is O(n3) for the Floyd-Warshall part, and O(k!n) for the all permutations part, and 100^3+(12!)(100) is practically peanuts unless you have some really restrictive constraints.]
run Djikstra's Algorithm to find the shortest paths between all of the critical nodes (start, end, and must-pass), then a depth-first traversal should tell you the shortest path through the resulting subgraph that touches all of the nodes start ... mustpasses ... end
This is two problems... Steven Lowe pointed this out, but didn't give enough respect to the second half of the problem.
You should first discover the shortest paths between all of your critical nodes (start, end, mustpass). Once these paths are discovered, you can construct a simplified graph, where each edge in the new graph is a path from one critical node to another in the original graph. There are many pathfinding algorithms that you can use to find the shortest path here.
Once you have this new graph, though, you have exactly the Traveling Salesperson problem (well, almost... No need to return to your starting point). Any of the posts concerning this, mentioned above, will apply.
Actually, the problem you posted is similar to the traveling salesman, but I think closer to a simple pathfinding problem. Rather than needing to visit each and every node, you simply need to visit a particular set of nodes in the shortest time (distance) possible.
The reason for this is that, unlike the traveling salesman problem, a corn maze will not allow you to travel directly from any one point to any other point on the map without needing to pass through other nodes to get there.
I would actually recommend A* pathfinding as a technique to consider. You set this up by deciding which nodes have access to which other nodes directly, and what the "cost" of each hop from a particular node is. In this case, it looks like each "hop" could be of equal cost, since your nodes seem relatively closely spaced. A* can use this information to find the lowest cost path between any two points. Since you need to get from point A to point B and visit about 12 inbetween, even a brute force approach using pathfinding wouldn't hurt at all.
Just an alternative to consider. It does look remarkably like the traveling salesman problem, and those are good papers to read up on, but look closer and you'll see that its only overcomplicating things. ^_^ This coming from the mind of a video game programmer who's dealt with these kinds of things before.
This is not a TSP problem and not NP-hard because the original question does not require that must-pass nodes are visited only once. This makes the answer much, much simpler to just brute-force after compiling a list of shortest paths between all must-pass nodes via Dijkstra's algorithm. There may be a better way to go but a simple one would be to simply work a binary tree backwards. Imagine a list of nodes [start,a,b,c,end]. Sum the simple distances [start->a->b->c->end] this is your new target distance to beat. Now try [start->a->c->b->end] and if that's better set that as the target (and remember that it came from that pattern of nodes). Work backwards over the permutations:
[start->a->b->c->end]
[start->a->c->b->end]
[start->b->a->c->end]
[start->b->c->a->end]
[start->c->a->b->end]
[start->c->b->a->end]
One of those will be shortest.
(where are the 'visited multiple times' nodes, if any? They're just hidden in the shortest-path initialization step. The shortest path between a and b may contain c or even the end point. You don't need to care)
Andrew Top has the right idea:
1) Djikstra's Algorithm
2) Some TSP heuristic.
I recommend the Lin-Kernighan heuristic: it's one of the best known for any NP Complete problem. The only other thing to remember is that after you expanded out the graph again after step 2, you may have loops in your expanded path, so you should go around short-circuiting those (look at the degree of vertices along your path).
I'm actually not sure how good this solution will be relative to the optimum. There are probably some pathological cases to do with short circuiting. After all, this problem looks a LOT like Steiner Tree: http://en.wikipedia.org/wiki/Steiner_tree and you definitely can't approximate Steiner Tree by just contracting your graph and running Kruskal's for example.
Considering the amount of nodes and edges is relatively finite, you can probably calculate every possible path and take the shortest one.
Generally this known as the travelling salesman problem, and has a non-deterministic polynomial runtime, no matter what the algorithm you use.
http://en.wikipedia.org/wiki/Traveling_salesman_problem
The question talks about must-pass in ANY order. I have been trying to search for a solution about the defined order of must-pass nodes. I found my answer but since no question on StackOverflow had a similar question I'm posting here to let maximum people benefit from it.
If the order or must-pass is defined then you could run dijkstra's algorithm multiple times. For instance let's assume you have to start from s pass through k1, k2 and k3 (in respective order) and stop at e. Then what you could do is run dijkstra's algorithm between each consecutive pair of nodes. The cost and path would be given by:
dijkstras(s, k1) + dijkstras(k1, k2) + dijkstras(k2, k3) + dijkstras(k3, 3)
How about using brute force on the dozen 'must visit' nodes. You can cover all the possible combinations of 12 nodes easily enough, and this leaves you with an optimal circuit you can follow to cover them.
Now your problem is simplified to one of finding optimal routes from the start node to the circuit, which you then follow around until you've covered them, and then find the route from that to the end.
Final path is composed of :
start -> path to circuit* -> circuit of must visit nodes -> path to end* -> end
You find the paths I marked with * like this
Do an A* search from the start node to every point on the circuit
for each of these do an A* search from the next and previous node on the circuit to the end (because you can follow the circuit round in either direction)
What you end up with is a lot of search paths, and you can choose the one with the lowest cost.
There's lots of room for optimization by caching the searches, but I think this will generate good solutions.
It doesn't go anywhere near looking for an optimal solution though, because that could involve leaving the must visit circuit within the search.
One thing that is not mentioned anywhere, is whether it is ok for the same vertex to be visited more than once in the path. Most of the answers here assume that it's ok to visit the same edge multiple times, but my take given the question (a path should not visit the same vertex more than once!) is that it is not ok to visit the same vertex twice.
So a brute force approach would still apply, but you'd have to remove vertices already used when you attempt to calculate each subset of the path.