There is great problem in one of the algo contest sites. I am trying to solve it for 5 days. I am not asking you to solve me this for me, as I am new to algorithms I would like to ask you help me with classification of this type problem, did anyone solved problems like this, what is the type of problem NP or not. Please do not think that I asking you to solve this for me, my purpose is just to learn algorithms and this is the problem which is enough difficult for me:
The goal of this puzzle is to determine where to place a set of gas
stations so that they are closest to airports. Airports make use of a
lot of gas for fueling planes, so placing gas stations close them is
of strategical importance.
Input Specification Your program should take one and only one command
line argument: the input file (passed in argv, args, arguments
depending on the language). The input file is formatted as follows:
the first line contains an integer: n the number of airports the n
following lines each contain 2 floating point values xi yi
representing the coordinates of the ith airport the following line
contains the number p of cases to analyze (p is always less than 5)
the following p lines each contain one integer gi giving the number of
required gas stations
Output Specifications:
You program should output
the result to the standard output (printf, print, echo, write): Your
output should contain p lines, each line providing the gj coordinates
xj,yj of the gas stations. Your solution score will be measured by the
quality of the solution. The quality of the solution is measured by
the total distance, the total distance D is the square root of the sum
of squared distances from each airport to its closest gas station. The
lower the total distance D, the higher your score will be.
This problem is the canonical unsupervised k-means classification problem. See here for the full details: http://en.wikipedia.org/wiki/K-means_clustering
For a quick hint (if you want to avoid complete spoilers) k-means simply starts by picking random locations for your gas stations. It improves the solution each iteration thereafter by reducing the cost of each individual gas station one at a time. It does this by moving a gas station with the goal of minimizing its cost for the set of airports that it currently fuels.
This seems to a be variant of the Facility location problem. Finding the optimal locations is NP-hard, but many approximation methods can be applied to find solutions within a certain guaranteed distance of the optimum. Alternatively, soft methods like clustering can be used, as proposed in other answers.
For the case gi=1. It is easy - you would just computer the center of gravity/mass (from all airports, heck you could even weigth each airport with the amount of fuel they consume, so it would place the fuel station closer to heavy consuming airports, but as this is not required you would give all the same weight). This would yield an optimal solution (this is also a good example that Nonlinear, global optimization does NOT necessary imply NP hard).
My idea would be to partition the set of airports into gi sets, and afterwards apply to each set the center-of-gravity/mass. This would be classified as a clustering problem (or maybe a partition, depends how you formulate it). (Practically I would apply a k-means clustering to solve this). (Here it gets indeed NP hard if you want perfect result, but maybe someone come up with another good solution)
Related
This is a problem I came across frequently and I'm searching a more effective way to solve it. Take a look at this pics:
Let's say you want to find the shortest distance from the red point to a line segment an. Assume you only know the start/end point (x,y) of the segments and the point. Now this can be done in O(n), where n are the line segments, by checking every distance from the point to a line segment. This is IMO not effective, because in the worst case there have to be n-1 distance checks till the right one is found.
This can be a real performance issue for n = 1000 f.e. (which is a likely number), especially if the distance calculation isn't just done in the euclidean space by the Pythagorean theorem but for example by a geodesic method like the haversine formula or Vincenty's.
This is a general problem in different situations:
Is the point inside a radius of the vertices?
Which set of vertices is nearest to the point?
Is the point surrounded by line segments?
To answer these questions, the only approach I know is O(n). Now I would like to know if there is a data structure or a different strategy to solve these problems more efficiently?
To make it short: I'm searching a way, where the line segments / vertices could be "filtered" somehow to get a set of potential candidates before I start my distance calculations. Something to reduce the complexity to O(m) where m < n.
Probably not an acceptable answer, but too long for a comment: The most appropriate answer here depends on details that you did not state in the question.
If you only want to perform this test once, then there will be no way avoid a linear search. However, if you have a fixed set of lines (or a set of lines that does not change too significantly over time), then you may employ various techniques for accelerating the queries. These are sometimes referred to as Spatial Indices, like a Quadtree.
You'll have to expect a trade-off between several factors, like the query time and the memory consumption, or the query time and the time that is required for updating the data structure when the given set of lines changes. The latter also depends on whether it is a structural change (lines being added or removed), or whether only the positions of the existing lines change.
My problem is that I need to compare URL paths and deduce if they are similar. Below I provide example data to process:
# GROUP 1
/robots.txt
# GROUP 2
/bot.html
# GROUP 3
/phpMyAdmin-2.5.6-rc1/scripts/setup.php
/phpMyAdmin-2.5.6-rc2/scripts/setup.php
/phpMyAdmin-2.5.6/scripts/setup.php
/phpMyAdmin-2.5.7-pl1/scripts/setup.php
/phpMyAdmin-2.5.7/scripts/setup.php
/phpMyAdmin-2.6.0-alpha/scripts/setup.php
/phpMyAdmin-2.6.0-alpha2/scripts/setup.php
# GROUP 4
//phpMyAdmin/
I tried Levenshtein distance to compare, but for me is not enough accurate. I do not need 100% accurate algorithm, but I think 90% and above is a must.
I think that I need some sort of classifier, but the problem is that each portion of new data can containt path that should be classified to the new unknown class.
Could you please direct me to the right thoutht?
Thanks
Levenshtein distance is best option, but tuned distance. You have to use weighted Edit distance and possibly split path on tokens - words and numbers. So for example version like "2.5.6-rc2 and 2.5.6" can be treated as 0 weight difference, but name token like phpMyAdmin and javaMyAdmin give 1 weight difference.
When checking #jakub.gieryluk suggestion I accidentally have found solution that satisfy me - "Hobohm clustering algorithm, originally devised to reduce redundancy of biological sequence data sets."
Tests of PERL library implemented by Bruno Vecchi gave me really good results. The only problem is that I need Python implementation, but I belive that I can either find one on the Internet or reimplement code by myself.
Next thing is that I have not checked active learning ability of this algorithm yet ;)
I know it's not the exact answer to your question, but are you familiar with k-means algorithm?
I guess even the Levenshtein can work here, the difficulty however is how to compute centroids with that approach.
Perhaps you can divide input set into disjoint subsets, then for each URL in each subset compute the distance to all the other URLs in the same subset, and the URL that has lowest sum of distances, should be the centroid (of course, it depends on how big is the input set; for huge sets it might be not a good idea to do so).
The good thing about k-means is that you can start with absolutely random division, and then iteratively make it better.
The bad thing about k-means is that you have to precise k before start. However, during the run (perhaps where the situation stabilized after first couple of iterations), you can measure intra-similarity of each set, and if it is low, you can divide the set into two subsets and go on with the same algorithm.
I have a need to take a 2D graph of n points and reduce it the r points (where r is a specific number less than n). For example, I may have two datasets with slightly different number of total points, say 1021 and 1001 and I'd like to force both datasets to have 1000 points. I am aware of a couple of simplification algorithms: Lang Simplification and Douglas-Peucker. I have used Lang in a previous project with slightly different requirements.
The specific properties of the algorithm I am looking for is:
1) must preserve the shape of the line
2) must allow me reduce dataset to a specific number of points
3) is relatively fast
This post is a discussion of the merits of the different algorithms. I will post a second message for advice on implementations in Java or Groovy (why reinvent the wheel).
I am concerned about requirement 2 above. I am not an expert enough in these algorithms to know whether I can dictate the exact number of output points. The implementation of Lang that I've used took lookAhead, tolerance and the array of Points as input, so I don't see how to dictate the number of points in the output. This is a critical requirement of my current needs. Perhaps this is due to the specific implementation of Lang we had used, but I have not seen a lot of information on Lang on the web. Alternatively we could use Douglas-Peucker but again I am not sure if the number of points in the output can be specified.
I should add I am not an expert on these types of algorithms or any kind of math wiz, so I am looking for mere mortal type advice :) How do I satisfy requirements 1 and 2 above? I would sacrifice performance for the right solution.
I think you can adapt Douglas-Pücker quite straightforwardly. Adapt the recursive algorithm so that rather than producing a list it produces a tree mirroring the structure of the recursive calls. The root of the tree will be the single-line approximation P0-Pn; the next level will represent the two-line approximation P0-Pm-Pn where Pm is the point between P0 and Pn which is furthest from P0-Pn; the next level (if full) will represent a four-line approximation, etc. You can then trim the tree either on the basis of depth or on the basis of distance of the inserted point from the parent line.
Edit: in fact, if you take the latter approach you don't need to build a tree. Instead you populate a priority queue where the priority is given by the distance of the inserted point from the parent line. Then when you've finished the queue tells you which points to remove (or keep, according to the order of the priorities).
You can find my C++ implementation and article on Douglas-Peucker simplification here and here. I also provide a modified version of the Douglas-Peucker simplification that allows you to specify the number of points of the resulting simplified line. It uses a priority queue as mentioned by 'Peter Taylor'. Its a lot slower though, so I don't know if it would satisfy the 'is relatively fast' requirement.
I'm planning on providing an implementation for Lang simplification (and several others). Currently I don't see any easy way how to adjust Lang to reduce to a fixed point count. If you
could live with a less strict requirement: 'must allow me reduce dataset to an approximate number of points', then you could use an iterative approach. Guess an initial value for lookahead: point count / desired point count. Then slowly increase the lookahead until you approximately hit the desired point count.
I hope this helps.
p.s.: I just remembered something, you could also try the Visvalingam-Whyatt algorithm. In short:
-compute the triangle area for each point with its direct neighbors
-sort these areas
-remove the point with the smallest area
-update the area of its neighbors
-resort
-continue until n points remain
Hey guys, I have a sort of speed dating type application (not used for dating, just a similar concept) that compares users and matches them in a round based event.
Currently I am storing each user to user comparison (using cosine similarity) and then finding a round in which both users are available. My current set up works fine for smaller scale but I seem to be missing a few matchings in larger data sets.
For example with a setup like so (assuming 6 users, 3 from each group)
Round (User1, User2)
----------------------------
1 (x1,y1) (x2,y2) (x3,y3)
2 (x1,y2) (x2,y3) (x3,y1)
3 (x1,y3) (x2,y1) (x3,y2)
My approach works well right now to ensure I have each user meeting the appropriate user without having overlaps so a user is left out, just not with larger data sets.
My current algorithm
I store a comparison of each user from x to each user from y like so
Round, user1, user2, similarity
And to build the event schedule I simply sort the comparisons by similarity and then iterate over the results, finding an open round for both users, like so:
event.user_maps.all(:order => 'similarity desc').each do |map|
(1..event.rounds).each do |round|
if user_free_in_round?(map.user1) and user_free_in_round?(map.user2)
#creates the pairing and breaks from the loop
end
end
end
This isn't exact code but the general algorithm to build the schedule. Does anyone know a better way of filling in a matrix of item pairings where no one item can be in more than one place in the same slot?
EDIT
For some clarification, the issue I am having is that in larger sets my algorithm of placing highest similarity matches first can sometimes result in collisions. What I mean by that is that the users are paired in such a way that they have no other user to meet with.
Like so:
Round (User1, User2)
----------------------------
1 (x1,y1) (x2,y2) (x3,y3)
2 (x1,y3) (x2,nil) (x3,y1)
3 (x1,y2) (x2,y1) (x3,y2)
I want to be able to prevent this from happening while preserving the need for higher similar users given higher priority in scheduling.
In real scenarios there are far more matches than there are available rounds and an uneven number of x users to y users and in my test cases instead of getting every round full I will only have about 90% or so of them filled while collisions like the above are causing problems.
I think the question still needs clarification even after edit, but I could be missing something.
As far as I can tell, what you want is that each new round should start with the best possible matching (defined as sum of the cosine similarities of all the matched pairs). After any pair (x_i,y_j) have been matched in a round, they are not eligible for the next round.
You could do this by building a bipartite graph where your Xs are nodes in one side and Ys are nodes in another side, and the edge weight is cosine similarity. Then you find the max weighted match in this graph. For the next rounds, you eliminate the edges that have already been used in previous round and run the matching algorithm again. For details on how to code max weight matching in bipartite graph, see here.
BTW, this solution is not optimum since we are proceeding from one round to next in a greedy fashion. I have a feeling that getting the optimum solution would be NP hard, but I don't have a proof so can't be sure.
I agree that the question still needs clarification. As Amit expressed, I have a gut feeling that this is an NP hard problem, so I am assuming that you are looking for an approximate solution.
That said, I would need more information on the tradeoffs you would be willing to make (and perhaps I'm just missing something in your question). What are the explicit goals of the algorithm?
Is there a lower threshold for similarity below which you don't want a pairing to happen? I'm still a bit confused as to why there would be individuals which could not be paired up at all during a given round...
Essentially, you are performing a search over the space of possible pairings, correct? Maybe you could use backtracking or some form of constraint-based algorithm to make sure that you can obtain a complete solution for a given round...?
My best shot so far:
A delivery vehicle needs to make a series of deliveries (d1,d2,...dn), and can do so in any order--in other words, all the possible permutations of the set D = {d1,d2,...dn} are valid solutions--but the particular solution needs to be determined before it leaves the base station at one end of the route (imagine that the packages need to be loaded in the vehicle LIFO, for example).
Further, the cost of the various permutations is not the same. It can be computed as the sum of the squares of distance traveled between di -1 and di, where d0 is taken to be the base station, with the caveat that any segment that involves a change of direction costs 3 times as much (imagine this is going on on a railroad or a pneumatic tube, and backing up disrupts other traffic).
Given the set of deliveries D represented as their distance from the base station (so abs(di-dj) is the distance between two deliveries) and an iterator permutations(D) which will produce each permutation in succession, find a permutation which has a cost less than or equal to that of any other permutation.
Now, a direct implementation from this description might lead to code like this:
function Cost(D) ...
function Best_order(D)
for D1 in permutations(D)
Found = true
for D2 in permutations(D)
Found = false if cost(D2) > cost(D1)
return D1 if Found
Which is O(n*n!^2), e.g. pretty awful--especially compared to the O(n log(n)) someone with insight would find, by simply sorting D.
My question: can you come up with a plausible problem description which would naturally lead the unwary into a worse (or differently awful) implementation of a sorting algorithm?
I assume you're using this question for an interview to see if the applicant can notice a simple solution in a seemingly complex question.
[This assumption is incorrect -- MarkusQ]
You give too much information.
The key to solving this is realizing that the points are in one dimension and that a sort is all that is required. To make this question more difficult hide this fact as much as possible.
The biggest clue is the distance formula. It introduces a penalty for changing directions. The first thing an that comes to my mind is minimizing this penalty. To remove the penalty I have to order them in a certain direction, this ordering is the natural sort order.
I would remove the penalty for changing directions, it's too much of a give away.
Another major clue is the input values to the algorithm: a list of integers. Give them a list of permutations, or even all permutations. That sets them up to thinking that a O(n!) algorithm might actually be expected.
I would phrase it as:
Given a list of all possible
permutations of n delivery locations,
where each permutation of deliveries
(d1, d2, ...,
dn) has a cost defined by:
Return permutation P such that the
cost of P is less than or equal to any
other permutation.
All that really needs to be done is read in the first permutation and sort it.
If they construct a single loop to compare the costs ask them what the big-o runtime of their algorithm is where n is the number of delivery locations (Another trap).
This isn't a direct answer, but I think more clarification is needed.
Is di allowed to be negative? If so, sorting alone is not enough, as far as I can see.
For example:
d0 = 0
deliveries = (-1,1,1,2)
It seems the optimal path in this case would be 1 > 2 > 1 > -1.
Edit: This might not actually be the optimal path, but it illustrates the point.
YOu could rephrase it, having first found the optimal solution, as
"Give me a proof that the following convination is the most optimal for the following set of rules, where optimal means the smallest number results from the sum of all stage costs, taking into account that all stages (A..Z) need to be present once and once only.
Convination:
A->C->D->Y->P->...->N
Stage costs:
A->B = 5,
B->A = 3,
A->C = 2,
C->A = 4,
...
...
...
Y->Z = 7,
Z->Y = 24."
That ought to keep someone busy for a while.
This reminds me of the Knapsack problem, more than the Traveling Salesman. But the Knapsack is also an NP-Hard problem, so you might be able to fool people to think up an over complex solution using dynamic programming if they correlate your problem with the Knapsack. Where the basic problem is:
can a value of at least V be achieved
without exceeding the weight W?
Now the problem is a fairly good solution can be found when V is unique, your distances, as such:
The knapsack problem with each type of
item j having a distinct value per
unit of weight (vj = pj/wj) is
considered one of the easiest
NP-complete problems. Indeed empirical
complexity is of the order of O((log
n)2) and very large problems can be
solved very quickly, e.g. in 2003 the
average time required to solve
instances with n = 10,000 was below 14
milliseconds using commodity personal
computers1.
So you might want to state that several stops/packages might share the same vj, inviting people to think about the really hard solution to:
However in the
degenerate case of multiple items
sharing the same value vj it becomes
much more difficult with the extreme
case where vj = constant being the
subset sum problem with a complexity
of O(2N/2N).
So if you replace the weight per value to distance per value, and state that several distances might actually share the same values, degenerate, some folk might fall in this trap.
Isn't this just the (NP-Hard) Travelling Salesman Problem? It doesn't seem likely that you're going to make it much harder.
Maybe phrasing the problem so that the actual algorithm is unclear - e.g. by describing the paths as single-rail railway lines so the person would have to infer from domain knowledge that backtracking is more costly.
What about describing the question in such a way that someone is tempted to do recursive comparisions - e.g. "can you speed up the algorithm by using the optimum max subset of your best (so far) results"?
BTW, what's the purpose of this - it sounds like the intent is to torture interviewees.
You need to be clearer on whether the delivery truck has to return to base (making it a round trip), or not. If the truck does return, then a simple sort does not produce the shortest route, because the square of the return from the furthest point to base costs so much. Missing some hops on the way 'out' and using them on the way back turns out to be cheaper.
If you trick someone into a bad answer (for example, by not giving them all the information) then is it their foolishness or your deception that has caused it?
How great is the wisdom of the wise, if they heed not their ego's lies?