I am currently working with multi-objective problem with 3 objectives and I am using weighted sum approach, where weights on all objectives together should sum up to 1, I am also using 0.1 as a step, for example if I have two objectives my weights would be:
0.9 0.1
0.8 0.2
0.7 0.3
But currently I'm stuck with 3 objectives and trying to figure out an algorithm that could make a similar distribution of weights across 3 objectives like this:
1obj 2obj 3obj
1.0 0.0 0.0
0.9 (0 ; 0.1) (0.1 ; 0)
0.8 (0 ; 0.1 ; 0.2) (0.2 ; 0.1 ; 0)
Can you pls suggest an algorithm that will get all possible combinations
The simplest way IMHO to get all combinations in this case is to model the problem as a graph and run an graph discovery algorithm, such as BFS.
The graph will be G=(V,E) where V = { (x1,x2,x3) | x1 + x2 + x3 = 1 } and E = { (v1,v2) | can get from v1 to v2 within a single change} (For example ((1,0,0.0,0.0),(0.9,0.0,0.1)) is an edge - because you can get from one to the other with a single change).
Now, since the graph is strongly connected (convince yourself why) - BFS will discover all vertices in this graph, and thus all possible combinations.
Caution: The number of possibilities is exponential, so for large number of objects or small step - discovering the entire graph will take a long time.
Small optimization: (Not needed if the graph size is not expected to be too large): One can optimize the space consumption of this solution by modeling the graph as a DAG (not allowing "back edges" as a structural restriction). This will also yield the correct result because all vertices are still accessible from a single (unique) source.
This saves space consumption because when running BFS in a DAG - a visited set does not have to be maintained.
Related
I have this question :
Airline company has N different planes and T pilots. Every pilot has a list of planes he can fly. Every flight needs 2 pilots. The company want to have as much flights simultaneously as possible. Find an algorithm that finds if you can have all the flights simultaneously.
This is the solution I thought about is finding max flow on this graph:
I am just not sure what the capacity should be. Can you help me with that?
Great idea to find the max flow.
For each edge from source --> pilot, assign a capacity of 1. Each pilot can only fly one plane at a time since they are running simultaneously.
For each edge from pilot --> plane, assign a capacity of 1. If this edge is filled with flow of 1, it represents that the given pilot is flying that plane.
For each edge from plane --> sink, assign a capacity of 2. This represents that each plane must be supplied by exactly 2 pilots.
Now, find a maximum flow. If the resulting maximum flow is two times the number of planes, then it's possible to satisfy the constraints. In this case, the edges between planes and pilots that are at capacity represent the matching.
The other answer is fine but you don't really need to involve flow as this can be reduced just as well to ordinary maximum bipartite matching:
For each plane, add another auxiliary plane to the plane partition with edges to the same pilots as the first plane.
Find a maximum bipartite matching M.
The answer is now true if and only if M = 2 N.
If you like, you can think of this as saying that each plane needs a pilot and a co-pilot, and the two vertices associated to each plane now represents those two roles.
The reduction to maximum bipartite matching is linear time, so using e.g. the Hopcroft–Karp algorithm to find the matching, you can solve the problem in O(|E| √|V|) where E is the number of edges between the partitions, and V = T + N.
In practice, the improvement over using a maximum flow based approach should depend on the quality of your implementations as well as the particular choice of representation of the graph, but chances are that you're better off this way.
Implementation example
To illustrate the last point, let's give an idea of how the two reductions could look in practice. One representation of a graph that's often useful due to its built-in memory locality is that of a CSR matrix, so let us assume that the input is such a matrix, whose rows correspond to the planes, and whose columns correspond to the pilots.
We will use the Python library SciPy which comes with algorithms for both maximum bipartite matching and maximum flow, and which works with CSR matrix representations for graphs under the hood.
In the algorithm given above, we will then need to construct the biadjacency matrix of the graph with the additional vertices added. This is nothing but the result of stacking the input matrix on top of itself, which is straightforward to phrase in terms of the CSR data structures: Following Wikipedia's notation, COL_INDEX should just be repeated, and ROW_INDEX should be replaced with ROW_INDEX concatenated with a copy of ROW_INDEX in which all elements are increased by the final element of ROW_INDEX.
In SciPy, a complete implementation which answers yes or no to the problem in OP would look as follows:
import numpy as np
from scipy.sparse.csgraph import maximum_bipartite_matching
def reduce_to_max_matching(a):
i, j = a.shape
data = np.ones(a.nnz * 2, dtype=bool)
indices = np.concatenate([a.indices, a.indices])
indptr = np.concatenate([a.indptr, a.indptr[1:] + a.indptr[-1]])
graph = csr_matrix((data, indices, indptr), shape=(2*i, j))
return (maximum_bipartite_matching(graph) != -1).sum() == 2 * i
In the maximum flow approach given by #HeatherGuarnera's answer, we will need to set up the full adjacency matrix of the new graph. This is also relatively straightforward; the input matrix will appear as a certain submatrix of the adjacency matrix, and we need to add a row for the source vertex and a column for the target. The example section of the documentation for SciPy's max flow solver actually contains an illustration of what this looks like in practice. Adopting this, a complete solution looks as follows:
import numpy as np
from scipy.sparse.csgraph import maximum_flow
def reduce_to_max_flow(a):
i, j = a.shape
n = a.nnz
data = np.concatenate([2*np.ones(i, dtype=int), np.ones(n + j, dtype=int)])
indices = np.concatenate([np.arange(1, i + 1),
a.indices + i + 1,
np.repeat(i + j + 1, j)])
indptr = np.concatenate([[0],
a.indptr + i,
np.arange(n + i + 1, n + i + j + 1),
[n + i + j]])
graph = csr_matrix((data, indices, indptr), shape=(2+i+j, 2+i+j))
flow = maximum_flow(graph, 0, graph.shape[0]-1)
return flow.flow_value == 2*i
Let us compare the timings of the two approaches on a single example consisting of 40 planes and 100 pilots, on a graph whose edge density is 0.1:
from scipy.sparse import random
inp = random(40, 100, density=.1, format='csr', dtype=bool)
%timeit reduce_to_max_matching(inp) # 191 µs ± 3.57 µs per loop
%timeit reduce_to_max_flow(inp) # 1.29 ms ± 20.1 µs per loop
The matching-based approach is faster, but not by a crazy amount. On larger problems, we'll start to see the advantages of using matching instead; with 400 planes and 1000 pilots:
inp = random(400, 1000, density=.1, format='csr', dtype=bool)
%timeit reduce_to_max_matching(inp) # 473 µs ± 5.52 µs per loop
%timeit reduce_to_max_flow(inp) # 68.9 ms ± 555 µs per loop
Again, this exact comparison relies on the use of specific predefined solvers from SciPy and how those are implemented, but if nothing else, this hints that simpler is better.
I am going through the following example on Problog probabilistic graph
I tried to computed probability of path from 1 to 5. Here are my manual computations
0.6*0.4+0.1*0.3*0.8 = 0.264
However, Problog returns P(path(1,5)) = 0.25824
Am I computing it correctly?
No, you can't just add up all the probabilities for the different paths. To see that, just assume that both paths from 1 to 5 had a probability of 0.7 each. You would get a probability of 1.4 which is clearly wrong (meaning that it is impossible that there is no path).
The way to calculate the probability for either of two events A and B is to get the probability of neither being true and then looking at the inverse of this event.
P(1->2->5) = 0.24
P(1->3->4->5) = 0.024
P(either) = 1 - (1 - 0.24) * (1 - 0.024)
= 1 - 0.74176
= 0.25824
Sorry for probably bad terminology, my statistics knowledge is a bit rusty.
I am trying to implement a stochastic ant colony optimisation algorithm, and I'm having trouble working out how to implement movement choices based on probabilities.
the standard (greedy) version that I have implemented so far is that an ant m at a vertex i on a graph G = (V,E) where E is the set of edges (i, j), will choose the next vertex j based on the following criteria:
j = argmax(<fitness function for j>)
such that j is connected to i
the problem I am having is in trying to implement a stochastic version of this, so that now the criteria for choosing a new vertex, j is:
P(j) = <fitness function for j>/sum(<fitness function for J>)
where P(j) is the probability of choosing vertex j,
such j is connected to i,
and J is the set of all vertices connected to i
I understand the mathematics behind it, I am just having trouble working out how i should actually implement it.
if, say, i have 3 vertices connected to i, each with a probability of 0.2, 0.3, 0.5 - what is the best way to make the selection? should I just randomly select a vertex j, then generate a random number r in the range (0,1) and if r >= P(j), select vertex j? or is there a better way?
Looking at the problem statement, I think you are not trying to visit all nodes (connected to i (say) ), but some of the nodes based on some probability distribution. Lets take an example:
You have a node i and connected to it are 5 nodes, a1...a5, with probabilities p1...p5, such that sum(p_i) = 1. No, say the precision of probabilities that you consider is 2 places after decimal. Also, you dont want to visit all 5 nodes, but only k of them. Lets say, in this example, k = 2. So, since 2 places of decimal is your probability precision, add 3 to it to increase normality of probability distribution in the random function. (You can change this 3 to any number of your choice, as far as performance is concerned) (Since you have not tagged any language, I'll take example of java's nextInt() function to generate random numbers.)
Lets give some values:
p1...p5 = {0.17, 0.11, 0.45, 0.03, 0.24}
Now, in a loop from 1 to k, generate a random number from (0...10^5). {5 = 2 + 3, ie. precision + 3}. If the generated number is from 0 to 16999, go with node a1, 17000 to 27999, go with a2, 28000 to 72999, go with a3...and so on. You get the idea.
What you're trying to implement is a weighted random choice depending on the probabilities for the components of the solution, or a random proportional selection rule on ACO terms. Here is an snippet of the implementation of this rule on the Isula Framework:
double value = random.nextDouble();
while (componentWithProbabilitiesIterator.hasNext()) {
Map.Entry<C, Double> componentWithProbability = componentWithProbabilitiesIterator
.next();
Double probability = componentWithProbability.getValue();
total += probability;
if (total >= value) {
nextNode = componentWithProbability.getKey();
getAnt().visitNode(nextNode);
return true;
}
}
You just need to generate a random value between 0 and 1 (stored in value), and start accumulating the probabilities of the components (on the total variable). When the total exceeds the threshold defined in value, we have found the component to add to the solution.
I have a set of points like:
(x , y , z , t)
(1 , 3 , 6 , 0.5)
(1.5 , 4 , 6.5 , 1)
(3.5 , 7 , 8 , 1.5)
(4 , 7.25 , 9 , 2)
I am looking to find the best linear fit on these points, let say a function like:
f(t) = a * x +b * y +c * z
This is Linear Regression problem. The "best fit" depends on the metric you define for being better.
One simple example is the Least Squares Metric, which aims to minimize the sum of squares: (f((x_i,y_i,z_i)) - w_i)^2 - where w_i is the measured value for the sample.
So, in least squares you are trying to minimize SUM{(a*x_i+b*y_i+c*z^i - w_i)^2 | per each i }. This function has a single global minimum at:
(a,b,c) = (X^T * X)^-1 * X^T * w
Where:
X is a 3xm matrix (m is the number of samples you have)
X^T - is the transposed of this matrix
w - is the measured results: `(w_1,w_2,...,w_m)`
The * operator represents matrix multiplication
There are more complex other methods, that use other distance metric, one example is the famous SVR with a linear kernel.
It seems that you are looking for the major axis of a point cloud.
You can work this out by finding the Eigenvector associated to the largest Eigenvalue of the covariance matrix. Could be an opportunity to use the power method (starting the iterations with the point farthest from the centroid, for example).
Can also be addressed by Singular Value Decomposition, preferably using methods that compute the largest values only.
If your data set contains outliers, then RANSAC could be a better choice: take two points at random and compute the sum of distances to the line they define. Repeat a number of times and keep the best fit.
Using the squared distances will answer your request for least-squares, but non-squared distances will be more robust.
You have a linear problem.
For example, my equation will be Y=ax1+bx2+c*x3.
In MATLAB do it:
B = [x1(:) x2(:) x3(:)] \ Y;
Y_fit = [x1(:) x2(:) x3(:)] * B;
In PYTHON do it:
import numpy as np
B, _, _, _ = np.linalg.lstsq([x1[:], x2[:], x3[:]], Y)
Y_fit = np.matmul([x1[:] x2[:] x3[:]], B)
I know that many algorithms are available for calculating the shortest path between two points in a graph or a grid, like breadth-first, all-pairs (Floyd's), Dijkstra's.
However, as I noticed, all of these algorithms compute all the paths in that graph or grid, not only those between the two points we are interested in.
MY QUESTION IS:
if I have a grid, i.e. a two dimensional array, and I'm interested in computing the shortest path between two points, say P1 and P2, and if there are restrictions on the way I can move on the grid (for example only diagonally, or only diagonally and upwards, etc.),
what algorithm can compute this?
Please notice here that if you have an answer, I would like you to post the name of the algorithm rather than the algorithm itself (of course, even better if you also post the algorithm); for example, whether it is Dijkstra's algorithm, or Floyd's, or whatever.
Please help me, I've been thinking about this for months!
okey guys i found this algorithm on TOPCODER.COM
here in the grid you can move only (diagonally and up)
but i can't understand what algorithm is this by any means could any one know?
#include<iostream>
#include <cmath>
using namespace std;
inline int Calc(int x,int y)
{
if(abs(x)>=abs(y)) return abs(x);
int z=(abs(x)+abs(y))/2;
return z+abs(abs(x)-z);
}
class SliverDistance
{
public:
int minSteps(int x1,int y1, int x2, int y2)
{
int ret=0;
if(((x1+y1)&1)!=((x2+y2)&1))y1++,ret++;
return ret+Calc(x2-x1,y2-y1);
}
};
Lee's algorithm: http://en.wikipedia.org/wiki/Lee_algorithm
It's essentially a BF search, here's an example: http://www.oop.rwth-aachen.de/documents/oop-2007/sss-oop-2007.pdf
To implement it effectively, check my answer here: Change FloodFill-Algorithm to get Voronoi Territory for two data points? - when I say mark, you mark it with the number on the position you came from + 1.
For example, if you have this grid, where a * = obstacle and you can move up, down, left and right, and you start from S and must go to D, and 0 = free position:
S 0 0 0
* * 0 *
* 0 0 *
0 0 * *
* 0 0 D
You put S in your queue, then "expand" it:
S 1 0 0
* * 0 *
* 0 0 *
0 0 * *
* 0 0 D
Then expand all of its neighbours:
S 1 2 0
* * 0 *
* 0 0 *
0 0 * *
* 0 0 D
And all of those neighbours' neighbours:
S 1 2 3
* * 3 *
* 0 0 *
0 0 * *
* 0 0 D
And so on, in the end you'll get:
S 1 2 3
* * 3 *
* 5 4 *
7 6 * *
* 7 8 9
So the distance from S to D is 9. The running time is O(NM), where N = number of lines and M = number of columns. I think this is the easiest algorithm to implement on grids, and it's also very efficient in practice. It should be faster than a classical dijkstra, although dijkstra might win if you implement it using heaps.
You may be disinformed. There exist different variants of Dijkstra's algorithm. One computes the shortest paths from each point to every other point (like Floyd's).
However, the typical Dijkstra algorithm is based on a priority queue and only computes your required shortest path. It does construct several paths during its execution, but those are all partial paths from A to some other nodes that might be on the final solution path.
Hence, you can easily interpret your grid as a graph (the restrictions like diagonals can then be taken into account accordingly) and run a Dijkstra search for the shortest path from A to B on that. It's really just a matter of modelling your problem, not that you need some fancy algorithm.
Use the A Star (A*) algorithm.
If your movement is restrictive enough (e.g. you can only move to the right, or up, or to the diagonal up and right), then you can exploit its overlapping subproblems and suboptimal substructure nature and use dynamic programming.
What I fail to understand is, if you want the shortest path between A and B, don't you still need to look at A to C and A to D if C and D point to B? Your shortest path could very well be A-C-B or A-D-B. You just need to throw out unconnected nodes. In one of my projects, I took points A and B, checked to see what other points were connected, and those that weren't were deleted from the entire graph. Then I proceeded with using Dijkstra's algorithm.
Here's a python implementation of shortest path in a matrix from (0,0) to (0,m-1) using BFS. You can change it to fit variable points.
n,m,k1,k2=[int(i) for i in input().split()]
arr=[[int(j) for j in input().split()] for i in range(n)]
x=[[-1 for i in range(m)] for j in range(n)]
x[0][0]=0
vis={}
q=[(0,0)]
while len(q)!=0:
curr=q[0]
rem=q.pop(0)
vis[curr]=True
r=curr[0]
c=curr[1]
if r-1>=0 and arr[r-1][c]==0:
if vis.get((r-1,c),-1)==-1 or vis[(r-1,c)]!=True:
q.append((r-1,c))
x[r-1][c]=x[r][c]+1
if r+1<n and arr[r+1][c]==0:
if vis.get((r+1,c),-1)==-1 or vis[(r+1,c)]!=True:
q.append((r+1,c))
x[r+1][c]=x[r][c]+1
if c-1>=0 and arr[r][c-1]==0:
if vis.get((r,c-1),-1)==-1 or vis[(r,c-1)]!=True:
q.append((r,c-1))
x[r][c-1]=x[r][c]+1
if c+1<m and arr[r][c+1]==0:
if vis.get((r,c+1),-1)==-1 or vis[(r,c+1)]!=True:
q.append((r,c+1))
x[r][c+1]=x[r][c]+1
#for i in x:
#print(i)
ans=x[0][m-1]
if ans==-1:
print(-1)
else:
print(ans)
input matrix should consist of 0's and 1's. 0 is for possible movement.
n is number of rows .
m is number of columns.
arr is the given matrix.
x is the distance matrix from (0,0).
vis is a dictionary giving a boolean if the node is visited.
output of -1 shows that there is no such path possible.
Your grid forms a graph (or at least can be viewed as a graph). Eliminating some directions of movement indicates that it's a directed graph. If you can't move from one node to another at all, that's an edge that isn't present in the graph.
Once you've encoded your grid into graph form, it's a simple matter of selecting among the well-known graph algorithms (of which you're apparently already aware) to traverse it for the type of result you want (e.g. shortest path).
Edit: I've looked at the answer you posted, but I'm not sure what that code is supposed to be/do. Just for example, it has: if(y>=0) max(abs(x),y);. This doesn't seem (at least to me) to make much sense -- the result from the max is simply thrown away. To accomplish something useful, it needs to be returned or assigned or something on that order. As it stands, the best you can hope is that the compiler spots it as dead code, and doesn't generate anything for it.
My guess is that the code doesn't really work quite as intended, and if it does anything useful, it's more by accident than design. It would take a fair amount of time and effort to be sure you've sorted out problems like this to the point that you were really sure what it did, and even more difficult to guess what was really intended.
use A* algorithm for finding the path between two points in a 2D grid.
http://theory.stanford.edu/~amitp/GameProgramming/ImplementationNotes.html