faster than O(n^4) algorithm for solving the following problem
This question is quite similar to the previous one I asked here. I recommend taking a look at it and the solution given by @platty before continuing.
Given a set of n inequalities each of the form ax+by+cz≤d for some a,b,c,d in Q, determine if there exists x, y and z in Q that satisfy all the inequalities.
Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).
I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks
geometry algorithms computational-complexity
asked Dec 11 '18 at 6:00
mathew
410215
add a comment |
This question is quite similar to the previous one I asked here. I recommend taking a look at it and the solution given by @platty before continuing.
Given a set of n inequalities each of the form ax+by+cz≤d for some a,b,c,d in Q, determine if there exists x, y and z in Q that satisfy all the inequalities.
Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).
I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks
geometry algorithms computational-complexity
asked Dec 11 '18 at 6:00
mathew
410215
In practice, the simplex method will usually work quickly.
– Greg Martin
Dec 11 '18 at 6:27
Note that you can actually have a feasible region that contains no intersection points of three planes. It could, for example, be an infinite prism stretching both ways.
– Todor Markov
Dec 11 '18 at 19:27
add a comment |
This question is quite similar to the previous one I asked here. I recommend taking a look at it and the solution given by @platty before continuing.
Given a set of n inequalities each of the form ax+by+cz≤d for some a,b,c,d in Q, determine if there exists x, y and z in Q that satisfy all the inequalities.
Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).
I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks
geometry algorithms computational-complexity
asked Dec 11 '18 at 6:00
mathew
410215
This question is quite similar to the previous one I asked here. I recommend taking a look at it and the solution given by @platty before continuing.
Given a set of n inequalities each of the form ax+by+cz≤d for some a,b,c,d in Q, determine if there exists x, y and z in Q that satisfy all the inequalities.
Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).
I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks
geometry algorithms computational-complexity
geometry algorithms computational-complexity
asked Dec 11 '18 at 6:00
mathew
410215
asked Dec 11 '18 at 6:00
mathew
410215
asked Dec 11 '18 at 6:00
mathew
410215
asked Dec 11 '18 at 6:00
mathew
410215
asked Dec 11 '18 at 6:00
mathew
410215
410215
In practice, the simplex method will usually work quickly.
– Greg Martin
Dec 11 '18 at 6:27
Note that you can actually have a feasible region that contains no intersection points of three planes. It could, for example, be an infinite prism stretching both ways.
– Todor Markov
Dec 11 '18 at 19:27
add a comment |
In practice, the simplex method will usually work quickly.
– Greg Martin
Dec 11 '18 at 6:27
Note that you can actually have a feasible region that contains no intersection points of three planes. It could, for example, be an infinite prism stretching both ways.
– Todor Markov
Dec 11 '18 at 19:27
In practice, the simplex method will usually work quickly.
– Greg Martin
Dec 11 '18 at 6:27
In practice, the simplex method will usually work quickly.
– Greg Martin
Dec 11 '18 at 6:27
Note that you can actually have a feasible region that contains no intersection points of three planes. It could, for example, be an infinite prism stretching both ways.
– Todor Markov
Dec 11 '18 at 19:27
Note that you can actually have a feasible region that contains no intersection points of three planes. It could, for example, be an infinite prism stretching both ways.
– Todor Markov
Dec 11 '18 at 19:27
add a comment |
1 Answer
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You can modify your algorithm a bit to make it $O(n^3)$.
If we have a line, we can check if there's a point on that line satisfying all inequalities in $O(n)$ time. If we denote one of the directions on this line "up", each plane gives you an upper or lower bound on the part of the line satisfying the inequalities. So we can compute the intersection points between the line and each plane, and check if the lowest upper bound is above the highest lower bound.
Like in your algorithm, candidate lines can be intersection lines between each pair of planes ($O(n^2)$ of them. So you get a total complexity of $O(n^3)$.
Note, care must be taken for the special case when all planes are parallel and so there's no intersection lines.
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
Thank you for this, if no better solutions are put forward I'll accept your answer however I'm still hoping a polynomial time algorithm will be found.
– mathew
Dec 12 '18 at 9:06
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
You can modify your algorithm a bit to make it $O(n^3)$.
If we have a line, we can check if there's a point on that line satisfying all inequalities in $O(n)$ time. If we denote one of the directions on this line "up", each plane gives you an upper or lower bound on the part of the line satisfying the inequalities. So we can compute the intersection points between the line and each plane, and check if the lowest upper bound is above the highest lower bound.
Like in your algorithm, candidate lines can be intersection lines between each pair of planes ($O(n^2)$ of them. So you get a total complexity of $O(n^3)$.
Note, care must be taken for the special case when all planes are parallel and so there's no intersection lines.
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
Thank you for this, if no better solutions are put forward I'll accept your answer however I'm still hoping a polynomial time algorithm will be found.
– mathew
Dec 12 '18 at 9:06
add a comment |
You can modify your algorithm a bit to make it $O(n^3)$.
If we have a line, we can check if there's a point on that line satisfying all inequalities in $O(n)$ time. If we denote one of the directions on this line "up", each plane gives you an upper or lower bound on the part of the line satisfying the inequalities. So we can compute the intersection points between the line and each plane, and check if the lowest upper bound is above the highest lower bound.
Like in your algorithm, candidate lines can be intersection lines between each pair of planes ($O(n^2)$ of them. So you get a total complexity of $O(n^3)$.
Note, care must be taken for the special case when all planes are parallel and so there's no intersection lines.
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
Thank you for this, if no better solutions are put forward I'll accept your answer however I'm still hoping a polynomial time algorithm will be found.
– mathew
Dec 12 '18 at 9:06
add a comment |
You can modify your algorithm a bit to make it $O(n^3)$.
If we have a line, we can check if there's a point on that line satisfying all inequalities in $O(n)$ time. If we denote one of the directions on this line "up", each plane gives you an upper or lower bound on the part of the line satisfying the inequalities. So we can compute the intersection points between the line and each plane, and check if the lowest upper bound is above the highest lower bound.
Like in your algorithm, candidate lines can be intersection lines between each pair of planes ($O(n^2)$ of them. So you get a total complexity of $O(n^3)$.
Note, care must be taken for the special case when all planes are parallel and so there's no intersection lines.
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
You can modify your algorithm a bit to make it $O(n^3)$.
If we have a line, we can check if there's a point on that line satisfying all inequalities in $O(n)$ time. If we denote one of the directions on this line "up", each plane gives you an upper or lower bound on the part of the line satisfying the inequalities. So we can compute the intersection points between the line and each plane, and check if the lowest upper bound is above the highest lower bound.
Like in your algorithm, candidate lines can be intersection lines between each pair of planes ($O(n^2)$ of them. So you get a total complexity of $O(n^3)$.
Note, care must be taken for the special case when all planes are parallel and so there's no intersection lines.
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
answered Dec 11 '18 at 19:36
Todor Markov
1,16817
1,16817
Thank you for this, if no better solutions are put forward I'll accept your answer however I'm still hoping a polynomial time algorithm will be found.
– mathew
Dec 12 '18 at 9:06
add a comment |
Thank you for this, if no better solutions are put forward I'll accept your answer however I'm still hoping a polynomial time algorithm will be found.
– mathew
Dec 12 '18 at 9:06
Thank you for this, if no better solutions are put forward I'll accept your answer however I'm still hoping a polynomial time algorithm will be found.
– mathew
Dec 12 '18 at 9:06
Thank you for this, if no better solutions are put forward I'll accept your answer however I'm still hoping a polynomial time algorithm will be found.
– mathew
Dec 12 '18 at 9:06
add a comment |
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In practice, the simplex method will usually work quickly.
– Greg Martin
Dec 11 '18 at 6:27
Note that you can actually have a feasible region that contains no intersection points of three planes. It could, for example, be an infinite prism stretching both ways.
– Todor Markov
Dec 11 '18 at 19:27