[algogeeks] Re: Good problem
It is not a solution for minimize maximum distance either, since |x(i)-x(j)|
= c does not hold for every pair of points, only for adjacent points.
2009/10/21 saltycookie saltycoo...@gmail.com
Yes, you are quite right. If I am not mistaken, you give a good solution
for finding the minimum maximum distance.
But what about the original problem where we want to find the maximum
minimum distance? I am not clear about the connection between the two
problems.
Thanks.
2009/10/21 Dave dave_and_da...@juno.com
林夏祥 , think again. If we are trying to minimize the maximum distance,
then we want to minimize the upper bound. That is what I specified:
letting c be the upper bound, find the smallest c such that all of the
distances do not exceed c. That gives rise to the inequalities
|x(i)-x(j)| = c.
If necessary, this can be written as two inequalities:
x(i) - x(j) = c and
x(j) - x(i) = c.
Since the relationship is and, we can just use the two inequalities
as part of the constraint conditions.
Dave
On Oct 21, 12:02 am, 林夏祥 saltycoo...@gmail.com wrote:
I don't think LP can solve it. We are to maximize c, not minimize c.
The formulas we have are:
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
The first inequality actually is combination of two linear equalities:
x(i)
- x(j) = c or x(i) - x(j) = -c. Notice the relation of the two is
or,
and we cannot put them together to get a system of linear inequalities.
2009/10/21 Dave dave_and_da...@juno.com
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
Dave
On Oct 5, 9:22 am, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a
range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--
此致
敬礼！
林夏祥- Hide quoted text -
- Show quoted text -
--
此致
敬礼！
林夏祥
--
此致
敬礼！
林夏祥
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[algogeeks] Re: Good problem
I don't think LP can solve it. We are to maximize c, not minimize c.
The formulas we have are:
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
The first inequality actually is combination of two linear equalities: x(i)
- x(j) = c or x(i) - x(j) = -c. Notice the relation of the two is or,
and we cannot put them together to get a system of linear inequalities.
2009/10/21 Dave dave_and_da...@juno.com
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
Dave
On Oct 5, 9:22 am, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--
此致
敬礼！
林夏祥
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[algogeeks] Re: Good problem
林夏祥 , think again. If we are trying to minimize the maximum distance,
then we want to minimize the upper bound. That is what I specified:
letting c be the upper bound, find the smallest c such that all of the
distances do not exceed c. That gives rise to the inequalities
|x(i)-x(j)| = c.
If necessary, this can be written as two inequalities:
x(i) - x(j) = c and
x(j) - x(i) = c.
Since the relationship is and, we can just use the two inequalities
as part of the constraint conditions.
Dave
On Oct 21, 12:02 am, 林夏祥 saltycoo...@gmail.com wrote:
I don't think LP can solve it. We are to maximize c, not minimize c.
The formulas we have are:
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
The first inequality actually is combination of two linear equalities: x(i)
- x(j) = c or x(i) - x(j) = -c. Notice the relation of the two is or,
and we cannot put them together to get a system of linear inequalities.
2009/10/21 Dave dave_and_da...@juno.com
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
Dave
On Oct 5, 9:22 am, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--
此致
敬礼！
林夏祥- Hide quoted text -
- Show quoted text -
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[algogeeks] Re: Good problem
Yes, you are quite right. If I am not mistaken, you give a good solution for
finding the minimum maximum distance.
But what about the original problem where we want to find the maximum
minimum distance? I am not clear about the connection between the two
problems.
Thanks.
2009/10/21 Dave dave_and_da...@juno.com
林夏祥 , think again. If we are trying to minimize the maximum distance,
then we want to minimize the upper bound. That is what I specified:
letting c be the upper bound, find the smallest c such that all of the
distances do not exceed c. That gives rise to the inequalities
|x(i)-x(j)| = c.
If necessary, this can be written as two inequalities:
x(i) - x(j) = c and
x(j) - x(i) = c.
Since the relationship is and, we can just use the two inequalities
as part of the constraint conditions.
Dave
On Oct 21, 12:02 am, 林夏祥 saltycoo...@gmail.com wrote:
I don't think LP can solve it. We are to maximize c, not minimize c.
The formulas we have are:
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
The first inequality actually is combination of two linear equalities:
x(i)
- x(j) = c or x(i) - x(j) = -c. Notice the relation of the two is or,
and we cannot put them together to get a system of linear inequalities.
2009/10/21 Dave dave_and_da...@juno.com
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
Dave
On Oct 5, 9:22 am, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a
range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--
此致
敬礼！
林夏祥- Hide quoted text -
- Show quoted text -
--
此致
敬礼！
林夏祥
--~--~-~--~~~---~--~~
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[algogeeks] Re: Good problem
to maximize the minimum distance between any two points:- to maximize the
minimum distance between adjacent points
- for this all points must be equally spaced.
hence, choose 'n' equally spaced points in the range (r1, r2) starting from
r1 and ending at r2.
2009/10/21 saltycookie saltycoo...@gmail.com
Yes, you are quite right. If I am not mistaken, you give a good solution
for finding the minimum maximum distance.
But what about the original problem where we want to find the maximum
minimum distance? I am not clear about the connection between the two
problems.
Thanks.
2009/10/21 Dave dave_and_da...@juno.com
林夏祥 , think again. If we are trying to minimize the maximum distance,
then we want to minimize the upper bound. That is what I specified:
letting c be the upper bound, find the smallest c such that all of the
distances do not exceed c. That gives rise to the inequalities
|x(i)-x(j)| = c.
If necessary, this can be written as two inequalities:
x(i) - x(j) = c and
x(j) - x(i) = c.
Since the relationship is and, we can just use the two inequalities
as part of the constraint conditions.
Dave
On Oct 21, 12:02 am, 林夏祥 saltycoo...@gmail.com wrote:
I don't think LP can solve it. We are to maximize c, not minimize c.
The formulas we have are:
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
The first inequality actually is combination of two linear equalities:
x(i)
- x(j) = c or x(i) - x(j) = -c. Notice the relation of the two is
or,
and we cannot put them together to get a system of linear inequalities.
2009/10/21 Dave dave_and_da...@juno.com
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
Dave
On Oct 5, 9:22 am, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a
range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--
此致
敬礼！
林夏祥- Hide quoted text -
- Show quoted text -
--
此致
敬礼！
林夏祥
--
Reduce, Reuse and Recycle
Regards,
Vivek.S
--~--~-~--~~~---~--~~
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[algogeeks] Re: Good problem
I take it from the problem description that you are given n intervals
[r1(i),r2(i)]_i=1^n and are to choose an x(i) in each interval so as
to minimize the maximum value of |x(i)-x(j)|. The intervals may
overlap, in which case you may be able to pick equally-spaced points,
or they may not overlap, in which the endpoints of the intervals
become constraints on the selection of the x(i), and may prevent you
from selecting equally-spaced points.
Dave
On Oct 21, 10:21 am, Vivek S s.vivek.ra...@gmail.com wrote:
to maximize the minimum distance between any two points:- to maximize the
minimum distance between adjacent points
- for this all points must be equally spaced.
hence, choose 'n' equally spaced points in the range (r1, r2) starting from
r1 and ending at r2.
2009/10/21 saltycookie saltycoo...@gmail.com
Yes, you are quite right. If I am not mistaken, you give a good solution
for finding the minimum maximum distance.
But what about the original problem where we want to find the maximum
minimum distance? I am not clear about the connection between the two
problems.
Thanks.
2009/10/21 Dave dave_and_da...@juno.com
林夏祥 , think again. If we are trying to minimize the maximum distance,
then we want to minimize the upper bound. That is what I specified:
letting c be the upper bound, find the smallest c such that all of the
distances do not exceed c. That gives rise to the inequalities
|x(i)-x(j)| = c.
If necessary, this can be written as two inequalities:
x(i) - x(j) = c and
x(j) - x(i) = c.
Since the relationship is and, we can just use the two inequalities
as part of the constraint conditions.
Dave
On Oct 21, 12:02 am, 林夏祥 saltycoo...@gmail.com wrote:
I don't think LP can solve it. We are to maximize c, not minimize c.
The formulas we have are:
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
The first inequality actually is combination of two linear equalities:
x(i)
- x(j) = c or x(i) - x(j) = -c. Notice the relation of the two is
or,
and we cannot put them together to get a system of linear inequalities.
2009/10/21 Dave dave_and_da...@juno.com
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
Dave
On Oct 5, 9:22 am, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a
range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--
此致
敬礼！
林夏祥- Hide quoted text -
- Show quoted text -
--
此致
敬礼！
林夏祥
--
Reduce, Reuse and Recycle
Regards,
Vivek.S- Hide quoted text -
- Show quoted text -
--~--~-~--~~~---~--~~
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[algogeeks] Re: Good problem
Hi,
Still waiting for solution...
On Wed, Oct 7, 2009 at 3:18 PM, monty 1987 1986mo...@gmail.com wrote:
The important thing is all the points do not lie in same range i.e.
x1 ,x2 ,x3 each of them have their own range.
On Wed, Oct 7, 2009 at 3:15 PM, monty 1987 1986mo...@gmail.com wrote:
The min. distance between two points i.e. the euclidean distance between
two points.
On Tue, Oct 6, 2009 at 5:52 PM, MrM maleki...@gmail.com wrote:
you can arrange them with equal distances !
if n=1 then, it does not matter where you put the point !
if n1 then, put them with distances = (r2i-r1i) / (n-1) !
it means ou put the first point on r1i and the last point on r2i, the
remaining point are distributed with equal distances !
On Oct 5, 5:22 pm, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a
range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--~--~-~--~~~---~--~~
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[algogeeks] Re: Good problem
Hi,
This problem is same as a Travelling salesman problemIn
travelling salesman we need to cover points in Min distance...here we need
to just opposite..
PS: Answer may be misleading ...if so Pls correct me... :)
-Mithun
On Tue, Oct 20, 2009 at 6:16 PM, monty 1987 1986mo...@gmail.com wrote:
Hi,
Still waiting for solution...
On Wed, Oct 7, 2009 at 3:18 PM, monty 1987 1986mo...@gmail.com wrote:
The important thing is all the points do not lie in same range i.e.
x1 ,x2 ,x3 each of them have their own range.
On Wed, Oct 7, 2009 at 3:15 PM, monty 1987 1986mo...@gmail.com wrote:
The min. distance between two points i.e. the euclidean distance between
two points.
On Tue, Oct 6, 2009 at 5:52 PM, MrM maleki...@gmail.com wrote:
you can arrange them with equal distances !
if n=1 then, it does not matter where you put the point !
if n1 then, put them with distances = (r2i-r1i) / (n-1) !
it means ou put the first point on r1i and the last point on r2i, the
remaining point are distributed with equal distances !
On Oct 5, 5:22 pm, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a
range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
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[algogeeks] Re: Good problem
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| = c for all i and j
r1(i) = x(i) = r2(i) for all i
Dave
On Oct 5, 9:22 am, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
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Algorithm Geeks group.
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[algogeeks] Re: Good problem
The min. distance between two points i.e. the euclidean distance between two
points.
On Tue, Oct 6, 2009 at 5:52 PM, MrM maleki...@gmail.com wrote:
you can arrange them with equal distances !
if n=1 then, it does not matter where you put the point !
if n1 then, put them with distances = (r2i-r1i) / (n-1) !
it means ou put the first point on r1i and the last point on r2i, the
remaining point are distributed with equal distances !
On Oct 5, 5:22 pm, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com
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[algogeeks] Re: Good problem
The important thing is all the points do not lie in same range i.e.
x1 ,x2 ,x3 each of them have their own range.
On Wed, Oct 7, 2009 at 3:15 PM, monty 1987 1986mo...@gmail.com wrote:
The min. distance between two points i.e. the euclidean distance between
two points.
On Tue, Oct 6, 2009 at 5:52 PM, MrM maleki...@gmail.com wrote:
you can arrange them with equal distances !
if n=1 then, it does not matter where you put the point !
if n1 then, put them with distances = (r2i-r1i) / (n-1) !
it means ou put the first point on r1i and the last point on r2i, the
remaining point are distributed with equal distances !
On Oct 5, 5:22 pm, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
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-~--~~~~--~~--~--~---
[algogeeks] Re: Good problem
you can arrange them with equal distances !
if n=1 then, it does not matter where you put the point !
if n1 then, put them with distances = (r2i-r1i) / (n-1) !
it means ou put the first point on r1i and the last point on r2i, the
remaining point are distributed with equal distances !
On Oct 5, 5:22 pm, monty 1987 1986mo...@gmail.com wrote:
We have to locate n points on the x-axis
For each point xi
the x co-ordinate of it lies between a range
[r1i,r2i]
Now we have to decide the location of points such that
minimum { distance between any two points } is maximum.
Any answer is welcomed.
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
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