Check for Point Collinearity and rearrange line
$begingroup$
imagine you are having a 2D object, which looks like this:
$$
l = (p_1,p_2,...,p_n)
$$
with $p_i = (x_i, y_i)$.
It may be possible that $p_1 = p_n$, or that $l$ only contains two points, which are not the same. (Am i correct that even if $p_1 = p_n$ the object is still considered a line-like object ? Since in this case it would basically represent the shell of a polygon without a surface.)
My Question is:
In $l$ there are points which are not collinear, meaning that $l$ is not just a straight line. Now my goal is to extract all straight lines from $l$, such that $l$ is seperated into a set of new lines. If you would plot the new lines all together, it would be equivalent to plotting $l$.
Does someone has an idea how to do that correctly, and how this process is adequately notated?
Greetings !
geometry
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 16:03
EkkoEkko
12
$endgroup$
add a comment |
$begingroup$
imagine you are having a 2D object, which looks like this:
$$
l = (p_1,p_2,...,p_n)
$$
with $p_i = (x_i, y_i)$.
It may be possible that $p_1 = p_n$, or that $l$ only contains two points, which are not the same. (Am i correct that even if $p_1 = p_n$ the object is still considered a line-like object ? Since in this case it would basically represent the shell of a polygon without a surface.)
My Question is:
In $l$ there are points which are not collinear, meaning that $l$ is not just a straight line. Now my goal is to extract all straight lines from $l$, such that $l$ is seperated into a set of new lines. If you would plot the new lines all together, it would be equivalent to plotting $l$.
Does someone has an idea how to do that correctly, and how this process is adequately notated?
Greetings !
geometry
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 16:03
EkkoEkko
12
$endgroup$
$begingroup$
If your set $l$ is ordered, and you're trying to draw the line segments from one point to the next, you could do each line segment like this: $(1-t)p_i+t p_{i+1},$ as $t$ ranges from $0$ to $1$.
$endgroup$
– Adrian Keister
Dec 11 '18 at 16:07
$begingroup$
How exactly do you want to seperate $l$ into new lines? Do you want to seperate the polygon in a minimal partition or do you want to draw only new segments of each line $[p_i,p_{i+1}]$?
$endgroup$
– bruderjakob17
Dec 11 '18 at 16:40
$begingroup$
If i use $(p_{i}, p_{i+1})$ to create new lines this would result in a giant set of lines. As an example, imagine $l = {(1,1), (2,2), (3,3), (2,4), (1,5) }$, in this case i want the result to be a set of two lines : {(1,1), (3,3)} and {(3,3),(1,5)} instead of 4 lines like in the approach of using $(p_i, p_{i+1})$.
$endgroup$
– Ekko
Dec 11 '18 at 16:49
add a comment |
$begingroup$
imagine you are having a 2D object, which looks like this:
$$
l = (p_1,p_2,...,p_n)
$$
with $p_i = (x_i, y_i)$.
It may be possible that $p_1 = p_n$, or that $l$ only contains two points, which are not the same. (Am i correct that even if $p_1 = p_n$ the object is still considered a line-like object ? Since in this case it would basically represent the shell of a polygon without a surface.)
My Question is:
In $l$ there are points which are not collinear, meaning that $l$ is not just a straight line. Now my goal is to extract all straight lines from $l$, such that $l$ is seperated into a set of new lines. If you would plot the new lines all together, it would be equivalent to plotting $l$.
Does someone has an idea how to do that correctly, and how this process is adequately notated?
Greetings !
geometry
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 16:03
EkkoEkko
12
$endgroup$
imagine you are having a 2D object, which looks like this:
$$
l = (p_1,p_2,...,p_n)
$$
with $p_i = (x_i, y_i)$.
It may be possible that $p_1 = p_n$, or that $l$ only contains two points, which are not the same. (Am i correct that even if $p_1 = p_n$ the object is still considered a line-like object ? Since in this case it would basically represent the shell of a polygon without a surface.)
My Question is:
In $l$ there are points which are not collinear, meaning that $l$ is not just a straight line. Now my goal is to extract all straight lines from $l$, such that $l$ is seperated into a set of new lines. If you would plot the new lines all together, it would be equivalent to plotting $l$.
Does someone has an idea how to do that correctly, and how this process is adequately notated?
Greetings !
geometry
geometry
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 16:03
EkkoEkko
12
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
asked Dec 11 '18 at 16:03
EkkoEkko
12
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
edited Dec 11 '18 at 16:11
Asaf Karagila♦
305k33435766
305k33435766
asked Dec 11 '18 at 16:03
EkkoEkko
12
asked Dec 11 '18 at 16:03
EkkoEkko
12
asked Dec 11 '18 at 16:03
EkkoEkko
12
12
$begingroup$
If your set $l$ is ordered, and you're trying to draw the line segments from one point to the next, you could do each line segment like this: $(1-t)p_i+t p_{i+1},$ as $t$ ranges from $0$ to $1$.
$endgroup$
– Adrian Keister
Dec 11 '18 at 16:07
$begingroup$
How exactly do you want to seperate $l$ into new lines? Do you want to seperate the polygon in a minimal partition or do you want to draw only new segments of each line $[p_i,p_{i+1}]$?
$endgroup$
– bruderjakob17
Dec 11 '18 at 16:40
$begingroup$
If i use $(p_{i}, p_{i+1})$ to create new lines this would result in a giant set of lines. As an example, imagine $l = {(1,1), (2,2), (3,3), (2,4), (1,5) }$, in this case i want the result to be a set of two lines : {(1,1), (3,3)} and {(3,3),(1,5)} instead of 4 lines like in the approach of using $(p_i, p_{i+1})$.
$endgroup$
– Ekko
Dec 11 '18 at 16:49
add a comment |
$begingroup$
If your set $l$ is ordered, and you're trying to draw the line segments from one point to the next, you could do each line segment like this: $(1-t)p_i+t p_{i+1},$ as $t$ ranges from $0$ to $1$.
$endgroup$
– Adrian Keister
Dec 11 '18 at 16:07
$begingroup$
How exactly do you want to seperate $l$ into new lines? Do you want to seperate the polygon in a minimal partition or do you want to draw only new segments of each line $[p_i,p_{i+1}]$?
$endgroup$
– bruderjakob17
Dec 11 '18 at 16:40
$begingroup$
If i use $(p_{i}, p_{i+1})$ to create new lines this would result in a giant set of lines. As an example, imagine $l = {(1,1), (2,2), (3,3), (2,4), (1,5) }$, in this case i want the result to be a set of two lines : {(1,1), (3,3)} and {(3,3),(1,5)} instead of 4 lines like in the approach of using $(p_i, p_{i+1})$.
$endgroup$
– Ekko
Dec 11 '18 at 16:49
$begingroup$
If your set $l$ is ordered, and you're trying to draw the line segments from one point to the next, you could do each line segment like this: $(1-t)p_i+t p_{i+1},$ as $t$ ranges from $0$ to $1$.
$endgroup$
– Adrian Keister
Dec 11 '18 at 16:07
$begingroup$
If your set $l$ is ordered, and you're trying to draw the line segments from one point to the next, you could do each line segment like this: $(1-t)p_i+t p_{i+1},$ as $t$ ranges from $0$ to $1$.
$endgroup$
– Adrian Keister
Dec 11 '18 at 16:07
$begingroup$
How exactly do you want to seperate $l$ into new lines? Do you want to seperate the polygon in a minimal partition or do you want to draw only new segments of each line $[p_i,p_{i+1}]$?
$endgroup$
– bruderjakob17
Dec 11 '18 at 16:40
$begingroup$
How exactly do you want to seperate $l$ into new lines? Do you want to seperate the polygon in a minimal partition or do you want to draw only new segments of each line $[p_i,p_{i+1}]$?
$endgroup$
– bruderjakob17
Dec 11 '18 at 16:40
$begingroup$
If i use $(p_{i}, p_{i+1})$ to create new lines this would result in a giant set of lines. As an example, imagine $l = {(1,1), (2,2), (3,3), (2,4), (1,5) }$, in this case i want the result to be a set of two lines : {(1,1), (3,3)} and {(3,3),(1,5)} instead of 4 lines like in the approach of using $(p_i, p_{i+1})$.
$endgroup$
– Ekko
Dec 11 '18 at 16:49
$begingroup$
If i use $(p_{i}, p_{i+1})$ to create new lines this would result in a giant set of lines. As an example, imagine $l = {(1,1), (2,2), (3,3), (2,4), (1,5) }$, in this case i want the result to be a set of two lines : {(1,1), (3,3)} and {(3,3),(1,5)} instead of 4 lines like in the approach of using $(p_i, p_{i+1})$.
$endgroup$
– Ekko
Dec 11 '18 at 16:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Set $V:= emptyset$.
Now, for $i = 1,dots, n-1$, do the following:
If $p_i = p_{i+1}$, continue.
Check, whether there is a tupel $(a,b,X)in V$ s.t.
$DeclareMathOperator{span}{span} p_i,p_{i+1} in a +
span{{b-a}}$.
If that is the case, write $p_i = a + lambda (b-a)$ and $p_{i+1} = a
+ mu (b-a)$. Without loss we may assume $lambda > mu$. Now, replace $(a,b,X)in V$ by $DeclareMathOperator{contract}{contract}
(a,b,contract{(Xcup {[mu, lambda]})})$, where $contract({Y_1, dots, Y_k}):=begin{cases}contract({Y_1,dots ,Y_k}backslash{Y_i,Y_j}cup{Y_i cup Y_j})& text{if there are $ineq j$ s.t. $Y_i cap Y_j neq emptyset$}\
{Y_1, dots , Y_k} & text{else}end{cases}$Else, add $(p_i,p_{i+1}, {[0,1]})$ to $V$.
Now, you can draw all lines you need as follows: For every $(a,b,X) in V$ and for all $[mu,lambda]in X$, draw $[a+mu (b-a), a+lambda (b-a)] := {a + xi (b-a) space | space xi in [mu, lambda]}$.
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
$endgroup$
add a comment |
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$begingroup$
Set $V:= emptyset$.
Now, for $i = 1,dots, n-1$, do the following:
If $p_i = p_{i+1}$, continue.
Check, whether there is a tupel $(a,b,X)in V$ s.t.
$DeclareMathOperator{span}{span} p_i,p_{i+1} in a +
span{{b-a}}$.
If that is the case, write $p_i = a + lambda (b-a)$ and $p_{i+1} = a
+ mu (b-a)$. Without loss we may assume $lambda > mu$. Now, replace $(a,b,X)in V$ by $DeclareMathOperator{contract}{contract}
(a,b,contract{(Xcup {[mu, lambda]})})$, where $contract({Y_1, dots, Y_k}):=begin{cases}contract({Y_1,dots ,Y_k}backslash{Y_i,Y_j}cup{Y_i cup Y_j})& text{if there are $ineq j$ s.t. $Y_i cap Y_j neq emptyset$}\
{Y_1, dots , Y_k} & text{else}end{cases}$Else, add $(p_i,p_{i+1}, {[0,1]})$ to $V$.
Now, you can draw all lines you need as follows: For every $(a,b,X) in V$ and for all $[mu,lambda]in X$, draw $[a+mu (b-a), a+lambda (b-a)] := {a + xi (b-a) space | space xi in [mu, lambda]}$.
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
$endgroup$
add a comment |
$begingroup$
Set $V:= emptyset$.
Now, for $i = 1,dots, n-1$, do the following:
If $p_i = p_{i+1}$, continue.
Check, whether there is a tupel $(a,b,X)in V$ s.t.
$DeclareMathOperator{span}{span} p_i,p_{i+1} in a +
span{{b-a}}$.
If that is the case, write $p_i = a + lambda (b-a)$ and $p_{i+1} = a
+ mu (b-a)$. Without loss we may assume $lambda > mu$. Now, replace $(a,b,X)in V$ by $DeclareMathOperator{contract}{contract}
(a,b,contract{(Xcup {[mu, lambda]})})$, where $contract({Y_1, dots, Y_k}):=begin{cases}contract({Y_1,dots ,Y_k}backslash{Y_i,Y_j}cup{Y_i cup Y_j})& text{if there are $ineq j$ s.t. $Y_i cap Y_j neq emptyset$}\
{Y_1, dots , Y_k} & text{else}end{cases}$Else, add $(p_i,p_{i+1}, {[0,1]})$ to $V$.
Now, you can draw all lines you need as follows: For every $(a,b,X) in V$ and for all $[mu,lambda]in X$, draw $[a+mu (b-a), a+lambda (b-a)] := {a + xi (b-a) space | space xi in [mu, lambda]}$.
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
$endgroup$
add a comment |
$begingroup$
Set $V:= emptyset$.
Now, for $i = 1,dots, n-1$, do the following:
If $p_i = p_{i+1}$, continue.
Check, whether there is a tupel $(a,b,X)in V$ s.t.
$DeclareMathOperator{span}{span} p_i,p_{i+1} in a +
span{{b-a}}$.
If that is the case, write $p_i = a + lambda (b-a)$ and $p_{i+1} = a
+ mu (b-a)$. Without loss we may assume $lambda > mu$. Now, replace $(a,b,X)in V$ by $DeclareMathOperator{contract}{contract}
(a,b,contract{(Xcup {[mu, lambda]})})$, where $contract({Y_1, dots, Y_k}):=begin{cases}contract({Y_1,dots ,Y_k}backslash{Y_i,Y_j}cup{Y_i cup Y_j})& text{if there are $ineq j$ s.t. $Y_i cap Y_j neq emptyset$}\
{Y_1, dots , Y_k} & text{else}end{cases}$Else, add $(p_i,p_{i+1}, {[0,1]})$ to $V$.
Now, you can draw all lines you need as follows: For every $(a,b,X) in V$ and for all $[mu,lambda]in X$, draw $[a+mu (b-a), a+lambda (b-a)] := {a + xi (b-a) space | space xi in [mu, lambda]}$.
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
$endgroup$
Set $V:= emptyset$.
Now, for $i = 1,dots, n-1$, do the following:
If $p_i = p_{i+1}$, continue.
Check, whether there is a tupel $(a,b,X)in V$ s.t.
$DeclareMathOperator{span}{span} p_i,p_{i+1} in a +
span{{b-a}}$.
If that is the case, write $p_i = a + lambda (b-a)$ and $p_{i+1} = a
+ mu (b-a)$. Without loss we may assume $lambda > mu$. Now, replace $(a,b,X)in V$ by $DeclareMathOperator{contract}{contract}
(a,b,contract{(Xcup {[mu, lambda]})})$, where $contract({Y_1, dots, Y_k}):=begin{cases}contract({Y_1,dots ,Y_k}backslash{Y_i,Y_j}cup{Y_i cup Y_j})& text{if there are $ineq j$ s.t. $Y_i cap Y_j neq emptyset$}\
{Y_1, dots , Y_k} & text{else}end{cases}$Else, add $(p_i,p_{i+1}, {[0,1]})$ to $V$.
Now, you can draw all lines you need as follows: For every $(a,b,X) in V$ and for all $[mu,lambda]in X$, draw $[a+mu (b-a), a+lambda (b-a)] := {a + xi (b-a) space | space xi in [mu, lambda]}$.
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
answered Dec 16 '18 at 10:32
bruderjakob17bruderjakob17
416110
416110
add a comment |
add a comment |
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$begingroup$
If your set $l$ is ordered, and you're trying to draw the line segments from one point to the next, you could do each line segment like this: $(1-t)p_i+t p_{i+1},$ as $t$ ranges from $0$ to $1$.
$endgroup$
– Adrian Keister
Dec 11 '18 at 16:07
$begingroup$
How exactly do you want to seperate $l$ into new lines? Do you want to seperate the polygon in a minimal partition or do you want to draw only new segments of each line $[p_i,p_{i+1}]$?
$endgroup$
– bruderjakob17
Dec 11 '18 at 16:40
$begingroup$
If i use $(p_{i}, p_{i+1})$ to create new lines this would result in a giant set of lines. As an example, imagine $l = {(1,1), (2,2), (3,3), (2,4), (1,5) }$, in this case i want the result to be a set of two lines : {(1,1), (3,3)} and {(3,3),(1,5)} instead of 4 lines like in the approach of using $(p_i, p_{i+1})$.
$endgroup$
– Ekko
Dec 11 '18 at 16:49