Mixing
Projection On Parallelogram
In my research, I encountered the problem of finding the projection of any point on given parallelogram with the following properties :
- If the point is in region 1, the projection should be vertex A.
- If the point is in region 2, the projection should be on segment AB parallel to Y-axis.
- If the point is in region 3, the projection should be vertex B.
And so on.
The equations of parallelogram sides are also given in the figure.
I am trying to find out a computational way to figure out this projection.
I tried to set up the problem as a minimum L1 distance to parallelogram from the given point. But, I realized that this is not the correct projection.
Can some one provide help on what would be the efficient computational way to figure out this kind of projection? I am totally new to this field - Any references are highly appreciated.
Thank you very much!
geometry reference-request coordinate-systems projective-geometry computational-geometry
edited Nov 27 '18 at 9:26
MBorg
1751114
asked Nov 15 '18 at 16:55
chandu1729
2,371821
add a comment |
In my research, I encountered the problem of finding the projection of any point on given parallelogram with the following properties :
- If the point is in region 1, the projection should be vertex A.
- If the point is in region 2, the projection should be on segment AB parallel to Y-axis.
- If the point is in region 3, the projection should be vertex B.
And so on.
The equations of parallelogram sides are also given in the figure.
I am trying to find out a computational way to figure out this projection.
I tried to set up the problem as a minimum L1 distance to parallelogram from the given point. But, I realized that this is not the correct projection.
Can some one provide help on what would be the efficient computational way to figure out this kind of projection? I am totally new to this field - Any references are highly appreciated.
Thank you very much!
geometry reference-request coordinate-systems projective-geometry computational-geometry
edited Nov 27 '18 at 9:26
MBorg
1751114
asked Nov 15 '18 at 16:55
chandu1729
2,371821
There are some conditions to make clear: a) you said that AB is parallel to y axis, but then what are the dotted lines out of B indicating ? b) the arrow indicating the projection, is it normal to the edge ?, parallel to the axes ? ... c) the drawing is so distorted that looks to be in 3D, but equations are in 2d (?) You should provide a much cleaner scheme , before thinking to equations
– G Cab
Nov 15 '18 at 17:24
@GCab: a) AB is not parallel to y-axis. I meant the projection is parallel to y-axis. For example, let X be the given point in region 2 then Y is projection of X such that XY is parallel to Y-axis. b) Parallel to axes. c) The diagram is 2D and the shape is parallelogram and the describing equations are also in the diagram. Thank you very much for any help!
– chandu1729
Nov 16 '18 at 5:50
add a comment |
In my research, I encountered the problem of finding the projection of any point on given parallelogram with the following properties :
- If the point is in region 1, the projection should be vertex A.
- If the point is in region 2, the projection should be on segment AB parallel to Y-axis.
- If the point is in region 3, the projection should be vertex B.
And so on.
The equations of parallelogram sides are also given in the figure.
I am trying to find out a computational way to figure out this projection.
I tried to set up the problem as a minimum L1 distance to parallelogram from the given point. But, I realized that this is not the correct projection.
Can some one provide help on what would be the efficient computational way to figure out this kind of projection? I am totally new to this field - Any references are highly appreciated.
Thank you very much!
geometry reference-request coordinate-systems projective-geometry computational-geometry
edited Nov 27 '18 at 9:26
MBorg
1751114
asked Nov 15 '18 at 16:55
chandu1729
2,371821
In my research, I encountered the problem of finding the projection of any point on given parallelogram with the following properties :
- If the point is in region 1, the projection should be vertex A.
- If the point is in region 2, the projection should be on segment AB parallel to Y-axis.
- If the point is in region 3, the projection should be vertex B.
And so on.
The equations of parallelogram sides are also given in the figure.
I am trying to find out a computational way to figure out this projection.
I tried to set up the problem as a minimum L1 distance to parallelogram from the given point. But, I realized that this is not the correct projection.
Can some one provide help on what would be the efficient computational way to figure out this kind of projection? I am totally new to this field - Any references are highly appreciated.
Thank you very much!
geometry reference-request coordinate-systems projective-geometry computational-geometry
geometry reference-request coordinate-systems projective-geometry computational-geometry
edited Nov 27 '18 at 9:26
MBorg
1751114
asked Nov 15 '18 at 16:55
chandu1729
2,371821
edited Nov 27 '18 at 9:26
MBorg
1751114
asked Nov 15 '18 at 16:55
chandu1729
2,371821
edited Nov 27 '18 at 9:26
MBorg
1751114
edited Nov 27 '18 at 9:26
MBorg
1751114
edited Nov 27 '18 at 9:26
MBorg
1751114
1751114
asked Nov 15 '18 at 16:55
chandu1729
2,371821
asked Nov 15 '18 at 16:55
chandu1729
2,371821
asked Nov 15 '18 at 16:55
chandu1729
2,371821
2,371821
There are some conditions to make clear: a) you said that AB is parallel to y axis, but then what are the dotted lines out of B indicating ? b) the arrow indicating the projection, is it normal to the edge ?, parallel to the axes ? ... c) the drawing is so distorted that looks to be in 3D, but equations are in 2d (?) You should provide a much cleaner scheme , before thinking to equations
– G Cab
Nov 15 '18 at 17:24
@GCab: a) AB is not parallel to y-axis. I meant the projection is parallel to y-axis. For example, let X be the given point in region 2 then Y is projection of X such that XY is parallel to Y-axis. b) Parallel to axes. c) The diagram is 2D and the shape is parallelogram and the describing equations are also in the diagram. Thank you very much for any help!
– chandu1729
Nov 16 '18 at 5:50
add a comment |
There are some conditions to make clear: a) you said that AB is parallel to y axis, but then what are the dotted lines out of B indicating ? b) the arrow indicating the projection, is it normal to the edge ?, parallel to the axes ? ... c) the drawing is so distorted that looks to be in 3D, but equations are in 2d (?) You should provide a much cleaner scheme , before thinking to equations
– G Cab
Nov 15 '18 at 17:24
@GCab: a) AB is not parallel to y-axis. I meant the projection is parallel to y-axis. For example, let X be the given point in region 2 then Y is projection of X such that XY is parallel to Y-axis. b) Parallel to axes. c) The diagram is 2D and the shape is parallelogram and the describing equations are also in the diagram. Thank you very much for any help!
– chandu1729
Nov 16 '18 at 5:50
There are some conditions to make clear: a) you said that AB is parallel to y axis, but then what are the dotted lines out of B indicating ? b) the arrow indicating the projection, is it normal to the edge ?, parallel to the axes ? ... c) the drawing is so distorted that looks to be in 3D, but equations are in 2d (?) You should provide a much cleaner scheme , before thinking to equations
– G Cab
Nov 15 '18 at 17:24
There are some conditions to make clear: a) you said that AB is parallel to y axis, but then what are the dotted lines out of B indicating ? b) the arrow indicating the projection, is it normal to the edge ?, parallel to the axes ? ... c) the drawing is so distorted that looks to be in 3D, but equations are in 2d (?) You should provide a much cleaner scheme , before thinking to equations
– G Cab
Nov 15 '18 at 17:24
@GCab: a) AB is not parallel to y-axis. I meant the projection is parallel to y-axis. For example, let X be the given point in region 2 then Y is projection of X such that XY is parallel to Y-axis. b) Parallel to axes. c) The diagram is 2D and the shape is parallelogram and the describing equations are also in the diagram. Thank you very much for any help!
– chandu1729
Nov 16 '18 at 5:50
@GCab: a) AB is not parallel to y-axis. I meant the projection is parallel to y-axis. For example, let X be the given point in region 2 then Y is projection of X such that XY is parallel to Y-axis. b) Parallel to axes. c) The diagram is 2D and the shape is parallelogram and the describing equations are also in the diagram. Thank you very much for any help!
– chandu1729
Nov 16 '18 at 5:50
add a comment |
2 Answers
2
active
oldest
votes
So, each couple of lines given by
$$
b_{,1,,1} x + b_{,1,,2} y = pm 1
$$
and
$$
b_{,2,,1} x + b_{,2,,2} y = pm 1
$$
denotes a couple of lines symmetrical wrt the origin.
The four lines defines a parallelogram centered at the origin.
The four vertices are given by
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
; : ; left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)left( {matrix{ x cr y cr } } right)
= left( {matrix{
1 & 1 & { - 1} & { - 1} cr
1 & { - 1} & 1 & { - 1} cr
} } right)
$$
i.e.:
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= left( {matrix{ {b_{,1,,1} } & {b_{,1,,2} } cr {b_{,2,,1} } & {b_{,2,,2} } cr } } right)^{, - ,1}
left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
If some of the $b$ values are negative the denomination of the A points will change
with respect to that shown in the figure: it is better to rearrange the denominations as to match
with that shown, i.e. lower, higher $x$, lower,higher $y$ coordinates.
Now, at each vertex attach three unitary vectors as shown.
Compute the projection of the vector $A_{p,m}P$ onto the unitary vectors $bf x_{pm}$ and $bf y_{pm}$ stemming from $A_{p,m}$.
If both projections are positive, then the point P lies in the $90^circ$ sector encompassed by those vectors
and $P$ projects onto the vertex itself.
Repeat for the other vertices and check if both projections are positive.
If P is not in any of the sector above, then express the vector $A_{p,m}P$ in the reference system given by $bf x_{pm}, , bf v_{pm}$.
If the coordinates in that system are both positive, then $P$ projects on the edge $A_{pm}A_{pp}$, and its
coordinate wrt $bf v_{pm}$ gives the distance from the vertex $A_{pm}.
You can see that the reference system $bf y_{pp}, , bf v_{pp}$ in $A_{pp}$ will cover the edge $A_{pp}A_{mp}$, and so on.
You can also see that $P$, if external to the parallelogram, can have both coordinates positive only in one of the four systems.
To translate the above into a computational algorithm we can proceed along the following lines.
a) Construct the matrix
$$
{bf B} = left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)
$$
ensure that its determinant is non null and compute $bf B^{-1}$.
b) Compute the four points A
$$
left( {matrix{ {A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= {bf B}^{, - ,1} left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
c) Rearrange points A
for instance by taking as pivot the point with minimal abscissa ($A_0$ in the sketch) and arranging the others
according to the tangent of the angle that they form with the pivot (if $Delta x = 0 ; to ; Delta x = epsilon $);
re-name the suffixes to $(0,1,2,3)$ so that they increase along the perimeter (e.g. ACW).
d) Construct unit vectors $bf u_n, ; v_n$
$$
eqalign{
& {bf v}_{,n} = overrightarrow {A_{,n} A_{,left( {n + 1} right)} } mathop {;/}limits_{} ;left| {overrightarrow {A_{,n} A_{,left( {n + 1} right)} } } right|quad left( {nbmod 4} right) cr
& {bf u}_{,n} = left( {matrix{
{cos left( {left( {n - 2} right)pi /2} right)} cr
{sin left( {left( {n - 2} right)pi /2} right)} cr
} } right) cr}
$$
e) Allocate the point P in the corresponding sector
construct the two lists
$$
L_{,x} = left[ {A_{,n,,x} ,P_{,x} } right]quad L_{,y} = left[ {A_{,n,,y} ,P_{,y} } right]
$$
and sort them in increasing order;
f) use the rank indices of P in the ordered lists to determine its position and projection
if (corner sector in $A_0$)
$P_{,x} < A_{,0,,x} ; wedge ;A_{,0,,y} < P_{,y} $ then $P to A_{,0} $ exit;
elif (lateral sector below $A_0$)
$P_{,x} < A_{,1,,x} ; wedge ;A_{,1,,y} < P_{,y} < A_{,0,,y} $ then
- take the matrix ${bf V}_{,0} = left( {matrix{ {{bf u}_{,0} } & {{bf v}_{,0} } cr } } right)$
- determine the coordinates of $P$ in that system
$$
left( {matrix{ {P'_x } cr {P'_y } cr } } right)
= {bf V}_{,0} ^{, - ,{bf 1}} ;left( {matrix{ {P_x } cr {P_y } cr } } right)
$$
- if $P'_y<0 ; to$ error;
- elif $P'_x<0 ; to$ the point is internal to the p., $to$ ?
- else $P; to ;A_{,0} + P'_y {bf v}_{,0} $
elif
....
(continue with the same path for the other sectors)
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
Thanks for the response. I am not sure if I can code this - Do you think it can be expressed in a computational way? For example: We would like to write a function which takes the matrix B and point P (x,y) as inputs and computes (X,Y) as projection? Also, I would like to extend this Algorithm to high dimensions - so checking all the vertices will be susceptible to curse of dimensionality! Please let me know. Thanks Much.
– chandu1729
Nov 17 '18 at 19:27
@chandu1729: well I developed the answer mainly mathematically, but keeping an eye on computability. Mathematically (and computationally) I cannot see a better way. And it can be extended to 3D and higher. I am quite sure it is tranformable in an algorithm.
– G Cab
Nov 17 '18 at 21:16
@chandu1729: the difficulty, is mainly on rotation /reflection of the points for negative signs on the $b$'s. If the $b$'s are positive it should be quite viable.
– G Cab
Nov 17 '18 at 21:20
If I understand correctly, you are looking/traversing at each vertex. In n-dimensions there will be 2^n vertices which makes it exponential complexity!
– chandu1729
Nov 17 '18 at 21:24
I would highly appreciate if you can express your answer in an algorithmic way for general n-dimensional matrix B and point P. That will be easier for me to understand as well.
– chandu1729
Nov 17 '18 at 21:26
|
show 5 more comments
In my solution $x_1, x_2$ are replaced by $x,y.$
The solution should be easy to code.
Assume the given straight lines are not horizontal neither vertical. In this case are all coefficients $b_{ij}neq 0.$
From the coefficients $pm 1$ in the equations follows that the parallelogram is centered in $O.$
Thus if $A(x_A,y_A), B(x_B,y_B),$ then necessarily $C(-x_A,-y_A), D(-x_B,-y_B).$
In the enclosed figure are the equations of the straight lines $color{blue}{blue},$ and located across the corresponding line.
Each region is determined by a system of inequations written in black.
The input point $P(x,y)$ has to be transformed to an output $P'(X,Y).$ These new coordinates $X,Y$ are written in the corresponding region in $color{red}{red}.$
Remark
If a pair of lines is horizontal or vertical (the product $prod_{i,j=1,2} b_{ij}=0$ ), the above solution has to be modified, but is easier.
answered Nov 24 '18 at 13:39
user376343
2,8582823
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
So, each couple of lines given by
$$
b_{,1,,1} x + b_{,1,,2} y = pm 1
$$
and
$$
b_{,2,,1} x + b_{,2,,2} y = pm 1
$$
denotes a couple of lines symmetrical wrt the origin.
The four lines defines a parallelogram centered at the origin.
The four vertices are given by
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
; : ; left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)left( {matrix{ x cr y cr } } right)
= left( {matrix{
1 & 1 & { - 1} & { - 1} cr
1 & { - 1} & 1 & { - 1} cr
} } right)
$$
i.e.:
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= left( {matrix{ {b_{,1,,1} } & {b_{,1,,2} } cr {b_{,2,,1} } & {b_{,2,,2} } cr } } right)^{, - ,1}
left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
If some of the $b$ values are negative the denomination of the A points will change
with respect to that shown in the figure: it is better to rearrange the denominations as to match
with that shown, i.e. lower, higher $x$, lower,higher $y$ coordinates.
Now, at each vertex attach three unitary vectors as shown.
Compute the projection of the vector $A_{p,m}P$ onto the unitary vectors $bf x_{pm}$ and $bf y_{pm}$ stemming from $A_{p,m}$.
If both projections are positive, then the point P lies in the $90^circ$ sector encompassed by those vectors
and $P$ projects onto the vertex itself.
Repeat for the other vertices and check if both projections are positive.
If P is not in any of the sector above, then express the vector $A_{p,m}P$ in the reference system given by $bf x_{pm}, , bf v_{pm}$.
If the coordinates in that system are both positive, then $P$ projects on the edge $A_{pm}A_{pp}$, and its
coordinate wrt $bf v_{pm}$ gives the distance from the vertex $A_{pm}.
You can see that the reference system $bf y_{pp}, , bf v_{pp}$ in $A_{pp}$ will cover the edge $A_{pp}A_{mp}$, and so on.
You can also see that $P$, if external to the parallelogram, can have both coordinates positive only in one of the four systems.
To translate the above into a computational algorithm we can proceed along the following lines.
a) Construct the matrix
$$
{bf B} = left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)
$$
ensure that its determinant is non null and compute $bf B^{-1}$.
b) Compute the four points A
$$
left( {matrix{ {A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= {bf B}^{, - ,1} left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
c) Rearrange points A
for instance by taking as pivot the point with minimal abscissa ($A_0$ in the sketch) and arranging the others
according to the tangent of the angle that they form with the pivot (if $Delta x = 0 ; to ; Delta x = epsilon $);
re-name the suffixes to $(0,1,2,3)$ so that they increase along the perimeter (e.g. ACW).
d) Construct unit vectors $bf u_n, ; v_n$
$$
eqalign{
& {bf v}_{,n} = overrightarrow {A_{,n} A_{,left( {n + 1} right)} } mathop {;/}limits_{} ;left| {overrightarrow {A_{,n} A_{,left( {n + 1} right)} } } right|quad left( {nbmod 4} right) cr
& {bf u}_{,n} = left( {matrix{
{cos left( {left( {n - 2} right)pi /2} right)} cr
{sin left( {left( {n - 2} right)pi /2} right)} cr
} } right) cr}
$$
e) Allocate the point P in the corresponding sector
construct the two lists
$$
L_{,x} = left[ {A_{,n,,x} ,P_{,x} } right]quad L_{,y} = left[ {A_{,n,,y} ,P_{,y} } right]
$$
and sort them in increasing order;
f) use the rank indices of P in the ordered lists to determine its position and projection
if (corner sector in $A_0$)
$P_{,x} < A_{,0,,x} ; wedge ;A_{,0,,y} < P_{,y} $ then $P to A_{,0} $ exit;
elif (lateral sector below $A_0$)
$P_{,x} < A_{,1,,x} ; wedge ;A_{,1,,y} < P_{,y} < A_{,0,,y} $ then
- take the matrix ${bf V}_{,0} = left( {matrix{ {{bf u}_{,0} } & {{bf v}_{,0} } cr } } right)$
- determine the coordinates of $P$ in that system
$$
left( {matrix{ {P'_x } cr {P'_y } cr } } right)
= {bf V}_{,0} ^{, - ,{bf 1}} ;left( {matrix{ {P_x } cr {P_y } cr } } right)
$$
- if $P'_y<0 ; to$ error;
- elif $P'_x<0 ; to$ the point is internal to the p., $to$ ?
- else $P; to ;A_{,0} + P'_y {bf v}_{,0} $
elif
....
(continue with the same path for the other sectors)
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
Thanks for the response. I am not sure if I can code this - Do you think it can be expressed in a computational way? For example: We would like to write a function which takes the matrix B and point P (x,y) as inputs and computes (X,Y) as projection? Also, I would like to extend this Algorithm to high dimensions - so checking all the vertices will be susceptible to curse of dimensionality! Please let me know. Thanks Much.
– chandu1729
Nov 17 '18 at 19:27
@chandu1729: well I developed the answer mainly mathematically, but keeping an eye on computability. Mathematically (and computationally) I cannot see a better way. And it can be extended to 3D and higher. I am quite sure it is tranformable in an algorithm.
– G Cab
Nov 17 '18 at 21:16
@chandu1729: the difficulty, is mainly on rotation /reflection of the points for negative signs on the $b$'s. If the $b$'s are positive it should be quite viable.
– G Cab
Nov 17 '18 at 21:20
If I understand correctly, you are looking/traversing at each vertex. In n-dimensions there will be 2^n vertices which makes it exponential complexity!
– chandu1729
Nov 17 '18 at 21:24
I would highly appreciate if you can express your answer in an algorithmic way for general n-dimensional matrix B and point P. That will be easier for me to understand as well.
– chandu1729
Nov 17 '18 at 21:26
|
show 5 more comments
So, each couple of lines given by
$$
b_{,1,,1} x + b_{,1,,2} y = pm 1
$$
and
$$
b_{,2,,1} x + b_{,2,,2} y = pm 1
$$
denotes a couple of lines symmetrical wrt the origin.
The four lines defines a parallelogram centered at the origin.
The four vertices are given by
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
; : ; left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)left( {matrix{ x cr y cr } } right)
= left( {matrix{
1 & 1 & { - 1} & { - 1} cr
1 & { - 1} & 1 & { - 1} cr
} } right)
$$
i.e.:
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= left( {matrix{ {b_{,1,,1} } & {b_{,1,,2} } cr {b_{,2,,1} } & {b_{,2,,2} } cr } } right)^{, - ,1}
left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
If some of the $b$ values are negative the denomination of the A points will change
with respect to that shown in the figure: it is better to rearrange the denominations as to match
with that shown, i.e. lower, higher $x$, lower,higher $y$ coordinates.
Now, at each vertex attach three unitary vectors as shown.
Compute the projection of the vector $A_{p,m}P$ onto the unitary vectors $bf x_{pm}$ and $bf y_{pm}$ stemming from $A_{p,m}$.
If both projections are positive, then the point P lies in the $90^circ$ sector encompassed by those vectors
and $P$ projects onto the vertex itself.
Repeat for the other vertices and check if both projections are positive.
If P is not in any of the sector above, then express the vector $A_{p,m}P$ in the reference system given by $bf x_{pm}, , bf v_{pm}$.
If the coordinates in that system are both positive, then $P$ projects on the edge $A_{pm}A_{pp}$, and its
coordinate wrt $bf v_{pm}$ gives the distance from the vertex $A_{pm}.
You can see that the reference system $bf y_{pp}, , bf v_{pp}$ in $A_{pp}$ will cover the edge $A_{pp}A_{mp}$, and so on.
You can also see that $P$, if external to the parallelogram, can have both coordinates positive only in one of the four systems.
To translate the above into a computational algorithm we can proceed along the following lines.
a) Construct the matrix
$$
{bf B} = left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)
$$
ensure that its determinant is non null and compute $bf B^{-1}$.
b) Compute the four points A
$$
left( {matrix{ {A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= {bf B}^{, - ,1} left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
c) Rearrange points A
for instance by taking as pivot the point with minimal abscissa ($A_0$ in the sketch) and arranging the others
according to the tangent of the angle that they form with the pivot (if $Delta x = 0 ; to ; Delta x = epsilon $);
re-name the suffixes to $(0,1,2,3)$ so that they increase along the perimeter (e.g. ACW).
d) Construct unit vectors $bf u_n, ; v_n$
$$
eqalign{
& {bf v}_{,n} = overrightarrow {A_{,n} A_{,left( {n + 1} right)} } mathop {;/}limits_{} ;left| {overrightarrow {A_{,n} A_{,left( {n + 1} right)} } } right|quad left( {nbmod 4} right) cr
& {bf u}_{,n} = left( {matrix{
{cos left( {left( {n - 2} right)pi /2} right)} cr
{sin left( {left( {n - 2} right)pi /2} right)} cr
} } right) cr}
$$
e) Allocate the point P in the corresponding sector
construct the two lists
$$
L_{,x} = left[ {A_{,n,,x} ,P_{,x} } right]quad L_{,y} = left[ {A_{,n,,y} ,P_{,y} } right]
$$
and sort them in increasing order;
f) use the rank indices of P in the ordered lists to determine its position and projection
if (corner sector in $A_0$)
$P_{,x} < A_{,0,,x} ; wedge ;A_{,0,,y} < P_{,y} $ then $P to A_{,0} $ exit;
elif (lateral sector below $A_0$)
$P_{,x} < A_{,1,,x} ; wedge ;A_{,1,,y} < P_{,y} < A_{,0,,y} $ then
- take the matrix ${bf V}_{,0} = left( {matrix{ {{bf u}_{,0} } & {{bf v}_{,0} } cr } } right)$
- determine the coordinates of $P$ in that system
$$
left( {matrix{ {P'_x } cr {P'_y } cr } } right)
= {bf V}_{,0} ^{, - ,{bf 1}} ;left( {matrix{ {P_x } cr {P_y } cr } } right)
$$
- if $P'_y<0 ; to$ error;
- elif $P'_x<0 ; to$ the point is internal to the p., $to$ ?
- else $P; to ;A_{,0} + P'_y {bf v}_{,0} $
elif
....
(continue with the same path for the other sectors)
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
Thanks for the response. I am not sure if I can code this - Do you think it can be expressed in a computational way? For example: We would like to write a function which takes the matrix B and point P (x,y) as inputs and computes (X,Y) as projection? Also, I would like to extend this Algorithm to high dimensions - so checking all the vertices will be susceptible to curse of dimensionality! Please let me know. Thanks Much.
– chandu1729
Nov 17 '18 at 19:27
@chandu1729: well I developed the answer mainly mathematically, but keeping an eye on computability. Mathematically (and computationally) I cannot see a better way. And it can be extended to 3D and higher. I am quite sure it is tranformable in an algorithm.
– G Cab
Nov 17 '18 at 21:16
@chandu1729: the difficulty, is mainly on rotation /reflection of the points for negative signs on the $b$'s. If the $b$'s are positive it should be quite viable.
– G Cab
Nov 17 '18 at 21:20
If I understand correctly, you are looking/traversing at each vertex. In n-dimensions there will be 2^n vertices which makes it exponential complexity!
– chandu1729
Nov 17 '18 at 21:24
I would highly appreciate if you can express your answer in an algorithmic way for general n-dimensional matrix B and point P. That will be easier for me to understand as well.
– chandu1729
Nov 17 '18 at 21:26
|
show 5 more comments
So, each couple of lines given by
$$
b_{,1,,1} x + b_{,1,,2} y = pm 1
$$
and
$$
b_{,2,,1} x + b_{,2,,2} y = pm 1
$$
denotes a couple of lines symmetrical wrt the origin.
The four lines defines a parallelogram centered at the origin.
The four vertices are given by
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
; : ; left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)left( {matrix{ x cr y cr } } right)
= left( {matrix{
1 & 1 & { - 1} & { - 1} cr
1 & { - 1} & 1 & { - 1} cr
} } right)
$$
i.e.:
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= left( {matrix{ {b_{,1,,1} } & {b_{,1,,2} } cr {b_{,2,,1} } & {b_{,2,,2} } cr } } right)^{, - ,1}
left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
If some of the $b$ values are negative the denomination of the A points will change
with respect to that shown in the figure: it is better to rearrange the denominations as to match
with that shown, i.e. lower, higher $x$, lower,higher $y$ coordinates.
Now, at each vertex attach three unitary vectors as shown.
Compute the projection of the vector $A_{p,m}P$ onto the unitary vectors $bf x_{pm}$ and $bf y_{pm}$ stemming from $A_{p,m}$.
If both projections are positive, then the point P lies in the $90^circ$ sector encompassed by those vectors
and $P$ projects onto the vertex itself.
Repeat for the other vertices and check if both projections are positive.
If P is not in any of the sector above, then express the vector $A_{p,m}P$ in the reference system given by $bf x_{pm}, , bf v_{pm}$.
If the coordinates in that system are both positive, then $P$ projects on the edge $A_{pm}A_{pp}$, and its
coordinate wrt $bf v_{pm}$ gives the distance from the vertex $A_{pm}.
You can see that the reference system $bf y_{pp}, , bf v_{pp}$ in $A_{pp}$ will cover the edge $A_{pp}A_{mp}$, and so on.
You can also see that $P$, if external to the parallelogram, can have both coordinates positive only in one of the four systems.
To translate the above into a computational algorithm we can proceed along the following lines.
a) Construct the matrix
$$
{bf B} = left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)
$$
ensure that its determinant is non null and compute $bf B^{-1}$.
b) Compute the four points A
$$
left( {matrix{ {A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= {bf B}^{, - ,1} left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
c) Rearrange points A
for instance by taking as pivot the point with minimal abscissa ($A_0$ in the sketch) and arranging the others
according to the tangent of the angle that they form with the pivot (if $Delta x = 0 ; to ; Delta x = epsilon $);
re-name the suffixes to $(0,1,2,3)$ so that they increase along the perimeter (e.g. ACW).
d) Construct unit vectors $bf u_n, ; v_n$
$$
eqalign{
& {bf v}_{,n} = overrightarrow {A_{,n} A_{,left( {n + 1} right)} } mathop {;/}limits_{} ;left| {overrightarrow {A_{,n} A_{,left( {n + 1} right)} } } right|quad left( {nbmod 4} right) cr
& {bf u}_{,n} = left( {matrix{
{cos left( {left( {n - 2} right)pi /2} right)} cr
{sin left( {left( {n - 2} right)pi /2} right)} cr
} } right) cr}
$$
e) Allocate the point P in the corresponding sector
construct the two lists
$$
L_{,x} = left[ {A_{,n,,x} ,P_{,x} } right]quad L_{,y} = left[ {A_{,n,,y} ,P_{,y} } right]
$$
and sort them in increasing order;
f) use the rank indices of P in the ordered lists to determine its position and projection
if (corner sector in $A_0$)
$P_{,x} < A_{,0,,x} ; wedge ;A_{,0,,y} < P_{,y} $ then $P to A_{,0} $ exit;
elif (lateral sector below $A_0$)
$P_{,x} < A_{,1,,x} ; wedge ;A_{,1,,y} < P_{,y} < A_{,0,,y} $ then
- take the matrix ${bf V}_{,0} = left( {matrix{ {{bf u}_{,0} } & {{bf v}_{,0} } cr } } right)$
- determine the coordinates of $P$ in that system
$$
left( {matrix{ {P'_x } cr {P'_y } cr } } right)
= {bf V}_{,0} ^{, - ,{bf 1}} ;left( {matrix{ {P_x } cr {P_y } cr } } right)
$$
- if $P'_y<0 ; to$ error;
- elif $P'_x<0 ; to$ the point is internal to the p., $to$ ?
- else $P; to ;A_{,0} + P'_y {bf v}_{,0} $
elif
....
(continue with the same path for the other sectors)
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
So, each couple of lines given by
$$
b_{,1,,1} x + b_{,1,,2} y = pm 1
$$
and
$$
b_{,2,,1} x + b_{,2,,2} y = pm 1
$$
denotes a couple of lines symmetrical wrt the origin.
The four lines defines a parallelogram centered at the origin.
The four vertices are given by
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
; : ; left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)left( {matrix{ x cr y cr } } right)
= left( {matrix{
1 & 1 & { - 1} & { - 1} cr
1 & { - 1} & 1 & { - 1} cr
} } right)
$$
i.e.:
$$
left( {matrix{
{A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= left( {matrix{ {b_{,1,,1} } & {b_{,1,,2} } cr {b_{,2,,1} } & {b_{,2,,2} } cr } } right)^{, - ,1}
left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
If some of the $b$ values are negative the denomination of the A points will change
with respect to that shown in the figure: it is better to rearrange the denominations as to match
with that shown, i.e. lower, higher $x$, lower,higher $y$ coordinates.
Now, at each vertex attach three unitary vectors as shown.
Compute the projection of the vector $A_{p,m}P$ onto the unitary vectors $bf x_{pm}$ and $bf y_{pm}$ stemming from $A_{p,m}$.
If both projections are positive, then the point P lies in the $90^circ$ sector encompassed by those vectors
and $P$ projects onto the vertex itself.
Repeat for the other vertices and check if both projections are positive.
If P is not in any of the sector above, then express the vector $A_{p,m}P$ in the reference system given by $bf x_{pm}, , bf v_{pm}$.
If the coordinates in that system are both positive, then $P$ projects on the edge $A_{pm}A_{pp}$, and its
coordinate wrt $bf v_{pm}$ gives the distance from the vertex $A_{pm}.
You can see that the reference system $bf y_{pp}, , bf v_{pp}$ in $A_{pp}$ will cover the edge $A_{pp}A_{mp}$, and so on.
You can also see that $P$, if external to the parallelogram, can have both coordinates positive only in one of the four systems.
To translate the above into a computational algorithm we can proceed along the following lines.
a) Construct the matrix
$$
{bf B} = left( {matrix{
{b_{,1,,1} } & {b_{,1,,2} } cr
{b_{,2,,1} } & {b_{,2,,2} } cr
} } right)
$$
ensure that its determinant is non null and compute $bf B^{-1}$.
b) Compute the four points A
$$
left( {matrix{ {A_{,p,,p} } & {A_{,p,,m} } & {A_{,m,,p} } & {A_{,m,,m} } cr } } right)
= {bf B}^{, - ,1} left( {matrix{ 1 & 1 & { - 1} & { - 1} cr 1 & { - 1} & 1 & { - 1} cr } } right)
$$
c) Rearrange points A
for instance by taking as pivot the point with minimal abscissa ($A_0$ in the sketch) and arranging the others
according to the tangent of the angle that they form with the pivot (if $Delta x = 0 ; to ; Delta x = epsilon $);
re-name the suffixes to $(0,1,2,3)$ so that they increase along the perimeter (e.g. ACW).
d) Construct unit vectors $bf u_n, ; v_n$
$$
eqalign{
& {bf v}_{,n} = overrightarrow {A_{,n} A_{,left( {n + 1} right)} } mathop {;/}limits_{} ;left| {overrightarrow {A_{,n} A_{,left( {n + 1} right)} } } right|quad left( {nbmod 4} right) cr
& {bf u}_{,n} = left( {matrix{
{cos left( {left( {n - 2} right)pi /2} right)} cr
{sin left( {left( {n - 2} right)pi /2} right)} cr
} } right) cr}
$$
e) Allocate the point P in the corresponding sector
construct the two lists
$$
L_{,x} = left[ {A_{,n,,x} ,P_{,x} } right]quad L_{,y} = left[ {A_{,n,,y} ,P_{,y} } right]
$$
and sort them in increasing order;
f) use the rank indices of P in the ordered lists to determine its position and projection
if (corner sector in $A_0$)
$P_{,x} < A_{,0,,x} ; wedge ;A_{,0,,y} < P_{,y} $ then $P to A_{,0} $ exit;
elif (lateral sector below $A_0$)
$P_{,x} < A_{,1,,x} ; wedge ;A_{,1,,y} < P_{,y} < A_{,0,,y} $ then
- take the matrix ${bf V}_{,0} = left( {matrix{ {{bf u}_{,0} } & {{bf v}_{,0} } cr } } right)$
- determine the coordinates of $P$ in that system
$$
left( {matrix{ {P'_x } cr {P'_y } cr } } right)
= {bf V}_{,0} ^{, - ,{bf 1}} ;left( {matrix{ {P_x } cr {P_y } cr } } right)
$$
- if $P'_y<0 ; to$ error;
- elif $P'_x<0 ; to$ the point is internal to the p., $to$ ?
- else $P; to ;A_{,0} + P'_y {bf v}_{,0} $
elif
....
(continue with the same path for the other sectors)
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
edited Nov 21 '18 at 17:29
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
answered Nov 17 '18 at 1:04
G Cab
17.9k31237
17.9k31237
Thanks for the response. I am not sure if I can code this - Do you think it can be expressed in a computational way? For example: We would like to write a function which takes the matrix B and point P (x,y) as inputs and computes (X,Y) as projection? Also, I would like to extend this Algorithm to high dimensions - so checking all the vertices will be susceptible to curse of dimensionality! Please let me know. Thanks Much.
– chandu1729
Nov 17 '18 at 19:27
@chandu1729: well I developed the answer mainly mathematically, but keeping an eye on computability. Mathematically (and computationally) I cannot see a better way. And it can be extended to 3D and higher. I am quite sure it is tranformable in an algorithm.
– G Cab
Nov 17 '18 at 21:16
@chandu1729: the difficulty, is mainly on rotation /reflection of the points for negative signs on the $b$'s. If the $b$'s are positive it should be quite viable.
– G Cab
Nov 17 '18 at 21:20
If I understand correctly, you are looking/traversing at each vertex. In n-dimensions there will be 2^n vertices which makes it exponential complexity!
– chandu1729
Nov 17 '18 at 21:24
I would highly appreciate if you can express your answer in an algorithmic way for general n-dimensional matrix B and point P. That will be easier for me to understand as well.
– chandu1729
Nov 17 '18 at 21:26
|
show 5 more comments
Thanks for the response. I am not sure if I can code this - Do you think it can be expressed in a computational way? For example: We would like to write a function which takes the matrix B and point P (x,y) as inputs and computes (X,Y) as projection? Also, I would like to extend this Algorithm to high dimensions - so checking all the vertices will be susceptible to curse of dimensionality! Please let me know. Thanks Much.
– chandu1729
Nov 17 '18 at 19:27
@chandu1729: well I developed the answer mainly mathematically, but keeping an eye on computability. Mathematically (and computationally) I cannot see a better way. And it can be extended to 3D and higher. I am quite sure it is tranformable in an algorithm.
– G Cab
Nov 17 '18 at 21:16
@chandu1729: the difficulty, is mainly on rotation /reflection of the points for negative signs on the $b$'s. If the $b$'s are positive it should be quite viable.
– G Cab
Nov 17 '18 at 21:20
If I understand correctly, you are looking/traversing at each vertex. In n-dimensions there will be 2^n vertices which makes it exponential complexity!
– chandu1729
Nov 17 '18 at 21:24
I would highly appreciate if you can express your answer in an algorithmic way for general n-dimensional matrix B and point P. That will be easier for me to understand as well.
– chandu1729
Nov 17 '18 at 21:26
Thanks for the response. I am not sure if I can code this - Do you think it can be expressed in a computational way? For example: We would like to write a function which takes the matrix B and point P (x,y) as inputs and computes (X,Y) as projection? Also, I would like to extend this Algorithm to high dimensions - so checking all the vertices will be susceptible to curse of dimensionality! Please let me know. Thanks Much.
– chandu1729
Nov 17 '18 at 19:27
Thanks for the response. I am not sure if I can code this - Do you think it can be expressed in a computational way? For example: We would like to write a function which takes the matrix B and point P (x,y) as inputs and computes (X,Y) as projection? Also, I would like to extend this Algorithm to high dimensions - so checking all the vertices will be susceptible to curse of dimensionality! Please let me know. Thanks Much.
– chandu1729
Nov 17 '18 at 19:27
@chandu1729: well I developed the answer mainly mathematically, but keeping an eye on computability. Mathematically (and computationally) I cannot see a better way. And it can be extended to 3D and higher. I am quite sure it is tranformable in an algorithm.
– G Cab
Nov 17 '18 at 21:16
@chandu1729: well I developed the answer mainly mathematically, but keeping an eye on computability. Mathematically (and computationally) I cannot see a better way. And it can be extended to 3D and higher. I am quite sure it is tranformable in an algorithm.
– G Cab
Nov 17 '18 at 21:16
@chandu1729: the difficulty, is mainly on rotation /reflection of the points for negative signs on the $b$'s. If the $b$'s are positive it should be quite viable.
– G Cab
Nov 17 '18 at 21:20
@chandu1729: the difficulty, is mainly on rotation /reflection of the points for negative signs on the $b$'s. If the $b$'s are positive it should be quite viable.
– G Cab
Nov 17 '18 at 21:20
If I understand correctly, you are looking/traversing at each vertex. In n-dimensions there will be 2^n vertices which makes it exponential complexity!
– chandu1729
Nov 17 '18 at 21:24
If I understand correctly, you are looking/traversing at each vertex. In n-dimensions there will be 2^n vertices which makes it exponential complexity!
– chandu1729
Nov 17 '18 at 21:24
I would highly appreciate if you can express your answer in an algorithmic way for general n-dimensional matrix B and point P. That will be easier for me to understand as well.
– chandu1729
Nov 17 '18 at 21:26
I would highly appreciate if you can express your answer in an algorithmic way for general n-dimensional matrix B and point P. That will be easier for me to understand as well.
– chandu1729
Nov 17 '18 at 21:26
|
show 5 more comments
In my solution $x_1, x_2$ are replaced by $x,y.$
The solution should be easy to code.
Assume the given straight lines are not horizontal neither vertical. In this case are all coefficients $b_{ij}neq 0.$
From the coefficients $pm 1$ in the equations follows that the parallelogram is centered in $O.$
Thus if $A(x_A,y_A), B(x_B,y_B),$ then necessarily $C(-x_A,-y_A), D(-x_B,-y_B).$
In the enclosed figure are the equations of the straight lines $color{blue}{blue},$ and located across the corresponding line.
Each region is determined by a system of inequations written in black.
The input point $P(x,y)$ has to be transformed to an output $P'(X,Y).$ These new coordinates $X,Y$ are written in the corresponding region in $color{red}{red}.$
Remark
If a pair of lines is horizontal or vertical (the product $prod_{i,j=1,2} b_{ij}=0$ ), the above solution has to be modified, but is easier.
answered Nov 24 '18 at 13:39
user376343
2,8582823
add a comment |
In my solution $x_1, x_2$ are replaced by $x,y.$
The solution should be easy to code.
Assume the given straight lines are not horizontal neither vertical. In this case are all coefficients $b_{ij}neq 0.$
From the coefficients $pm 1$ in the equations follows that the parallelogram is centered in $O.$
Thus if $A(x_A,y_A), B(x_B,y_B),$ then necessarily $C(-x_A,-y_A), D(-x_B,-y_B).$
In the enclosed figure are the equations of the straight lines $color{blue}{blue},$ and located across the corresponding line.
Each region is determined by a system of inequations written in black.
The input point $P(x,y)$ has to be transformed to an output $P'(X,Y).$ These new coordinates $X,Y$ are written in the corresponding region in $color{red}{red}.$
Remark
If a pair of lines is horizontal or vertical (the product $prod_{i,j=1,2} b_{ij}=0$ ), the above solution has to be modified, but is easier.
answered Nov 24 '18 at 13:39
user376343
2,8582823
add a comment |
In my solution $x_1, x_2$ are replaced by $x,y.$
The solution should be easy to code.
Assume the given straight lines are not horizontal neither vertical. In this case are all coefficients $b_{ij}neq 0.$
From the coefficients $pm 1$ in the equations follows that the parallelogram is centered in $O.$
Thus if $A(x_A,y_A), B(x_B,y_B),$ then necessarily $C(-x_A,-y_A), D(-x_B,-y_B).$
In the enclosed figure are the equations of the straight lines $color{blue}{blue},$ and located across the corresponding line.
Each region is determined by a system of inequations written in black.
The input point $P(x,y)$ has to be transformed to an output $P'(X,Y).$ These new coordinates $X,Y$ are written in the corresponding region in $color{red}{red}.$
Remark
If a pair of lines is horizontal or vertical (the product $prod_{i,j=1,2} b_{ij}=0$ ), the above solution has to be modified, but is easier.
answered Nov 24 '18 at 13:39
user376343
2,8582823
In my solution $x_1, x_2$ are replaced by $x,y.$
The solution should be easy to code.
Assume the given straight lines are not horizontal neither vertical. In this case are all coefficients $b_{ij}neq 0.$
From the coefficients $pm 1$ in the equations follows that the parallelogram is centered in $O.$
Thus if $A(x_A,y_A), B(x_B,y_B),$ then necessarily $C(-x_A,-y_A), D(-x_B,-y_B).$
In the enclosed figure are the equations of the straight lines $color{blue}{blue},$ and located across the corresponding line.
Each region is determined by a system of inequations written in black.
The input point $P(x,y)$ has to be transformed to an output $P'(X,Y).$ These new coordinates $X,Y$ are written in the corresponding region in $color{red}{red}.$
Remark
If a pair of lines is horizontal or vertical (the product $prod_{i,j=1,2} b_{ij}=0$ ), the above solution has to be modified, but is easier.
answered Nov 24 '18 at 13:39
user376343
2,8582823
edited Nov 28 '18 at 8:32
answered Nov 24 '18 at 13:39
user376343
2,8582823
answered Nov 24 '18 at 13:39
user376343
2,8582823
answered Nov 24 '18 at 13:39
user376343
2,8582823
2,8582823
add a comment |
add a comment |
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There are some conditions to make clear: a) you said that AB is parallel to y axis, but then what are the dotted lines out of B indicating ? b) the arrow indicating the projection, is it normal to the edge ?, parallel to the axes ? ... c) the drawing is so distorted that looks to be in 3D, but equations are in 2d (?) You should provide a much cleaner scheme , before thinking to equations
– G Cab
Nov 15 '18 at 17:24
@GCab: a) AB is not parallel to y-axis. I meant the projection is parallel to y-axis. For example, let X be the given point in region 2 then Y is projection of X such that XY is parallel to Y-axis. b) Parallel to axes. c) The diagram is 2D and the shape is parallelogram and the describing equations are also in the diagram. Thank you very much for any help!
– chandu1729
Nov 16 '18 at 5:50