Mixing
![Is it possible to generate random numbers out of nonrandom numbers? [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Simultaneous movement toward barycenters - what can be guaranteed
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Suppose a tiling is given in 2D (an embedding of a planar triangulated graph), with all faces convex.
Now suppose one moves each point, one by one, to the barycenter of its neighbors. I think that with such an update one can guarantee that the planarity is preserved at each step (after each point is moved). But could planarity be guaranteed in case whole configuration moves simultaneously?
I understood that once such simultaneous position update converges (in case the positions of the points of the outer convex face are fixed), it can be guaranteed that the new tiling is also planar (Tutte's embedding). My question is: given no fixed positions, what happens if one performs a single simultaneous coordinate update from the initial configuration that is already planar (all faces convex)?
In case no guarantees can be made, are there special cases with which the simultaneous update would preserve planarity, i.e., special kinds of initial configurations, etc?
** In mathematical terms:
The initial configuration is $Y_0$, and can be written as $Y_0 = T_0Y_0$, where $T_0$ is a transition matrix (each row/column having weight coefficients expressing each coordinate as a convex combination of its neighbors). The update is achieved as $Y_1 = T_bY_0$, where $T_b$ is a transition matrix that moves a configuration to the barycenter (a simultaneous update). Obviously, $Y_1 = T'Y_0$, where $T'$ is also a transition matrix $T' = T_bT_0$. But is that sufficient to guarantee that $Y_1 = T''Y_1$, for some transition matrix $T''$, i.e., each coordinate is a convex combination of its neighbors?
linear-algebra geometry polygons triangulation barycentric-coordinates
asked Jul 27 at 14:06
kevin811
62
add a comment |Â
up vote
1
down vote
favorite
Suppose a tiling is given in 2D (an embedding of a planar triangulated graph), with all faces convex.
Now suppose one moves each point, one by one, to the barycenter of its neighbors. I think that with such an update one can guarantee that the planarity is preserved at each step (after each point is moved). But could planarity be guaranteed in case whole configuration moves simultaneously?
I understood that once such simultaneous position update converges (in case the positions of the points of the outer convex face are fixed), it can be guaranteed that the new tiling is also planar (Tutte's embedding). My question is: given no fixed positions, what happens if one performs a single simultaneous coordinate update from the initial configuration that is already planar (all faces convex)?
In case no guarantees can be made, are there special cases with which the simultaneous update would preserve planarity, i.e., special kinds of initial configurations, etc?
** In mathematical terms:
The initial configuration is $Y_0$, and can be written as $Y_0 = T_0Y_0$, where $T_0$ is a transition matrix (each row/column having weight coefficients expressing each coordinate as a convex combination of its neighbors). The update is achieved as $Y_1 = T_bY_0$, where $T_b$ is a transition matrix that moves a configuration to the barycenter (a simultaneous update). Obviously, $Y_1 = T'Y_0$, where $T'$ is also a transition matrix $T' = T_bT_0$. But is that sufficient to guarantee that $Y_1 = T''Y_1$, for some transition matrix $T''$, i.e., each coordinate is a convex combination of its neighbors?
linear-algebra geometry polygons triangulation barycentric-coordinates
asked Jul 27 at 14:06
kevin811
62
If all points lie in the same plane then all barycenters lie in that plane too. How could the resulting configuration be non-planar?
– Aretino
Jul 27 at 16:56
@Aretino Your argument is not enough for a guarantee (or do we have different understandings of the word "planar"?). Suppose the initial configuration is random, i.e., with intersections; does the update guarantee an intersection-free drawing?
– kevin811
Jul 29 at 18:32
Sorry, I thought planar meant "all in a plane" but I see now that the real meaning is more complex.
– Aretino
Jul 29 at 19:55
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose a tiling is given in 2D (an embedding of a planar triangulated graph), with all faces convex.
Now suppose one moves each point, one by one, to the barycenter of its neighbors. I think that with such an update one can guarantee that the planarity is preserved at each step (after each point is moved). But could planarity be guaranteed in case whole configuration moves simultaneously?
I understood that once such simultaneous position update converges (in case the positions of the points of the outer convex face are fixed), it can be guaranteed that the new tiling is also planar (Tutte's embedding). My question is: given no fixed positions, what happens if one performs a single simultaneous coordinate update from the initial configuration that is already planar (all faces convex)?
In case no guarantees can be made, are there special cases with which the simultaneous update would preserve planarity, i.e., special kinds of initial configurations, etc?
** In mathematical terms:
The initial configuration is $Y_0$, and can be written as $Y_0 = T_0Y_0$, where $T_0$ is a transition matrix (each row/column having weight coefficients expressing each coordinate as a convex combination of its neighbors). The update is achieved as $Y_1 = T_bY_0$, where $T_b$ is a transition matrix that moves a configuration to the barycenter (a simultaneous update). Obviously, $Y_1 = T'Y_0$, where $T'$ is also a transition matrix $T' = T_bT_0$. But is that sufficient to guarantee that $Y_1 = T''Y_1$, for some transition matrix $T''$, i.e., each coordinate is a convex combination of its neighbors?
linear-algebra geometry polygons triangulation barycentric-coordinates
asked Jul 27 at 14:06
kevin811
62
Suppose a tiling is given in 2D (an embedding of a planar triangulated graph), with all faces convex.
Now suppose one moves each point, one by one, to the barycenter of its neighbors. I think that with such an update one can guarantee that the planarity is preserved at each step (after each point is moved). But could planarity be guaranteed in case whole configuration moves simultaneously?
I understood that once such simultaneous position update converges (in case the positions of the points of the outer convex face are fixed), it can be guaranteed that the new tiling is also planar (Tutte's embedding). My question is: given no fixed positions, what happens if one performs a single simultaneous coordinate update from the initial configuration that is already planar (all faces convex)?
In case no guarantees can be made, are there special cases with which the simultaneous update would preserve planarity, i.e., special kinds of initial configurations, etc?
** In mathematical terms:
The initial configuration is $Y_0$, and can be written as $Y_0 = T_0Y_0$, where $T_0$ is a transition matrix (each row/column having weight coefficients expressing each coordinate as a convex combination of its neighbors). The update is achieved as $Y_1 = T_bY_0$, where $T_b$ is a transition matrix that moves a configuration to the barycenter (a simultaneous update). Obviously, $Y_1 = T'Y_0$, where $T'$ is also a transition matrix $T' = T_bT_0$. But is that sufficient to guarantee that $Y_1 = T''Y_1$, for some transition matrix $T''$, i.e., each coordinate is a convex combination of its neighbors?
linear-algebra geometry polygons triangulation barycentric-coordinates
asked Jul 27 at 14:06
kevin811
62
edited Jul 30 at 12:04
asked Jul 27 at 14:06
kevin811
62
asked Jul 27 at 14:06
kevin811
62
asked Jul 27 at 14:06
kevin811
62
62
If all points lie in the same plane then all barycenters lie in that plane too. How could the resulting configuration be non-planar?
– Aretino
Jul 27 at 16:56
@Aretino Your argument is not enough for a guarantee (or do we have different understandings of the word "planar"?). Suppose the initial configuration is random, i.e., with intersections; does the update guarantee an intersection-free drawing?
– kevin811
Jul 29 at 18:32
Sorry, I thought planar meant "all in a plane" but I see now that the real meaning is more complex.
– Aretino
Jul 29 at 19:55
add a comment |Â
If all points lie in the same plane then all barycenters lie in that plane too. How could the resulting configuration be non-planar?
– Aretino
Jul 27 at 16:56
@Aretino Your argument is not enough for a guarantee (or do we have different understandings of the word "planar"?). Suppose the initial configuration is random, i.e., with intersections; does the update guarantee an intersection-free drawing?
– kevin811
Jul 29 at 18:32
Sorry, I thought planar meant "all in a plane" but I see now that the real meaning is more complex.
– Aretino
Jul 29 at 19:55
If all points lie in the same plane then all barycenters lie in that plane too. How could the resulting configuration be non-planar?
– Aretino
Jul 27 at 16:56
If all points lie in the same plane then all barycenters lie in that plane too. How could the resulting configuration be non-planar?
– Aretino
Jul 27 at 16:56
@Aretino Your argument is not enough for a guarantee (or do we have different understandings of the word "planar"?). Suppose the initial configuration is random, i.e., with intersections; does the update guarantee an intersection-free drawing?
– kevin811
Jul 29 at 18:32
@Aretino Your argument is not enough for a guarantee (or do we have different understandings of the word "planar"?). Suppose the initial configuration is random, i.e., with intersections; does the update guarantee an intersection-free drawing?
– kevin811
Jul 29 at 18:32
Sorry, I thought planar meant "all in a plane" but I see now that the real meaning is more complex.
– Aretino
Jul 29 at 19:55
Sorry, I thought planar meant "all in a plane" but I see now that the real meaning is more complex.
– Aretino
Jul 29 at 19:55
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864419%2fsimultaneous-movement-toward-barycenters-what-can-be-guaranteed%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
If all points lie in the same plane then all barycenters lie in that plane too. How could the resulting configuration be non-planar?
– Aretino
Jul 27 at 16:56
@Aretino Your argument is not enough for a guarantee (or do we have different understandings of the word "planar"?). Suppose the initial configuration is random, i.e., with intersections; does the update guarantee an intersection-free drawing?
– kevin811
Jul 29 at 18:32
Sorry, I thought planar meant "all in a plane" but I see now that the real meaning is more complex.
– Aretino
Jul 29 at 19:55