What's wrong with this Penrose pattern?












16












$begingroup$


I programmed the Penrose tiling by projecting a portion of 5D lattice to 2D space, by the "cut and project" method described in




  1. Quasicrystals: projections of 5-D lattice into 2 and 3 dimensions, H. Au-Yang and J. Perk.

  2. Generalised 2D Penrose tilings, A. Pavlovitch and M. Kléman


The orthonormal basis is chosen as
$$
M=sqrt{frac{2}{5}} begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$

Each row presents a basis vector, i.e.
$$
M_icdot M_j=0, ;;textrm{for } ineq j.$$

and
$$||M_i||=1, ;;textrm{for } 1leq i leq 5.
$$



$M$ consists of the parallel operator (representing the physical space)
$$
A=begin{bmatrix}
M_1\
M_2 \
end{bmatrix}=
begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
end{bmatrix}
$$

and the perpendicular operator
$$
B=begin{bmatrix}
M_2\
M_3 \
M_4 \
end{bmatrix}=begin{bmatrix}
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$



The 5D lattice points are integer combinations of basis such as
$$
p=i begin{bmatrix}
1\
0\
0\
0\
0\
end{bmatrix}
+ jbegin{bmatrix}
0\
1\
0\
0\
0\
end{bmatrix}
+dots, ;; i,j,dots in mathbb{Z}
$$



A 5D cube (centered at origin) is projected into 3D as polytope
$$
v'= B v, ;; vin hypercube
$$

so that I can check whether a $p$ is inside this polytope (20 faces). This is called "cutting" the 5D lattice points.



The resultant 2d projection $Ap$ is
enter image description here



Everything works fine, however, my result differs from the "standard" one (e.g. in wiki page) as follows



enter image description here



Is this a mistake or an alternative view of the same tiling?



Finally, I find this image (from Vertex Frequencies in Generalized Penrose Patterns, by E. Zobetz and A. Preisinger)



enter image description here



where the center of standard tiling exhibits the "S" pattern, while the center of my version has the "ST" pattern. But what does it mean exactly?










share|cite|improve this question











$endgroup$












  • $begingroup$
    One thing you might want to do is check how your trig functions are rendered in the presentation. $cos(frac{2}{5pi})$ looks funky in this context, probably should be $cos(frac{2pi}{5})$. Likewise for the other trig functions.
    $endgroup$
    – Oscar Lanzi
    Jan 1 at 1:02






  • 1




    $begingroup$
    @oscar Lanzi, thanks! That's a typo. I have corrected that.
    $endgroup$
    – whitegreen
    Jan 3 at 3:32
















16












$begingroup$


I programmed the Penrose tiling by projecting a portion of 5D lattice to 2D space, by the "cut and project" method described in




  1. Quasicrystals: projections of 5-D lattice into 2 and 3 dimensions, H. Au-Yang and J. Perk.

  2. Generalised 2D Penrose tilings, A. Pavlovitch and M. Kléman


The orthonormal basis is chosen as
$$
M=sqrt{frac{2}{5}} begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$

Each row presents a basis vector, i.e.
$$
M_icdot M_j=0, ;;textrm{for } ineq j.$$

and
$$||M_i||=1, ;;textrm{for } 1leq i leq 5.
$$



$M$ consists of the parallel operator (representing the physical space)
$$
A=begin{bmatrix}
M_1\
M_2 \
end{bmatrix}=
begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
end{bmatrix}
$$

and the perpendicular operator
$$
B=begin{bmatrix}
M_2\
M_3 \
M_4 \
end{bmatrix}=begin{bmatrix}
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$



The 5D lattice points are integer combinations of basis such as
$$
p=i begin{bmatrix}
1\
0\
0\
0\
0\
end{bmatrix}
+ jbegin{bmatrix}
0\
1\
0\
0\
0\
end{bmatrix}
+dots, ;; i,j,dots in mathbb{Z}
$$



A 5D cube (centered at origin) is projected into 3D as polytope
$$
v'= B v, ;; vin hypercube
$$

so that I can check whether a $p$ is inside this polytope (20 faces). This is called "cutting" the 5D lattice points.



The resultant 2d projection $Ap$ is
enter image description here



Everything works fine, however, my result differs from the "standard" one (e.g. in wiki page) as follows



enter image description here



Is this a mistake or an alternative view of the same tiling?



Finally, I find this image (from Vertex Frequencies in Generalized Penrose Patterns, by E. Zobetz and A. Preisinger)



enter image description here



where the center of standard tiling exhibits the "S" pattern, while the center of my version has the "ST" pattern. But what does it mean exactly?










share|cite|improve this question











$endgroup$












  • $begingroup$
    One thing you might want to do is check how your trig functions are rendered in the presentation. $cos(frac{2}{5pi})$ looks funky in this context, probably should be $cos(frac{2pi}{5})$. Likewise for the other trig functions.
    $endgroup$
    – Oscar Lanzi
    Jan 1 at 1:02






  • 1




    $begingroup$
    @oscar Lanzi, thanks! That's a typo. I have corrected that.
    $endgroup$
    – whitegreen
    Jan 3 at 3:32














16












16








16


5



$begingroup$


I programmed the Penrose tiling by projecting a portion of 5D lattice to 2D space, by the "cut and project" method described in




  1. Quasicrystals: projections of 5-D lattice into 2 and 3 dimensions, H. Au-Yang and J. Perk.

  2. Generalised 2D Penrose tilings, A. Pavlovitch and M. Kléman


The orthonormal basis is chosen as
$$
M=sqrt{frac{2}{5}} begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$

Each row presents a basis vector, i.e.
$$
M_icdot M_j=0, ;;textrm{for } ineq j.$$

and
$$||M_i||=1, ;;textrm{for } 1leq i leq 5.
$$



$M$ consists of the parallel operator (representing the physical space)
$$
A=begin{bmatrix}
M_1\
M_2 \
end{bmatrix}=
begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
end{bmatrix}
$$

and the perpendicular operator
$$
B=begin{bmatrix}
M_2\
M_3 \
M_4 \
end{bmatrix}=begin{bmatrix}
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$



The 5D lattice points are integer combinations of basis such as
$$
p=i begin{bmatrix}
1\
0\
0\
0\
0\
end{bmatrix}
+ jbegin{bmatrix}
0\
1\
0\
0\
0\
end{bmatrix}
+dots, ;; i,j,dots in mathbb{Z}
$$



A 5D cube (centered at origin) is projected into 3D as polytope
$$
v'= B v, ;; vin hypercube
$$

so that I can check whether a $p$ is inside this polytope (20 faces). This is called "cutting" the 5D lattice points.



The resultant 2d projection $Ap$ is
enter image description here



Everything works fine, however, my result differs from the "standard" one (e.g. in wiki page) as follows



enter image description here



Is this a mistake or an alternative view of the same tiling?



Finally, I find this image (from Vertex Frequencies in Generalized Penrose Patterns, by E. Zobetz and A. Preisinger)



enter image description here



where the center of standard tiling exhibits the "S" pattern, while the center of my version has the "ST" pattern. But what does it mean exactly?










share|cite|improve this question











$endgroup$




I programmed the Penrose tiling by projecting a portion of 5D lattice to 2D space, by the "cut and project" method described in




  1. Quasicrystals: projections of 5-D lattice into 2 and 3 dimensions, H. Au-Yang and J. Perk.

  2. Generalised 2D Penrose tilings, A. Pavlovitch and M. Kléman


The orthonormal basis is chosen as
$$
M=sqrt{frac{2}{5}} begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$

Each row presents a basis vector, i.e.
$$
M_icdot M_j=0, ;;textrm{for } ineq j.$$

and
$$||M_i||=1, ;;textrm{for } 1leq i leq 5.
$$



$M$ consists of the parallel operator (representing the physical space)
$$
A=begin{bmatrix}
M_1\
M_2 \
end{bmatrix}=
begin{bmatrix}
cos 0 & cos frac{2pi}{5} & cos frac{4pi}{5}& cos frac{6pi}{5}& cos frac{8pi}{5} \
sin 0 & sin frac{2pi}{5} & sin frac{4pi}{5}& sin frac{6pi}{5}& sin frac{8pi}{5} \
end{bmatrix}
$$

and the perpendicular operator
$$
B=begin{bmatrix}
M_2\
M_3 \
M_4 \
end{bmatrix}=begin{bmatrix}
cos 0 & cos frac{4pi}{5} & cos frac{8pi}{5}& cos frac{12pi}{5}& cos frac{16pi}{5} \
sin 0 & sin frac{4pi}{5} & sin frac{8pi}{5}& sin frac{12pi}{5}& sin frac{16pi}{5} \
frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}} & frac{1}{sqrt {2}}\
end{bmatrix}
$$



The 5D lattice points are integer combinations of basis such as
$$
p=i begin{bmatrix}
1\
0\
0\
0\
0\
end{bmatrix}
+ jbegin{bmatrix}
0\
1\
0\
0\
0\
end{bmatrix}
+dots, ;; i,j,dots in mathbb{Z}
$$



A 5D cube (centered at origin) is projected into 3D as polytope
$$
v'= B v, ;; vin hypercube
$$

so that I can check whether a $p$ is inside this polytope (20 faces). This is called "cutting" the 5D lattice points.



The resultant 2d projection $Ap$ is
enter image description here



Everything works fine, however, my result differs from the "standard" one (e.g. in wiki page) as follows



enter image description here



Is this a mistake or an alternative view of the same tiling?



Finally, I find this image (from Vertex Frequencies in Generalized Penrose Patterns, by E. Zobetz and A. Preisinger)



enter image description here



where the center of standard tiling exhibits the "S" pattern, while the center of my version has the "ST" pattern. But what does it mean exactly?







tiling crystallography






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 3:35







whitegreen

















asked Dec 5 '18 at 10:02









whitegreenwhitegreen

477314




477314












  • $begingroup$
    One thing you might want to do is check how your trig functions are rendered in the presentation. $cos(frac{2}{5pi})$ looks funky in this context, probably should be $cos(frac{2pi}{5})$. Likewise for the other trig functions.
    $endgroup$
    – Oscar Lanzi
    Jan 1 at 1:02






  • 1




    $begingroup$
    @oscar Lanzi, thanks! That's a typo. I have corrected that.
    $endgroup$
    – whitegreen
    Jan 3 at 3:32


















  • $begingroup$
    One thing you might want to do is check how your trig functions are rendered in the presentation. $cos(frac{2}{5pi})$ looks funky in this context, probably should be $cos(frac{2pi}{5})$. Likewise for the other trig functions.
    $endgroup$
    – Oscar Lanzi
    Jan 1 at 1:02






  • 1




    $begingroup$
    @oscar Lanzi, thanks! That's a typo. I have corrected that.
    $endgroup$
    – whitegreen
    Jan 3 at 3:32
















$begingroup$
One thing you might want to do is check how your trig functions are rendered in the presentation. $cos(frac{2}{5pi})$ looks funky in this context, probably should be $cos(frac{2pi}{5})$. Likewise for the other trig functions.
$endgroup$
– Oscar Lanzi
Jan 1 at 1:02




$begingroup$
One thing you might want to do is check how your trig functions are rendered in the presentation. $cos(frac{2}{5pi})$ looks funky in this context, probably should be $cos(frac{2pi}{5})$. Likewise for the other trig functions.
$endgroup$
– Oscar Lanzi
Jan 1 at 1:02




1




1




$begingroup$
@oscar Lanzi, thanks! That's a typo. I have corrected that.
$endgroup$
– whitegreen
Jan 3 at 3:32




$begingroup$
@oscar Lanzi, thanks! That's a typo. I have corrected that.
$endgroup$
– whitegreen
Jan 3 at 3:32










2 Answers
2






active

oldest

votes


















4












$begingroup$

The Wikipedia article states "Therefore, a finite patch cannot differentiate between the uncountably many Penrose tilings, nor even determine which position within the tiling is being shown". Thus they could be the same tiling but differently centered.



In particular, aside from the two dimensional shifting of Penrose tilings, if a higher dimensional lattice is shifted before projecting to two dimensions, then a similar situation holds.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I do agree with this argument, however, I guess that's not the end of the story. I wrote to the authors of the two papers above, the problem turns out be that the lattice $p$ should be carefully shifted to create the standard Penrose tiling. But I have not figured out the specific translation to the lattice.
    $endgroup$
    – whitegreen
    Jan 3 at 3:39












  • $begingroup$
    @whitegreen Perhaps write up your findings as another answer?
    $endgroup$
    – Travis
    Jan 3 at 3:40



















0












$begingroup$

I wrote to Prof. J.H.H.Perk, this is a quote from his email:



You want to make a different choice (the translation on lattice $p$), so that no three grid lines (3) pass through the same point. This requires that at least three of the five gammas are nonzero. Generically then you get a regular Penrose tiling, but you can still get singular Penrose tilings like the one for all gammas zero. Those choices are everywhere dense, like rationals are dense within the reals, but only infinitesimal part of them.



Which means the 5D lattice points are
$$
p=gamma_1 begin{bmatrix}
1\
0\
0\
0\
0\
end{bmatrix}
+ gamma_2 begin{bmatrix}
0\
1\
0\
0\
0\
end{bmatrix}
+ gamma_3 begin{bmatrix}
0\
0\
1\
0\
0\
end{bmatrix}
+ gamma_4 begin{bmatrix}
0\
0\
0\
1\
0\
end{bmatrix}
+ gamma_5 begin{bmatrix}
0\
0\
0\
0\
1\
end{bmatrix}
$$

However, I have not figured out the correct $gamma_1,dots,gamma_5 in mathbb{R}$.



If $gamma_i=0$ (the lattice is not shifted), the 3D window is as follows,
enter image description here



which contains ten points. This is against the case of Penrose pattern whose window should contain five points.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026892%2fwhats-wrong-with-this-penrose-pattern%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    The Wikipedia article states "Therefore, a finite patch cannot differentiate between the uncountably many Penrose tilings, nor even determine which position within the tiling is being shown". Thus they could be the same tiling but differently centered.



    In particular, aside from the two dimensional shifting of Penrose tilings, if a higher dimensional lattice is shifted before projecting to two dimensions, then a similar situation holds.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer. I do agree with this argument, however, I guess that's not the end of the story. I wrote to the authors of the two papers above, the problem turns out be that the lattice $p$ should be carefully shifted to create the standard Penrose tiling. But I have not figured out the specific translation to the lattice.
      $endgroup$
      – whitegreen
      Jan 3 at 3:39












    • $begingroup$
      @whitegreen Perhaps write up your findings as another answer?
      $endgroup$
      – Travis
      Jan 3 at 3:40
















    4












    $begingroup$

    The Wikipedia article states "Therefore, a finite patch cannot differentiate between the uncountably many Penrose tilings, nor even determine which position within the tiling is being shown". Thus they could be the same tiling but differently centered.



    In particular, aside from the two dimensional shifting of Penrose tilings, if a higher dimensional lattice is shifted before projecting to two dimensions, then a similar situation holds.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer. I do agree with this argument, however, I guess that's not the end of the story. I wrote to the authors of the two papers above, the problem turns out be that the lattice $p$ should be carefully shifted to create the standard Penrose tiling. But I have not figured out the specific translation to the lattice.
      $endgroup$
      – whitegreen
      Jan 3 at 3:39












    • $begingroup$
      @whitegreen Perhaps write up your findings as another answer?
      $endgroup$
      – Travis
      Jan 3 at 3:40














    4












    4








    4





    $begingroup$

    The Wikipedia article states "Therefore, a finite patch cannot differentiate between the uncountably many Penrose tilings, nor even determine which position within the tiling is being shown". Thus they could be the same tiling but differently centered.



    In particular, aside from the two dimensional shifting of Penrose tilings, if a higher dimensional lattice is shifted before projecting to two dimensions, then a similar situation holds.






    share|cite|improve this answer











    $endgroup$



    The Wikipedia article states "Therefore, a finite patch cannot differentiate between the uncountably many Penrose tilings, nor even determine which position within the tiling is being shown". Thus they could be the same tiling but differently centered.



    In particular, aside from the two dimensional shifting of Penrose tilings, if a higher dimensional lattice is shifted before projecting to two dimensions, then a similar situation holds.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 at 3:45

























    answered Dec 31 '18 at 22:55









    SomosSomos

    13.6k11135




    13.6k11135












    • $begingroup$
      Thanks for the answer. I do agree with this argument, however, I guess that's not the end of the story. I wrote to the authors of the two papers above, the problem turns out be that the lattice $p$ should be carefully shifted to create the standard Penrose tiling. But I have not figured out the specific translation to the lattice.
      $endgroup$
      – whitegreen
      Jan 3 at 3:39












    • $begingroup$
      @whitegreen Perhaps write up your findings as another answer?
      $endgroup$
      – Travis
      Jan 3 at 3:40


















    • $begingroup$
      Thanks for the answer. I do agree with this argument, however, I guess that's not the end of the story. I wrote to the authors of the two papers above, the problem turns out be that the lattice $p$ should be carefully shifted to create the standard Penrose tiling. But I have not figured out the specific translation to the lattice.
      $endgroup$
      – whitegreen
      Jan 3 at 3:39












    • $begingroup$
      @whitegreen Perhaps write up your findings as another answer?
      $endgroup$
      – Travis
      Jan 3 at 3:40
















    $begingroup$
    Thanks for the answer. I do agree with this argument, however, I guess that's not the end of the story. I wrote to the authors of the two papers above, the problem turns out be that the lattice $p$ should be carefully shifted to create the standard Penrose tiling. But I have not figured out the specific translation to the lattice.
    $endgroup$
    – whitegreen
    Jan 3 at 3:39






    $begingroup$
    Thanks for the answer. I do agree with this argument, however, I guess that's not the end of the story. I wrote to the authors of the two papers above, the problem turns out be that the lattice $p$ should be carefully shifted to create the standard Penrose tiling. But I have not figured out the specific translation to the lattice.
    $endgroup$
    – whitegreen
    Jan 3 at 3:39














    $begingroup$
    @whitegreen Perhaps write up your findings as another answer?
    $endgroup$
    – Travis
    Jan 3 at 3:40




    $begingroup$
    @whitegreen Perhaps write up your findings as another answer?
    $endgroup$
    – Travis
    Jan 3 at 3:40











    0












    $begingroup$

    I wrote to Prof. J.H.H.Perk, this is a quote from his email:



    You want to make a different choice (the translation on lattice $p$), so that no three grid lines (3) pass through the same point. This requires that at least three of the five gammas are nonzero. Generically then you get a regular Penrose tiling, but you can still get singular Penrose tilings like the one for all gammas zero. Those choices are everywhere dense, like rationals are dense within the reals, but only infinitesimal part of them.



    Which means the 5D lattice points are
    $$
    p=gamma_1 begin{bmatrix}
    1\
    0\
    0\
    0\
    0\
    end{bmatrix}
    + gamma_2 begin{bmatrix}
    0\
    1\
    0\
    0\
    0\
    end{bmatrix}
    + gamma_3 begin{bmatrix}
    0\
    0\
    1\
    0\
    0\
    end{bmatrix}
    + gamma_4 begin{bmatrix}
    0\
    0\
    0\
    1\
    0\
    end{bmatrix}
    + gamma_5 begin{bmatrix}
    0\
    0\
    0\
    0\
    1\
    end{bmatrix}
    $$

    However, I have not figured out the correct $gamma_1,dots,gamma_5 in mathbb{R}$.



    If $gamma_i=0$ (the lattice is not shifted), the 3D window is as follows,
    enter image description here



    which contains ten points. This is against the case of Penrose pattern whose window should contain five points.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      I wrote to Prof. J.H.H.Perk, this is a quote from his email:



      You want to make a different choice (the translation on lattice $p$), so that no three grid lines (3) pass through the same point. This requires that at least three of the five gammas are nonzero. Generically then you get a regular Penrose tiling, but you can still get singular Penrose tilings like the one for all gammas zero. Those choices are everywhere dense, like rationals are dense within the reals, but only infinitesimal part of them.



      Which means the 5D lattice points are
      $$
      p=gamma_1 begin{bmatrix}
      1\
      0\
      0\
      0\
      0\
      end{bmatrix}
      + gamma_2 begin{bmatrix}
      0\
      1\
      0\
      0\
      0\
      end{bmatrix}
      + gamma_3 begin{bmatrix}
      0\
      0\
      1\
      0\
      0\
      end{bmatrix}
      + gamma_4 begin{bmatrix}
      0\
      0\
      0\
      1\
      0\
      end{bmatrix}
      + gamma_5 begin{bmatrix}
      0\
      0\
      0\
      0\
      1\
      end{bmatrix}
      $$

      However, I have not figured out the correct $gamma_1,dots,gamma_5 in mathbb{R}$.



      If $gamma_i=0$ (the lattice is not shifted), the 3D window is as follows,
      enter image description here



      which contains ten points. This is against the case of Penrose pattern whose window should contain five points.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        I wrote to Prof. J.H.H.Perk, this is a quote from his email:



        You want to make a different choice (the translation on lattice $p$), so that no three grid lines (3) pass through the same point. This requires that at least three of the five gammas are nonzero. Generically then you get a regular Penrose tiling, but you can still get singular Penrose tilings like the one for all gammas zero. Those choices are everywhere dense, like rationals are dense within the reals, but only infinitesimal part of them.



        Which means the 5D lattice points are
        $$
        p=gamma_1 begin{bmatrix}
        1\
        0\
        0\
        0\
        0\
        end{bmatrix}
        + gamma_2 begin{bmatrix}
        0\
        1\
        0\
        0\
        0\
        end{bmatrix}
        + gamma_3 begin{bmatrix}
        0\
        0\
        1\
        0\
        0\
        end{bmatrix}
        + gamma_4 begin{bmatrix}
        0\
        0\
        0\
        1\
        0\
        end{bmatrix}
        + gamma_5 begin{bmatrix}
        0\
        0\
        0\
        0\
        1\
        end{bmatrix}
        $$

        However, I have not figured out the correct $gamma_1,dots,gamma_5 in mathbb{R}$.



        If $gamma_i=0$ (the lattice is not shifted), the 3D window is as follows,
        enter image description here



        which contains ten points. This is against the case of Penrose pattern whose window should contain five points.






        share|cite|improve this answer











        $endgroup$



        I wrote to Prof. J.H.H.Perk, this is a quote from his email:



        You want to make a different choice (the translation on lattice $p$), so that no three grid lines (3) pass through the same point. This requires that at least three of the five gammas are nonzero. Generically then you get a regular Penrose tiling, but you can still get singular Penrose tilings like the one for all gammas zero. Those choices are everywhere dense, like rationals are dense within the reals, but only infinitesimal part of them.



        Which means the 5D lattice points are
        $$
        p=gamma_1 begin{bmatrix}
        1\
        0\
        0\
        0\
        0\
        end{bmatrix}
        + gamma_2 begin{bmatrix}
        0\
        1\
        0\
        0\
        0\
        end{bmatrix}
        + gamma_3 begin{bmatrix}
        0\
        0\
        1\
        0\
        0\
        end{bmatrix}
        + gamma_4 begin{bmatrix}
        0\
        0\
        0\
        1\
        0\
        end{bmatrix}
        + gamma_5 begin{bmatrix}
        0\
        0\
        0\
        0\
        1\
        end{bmatrix}
        $$

        However, I have not figured out the correct $gamma_1,dots,gamma_5 in mathbb{R}$.



        If $gamma_i=0$ (the lattice is not shifted), the 3D window is as follows,
        enter image description here



        which contains ten points. This is against the case of Penrose pattern whose window should contain five points.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 3:56

























        answered Jan 3 at 3:45









        whitegreenwhitegreen

        477314




        477314






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026892%2fwhats-wrong-with-this-penrose-pattern%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Plaza Victoria

            Puebla de Zaragoza

            Musa