• Re the Kenyon 17 two part conjecture

    From Roger Bagula@21:1/5 to All on Thu Jan 19 09:27:05 2017
    A two part triangle space fill that isn't one of the Kenyon 17: http://www.math.brown.edu/~rkenyon/gallery/all17.pdf
    As you know I spent some time making Mathematica code for all 17
    tiles. The triangle space-fill in the Kenyon 17 is symmetrical
    and is actually related to the von Koch set.
    This affine with the 1 to Sqrt[6] affine scaling is new. ( at least to me).
    The result was found experimentally in the area of simplex affine triangle IFS’s.
    Steward Robert Hinsley would be happy. https://lh3.googleusercontent.com/-B0J8eYkg750/WIDyOVnIl0I/AAAAAAABFO8/gkf6B8b-HX8KGv9oSbZ5botXPRhZpBqJgCL0B/w506-h365/.10_affine_triaqngle_spacefill3000000_crs2.png
    (*begin of mathematica code*)

    Clear[cr, cols, cr2, cr3, cr4]
    allColors = ColorData["Legacy"][[3, 1]];
    firstCols = {"LightBlue", "DodgerBlue", "Cyan", "Blue", "White",
    "Yellow", "DeepNaplesYellow", "LightYellow", "LightPink", "White",
    "Red", "Tomato", "Pink", "LightPink", "Purple", "DarkOrchid",
    "Magenta", "ManganeseBlue", "DeepNaplesYellow", "Orange",
    "DarkOrange", "Tomato", "GoldOchre", "LightPink", "Magenta",
    "Green", "DarkOrchid", "LightSalmon", "LightPink", "Sienna",
    "Green", "Mint", "DarkSlateGray", "ManganeseBlue", "SlateGray",
    "DarkOrange", "MistyRose", "DeepNaplesYellow", "GoldOchre",
    "SapGreen", "Yellow", "Yellow", "Tomato", "DeepNaplesYellow",
    "DodgerBlue", "Cyan", "Red", "Blue", "DeepNaplesYellow", "Green",
    "Magenta", "DarkOrchid", "LightSalmon", "LightPink", "Sienna",
    "Green", "Mint", "DarkSlateGray", "ManganeseBlue", "SlateGray",
    "DarkOrange", "MistyRose", "DeepNaplesYellow", "GoldOchre",
    "SapGreen", "Yellow", "LimeGreen"};
    cols = ColorData["Legacy", #] & /@
    Join[firstCols, Complement[allColors, firstCols]];
    rotate[theta_] := {{Cos[theta], -Sin[theta]}, {Sin[theta],
    Cos[theta]}};
    cr[n_] := cr[n] = cols[[n]];
    cr2[n_] := cr2[n] = cols[[n + 4]];
    cr3[n_] := cr3[n] = cols[[n + 8]];
    cr4[n_] := cr4[n] = cols[[n + 12]];
    cr5[n_] := cr5[n] = cols[[n + 16]]
    Clear[f, dlst, pt, cr, ptlst, M, an]
    rotate[theta_] := {{Cos[theta], -Sin[theta]}, {Sin[theta],
    Cos[theta]}};
    n0 = 2;
    dlst = Table[ Random[Integer, {1, n0}], {n, 3000000}];


    pt = {0.5, 0.5};
    n1 = 1; m1 = Sqrt[6];
    r = N[n1 + m1];
    f[1, {x_, y_}] := N[{n1*x/r - m1*y/r, -m1*x/r - n1*y/r}] + {1/2, 1/2}

    f[2, {x_, y_}] :=
    N[{-n1*x/r - m1*y/r, -m1*x/r - n1*y/r}] + {1/2, 1/2}
    ptlst = Point[
    Developer`ToPackedArray[
    Table[pt = f[dlst[[j]], pt], {j, Length[dlst]}]],
    VertexColors -> Developer`ToPackedArray[cr2 /@ dlst]]; Graphics[{PointSize[.001], ptlst}, AspectRatio -> Automatic,
    PlotRange -> All, ImageSize -> 1000, Background -> Black]
    (* end*)

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  • From Stewart Robert Hinsley@21:1/5 to Roger Bagula on Fri Jan 20 11:59:41 2017
    On 19/01/2017 17:27, Roger Bagula wrote:
    A two part triangle space fill that isn't one of the Kenyon 17: http://www.math.brown.edu/~rkenyon/gallery/all17.pdf

    The "Kenyon 17" are the 2-part directly self-similar tiles. They don't
    include 2-part self-similar tiles involving reflections, which would add
    (at least) the golden bee and half-a-dozen tiles sort of associated with
    the Perron number 1+i. Nor do they include self-affine but not
    self-similar tiles, as here.

    As you know I spent some time making Mathematica code for all 17
    tiles. The triangle space-fill in the Kenyon 17 is symmetrical
    and is actually related to the von Koch set.
    This affine with the 1 to Sqrt[6] affine scaling is new. ( at least to me). The result was found experimentally in the area of simplex affine triangle IFS’s.

    The well known result is that the right-isoceles triangle is an order 2 self-similar tile with one element congruent to the other, all right
    angled triangles are order 2 self-similar tiles with one element similar
    to the other, and all triangles are order 2 self-affine tiles.

    ifstile.com has a page on Robinson triangles (and similarly constructed figures) which are the attractors of graph directed IFSs composed of
    (IIUC) similarity triangle. When I looked at them I reckoned that I
    could reverse engineer the order 3 versions, but the order 2 versions
    would need more study. (My program doesn't currently support graph
    directed IFSs.)

    Steward Robert Hinsley would be happy. https://lh3.googleusercontent.com/-B0J8eYkg750/WIDyOVnIl0I/AAAAAAABFO8/gkf6B8b-HX8KGv9oSbZ5botXPRhZpBqJgCL0B/w506-h365/.10_affine_triaqngle_spacefill3000000_crs2.png

    I haven't managed to convert your Mathematica code to an IFS, so I don't
    know whether there's anything special about this particular triangle,
    but sample code to generate a 2-part IFS for any triangle (with equal
    areas for each part) is

    Affine2D scalenize = new Affine2D(1, skew, 0, 0, ystretch, 0); AffineTransformList atl = new AffineTransformList(
    Affine2D.atReptile(2, 2) + Affine2D.atRotate90 - 1,
    Affine2D.atReptile(2, 2) + Affine2D.atRotate180 + 1); atl.Prefix(Affine2D.atInverted(scalenize)).Suffix(scalenize).Render("scalene");

    For each such triangle there is an uncountably infinite number of
    dissections parameterised by the ratio of the areas of the two parts. I
    don't think I ever coded this up, but the algorithm to do so is
    conceptually simple.

    1) start with the right isoceles triangle
    2) apply the transform x -> (1+s)x to one part
    3) apply the transform x -> (1-s)x to the other part.
    4) move the transforms so that their bases fill the base of the right
    isoceles triangle
    5) skew the transforms so that the remaining vertex of each coincides
    with the right angled vertex of the right isoceles triangle.
    6) apply the sample code above with appropriate values of ystretch and
    skew to get any desired triangle.

    Your Mathmatica code appears to be doing something on those lines.

    --
    SRH

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  • From Roger Bagula@21:1/5 to All on Sat Jan 20 13:45:46 2018
    See above post:
    at least one tile does exist that isn't one of the Kenyon 17.

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  • From Roger Bagula@21:1/5 to All on Tue Feb 20 07:26:07 2018
    Link to a picture: https://i.pinimg.com/564x/b9/06/a2/b906a22584f54844162fb2e98723d656.jpg

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  • From Roger Bagula@21:1/5 to All on Tue Feb 20 07:23:32 2018
    Broken Rule Tiling:
    Tiling with self-similar overlap:
    broken rule tiling;
    This tiling is two transforms
    and not any kind that Kenyon ever contemplated in his 17.
    (*Mathematica program*)
    Clear[f, dlst, pt, cr, ptlst, M, r, p, rotate]
    allColors = ColorData["Legacy"][[3, 1]];
    firstCols = {"Red", "Blue", "Magenta", "Green", "DarkOrchid", "LightSalmon",
    "LightPink", "Sienna", "Green", "Mint", "DarkSlateGray", "ManganeseBlue",
    "SlateGray", "DarkOrange", "MistyRose", "DeepNaplesYellow", "GoldOchre",
    "SapGreen", "Yellow"};
    cols = ColorData["Legacy", #] & /@
    Join[firstCols, Complement[allColors, firstCols]];
    Length[cols];
    (*IFS broken rule fractal tiling by Roger L.Bagula 19 Feb 2018©*)
    (*IFS \
    program type*)
    {m1,
    m2} = {{{0.7414178834516808`, 0}, {0,
    0.7414178834516808`}}, {{0.671043606703789`, 0}, {0,
    0.671043606703789`}}};
    A = {{0, 1}, {-1, 0}}
    a1 = Expand[(A.m1.{x, y}).A.m1.{x, y}]
    a2 = Expand[(A.m2.{-x, -y}).(A.m2.{-x, -y})]
    sc = Sqrt[1/FullSimplify[ExpandAll[(a1 + a2)/(x^2 + y^2)]]]

    f[1, {x_, y_}] = Expand[A.m1.{x, y} + {-1, 0}];
    f[2, {x_, y_}] = Expand[A.m2.{-x, -y} + {1, 0}];
    pt = {0.5, 0.5};
    {p1, p2} = {Det[sc*A.m1], Det[sc*A.m2]}

    dlst = Table[ Which[(r = Random[]) <= p1, 1, r <= 1, 2]
    , {n, 3000000}];
    cr[n_] := cr[n] = cols[[n ]];
    {0.5497004779019702`, 0.45029952209802954`}
    ptlst = Point[
    Developer`ToPackedArray[Table[pt = f[dlst[[j]], pt], {j, Length[dlst]}]],
    VertexColors -> Developer`ToPackedArray[cr /@ dlst]]; Graphics[{PointSize[.005], ptlst}, AspectRatio -> 1, ImageSize -> 1000]
    (*end program*)

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  • From Roger Bagula@21:1/5 to All on Sun Mar 4 08:17:00 2018
    Finding new roots to a triple pisot polynomial:

    I discovered this set of polynomials a couple of days ago:
    Special roots:
    {-0.662358978622373` - 0.5622795120623013` I, -0.662358978622373` +
    0.5622795120623013` I, -0.1557688521622851` + 0.8547594579428396` I,
    0.8181278307846581` + 0.29247994588053833` I}
    I’ve looked at the special root {1,5}, and {2,3} Moran solutions
    and they gives s=2 fractals( overlapping). I haven’t found a tiling.
    These n=3 120 degree solutions aren’t the only ones:
    n=6 60 degrees also gives special roots.
    in general:
    p[x_, n_] =
    FullSimplify[
    ExpandAll[(x^3 - x - 1)*(x^3 - Exp[I*2*Pi/n]*x - 1)*(x^3 +
    Exp[I*2*Pi/n]*x - 1)]]
    Table[ListPlot[{Re[x], Im[x]} /. NSolve[p[x, n] == 0, x],
    PlotStyle -> Red], {n, 1, 15, 1/3}];
    Show[%, PlotRange -> All]


    The roots form three circles with radius of about 1/3.

    (* mathematica*)
    (* special case tripisot*)
    FullSimplify[
    ExpandAll[(x^3 - x - 1)*(x^3 - Exp[I*2*Pi/3]*x - 1)*(x^3 + Exp[I*2*Pi/3]*x -
    1)]]
    (* solved*)
    ll = {Re[x], Im[x]} /.
    NSolve[(-1 - x + x^3) (1 + x^2 ((-1)^(1/3) - 2 x + x^4)) == 0, x] ListPlot[ll, PlotStyle -> Red]
    Table[Sqrt[ll[[i]].ll[[i]]], {i, Length[ll]}]
    (* special roots*)
    {{-0.662358978622373`, -0.5622795120623013`} \
    {-0.662358978622373`, 0.5622795120623013`}, {-0.1557688521622851`,
    0.8547594579428396`}, {0.8181278307846581`, 0.29247994588053833`}}
    (* forming matrices of all the roots*)
    m = {{Re[x], Im[x]}, {-Im[x],
    Re[x]}} /.
    NSolve[(-1 - x + x^3) (1 + x^2 ((-1)^(1/3) - 2 x + x^4)) == 0, x] Table[Det[m[[i]]], {i, 9}]
    (* doing Moran powers for all the roots*)
    Table[
    Table[{n, k, i,
    Det[MatrixPower[m[[i]], n]] + Det[MatrixPower[m[[i]], k]]}, {i, 9}], {n, 1,
    5}, {k, 1, n}]
    (* finding roots that have unitary ( one) two matrix Morans*)
    Delete[Union[
    Flatten[
    Table[Union[
    Table[
    If[Abs[
    Det[MatrixPower[m[[i]], n]] + Det[MatrixPower[m[[i]], k]] - 1.0] <
    0.01, {n, k, i, m[[i]]}, {}], {i, 9}]], {n, 1, 5}, {k, 1, n}],
    2]], 1]
    c = {3, 4, 6, 7}
    (* showing the spacial roots*)
    w =
    Table[x /.
    NSolve[(-1 - x + x^3) (1 + x^2 ((-1)^(1/3) - 2 x + x^4)) == 0,
    x][[c[[i]]]], {i, 4}]
    {-0.662358978622373` - 0.5622795120623013` I, -0.662358978622373` +
    0.5622795120623013` I, -0.1557688521622851` + 0.8547594579428396` I,
    0.8181278307846581` + 0.29247994588053833` I}
    (* end*)

    IFS program:
    (*Mathematica program*)Clear[f, dlst, pt, cr, ptlst, M, r, p, rotate]
    allColors = ColorData["Legacy"][[3, 1]];
    firstCols = {"Red", "Blue", "Magenta", "Green", "DarkOrchid", "LightSalmon",
    "LightPink", "Sienna", "Green", "Mint", "DarkSlateGray", "ManganeseBlue",
    "SlateGray", "DarkOrange", "MistyRose", "DeepNaplesYellow", "GoldOchre",
    "SapGreen", "Yellow"};
    cols = ColorData["Legacy", #] & /@
    Join[firstCols, Complement[allColors, firstCols]];
    Length[cols];
    (*IFS by Roger L.Bagula 4 Mar 2018©*)
    (*IFS program type*)
    {m1, m2} =
    Table[MatrixPower[{{-0.1557688521622851`,
    0.8547594579428396`}, {-0.8547594579428396`, -0.1557688521622851`}},
    i], {i, 1, 5, 4}]
    A = {{0, 1}, {1, 0}}
    a1 = Expand[(A.m1.{x, y}).A.m1.{x, y}] // Chop
    a2 = Expand[(A.m2.{-x, -y}).(A.m2.{-x, -y})] // Chop
    sc = FullSimplify[
    ExpandAll[Sqrt[1/FullSimplify[ExpandAll[(a1 + a2)/(x^2 + y^2)]]]]]

    f[1, {x_, y_}] = m1.{x, y} + {-1, 0}
    f[2, {x_, y_}] = A.m2.{x, y} + {1, 0}
    pt = {0.5, 0.5};
    {p1, p2} = {Det[m1], Det[m2]}

    dlst = Table[ Which[(r = Random[]) <= p1, 1, r <= 1, 2]
    , {n, 90000}];
    cr[n_] := cr[n] = cols[[n ]];

    ptlst = Point[
    Developer`ToPackedArray[Table[pt = f[dlst[[j]], pt], {j, Length[dlst]}]],
    VertexColors -> Developer`ToPackedArray[cr /@ dlst]]; Graphics[{PointSize[.001], ptlst}, AspectRatio -> 1, ImageSize -> 1000,
    Background -> Black]
    (*end program*) https://i.pinimg.com/564x/45/34/d5/4534d548d14c6df840f4e14158f5dfc1.jpg

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  • From Roger Bagula@21:1/5 to All on Sun Mar 4 10:41:15 2018
    They actually come in pairs of cubic polynomials to give 6 special roots on the polynomial:
    ( nearly symmetric)
    1 - 6 x^3 + 14 x^6 - 20 x^9 + 15 x^12 - 6 x^15 + x^18
    Where the roots form 3 hexagons of relative radius of about 1/3.
    Actually:
    r=(1.32472-0.682328)/2=0.321196
    p[x_] = FullSimplify[
    ExpandAll[(x^3 + x - 1)*(x^3 - x - 1)*(x^3 - Exp[I*2*Pi/3]*x - 1)*(x^3 +
    Exp[I*2*Pi/3]*x - 1)*(x^3 - Exp[-I*2*Pi/3]*x - 1)*(x^3 +
    Exp[-I*2*Pi/3]*x - 1)]]
    1 - 6 x^3 + 14 x^6 - 20 x^9 + 15 x^12 - 6 x^15 + x^18
    ll = {Re[x], Im[x]} /. NSolve[p[x] == 0, x]
    Out[596]= {{-0.835342, -0.876227}, {-0.835342,
    0.876227}, {-0.662359, -1.14724}, {-0.662359,
    1.14724}, {-0.662359, -0.56228}, {-0.662359,
    0.56228}, {-0.341164, -0.590913}, {-0.341164,
    0.590913}, {-0.341164, -1.16154}, {-0.341164,
    1.16154}, {-0.155769, -0.854759}, {-0.155769, 0.854759}, {0.682328,
    0}, {0.818128, -0.29248}, {0.818128,
    0.29248}, {1.17651, -0.285314}, {1.17651, 0.285314}, {1.32472, 0}}
    ListPlot[ll, PlotStyle -> Red] https://i.pinimg.com/564x/97/fe/50/97fe5018891f7cb8c28a73aac528710e.jpg

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  • From Stewart Robert Hinsley@21:1/5 to Roger Bagula on Tue Mar 6 19:46:59 2018
    On 20/02/2018 15:23, Roger Bagula wrote:
    Broken Rule Tiling:
    Tiling with self-similar overlap:
    broken rule tiling;
    This tiling is two transforms
    and not any kind that Kenyon ever contemplated in his 17.

    As it's not self similar it's not relevant to the Kenyon conjecture. (If
    you allow reflections you get half-a-dozen or so more additional
    rep-tiles, as well as the golden bee. If you allow self-affine
    dissections you get at least all triangles and parallelograms, rather
    than a restricted set. If you allow graph-directed IFSs you get
    additional figures - see ifstile.com.)

    I wouldn't be surprised that if you allowed overlaps that gives you
    other figured which are tiles. But while tiles can be remarkably cryptic
    (e.g. the Levy curve) this doesn't look like a tile to me - if you try
    to fill in the white lines with smaller copies you just get smaller
    white lines. I'd guess that you could fix that point by moving the
    copies together so that you have both a squarish overlap and a linear
    overlap (replacing the linear gap), but that doesn't guarantee that the
    figure obtained is a tile - for self-similar tiles there is a
    constructive proof (dissect and expand) that they tile the plane*, but
    that proof fails if the attractor is not a dissection of itself, so it
    is necessary to provide an argument that it is a tile.

    *If I recall correctly, from consideration of which polyomino IFS
    attractors are fixed and which have additional degrees of freedom, this procedure only works for self-affine figures if the difference in
    orientation between copies is 0 or pi.

    Some els (rectilinear hexagons) have rectangular tilings with unit cells composed of two copies (e.g. L-triomino, L-tetramino, P- and
    L-pentaminos, etc). I believe that this also have herringbone tilings
    with a single copy in the unit cell (as well as aperiodic tilings).

    I'm wondering whether all els have herringbone tilings. I'll have to
    think a little more on that point.

    (*Mathematica program*)
    Clear[f, dlst, pt, cr, ptlst, M, r, p, rotate]
    allColors = ColorData["Legacy"][[3, 1]];
    firstCols = {"Red", "Blue", "Magenta", "Green", "DarkOrchid", "LightSalmon",
    "LightPink", "Sienna", "Green", "Mint", "DarkSlateGray", "ManganeseBlue",
    "SlateGray", "DarkOrange", "MistyRose", "DeepNaplesYellow", "GoldOchre",
    "SapGreen", "Yellow"};
    cols = ColorData["Legacy", #] & /@
    Join[firstCols, Complement[allColors, firstCols]];
    Length[cols];
    (*IFS broken rule fractal tiling by Roger L.Bagula 19 Feb 2018©*)
    (*IFS \
    program type*)
    {m1,
    m2} = {{{0.7414178834516808`, 0}, {0,
    0.7414178834516808`}}, {{0.671043606703789`, 0}, {0,
    0.671043606703789`}}};
    A = {{0, 1}, {-1, 0}}
    a1 = Expand[(A.m1.{x, y}).A.m1.{x, y}]
    a2 = Expand[(A.m2.{-x, -y}).(A.m2.{-x, -y})]
    sc = Sqrt[1/FullSimplify[ExpandAll[(a1 + a2)/(x^2 + y^2)]]]

    f[1, {x_, y_}] = Expand[A.m1.{x, y} + {-1, 0}];
    f[2, {x_, y_}] = Expand[A.m2.{-x, -y} + {1, 0}];
    pt = {0.5, 0.5};
    {p1, p2} = {Det[sc*A.m1], Det[sc*A.m2]}

    dlst = Table[ Which[(r = Random[]) <= p1, 1, r <= 1, 2]
    , {n, 3000000}];
    cr[n_] := cr[n] = cols[[n ]];
    {0.5497004779019702`, 0.45029952209802954`}
    ptlst = Point[
    Developer`ToPackedArray[Table[pt = f[dlst[[j]], pt], {j, Length[dlst]}]],
    VertexColors -> Developer`ToPackedArray[cr /@ dlst]]; Graphics[{PointSize[.005], ptlst}, AspectRatio -> 1, ImageSize -> 1000]
    (*end program*)


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  • From Roger Bagula@21:1/5 to All on Fri Mar 9 10:29:55 2018
    Steward Robert Hinsley
    Well darn , you are still alive. LOL.
    I thought the "demon" had declared you dead when they deXed your great web site without half a thought.
    Your {1,3} triangle tile on the x^3-x^2-1 Pisot
    is being called a "cow tile" by Ed Pegg.
    I've started three tiling interest groups besides the older yahoo one.
    Yahoo has big problems.
    It would be nice with four groups if you'd join one.
    Linked in:
    https://www.linkedin.com/groups/13577869
    Facebook:
    https://www.facebook.com/groups/391950357895182/?pnref=story
    Google+:
    https://plus.google.com/communities/115723411672694466822
    And Yahoo: https://groups.yahoo.com/neo/groups/true_tile/info;_ylc=X3oDMTJmNHRtNDdvBF9TAzk3MzU5NzE0BGdycElkAzEzOTk3MzMzBGdycHNwSWQDMTcwNTA4MzM4OARzZWMDdnRsBHNsawN2Z2hwBHN0aW1lAzE1MTQ3OTU3NzQ-
    Since we first got definitions for the Rauzy and Akiyama tilings about 2000, things in tiling have really taken off. Their is an encyclopedia that Dieter Steemann quotes in his Wolfram demos of the tiling Kenyon program.
    You should printout your tiling web pages ( the original ones) and try to get them published. The web isn't a very sure place these days.
    Roger Bagula

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  • From Roger Bagula@21:1/5 to All on Sat Mar 10 09:01:25 2018
    Ed Pegg posted this interesting link: http://tds.math.msu.su/.../f/fe/Poster_Taran_Rauzy3D.pdf

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  • From Roger Bagula@21:1/5 to All on Sat Mar 10 09:03:16 2018
    Ed Pegg posted this link at facebook; http://tds.math.msu.su/wiki/images/f/fe/Poster_Taran_Rauzy3D.pdf

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