Larry Riddle, Agnes Scott College

## Flowsnake IFS Details

 $${f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { -1/ \sqrt 14} & {3\sqrt 3 /14 } \\ {-3\sqrt 3 /14} & { -1/ \sqrt 14} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/14 } \\ {3\sqrt 3 /14} \\ \end{array}} \right]$$ scale by $$1/ \sqrt{7}$$, rotate by A-120° $${f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 5/14} & {- \sqrt 3 / 14} \\ { \sqrt 3 / 14} & { 5/14} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/14 } \\ {3\sqrt 3 /14} \\ \end{array}} \right]$$ scale by $$1/ \sqrt{7}$$, rotate by A $${f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 5/14} & {- \sqrt 3 / 14} \\ { \sqrt 3 / 14} & { 5/14} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {3/7} \\ {2 \sqrt 3 / 7} \\ \end{array}} \right]$$ scale by $$1/ \sqrt{7}$$, rotate by A $${f_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { -1/ \sqrt 14} & {3\sqrt 3 /14 } \\ {-3\sqrt 3 /14} & { -1/ \sqrt 14} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {11/14} \\ {5 \sqrt 3 / 14} \\ \end{array}} \right]$$ scale by $$1/ \sqrt{7}$$, rotate by A-120° $${f_5}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 5/14} & {- \sqrt 3 / 14} \\ { \sqrt 3 / 14} & { 5/14} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {5/14} \\ {\sqrt 3 /14} \\ \end{array}} \right]$$ scale by $$1/ \sqrt{7}$$, rotate by A $${f_6}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { -2/7} & {- \sqrt 3 /7} \\ { \sqrt 3 /7} & { -2/7} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {9/14} \\ {- \sqrt 3 / 14} \\ \end{array}} \right]$$ scale by $$1/ \sqrt{7}$$, rotate by A+120° $${f_7}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 5/14} & {- \sqrt 3 / 14} \\ { \sqrt 3 / 14} & { 5/14} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {9/14} \\ {- \sqrt 3 / 14} \\ \end{array}} \right]$$ scale by $$1/ \sqrt{7}$$, rotate by A

Start with a pattern of seven regular hexagons. We will place those hexagons so that they lie with two vertices at (0,0) and (1,0). Eight vertices of the smaller hexagons are connected as shown above to create the basic red motif. We can think of each red segment as having been scaled by a factor r from the horizontal unit line from (0,0) to (1,0), rotated by an appropriate angle, then translated. To find these values, we need to first determine the scaling factor r and the rotation for the first line segment starting at (0,0).

Consider the blue triangle shown in the figure above. Let A be the angle at the origin. The upper angle is 120° by properties of regular hexagons. We can use the Law of Cosines to get

\begin{align} {\left( {\frac{1}{2}} \right)^2} &= {r^2} + {\left( {\frac{r}{2}} \right)^2} - 2 \cdot r \cdot \frac{r}{2} \cdot \cos {120^ \circ } \\ &= {r^2} + \frac{{{r^2}}}{4} + \frac{{{r^2}}}{2} = \frac{{7{r^2}}}{4} \\ \end{align}

This shows that $$r = \frac{1}{\sqrt 7}$$. Now by the Law of Sines,

\begin{align} \frac{{\sin A}}{{\sin {{120}^ \circ }}} = \frac{{r/2}}{{1/2}} = \frac{1}{{\sqrt 7 }} \quad &\Rightarrow \sin A = \frac{1}{{\sqrt 7 }}\sin {120^ \circ } = \frac{{\sqrt 3 }}{{2\sqrt 7 }} \\ \\ &\Rightarrow A = \arcsin \left( {\frac{{\sqrt 3 }}{{2\sqrt 7 }}} \right) \approx {19.1066^ \circ } \\ \end{align}

The red line segment starting at the origin makes an angle of A+60° with the horizontal axis. But because of the orientation of that first red segment, we must rotate this segment by A+60°−180° = A−120° (this will be a negative value because the rotation will be in the clockwise direction.)

We can now read off the rest of the rotation angles from the figure above. They are respectively, A, A, A-120°, A, A+120°, and A.

The matrices in the IFS involve the sine and cosine values for the rotation angles. Since $$A = \arcsin \left( {\frac{{\sqrt 3 }}{{2\sqrt 7 }}} \right)$$, we have

$$\sin (A) = \dfrac{{\sqrt 3 }}{{2\sqrt 7 }} = \dfrac{{\sqrt 3 \cdot \sqrt 7 }}{{14}}$$

$$\cos (A) = \sqrt {1 - {{\sin }^2}(A)} = \sqrt {1 - \dfrac{{21}}{{{{14}^2}}}} = \dfrac{{5\sqrt 7 }}{{14}}$$

$$\sin (A - {120^ \circ }) = \sin (A)\cos ({120^ \circ }) - \sin ({120^ \circ })\cos (A) = - \dfrac{{\sqrt 3 \sqrt 7 }}{{28}} - \dfrac{{5\sqrt 3 \sqrt 7 }}{{28}} = - \dfrac{{3\sqrt 3 \sqrt 7 }}{{14}}$$

$$\cos (A - {120^ \circ }) = \cos (A)\cos ({120^ \circ }) + \sin (A)\sin ({120^ \circ }) = - \dfrac{{5\sqrt 7 }}{{28}} + \dfrac{{3\sqrt 7 }}{{28}} = - \dfrac{{\sqrt 7 }}{{14}}$$

$$\sin (A + {120^\circ }) = \sin (A)\cos ({120^\circ }) + \sin ({120^\circ })\cos (A) = - \dfrac{{\sqrt 3 \sqrt 7 }}{{28}} + \dfrac{{5\sqrt 3 \sqrt 7 }}{{28}} = - \dfrac{{\sqrt 3 \sqrt 7 }}{7}$$

$$\cos (A + {120^ \circ }) = \cos (A)\cos ({120^ \circ }) - \sin (A)\sin ({120^ \circ }) = - \dfrac{{5\sqrt 7 }}{{28}} - \dfrac{{3\sqrt 7 }}{{28}} = - \dfrac{{2\sqrt 7 }}{7}$$

The segments must also be translated to the appropriate points after being scaled and rotated, as shown in the figure below.

We just need to compute the coordinates of P1, P2, P3, P4, and P5.

$$P1 = \left[ {\begin{array}{*{20}{c}} {r\cos (60^ \circ + A)} \\ {r\sin (60^ \circ + A)} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1/14} \\ {3 \sqrt 3 / 14} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.07143} \\ {0.37115} \\ \end{array}} \right]$$

$$P2 = P1 + \left[ {\begin{array}{*{20}{c}} {r\cos (A)} \\ {r\sin (A))} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3/7} \\ {2 \sqrt 3 / 7} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.42857} \\ {0.49487} \\ \end{array}} \right]$$

$$P3 = P2 + \left[ {\begin{array}{*{20}{c}} {r\cos (A)} \\ {r\sin (A)} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11/14} \\ {5 \sqrt 3 / 14} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.78571} \\ {0.61859} \\ \end{array}} \right]$$

$$P4 = \left[ {\begin{array}{*{20}{c}} { r\cos ( A)} \\ { r\sin (A)} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {5/14} \\ { \sqrt 3 / 14} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.35714} \\ {0.12372} \\ \end{array}} \right]$$

$$P5 = \left[ {\begin{array}{*{20}{c}} {1-r\cos (A)} \\ { - r\sin (A)} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {9/14} \\ {- \sqrt 3 / 14} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.64286} \\ { - 0.12372} \\ \end{array}} \right]$$