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  <front>
    <journal-meta>
      <journal-title-group>
        <journal-title>International Congress on Education and Technology in Sciences, December</journal-title>
      </journal-title-group>
    </journal-meta>
    <article-meta>
      <title-group>
        <article-title>An algorithm since the vector approach in Mathematica to generate the Von Koch curve and islands in a general way</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Segundo B. Correa-Erazo</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Ricardo Velezmoro-León</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Robert Ipanaqué-Chero</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>Universidad Nacional de Piura</institution>
          ,
          <addr-line>Urbanización Miraflores s/n, Piura</addr-line>
          ,
          <country country="PE">Perú</country>
        </aff>
      </contrib-group>
      <pub-date>
        <year>2023</year>
      </pub-date>
      <volume>0</volume>
      <fpage>4</fpage>
      <lpage>06</lpage>
      <abstract>
        <p>In this paper, the Von Koch curve is generated in a general way from a vector approach without using the classic system of iterated functions. This approach also allows constructing diferent Koch islands based on any regular polygon; to do this, a Koch curve is drawn on each of the sides of the regular polygon, taking into account that if the vertices of the polygon are traversed counterclockwise, An island will be generated inside the polygon, but if the vertices are traversed clockwise, an island outside the polygon will be developed. Given the versatility of the method, an interior island and an exterior island are also united, giving origin to a new fractal. To visualize the results, Mathematica scientific software was used.</p>
      </abstract>
      <kwd-group>
        <kwd>eol&gt;Von Koch curve</kwd>
        <kwd>Von Koch Island</kwd>
        <kwd>algorithm</kwd>
        <kwd>new focus</kwd>
      </kwd-group>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>1. Introduction</title>
      <p>The system of iterated functions that generates the classical Von Koch curve is given by:
⎧ 1(, ) = ( 13 , 13 )
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎪⎪ 2(, ) = ( 16  −
⎨
⎪⎪ 3(, ) = ( 16  +
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎩ 4(, ) = ( 13  + 23 , 13 )
√
63  + 12 , −
√ √
63  + 13 , 63  + 16 )
√
63  + 16  +
√3
6 )
(d) Fourth iteration.</p>
      <p>
        (e) Seventh iteration.
contractive operator with the Hausdorf metric and has a single fixed point [
        <xref ref-type="bibr" rid="ref2">2</xref>
        ].
This means that if we take any arbitrary ends segment of the plane and apply the above IFS to
it, is obtained as a final result the classic Von Koch curve as can be seen in the figure 1.
      </p>
      <p>
        As described above, numerically, it can be observed that efectively when evaluating the
transformations 1, 2, 3 and 4 starting at any pair of points in the plane, the values converge
to the points of the classical Von Koch curve defined in the interval [
        <xref ref-type="bibr" rid="ref1">0, 1</xref>
        ]. In table 1 it can be
seen that if we recursively evaluate the transformations 1 and 4 at the point (2, 3) using ten
iterations, the first transformation converges to (0, 0), while 4 does so towards (1, 0) which
are the extremes of the classical Von Koch curve.
      </p>
      <p>
        The big question is how to generate the Von Koch curve by taking a segment with arbitrary
ends {, } of the plane. We want the Koch curve obtained to be located on the segment
 . The interest in searching for diferent variants of the Von-Koch curve is because there
are exciting applications in the field of communications because, due to its shape, multiple
resonances can be obtained when the angle is modified from 600 to a diferent one [
        <xref ref-type="bibr" rid="ref3">3</xref>
        ].
In this article, an algorithm is built to generate the Von Koch curve generally based on three
operations, obtaining the five necessary points to develop the first iteration. This algorithm
has the power to draw the Koch curve on any segment of the plane that is taken; in addition, it
allows the generation of Koch islands based on any regular polygon, and the versatility of the
algorithm ofers the possibility of generating an island inside the polygon and another outside
the polygon depending on whether the vertices are taken in a counterclockwise and clockwise
direction respectively.
      </p>
    </sec>
    <sec id="sec-2">
      <title>2. Mathematical and computational foundations</title>
      <sec id="sec-2-1">
        <title>2.1. About the Von Koch curve classical</title>
        <p>
          The Von Koch curve classical is obtained from the following description algorithm [
          <xref ref-type="bibr" rid="ref4">4</xref>
          ]:
• step 1: In the first iteration, we divide the unit segment into three equal parts;
• Step 2: Above the central part, we build an equilateral triangle;
• Step 3: Now, we eliminate the base of the triangle, thus obtaining four segments of
length 13 ; that is, the length of the curve is now 43 ;
• Step 4: For the second iteration, we construct equilateral triangles in each of the four
segments and apply step 3.
        </p>
        <sec id="sec-2-1-1">
          <title>From this algorithm, the four transformations are obtained using the scheme in figure 2:</title>
          <p>The first transformation contracts the segment of length one to its third part; that is to say,
1(, ) = (︀ 13 , 1 )︀ . To obtain the second transformation rotate 1 an angle of 600 and we
3
move it to the point (︀ 31 , 0)︀ . Therefore:</p>
          <p>⎛cos(600) − sin(600)⎞ ⎛ 13 ⎞
2(, ) = ⎝ ⎠ . ⎝
sin(600)
cos(600)
⎛ 1 ⎞</p>
          <p>3
⎠ + ⎝ ⎠ = ⎜
0 ⎝
⎛ 61  −
√
63  + 13 ⎞
⎟
⎠</p>
          <p>Similarly, to determine the transformation 3 we must rotate 1 an angle of − 600 and
translate it to the point ︁( 12 , √63 )︁ . Therefore:</p>
          <p>⎛cos(− 600) − sin(− 600)⎞ ⎛ 13 ⎞
3(, ) = ⎝ ⎠ . ⎝</p>
          <p>Among the peculiarities of this curve is that it has infinite length because if we calculate its
length in stages, it has that in the first stage, its length will be 1 = 4( 31 ), in the second stage is
have 2 = 16( 19 ); and in the − th iteration it result  = (︀ 4 )︀ . Therefore, the length of the
3</p>
        </sec>
        <sec id="sec-2-1-2">
          <title>Von Koch curve will be:</title>
          <p>lim
→∞
︂( 4 )︂ 
3
= ∞</p>
          <p>
            On the other hand, it is a continuous curve everywhere but diferentiable in none of them [
            <xref ref-type="bibr" rid="ref5">5</xref>
            ]
since it has a corner at each of its points and can be used as an example in the classroom of the
diferential calculus course to explain the characteristics of a curve that is not diferentiable at
any of its points. When a cognitive conflict is created when explaining diferentiability, a test is
ofered in [
            <xref ref-type="bibr" rid="ref6">6</xref>
            ]. Its dimension is calculated using the formula:
dim =
          </p>
          <p>log 
log (︀ 1 )︀</p>
          <p>
            Where  is the number of boxes with which the curve is covered and  is the scale
that, in this case, will be 13 . Therefore the dimension of the Von Koch curve will be
dim = log 4 ︂) = 1 · 26186 [
            <xref ref-type="bibr" rid="ref3">3</xref>
            ] and can be interpreted as a dimension of capacity; That is, the
log︂( 11
          </p>
          <p>3
Koch curve has a capacity of 1 · 26186 to occupy the space R2.</p>
        </sec>
        <sec id="sec-2-1-3">
          <title>One of the variants ofered by the Koch curve is that a significant mathematical curve called a</title>
          <p>
            snowflake or Koch island can be constructed from it. This curve has acquired a lot of importance
because it is being used as an antenna in wireless communications [
            <xref ref-type="bibr" rid="ref7">7</xref>
            ] in antennas for satellites
[8]. It is constructed by starting with an equilateral triangle, removing the inner third of each
side, making another equilateral triangle where the side was removed, and then repeating the
process indefinitely [9]. The diferent stages of the construction are shown in the figure 3.
          </p>
          <p>In the same way, the snowflake can be generated starting with a square and, in general, from
any polygon. This range of possibilities leads us to think of an algorithm that is simple to handle,
given that if we use a system of iterated functions, twelve transformations would be needed
when we start with a triangle. In the case of starting with a square sixteen and generally for a
polygon with  sides, 4 transformations with  ≥ 3 are required.
2.2. Standard Mathematica tools for generating the Von Koch curve classical
Mathematica incorporates the KochCurve function to generate the Von Koch curve classical.</p>
        </sec>
        <sec id="sec-2-1-4">
          <title>The variants of this function are:</title>
          <p>which gives the line segments representing the (ℎ)-step Koch curve,
which takes a series of steps of unit length at successive relative angles  , and
KochCurve[]</p>
          <p>KochCurve[, { 1,  2, . . .}]</p>
          <p>KochCurve[, {{1,  1}, {2,  2}, . . .}]
which takes successive steps of lengths proportional to .</p>
        </sec>
        <sec id="sec-2-1-5">
          <title>Below are examples for each variant.</title>
          <p>
            In[
            <xref ref-type="bibr" rid="ref1">1</xref>
            ]:= Graphics[KochCurve[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]]
Out[
            <xref ref-type="bibr" rid="ref1">1</xref>
            ]:= See figure 4.
          </p>
          <p>
            In[
            <xref ref-type="bibr" rid="ref2">2</xref>
            ]:= Graphics[KochCurve[3,{0,85 Degree,-85 Degree,
-85 Degree,85 Degree}]]
Out[
            <xref ref-type="bibr" rid="ref2">2</xref>
            ]:= See figure 5.
          </p>
          <p>
            In[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]:= Graphics[KochCurve[3,{{1,0},{1,90 Degree},{1,-90 Degree},
{2,-90 Degree},{1,90 Degree},{1,90 Degree},
{1,-90 Degree}}]]
          </p>
        </sec>
      </sec>
      <sec id="sec-2-2">
        <title>2.3. Algorithm</title>
        <p>It is known that the IFS that generates the Koch curve has four transformations, and if we
start with a segment of extremes  (1, 2) and (1, 2) when evaluating the IFS in the first
iteration will be obtained 8 points; for the second 32 points; for the third 128 and so on, which
would require more computational time.</p>
        <p>This algorithm consists of three operations for which 5 points are obtained in the first iteration,
20 points for the second, 80 for the third, and so on. The tools that will be used to describe the
algorithm are obtained from the following analysis.</p>
        <p>Consider the scheme proposed in figure 7. The points  and  remain fixed, and the only
thing we would need are the points 1, 2 and 3. Indeed:
1 is obtained through:</p>
        <p>1 2 + 
1 =  + 3 ( −  ) ⇒ 1 = 3</p>
        <p>To get to 2 we use the complex transformation  : R2 → R2 defined by  (1, 2) =
(− 2, 1) [10], given that the segment  2 is perpendicular to the segment  , where  is
the midpoint of the segment  . Therefore:
• step 1: To consider two arbitrarily diferent points of the plane;
• Step 2: To calculate the points {1, 2, 3} and add the points  y , obtaining the
list {, 1, 2, 3, }, which corresponds to the first iteration;
• Step 3: Partition the previous list two by two, leaving:</p>
        <p>{{, 1}, {1, 2}, {2, 3}, {3, }}</p>
        <sec id="sec-2-2-1">
          <title>And to apply step 2 to each of the partitions;</title>
          <p>• Step 4: Re-partition the resulting list and apply step 2 again, as often as the desired
iteration indicates;
• Step 5: Draw the traverse from the points obtained in the required iteration; this
traverse will be the Von Koch curve at the desired stage.</p>
        </sec>
      </sec>
      <sec id="sec-2-3">
        <title>2.4. About code in Mathematica</title>
        <p>The coding of the above algorithm in the scientific software Mathematica is based on simple
commands ofered by the software [11], which are detailed below:</p>
        <p>The Flatten command removes the grouping signs corresponding to the braces from any list.</p>
        <sec id="sec-2-3-1">
          <title>For example:</title>
          <p>The Partition command partitions a list into a set of lists according to the user’s convenience.</p>
        </sec>
        <sec id="sec-2-3-2">
          <title>It can be two by two, three by three, etc. For example:</title>
          <p>
            In[
            <xref ref-type="bibr" rid="ref2">2</xref>
            ]:= Partition[{2,4,6,2,1,8},2]
Out[
            <xref ref-type="bibr" rid="ref2">2</xref>
            ]:= {{2,4},{6,2},{1,8}}
          </p>
          <p>
            If in the previous entry after option 2, we add option 1; it results in a consecutive partition
increasing the number of lists. This is:
In[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]:= Partition[{2,4,6,2,1,8},2,1]
Out[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]:= {{2,4},{4,6},{6,2},{2,1},{1,8}}
          </p>
        </sec>
        <sec id="sec-2-3-3">
          <title>Using the previous commands defined the command IniKoch as follows:</title>
        </sec>
        <sec id="sec-2-3-4">
          <title>Command1</title>
          <p>
            In[
            <xref ref-type="bibr" rid="ref4">4</xref>
            ]:= IniKoch[P:{_,_},Q:{_,_}]:=Module[{J,PuntP1,PuntP2,PuntP3};
J[x1_,x2_]:={-x2,x1};
PuntP1[{P_,Q_}]:=(2P+Q)/3;
PuntP2[{P_,Q_}]:=(P+Q)/2+(sqrt3/6)J[Q-P];
          </p>
          <p>PuntP3[{P_,Q_}]:=(P+2Q)/3;
{P,PuntP1[{P,Q}],PuntP2[{P,Q}],PuntP3[{P,Q}],Q}]</p>
        </sec>
        <sec id="sec-2-3-5">
          <title>The command IniKoch defined above has the mission of generating the five points</title>
          <p>{, 1, 2, 3, } corresponding to the first iteration of the curve Von Koch. It should be
noted that the command input corresponds to any two points of the plane {, } and the points
{1, 2, 3} are calculated based on as mentioned in the description of the algorithm. Using
the IniKoch command, the KochCurveGeneral command is defined that will generate the</p>
        </sec>
        <sec id="sec-2-3-6">
          <title>Von Koch curve in a general way as follows:</title>
        </sec>
        <sec id="sec-2-3-7">
          <title>Command2</title>
          <p>
            In[
            <xref ref-type="bibr" rid="ref5">5</xref>
            ]:= CurvaVonKochGeneral[P:{_,_},Q:{_,_},n_,Opts___]:=Module[{},
vk[
            <xref ref-type="bibr" rid="ref1">1</xref>
            ]:={P,Q};
vk[
            <xref ref-type="bibr" rid="ref2">2</xref>
            ]:=N[IniKoch[vk[
            <xref ref-type="bibr" rid="ref1">1</xref>
            ],6];
vk[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]:=IniKoch/@Partition[vk[
            <xref ref-type="bibr" rid="ref2">2</xref>
            ],2,1];
vk[i_]:=IniKoch/@Partition[Partition[Flatten[vk[i-1]],2],2,1];
Graphics[{Hue[color],Line[vk[n]]},Opts]
          </p>
          <p>To increase the variants of the Koch curve, we can think about applying the previous
commands to the sides of a regular polygon; however, determining the vertices of a regular polygon
without computational help is very laborious. This dificulty is resolved by the command in
Mathematica PoliRG2P detailed below. Let’s think of a point  on the plane as the center of
the polygon; then, from any other point  on the plane, we can construct any regular polygon
using a rotation matrix as indicated in the figure 8. From this idea, the command that will
generate the vertices of the regular polygon is:</p>
        </sec>
        <sec id="sec-2-3-8">
          <title>Command3</title>
          <p>If we execute the previous command for the points (0, 2) and  (3, 4) for  = 6 is obtained:
(c) Second iteration.</p>
          <p>(d) Fifth iteration.</p>
          <p>(a) First iteration.</p>
          <p>
            Out[
            <xref ref-type="bibr" rid="ref6">6</xref>
            ]:= {{3,4},{3/2-sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ],3/2(2+sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ])},
{-3/2-sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ],1+3sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]/2},{-3,0},{-3/2+sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ],
1-3sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]/2},{3/2+sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ],3-3sqrt[
            <xref ref-type="bibr" rid="ref3">3</xref>
            ]/2}}
(c) Second iteration.
          </p>
          <p>(d) Sixth iteration.</p>
          <p>the which are the vertices of a regular hexagon, see figure 9.</p>
          <p>By combining commands 2 and 3, we can create a new command IslaKochGeneralInt that
will draw the Koch curve internally on each side of the regular polygon. The command is
defined as follows:</p>
        </sec>
        <sec id="sec-2-3-9">
          <title>Command4</title>
          <p>
            Out[
            <xref ref-type="bibr" rid="ref7">7</xref>
            ]:= IslaKochGeneralInt[Cen:{_,_},P:{_, _},nlados_,iter_,
Opts___]:= Module[{vv,ll},
vv=PoliRG2P[Cen,P,nlados];
ll=Append[vv,vv[[
            <xref ref-type="bibr" rid="ref1">1</xref>
            ]]];
Show[Table[CurvaVonKochGeneral[ll[[i]], ll[[i + 1]],
iter, 0.65], i, 1,Length[ll] - 1], Axes -&gt; True,
AxesLabel -&gt; {X,Y}] ]
(c) Second iteration.
          </p>
          <p>(d) Seventh iteration.</p>
          <p>To draw the Koch curve externally on the sides of the polygon, we reverse the list of vertices
using the Mathematica command Reverse; this will cause the command to traverse the vertices
of the polygon clockwise; thus, command 4 is slightly modified by internally adding the option
to invert the list, being defined as follows:</p>
        </sec>
        <sec id="sec-2-3-10">
          <title>Command5</title>
          <p>
            Out[8]:= IslaKochGeneralExt[Cen:{_,_},P:{_, _},nlados_,iter_,
Opts___]:= Module[{vv,ll},
vv=PoliRG2P[Cen,P,nlados];
vvr= Reverse[vv];
ll=Append[vvr,vvr[[
            <xref ref-type="bibr" rid="ref1">1</xref>
            ]]];
Show[Table[CurvaVonKochGeneral[ll[[i]], ll[[i + 1]],
iter, 0.65], i, 1,Length[ll] - 1], Axes -&gt; True,
          </p>
          <p>AxesLabel -&gt; {X,Y}] ]</p>
          <p>Another advantage of this method is that the input list of command 2 consists of only two
points, namely  and . However, it can be generalized so that the input is more than two
(c) Fourth iteration.</p>
          <p>(d) Seventh iteration.
points so that Koch curves can be graphed in one polygonal. Having said this, the command
will be:</p>
        </sec>
        <sec id="sec-2-3-11">
          <title>Command6</title>
        </sec>
      </sec>
    </sec>
    <sec id="sec-3">
      <title>3. Results</title>
      <sec id="sec-3-1">
        <title>From the code proposed above, the following results are obtained:</title>
        <p>(c) Fifth iteration.</p>
        <p>(d) Seventh iteration.</p>
        <p>Example 1. for the segment whose ends are {(0, 1), (2, 0)}, is obtain the results of figure 10.</p>
        <p>Example 2. If we want the rotation to be downwards, we change the order of the points;
that is, we take in the order {(2, 0), (0, 1)} and obtain the graphic of the figure 11.</p>
        <p>Example 3. For the segment of extremes {(2, 1), (5, 5)}, is obtained (see figure 12):
Example 4. If we apply command 4 to the points (0, 2) y (3, 4) for an equilateral triangle
is obtain (see figure 13):</p>
        <p>Example 5. If in the previous example we apply command 5 we obtain the island in the
ifgure 14.</p>
        <p>Example 6. If we apply commands 5 and 6 for the same center and vertex and then join
them, we obtain a fascinating island, the same one formed by drawing the internal and external</p>
      </sec>
      <sec id="sec-3-2">
        <title>Koch curves on the sides of a polygon. For (2, 3) y  (4, 1), is obtained (see figure 15):</title>
        <p>Ejemplo 7. For the comannd 6, when entering points {{0, 0}, {2, 2}, {4, 2}, {5, 0}} is
ob(c) Second iteration.</p>
        <p>(d) Sixth iteration.
tained the images of the figure 16:</p>
        <p>Ejemplo 8. If we apply command 6 to the list of points {{0, 0}, {2, 2}, {4, 0}, {0, 0}}, That
is, to the vertices of a triangle traveled counterclockwise, is obtained the images in the figure 17:</p>
      </sec>
    </sec>
    <sec id="sec-4">
      <title>4. Conclusions</title>
      <p>This article shows an algorithm described based on simple computational operations extracted
from a vector point of view. The importance lies in the versatility of the algorithm since it
is possible to construct a Koch curve simply by taking any initial segment without the need
to build a system of iterated functions for each case. It was also shown that from this vector
approach, it is possible to construct two Koch islands, one interior, and one exterior, from any
regular polygon and generate a new fractal as the union of the two islands inherent to each
polygon. In addition to this, a command was also implemented with the power to graph a Koch
curve in each of the segments of a polygonal, giving rise to a new generalized Koch curve; it is
recommended to computationally implement the approach proposed in this article for when
the points belong to a sphere given that the world in which we live has a spherical shape.</p>
    </sec>
    <sec id="sec-5">
      <title>Acknowledgments</title>
      <p>Thanks to the National University of Piura authorities for purchasing the Mathematica scientific
software license with which the diferent results were obtained.
(c) Second iteration.</p>
      <p>(d) Sixth iteration.
[8] O. F. G. Palacios, R. E. D. Vargas, J. A. H. Perez, S. B. C. Erazo, S-band koch snowflake
fractal antenna for cubesats, in: 2016 IEEE ANDESCON, IEEE, 2016, pp. 1–4. doi:10.1109/
andescon.2016.7836227.</p>
      <p>[9] E. W. Weisstein, Koch snowflake, https://mathworld. wolfram. com/ (2008).
[10] E. Abbena, S. Salamon, A. Gray, Modern diferential geometry of curves and surfaces with</p>
      <sec id="sec-5-1">
        <title>Mathematica, CRC press, 2017.</title>
        <p>[11] J. W. Gray, Mastering Mathematica®: Programming Methods and Applications, Academic
Press, 2014.</p>
      </sec>
    </sec>
  </body>
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