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0

$begingroup$

A set of equations is given as follows:

$$ x^Omega+y^Omega=1 $$

$$ (1-x)^Omega+y^Omega=1 $$

$$ x^Omega+(1-y)^Omega=1 $$

$$ (1-x)^Omega+(1-y)^Omega=1. $$

$(OmegasubsetBbb Q) =1.1,1.2,1.3,...,N,$ where $NinBbb N$. In the equations, $x,y in Bbb A(0,1),$ where A is the set of algebraic numbers.

I was able to calculate the cardinality of $Omega:$

$|Omega|=(20(N-1))^2.$ This is the number of intersection points in the inner grid.

Copy the structure $K$ times, which includes the original, and rotate each structure such that all corner points are equally spaced apart. For example when $K=2,$ you would rotate the copied structure by $pi/4$ radians. When $K=3,$ you would rotate the first copy by $pi/4$ and the second copy by $pi/8.$ When $K=3$ you would rotate the first copy by $pi/4$ the second copy by $pi/8$ and the third copy by $pi/16.$

The rotation scheme is as follows:

$$ A=(y-frac12)cos(z)+(x-frac12)sin(z)+frac12 $$

$$ B=(x-frac12)cos(z)-(y-frac12)sin(z)+frac12. $$

Then we have the first copy of the original structure as: $$ A^Omega +B^Omega=1 $$

$$ (1-A)^Omega +B^Omega=1 $$

$$ A^Omega +(1-B)^Omega=1 $$

$$ (1-A)^Omega +(1-B)^Omega=1. $$

The amounts of rotation for each structure can be expressed by the series:

$$ pisum_s=2^K frac12^s=pi (frac12^2+frac12^3+...+frac12^K).$$

As you can see, for $K=2$ total structures, the sum is $fracpi4$ which tells you to rotate the first copy by that amount.

Taking infinite structures, the angle of rotation goes to zero and the sum of all rotations converges to $fracpi2:$

$$ pisum_s=2^infty frac12^s=fracpi2. $$

How many total intersections occur for a given $K$ and $N$?

How many intersections of degree $D$ are there for a given $K$ and $N$? What is $maxD$ for a given $K$ and $N?$

This is $K=4$ and the number of curves is about $392.$

This is $K=1$ and the number of curves is about $60.$

combinatorics permutations combinations computability

share|cite|improve this question

edited Mar 11 at 5:53

Ultradark

asked Mar 10 at 21:09

UltradarkUltradark

2951518

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What structures are we talking about? What is the "inner grid" you are referring to? If $Omega$ is a set, what does $x^Omega$ mean?
  $endgroup$
  – Morgan Rodgers
  Mar 10 at 23:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $Omega$ is a set of $N$ elements then there are $N$ equations of that form
  $endgroup$
  – Ultradark
  Mar 10 at 23:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A structure is a set of the four equations
  $endgroup$
  – Ultradark
  Mar 10 at 23:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the structure is the set of equations (four? it seems like there are $4cdot|Omega|$ equations), what does it mean to "rotate" a structure? Do you mean to say that a structure is a set of points in a plane whose coordinates satisfy a set of equations?
  $endgroup$
  – Morgan Rodgers
  Mar 10 at 23:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how did you get that $|Omega| = (20(N-1))^2$ when $| 1.1, 1.2, 1.3, ldots, N| = 10(N-1)$?
  $endgroup$
  – Morgan Rodgers
  Mar 10 at 23:31
  
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

A set of equations is given as follows:

$$ x^Omega+y^Omega=1 $$

$$ (1-x)^Omega+y^Omega=1 $$

$$ x^Omega+(1-y)^Omega=1 $$

$$ (1-x)^Omega+(1-y)^Omega=1. $$

$(OmegasubsetBbb Q) =1.1,1.2,1.3,...,N,$ where $NinBbb N$. In the equations, $x,y in Bbb A(0,1),$ where A is the set of algebraic numbers.

I was able to calculate the cardinality of $Omega:$

$|Omega|=(20(N-1))^2.$ This is the number of intersection points in the inner grid.

Copy the structure $K$ times, which includes the original, and rotate each structure such that all corner points are equally spaced apart. For example when $K=2,$ you would rotate the copied structure by $pi/4$ radians. When $K=3,$ you would rotate the first copy by $pi/4$ and the second copy by $pi/8.$ When $K=3$ you would rotate the first copy by $pi/4$ the second copy by $pi/8$ and the third copy by $pi/16.$

The rotation scheme is as follows:

$$ A=(y-frac12)cos(z)+(x-frac12)sin(z)+frac12 $$

$$ B=(x-frac12)cos(z)-(y-frac12)sin(z)+frac12. $$

Then we have the first copy of the original structure as: $$ A^Omega +B^Omega=1 $$

$$ (1-A)^Omega +B^Omega=1 $$

$$ A^Omega +(1-B)^Omega=1 $$

$$ (1-A)^Omega +(1-B)^Omega=1. $$

The amounts of rotation for each structure can be expressed by the series:

$$ pisum_s=2^K frac12^s=pi (frac12^2+frac12^3+...+frac12^K).$$

As you can see, for $K=2$ total structures, the sum is $fracpi4$ which tells you to rotate the first copy by that amount.

Taking infinite structures, the angle of rotation goes to zero and the sum of all rotations converges to $fracpi2:$

$$ pisum_s=2^infty frac12^s=fracpi2. $$

How many total intersections occur for a given $K$ and $N$?

How many intersections of degree $D$ are there for a given $K$ and $N$? What is $maxD$ for a given $K$ and $N?$

This is $K=4$ and the number of curves is about $392.$

This is $K=1$ and the number of curves is about $60.$

combinatorics permutations combinations computability

share|cite|improve this question

edited Mar 11 at 5:53

Ultradark

asked Mar 10 at 21:09

UltradarkUltradark

2951518

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What structures are we talking about? What is the "inner grid" you are referring to? If $Omega$ is a set, what does $x^Omega$ mean?
  $endgroup$
  – Morgan Rodgers
  Mar 10 at 23:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $Omega$ is a set of $N$ elements then there are $N$ equations of that form
  $endgroup$
  – Ultradark
  Mar 10 at 23:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A structure is a set of the four equations
  $endgroup$
  – Ultradark
  Mar 10 at 23:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the structure is the set of equations (four? it seems like there are $4cdot|Omega|$ equations), what does it mean to "rotate" a structure? Do you mean to say that a structure is a set of points in a plane whose coordinates satisfy a set of equations?
  $endgroup$
  – Morgan Rodgers
  Mar 10 at 23:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how did you get that $|Omega| = (20(N-1))^2$ when $| 1.1, 1.2, 1.3, ldots, N| = 10(N-1)$?
  $endgroup$
  – Morgan Rodgers
  Mar 10 at 23:31
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

A set of equations is given as follows:

$$ x^Omega+y^Omega=1 $$

$$ (1-x)^Omega+y^Omega=1 $$

$$ x^Omega+(1-y)^Omega=1 $$

$$ (1-x)^Omega+(1-y)^Omega=1. $$

$(OmegasubsetBbb Q) =1.1,1.2,1.3,...,N,$ where $NinBbb N$. In the equations, $x,y in Bbb A(0,1),$ where A is the set of algebraic numbers.

I was able to calculate the cardinality of $Omega:$

$|Omega|=(20(N-1))^2.$ This is the number of intersection points in the inner grid.

Copy the structure $K$ times, which includes the original, and rotate each structure such that all corner points are equally spaced apart. For example when $K=2,$ you would rotate the copied structure by $pi/4$ radians. When $K=3,$ you would rotate the first copy by $pi/4$ and the second copy by $pi/8.$ When $K=3$ you would rotate the first copy by $pi/4$ the second copy by $pi/8$ and the third copy by $pi/16.$

The rotation scheme is as follows:

$$ A=(y-frac12)cos(z)+(x-frac12)sin(z)+frac12 $$

$$ B=(x-frac12)cos(z)-(y-frac12)sin(z)+frac12. $$

Then we have the first copy of the original structure as: $$ A^Omega +B^Omega=1 $$

$$ (1-A)^Omega +B^Omega=1 $$

$$ A^Omega +(1-B)^Omega=1 $$

$$ (1-A)^Omega +(1-B)^Omega=1. $$

The amounts of rotation for each structure can be expressed by the series:

$$ pisum_s=2^K frac12^s=pi (frac12^2+frac12^3+...+frac12^K).$$

As you can see, for $K=2$ total structures, the sum is $fracpi4$ which tells you to rotate the first copy by that amount.

Taking infinite structures, the angle of rotation goes to zero and the sum of all rotations converges to $fracpi2:$

$$ pisum_s=2^infty frac12^s=fracpi2. $$

How many total intersections occur for a given $K$ and $N$?

How many intersections of degree $D$ are there for a given $K$ and $N$? What is $maxD$ for a given $K$ and $N?$

This is $K=4$ and the number of curves is about $392.$

This is $K=1$ and the number of curves is about $60.$

combinatorics permutations combinations computability

share|cite|improve this question

edited Mar 11 at 5:53

Ultradark

asked Mar 10 at 21:09

UltradarkUltradark

2951518

$endgroup$

share|cite|improve this question

edited Mar 11 at 5:53

Ultradark

asked Mar 10 at 21:09

UltradarkUltradark

2951518

share|cite|improve this question

edited Mar 11 at 5:53

Ultradark

asked Mar 10 at 21:09

UltradarkUltradark

2951518

asked Mar 10 at 21:09

UltradarkUltradark

2951518

asked Mar 10 at 21:09

UltradarkUltradark

2951518