STATUS: Draft

---

import numpy as np
import sympy as sp
import pickle
from IPython.display import HTML
import ipywidgets as widgets
import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
import pandas as pd
import itertools
pd.set_option('display.max_colwidth', None)

#

# function to print latex
def renderListToLatex(e):
    latex_rendering = []

    for i in range(len(e)):
        latex_rendering.append("$$" + sp.latex(e[i]) + "$$<br/>")
    
    return(HTML("".join(latex_rendering[0:])))

Solving Polynomial Equations (20)

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Aim: Understand the bridge betwen polygon subdivisions and polynomial equations

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Observe consider the operation $\bar{\triangledown}$ and its use as an operation between multisets of polygonal subdivisions

Observe It is possible to use this operation to operate on multisets:

$$ [\text{all polygons that can be divided into triangles}] \bar{\triangledown} [\text{all polygons that can be divided into triangles}] $$

Observe: It is to find these combinations, it is possible to use concatenation of trees and come at this from a computational perspective.

Let $F1$ be a function to calculate all combinations of rooted planar binary trees

def F1(s):
    if len(s) == 1:
        yield s
    else:
        for i in range(1, len(s), 2):
            for l in F1(s[:i]):
                for r in F1(s[i+1:]):
                    yield '({}{}{})'.format(l, s[i], r)

allComb = []
for t in F1('x:x:x:x'):
    allComb.append(t)
#allComb

Let $F2$ be a function to multiply rooted binary trees.

def F2(t1, t2):
    treeStructure = ['x:' for i in range(t1 + t2)]
    t = ''.join(treeStructure)[:-1]
    allComb = []
    for i in F1(t):
        allComb.append(i)
    return(allComb)

Observe: The nature of this multiplication on multisets yields all combinations of multiples.

print("Multiplication (2, 0):\n", F2(2, 0))
print("\nMultiplication (0, 2):\n", F2(0,3))
print("\nMultiplication (2, 3):\n",F2(2, 3))

Multiplication (2, 0):
 ['(x:x)']

Multiplication (0, 2):
 ['(x:(x:x))', '((x:x):x)']

Multiplication (2, 3):
 ['(x:(x:(x:(x:x))))', '(x:(x:((x:x):x)))', '(x:((x:x):(x:x)))', '(x:((x:(x:x)):x))', '(x:(((x:x):x):x))', '((x:x):(x:(x:x)))', '((x:x):((x:x):x))', '((x:(x:x)):(x:x))', '(((x:x):x):(x:x))', '((x:(x:(x:x))):x)', '((x:((x:x):x)):x)', '(((x:x):(x:x)):x)', '(((x:(x:x)):x):x)', '((((x:x):x):x):x)']

Observe: Because of the way the $\bar{\triangledown}$ operator works, if $A$ is the sum of all polygonal triangle subdivisions it will be the case that $ A = | + A \bar{\triangledown} A $

Observe: Recall the relationship between the number of leaves in a rooted planar binary tree and the number of polygon sizes.

Number of leaves	Number of ways to arrange leaves in a rooted binary tree	Number of Polygon sides
3	2	4
4	5	5
5	14	6
6	42	7
7	132	8

Observe: Carrying out the multiplication fo $A$ with $A$ will result in:

$$ A = | + A \bar{\triangledown} A $$

Observe: It is simpler to move back to a more usual algebra, in order to map back to polynomials

$$\Phi : \text{Rooted subdivided polygons} \rightarrow \text{Polynomials in } t$$

Observe: This can be done as the following:

$$ \Phi (P) = t^n $$

where $n$ is the number of triangles in the subdivision $P$

Example: If $P, Q$ are both subdivided polygons (into triangles) then:

$$ \Phi(P \bar{\triangledown} Q) $$

Observe that $P$ and $Q$ will have the same number of triangles, however there will anoter $|$ that arises

Observe This can be written as:

$$ \Phi(P \bar{\triangledown} Q) = t \Phi (P) \Phi(Q) $$

Observe Possible to apply this homeomorphism to the fundamental equation of the multiset of subdivided polygons:

$$ A = | + A \bar{\triangledown} A $$

$$\Phi (A) = [\Phi(P): P \in A] $$

# TO DO