User blog:Ynought/E function v.1

E function
\(E\) function takes place on a plane with starting point \((0|0)\) at the bottom left corner and the thick ness of the line is \(n^{-n}\) and the lenght is \(n^{-1}\).And the size of the units doesn't matter

Logic and symbols
\(\alpha_n\) is the angle that line nr.\(n\) deviates from the x plane

\(G_1\) is the starting graph.And \(G_i\) is the graph that appears if you repeat the rules for the transformation of \(G_1\) for the \(i\)-th time

\(\star\) denotes homeomorphical embeding lets say i.e. \(G_i\star G_j\) means that \(G_i\) is homeorphically embedable into \(G_j\)

\(\omega=\text{sup}(0,1,2,3...)\)

\(\text{int}(n)=\) n rounded too the nearest integer.(you start rounding up from n.5)

\(L_i^{G_n}[x,y,\alpha]\) means that the coordinates of \(L_i^{G_n}\) are \((x|y)\) and its angle is \(\alpha\)

\(a\Rightarrow b\) means \(a\) turns into \(b\) under the rules specified

\(L_i\| L_j\) means that \(L_i\) is parallel too \(L_j\) but i will restrict parallel such that they are only parallel if they have a point where either the \(x\) or \(y\) axis's are equall at some point of the line

\(L_i\nparallel L_j\) means that \(L_i\) is not parallel too \(L_j\) but i will restrict parallel such that they are only parallel if they have a point where either the \(x\) or \(y\) axis's are equall at some point of the line

The definition of line number k:
You start looking for the line with the bottom most point of that line has the least \(y\) value.If that is a tie then you choose the line with the least \(x\) of those lines.call that one line 1.Repeat the rule above.Call that one line 2.Keep repeating that rule until you have every line numbered.And \(L_n\) is the n-th line,and this has to hold \(n\in\mathbb{N}_{>0}\).And \(L_n^{G_i}\) is the \(n\)-th line in \(G_i\)

Termination and the value of E(n)
\(E(n)=\text{max}(k(\nexists i(\omega>k> i\geqslant 1 \land G_i\star G_k))\) where \(G_1\) has at most \(n\) lines.And in case that case never apears then use this \(E(n)=\text{max}(k(\nexists i(\omega>k> i\geqslant 1 \land G_i\star G_k\land \nexists a(\nexists b(L_a^{G_n}\| L_b^{G_n}(n\in\mathbb{N}_{\geqslant 1})))))\) \(G_1\) has at most \(n\) lines.if even that doesn't terminate then use this \(E(n)=\text{max}(k(\nexists i(\omega>k> i\geqslant 1 \land G_i\star G_k\land \nexists a(\nexists b(L_a^{G_n}\nparallel L_b^{G_n}(n\in\mathbb{N}_{\geqslant 1})))))\) \(G_1\) has at most \(n\) lines.If even that doesn't then i give up.If this is the only one that terminates of the 3,then i define function \(E_+(n)\) as the number you need to have as an input on \(E(n)\) such that it increases.

1st attempt
\(G_i\Rightarrow G_{i+1}\) with the rules:

\(G_1\) has at most \(n\) lines that have their bottom left point inside a point of a circle with \(r=n\) ( \(n\) from \(E(n)\)) Start with \(L_h\) where \(h\) is the number of lines that have been affected by this rule +1 and \(?\) is the new nr of the line and \(L_h^{G_i}[x,y,\alpha]\Rightarrow L_?^{G_{i+1}}[x^2,\frac{2x}{x^2},\text{int}(\alpha+sin(\alpha))]\).

2nd attempt
\(G_i\Rightarrow G_{i+1}\) with the rules:

\(G_1\) has at most \(n\) lines that have their bottom left point inside of a circle with \(r=n\) Start with \(L_h\) where \(h\) is the number of lines that have been affected by this rule +1 and \(?\) is the new number of the line:

1.If \(x\) and \(y\) are \(<1\) and \(>0\) then \(L_h^{G_i}[x,y,\alpha]\Rightarrow L_?^{G_{i+1}}[x+1,y,\alpha]\)

2.If \(x\geqslant 1\) and \(y\leqslant 1\) then \(L_h^{G_i}[x,y,\alpha]\Rightarrow L_?^{G_{i+1}}[x-0.1,y+10^x,\text{int}(\alpha+\text{sin}(\alpha))+1]\)

3.If \(x\leqslant 1\) and \(y\geqslant 1\) then \(L_h^{G_i}[x,y,\alpha]\Rightarrow L_?^{G_{i+1}}[x+9.99...9^y,y-0.1,\text{int}(\alpha+\text{sin}(\alpha))+1]\) with \(h\) \(9\)'s after the \(.\)

4.If \(x\geqslant 1\) and \(y\geqslant 1\) and after the transormation of the whole graph there won't be a terminating case (with this rule) then \(L_h^{G_i}[x,y,\alpha]\Rightarrow L_?^{G_{i+1}}[\frac{(x+y)^2}{2x+3y},y-0.1...1,\alpha]\) with \(h\) \(1\)'s

5..If \(x\geqslant 1\) and \(y\geqslant 1\) and after the transormation of the whole graph (with rule 4.)there will be a terminating case then \(L_h^{G_i}[x,y,\alpha]\Rightarrow L_?^{G_{i+1}}[\frac{(x+y)^2}{2x+3y},y-0.1...1,\text{int}(\alpha+\text{sin}(\alpha))+1]\) with \(h\) \(1\)'s

What to use?
You use the ruleset that produces the fastest growing \(E\) function.