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= Welcome to MWiki =
 
= Welcome to MWiki =
== Theorems of the month ==
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== Theorem of the month ==
=== Three-Cube Theorem ===
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The centre method solves every solvable LP in <math>\mathcal{O}(\omega{\vartheta}^{2})</math>.
  
By Fermat’s little theorem, <math>k \in {}^{\omega }{\mathbb{Z}}</math> is sum of three cubes if and only if
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== Proof and algorithm ==
 +
Let <math>z := \grave{m} + n</math> and <math>d \in [0, 1]</math> the density of <math>A</math>. First, normalise and scale <math>{b}^{T}y - {c}^{T}x \le 0, Ax \le b</math> as well as <math>{A}^{T}y \ge c</math>. Let <math>P_r := \{(x, y)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{z} : {b}^{T}y - {c}^{T}x \le r \in [0, \check{r}], Ax - b \le \underline{r}_m, c - {A}^{T}y \le \underline{r}_n\}</math> have the radius <math>\check{r} := s|\min \; \{b_1, ..., b_m, -c_1, ..., -c_n\}|</math> and the scaling factor <math>s \in [1, 2]</math>. It follows <math>\underline{0}_{z} \in \partial P_{\check{r}}</math>. By the strong duality theorem, the LP min <math>\{ r \in [0, \check{r}] : (x, y)^T \in P_r\}</math> solves the LPs max <math>\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\}</math> and min <math>\{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\}</math>.
  
<div style="text-align:center;"><math>k=(n - a)^3 + n^3 + (n + b)^3 = 3n^3 - a^3 + b^3+ 3c \ne \pm 4\mod 9</math></div>
 
  
and <math>a, b, c, d, m, n \in {}^{\omega }{\mathbb{Z}}</math> implies both <math>(a^2 + b^2)n - (a - b)n^2 = c =: dn</math> and <math>m^2 = n^2 - 4(b^2 - bn + d)</math> for <math>2a_{1,2} = n \pm m.\square</math>
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Its solution is the geometric centre <math>g</math> of the polytope <math>P_0</math>. For <math>p_k^* := (\text{min}\,p_k + \text{max}\,p_k)/2</math> and <math>k = 1, ..., \grave{z}</math> approximate <math>g</math> by <math>p_0 := (x_0, y_0, r_0)^T</math> until <math>||\Delta p||_1</math> is sufficiently small. The solution <math>t^o(x^o, y^o, r^o)^T</math> of the two-dimensional LP min <math>\{ r \in [0, \check{r}] : t \in {}^{\omega}\mathbb{R}_{&gt; 0}, t(x_0, y_0)^T \in P_r\}</math> approximates <math>g</math> better and achieves <math>r \le \check{r}/\sqrt{\grave{z}}</math>. Repeat this for <math>t^o(x^o, y^o)^T</math> until <math>g \in P_0</math> is computed in <math>\mathcal{O}({}_z\check{r} {}_e\check{r}dmn)</math> if it exists. Numbers of length <math>\mathcal{O}({\omega})</math> can only be processed in <math>\mathcal{O}(\vartheta)</math> as is generally known.
  
=== Fickett's Theorem ===
 
  
For any relative positions of two overlapping congruent rectangular <math>n</math>-prisms <math>Q</math> and <math>R</math> with <math>n \in {}^{\omega }\mathbb{N}_{\ge 2}</math> and <math>m := 2n - 1</math>, it can be stated for the exact standard measure <math>\mu</math>, where <math>\mu</math> for <math>n = 2</math> needs to be replaced by the Euclidean path length <math>L</math>, that:
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Solving all two-dimensional LPs <math>\text{min}_k r_k</math> by bisection methods for <math>r_k \in {}^{\omega}\mathbb{R}_{\ge 0}</math> and <math>k = 1, ..., z</math> in <math>\mathcal{O}({\vartheta}^2)</math> each time determines <math>q \in {}^{\omega}\mathbb{R}^k</math> where <math>q_k := \Delta p_k \Delta r_k/r</math> and <math>r := \text{min}_k \Delta r_k</math>. Let simplified <math>|\Delta p_1| = … = |\Delta p_{z}|</math>. Here min <math>r_z</math> for <math>p^* := p + wq</math> and <math>w \in {}^{\omega}\mathbb{R}_{\ge 0}</math> would be also to solve. If <math>\text{min}_k \Delta r_k r = 0</math> follows, stop, otherwise repeat until min <math>r = 0</math> or min <math>r &gt; 0</math> is sure. If necessary, constraints are temporarily relaxed by the same small modulus.<math>\square</math>
 
 
<div style="text-align:center;"><math>\hat{m} &lt; r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) &lt; m.</math></div>
 
 
 
==== Proof: ====
 
Since the underlying extremal problem has its maximum for rectangles with the side lengths <math>s</math> and <math>s + 2d0</math>, min <math>r = s/(3s - 2d0) \le r \le</math> max <math>r = (3s - 2d0)/s</math> holds. The proof for <math>n &gt; 2</math> is analogous.<math>\square</math>
 
 
 
== Recommended reading ==
 
  
 +
== Recommended readings ==
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
  
 
[[de:Hauptseite]]
 
[[de:Hauptseite]]

Revision as of 01:55, 1 January 2022

Welcome to MWiki

Theorem of the month

The centre method solves every solvable LP in [math]\displaystyle{ \mathcal{O}(\omega{\vartheta}^{2}) }[/math].

Proof and algorithm

Let [math]\displaystyle{ z := \grave{m} + n }[/math] and [math]\displaystyle{ d \in [0, 1] }[/math] the density of [math]\displaystyle{ A }[/math]. First, normalise and scale [math]\displaystyle{ {b}^{T}y - {c}^{T}x \le 0, Ax \le b }[/math] as well as [math]\displaystyle{ {A}^{T}y \ge c }[/math]. Let [math]\displaystyle{ P_r := \{(x, y)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{z} : {b}^{T}y - {c}^{T}x \le r \in [0, \check{r}], Ax - b \le \underline{r}_m, c - {A}^{T}y \le \underline{r}_n\} }[/math] have the radius [math]\displaystyle{ \check{r} := s|\min \; \{b_1, ..., b_m, -c_1, ..., -c_n\}| }[/math] and the scaling factor [math]\displaystyle{ s \in [1, 2] }[/math]. It follows [math]\displaystyle{ \underline{0}_{z} \in \partial P_{\check{r}} }[/math]. By the strong duality theorem, the LP min [math]\displaystyle{ \{ r \in [0, \check{r}] : (x, y)^T \in P_r\} }[/math] solves the LPs max [math]\displaystyle{ \{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\} }[/math] and min [math]\displaystyle{ \{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\} }[/math].


Its solution is the geometric centre [math]\displaystyle{ g }[/math] of the polytope [math]\displaystyle{ P_0 }[/math]. For [math]\displaystyle{ p_k^* := (\text{min}\,p_k + \text{max}\,p_k)/2 }[/math] and [math]\displaystyle{ k = 1, ..., \grave{z} }[/math] approximate [math]\displaystyle{ g }[/math] by [math]\displaystyle{ p_0 := (x_0, y_0, r_0)^T }[/math] until [math]\displaystyle{ ||\Delta p||_1 }[/math] is sufficiently small. The solution [math]\displaystyle{ t^o(x^o, y^o, r^o)^T }[/math] of the two-dimensional LP min [math]\displaystyle{ \{ r \in [0, \check{r}] : t \in {}^{\omega}\mathbb{R}_{> 0}, t(x_0, y_0)^T \in P_r\} }[/math] approximates [math]\displaystyle{ g }[/math] better and achieves [math]\displaystyle{ r \le \check{r}/\sqrt{\grave{z}} }[/math]. Repeat this for [math]\displaystyle{ t^o(x^o, y^o)^T }[/math] until [math]\displaystyle{ g \in P_0 }[/math] is computed in [math]\displaystyle{ \mathcal{O}({}_z\check{r} {}_e\check{r}dmn) }[/math] if it exists. Numbers of length [math]\displaystyle{ \mathcal{O}({\omega}) }[/math] can only be processed in [math]\displaystyle{ \mathcal{O}(\vartheta) }[/math] as is generally known.


Solving all two-dimensional LPs [math]\displaystyle{ \text{min}_k r_k }[/math] by bisection methods for [math]\displaystyle{ r_k \in {}^{\omega}\mathbb{R}_{\ge 0} }[/math] and [math]\displaystyle{ k = 1, ..., z }[/math] in [math]\displaystyle{ \mathcal{O}({\vartheta}^2) }[/math] each time determines [math]\displaystyle{ q \in {}^{\omega}\mathbb{R}^k }[/math] where [math]\displaystyle{ q_k := \Delta p_k \Delta r_k/r }[/math] and [math]\displaystyle{ r := \text{min}_k \Delta r_k }[/math]. Let simplified [math]\displaystyle{ |\Delta p_1| = … = |\Delta p_{z}| }[/math]. Here min [math]\displaystyle{ r_z }[/math] for [math]\displaystyle{ p^* := p + wq }[/math] and [math]\displaystyle{ w \in {}^{\omega}\mathbb{R}_{\ge 0} }[/math] would be also to solve. If [math]\displaystyle{ \text{min}_k \Delta r_k r = 0 }[/math] follows, stop, otherwise repeat until min [math]\displaystyle{ r = 0 }[/math] or min [math]\displaystyle{ r > 0 }[/math] is sure. If necessary, constraints are temporarily relaxed by the same small modulus.[math]\displaystyle{ \square }[/math]

Recommended readings

Nonstandard Mathematics