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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorem of the month ==
 
== Theorem of the month ==
The intex method solves every solvable LP in <math>\mathcal{O}({\vartheta}^{3})</math>.
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The centre method solves every solvable LP in <math>\mathcal{O}(\omega{\vartheta}^{2})</math>.
  
 
== Proof and algorithm ==
 
== Proof and algorithm ==
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 the <em>height</em> <math>h</math> have the initial value <math>h_0 := s |\min \; \{b_1, ..., b_m, -d_1, ..., -d_n\}|</math> for the <em>elongation factor</em> <math>s \in \, ]1, 2]</math>.</br>
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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>.
The LP min <math>\{h \in [0, h_0] : x \in {}^{\omega}\mathbb{R}_{\ge 0}^{n}, y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m},{b}^{T}y - {c}^{T}x \le h, Ax - b \le (h, ..., h)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, c - {A}^{T}y \le (h, ..., h)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{n}\}</math> has <math>k</math> constraints and the feasible starting point <math>(x_0, y_0, h_0/s)^{T} \in {}^{\omega}\mathbb{R}_{\ge 0}^{m+n+1}</math>, e.g. <math>(0, 0, h_0/s)^{T}</math>.</br>
 
It identifies the mutually dual 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>.
 
  
Let the point <math>p := (x, y, h)^T</math> approximate the subpolytope’s centre of gravity <math>P^*</math> as <math>p_k^* := (\min p_k + \max p_k)/2</math> until <math>{|| \Delta p ||}_{1}</math> is sufficiently small. Here <math>x</math> takes precedence over <math>y</math>. Then extrapolate <math>p</math> via <math>{p}^{*}</math> into <math>\partial P^*</math> as <math>u</math>. Put <math>p := p^* + (u - p^*)/s</math> to shun <math>\partial P^*</math>. Hereon approximate <math>p</math> more deeply again as centre of gravity. After optionally solving all LPs min<math>{}_{k} {h}_{k}</math> by bisection methods for <math>{h}_{k} \in {}^{\omega}\mathbb{R}_{\ge 0}</math> in <math>\mathcal{O}({\vartheta}^{2})</math> each time, <math>v \in {}^{\omega}\mathbb{R}^{k}</math> may be determined such that <math>v_k := \Delta{p}_{k} \Delta{h}_{k}/r</math> and <math>r :=</math> min<math>{}_{k} \Delta{h}_{k}</math>. Simplified let <math>|\Delta{p}_{1}| = ... = |\Delta{p}_{m+n}|</math>.
 
  
Here min <math>{h}_{m+n+1}</math> may be solved for <math>p^* := p + tv</math> where <math>t \in {}^{\omega}\mathbb{R}_{\ge 0}</math> and <math>{v}_{m+n+1} = 0</math>. If min<math>{}_{k} {h}_{k} r = 0</math> follows, end, otherwise start over until min <math>h = 0</math> or min <math>h &gt; 0</math> is certain. If necessary, relax the constraints temporarily by the same small modulus.</br>
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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.
Since almost every iteration step in <math>\mathcal{O}({\omega\vartheta}^{2})</math> halves <math>h</math> at least, the strong duality theorem yields the result.<math>\square</math>
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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>
  
 
== Recommended readings ==
 
== Recommended readings ==

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