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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorem of the month ==
 
== Theorem of the month ==
=== RU method ===
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The centre method solves every solvable LP in <math>\mathcal{O}(\omega{\vartheta}^{2})</math>.
If the linear system (LS) <math>Ax = b \in {}^{\nu}\mathbb{Q}^{n}</math> can be uniquely solved for <math>n \in {}^{\nu}\mathbb{N}^*</math>, the ''root of unity method (<math>RU</math> method)'' computes <math>x \in {}^{\nu}\mathbb{Q}^{n}</math> for <math>A \in {}^{\nu}\mathbb{Q}^{n \times n}</math> in <math>\mathcal{O}(n^2)</math>.
 
  
=== Proof and algorithm ===
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== Proof and algorithm ==
Let <math>R_1 := (r_{1jk}) = (r_{1kj}) = R_1^T \in {}^{\nu}\mathbb{C}^{n \times n}, n \in {}^{\nu}2\mathbb{N}^*, r_{11k} := 1</math> and for <math>j &gt; 1</math> as well as <math>n_{jk} := j + k - 3</math> both <math>r_{1jk} := \hat{n}e^{i\tau n_{jk}/n}</math> for <math>n_{jk} &lt; n</math> and <math>r_{1jk} := \hat{n}e^{i\tau(n_{jk} - \acute{n})/n}</math> for <math>n_{jk} \ge n</math>. Interchanging the first and <math>j</math>-th row resp. column position and correspondingly interchanging the remaining row and column positions yields matrices <math>R_j = R_j^T</math> for <math>j &gt; 1</math>. Let <math>\delta_{jk}</math> be the Kronecker delta and <math>A := (a_{jk})</math>.
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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>.
  
If <math>a_{jk} \le 0</math> is given for at least one couple <math>(j, k)</math>, then compute the sums <math>s_0 := \sum\limits_{j=1}^m{b_j\varepsilon^j}</math> for an arbitrary transcendental number <math>\varepsilon</math> and <math>s_k := \sum\limits_{j=1}^m{a_{jk}\varepsilon^j} \ne 0</math> for all <math>k</math>. Replace <math>x_k</math> by <math>-x_k</math> for <math>s_k &lt; 0</math>. Add a multiple of <math>s^Tx</math> resp. <math>s_0</math> to <math>Ax = b</math>, such that <math>a_{jk} &gt; 0</math> holds for all <math>(j, k)</math>. Let <math>b_j = 1</math> for all <math>j</math> wlog. For <math>D_j := (d_{jk}), d_{jk} = \delta_{jk}⁄a_{jk}, C_j := D_j R_j</math> and <math>x_k^{(0)} := \hat{n}/ \max_j a_{jk}</math>, let <math>x^{(\grave{m})} = x^{(m)} + C_j^{-1}(b - Ax^{(m)}).\square</math>
 
  
=== Corollary ===
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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.
The RU method allows to determine every eigenvalue and -vector of <math>Ax = \lambda x \in {}^{\nu}\mathbb{Q}^{n} + {}^{\nu}\mathbb{Q}^{n}</math> for <math>n \in {}^{\nu}2\mathbb{N}^*, \lambda \in {}^{\nu}\mathbb{Q}+ {i}^{\nu}\mathbb{Q}</math> and <math>\in {}^{\nu}\mathbb{Q}^{n \times n}</math> in <math>\mathcal{O}(n^2)</math> by putting <math>x^{\prime(\grave{m})} = C_j^{-1}AC_j x^{\prime(m)}</math>.
 
  
'''Remark:''' Extending the theorem to complex <math>A</math> and <math>b</math> is easy.
 
== Recommended reading ==
 
  
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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 ==
 
[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