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# Linear Programming

In the following section, Set Theory, Topology and Nonstandard Analysis are presupposed. Simplex and intex method can solve an LP mostly in cubic and quadratic runtime respectively.

Diameter theorem for polytopes: Every diameter of an $$n$$-dimensional polytope defined by $$m$$ constraints for $$m, n \in {}^{\omega}\mathbb{N}_{\ge 2}$$ is at most $$2(m + n – 3)$$.

Proof: At most $$\acute{m}$$ hyperplanes can be assembled into an incomplete cycle of dimension 2 and there are at most $$n – 2$$ alternatives sidewards in all remaining dimensions. Overcoming every minimal distance requires at most two edges and yields the factor 2.$$\square$$

Remark: Dropping the finite limitation allows the analogous transfer of the theorem to polyhedra.

Definition: Let $$\vartheta := \omega\,{}_{e}\omega$$ and $${|| \cdot ||}_{1}$$ the 1-norm. If $$x \in X \subseteq {}^{\omega}\mathbb{R}^{\omega}$$ passes to the next step, it is followed by $${}^{*}$$ where $$\Delta x := {x}^{*} – x$$. The unit vector of $$x \in X^{*}$$ is $${_1}{x} := x/||x||$$. Let $$x, y \in X$$ in $$(x, y, …)^T$$ be row vectors. If a method requires computation time in seconds and memory in bits of $$\mathcal{O}({\omega}^{\mathcal{O}(1)})$$, it is polynomial. If one of both quantities is $$\mathcal{O}({e}^{|\mathcal{O}(\omega)|})$$, the method is exponential. Let the eigenproduct (yet: determinant) of a square matrix be the product of its eigenvalues.$$\triangle$$

Theorem: The simplex method1Dantzig, George B.: Lineare Programmierung und Erweiterungen; 1. Aufl.; 1966; Springer; Berlin. is exponential.

Proof and algorithm: Let LP max $$\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in P\}$$ have the feasible domain $$P := \{x \in {}^{\omega}\mathbb{R}^{n} : Ax \le b, b \in {}^{\omega}\mathbb{R}^{m}, A \in {}^{\omega}\mathbb{R}^{m \times n}, m, n \in {}^{\omega}\mathbb{N}^{*}\}$$. Its dual gives $${x}^{*} \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}$$. A (big) $$x^+ \in {}^{\omega}\mathbb{R}_{\geq0}$$ achieves $$x^* \in {}^{\omega}\mathbb{R}_{\geq0}^n$$ for $$x := x^* – {\underline{x}}_n^+$$. Solving max $$\{-h \in {}^{\omega}\mathbb{R}_{\le 0} : Ax – b \le \underline{h}_n\}$$ yields $$x \in P_{\ge 0}$$ if $$b \ge 0$$ does not hold. Let $$|\text{min } \{{b}_{1}, …, {b}_{m}\}|$$ be the initial value for $$h$$ and 0 its target value. Start with $$x := 0$$. Only one pivot step causes that $${b}^{*} \ge 0$$. Let $$i, j, k \in {}^{\omega}\mathbb{N}^{*}$$ and let $${a}_{i}^{T}$$ the $$i$$-th row vector of $$A$$. If $${c}_{j} \le 0$$ for all $$j$$, the LP is solved.

If for some $${c}_{j} > 0$$ also $${a}_{ij} \le 0$$ for all $$i$$, the LP is positively unbounded. If for some $${c}_{j} = 0$$ also $${a}_{ij} \le 0$$ for all $$i$$, drop $${c}_{j}$$ and $${A}_{.j}$$ as well as $${b}_{i}$$ and $${a}_{i}$$, but only when $${a}_{ij} < 0$$ holds. The inequality $${a}_{ij}{x}_{j} \ge 0 > {b}_{i}$$ for all $$j$$ has no solution, too. Dividing all $${a}_{i}^{T}x \le {b}_{i}$$ by $$||{a}_{i}||$$ and all $${c}_{j}$$ and $${a}_{ij}$$ by the minimum of $$|{a}_{ij}|$$ such that $${a}_{ij} \ne 0$$ for each $$j$$ maybe repeated and reversed later. Constraints with $${a}_{i} \le 0$$ may always be removed.

For each $${c}_{j} > 0$$ and non-base variable $${x}_{j}$$ select $${b_k}/{a}_{kj} :=$$ min $$\{{b_i}/{a}_{ij} : {a}_{ij} > 0\}$$ to obtain a next $${x}^{*} \in P_{\ge 0}$$ where $${x}_{j}^{*} = {x}_{j} + {b}_{k}/{a}_{kj}$$. To select the steepest edge, pick the pivot $${a}_{kj}$$ corresponding to $${x}_{j}$$ that maximises $${c}^{T}{_1}{\Delta x}$$ or $${c}_{j}^{2}/\grave{v}_{j}^{2}$$ for $$v_j := ||{A}_{.j}||$$ in the $$k$$-th constraint. Multiple maxima allow to use the rule of best pivot value max$${}_{k,j} {c}_{j}{b}_{k}/{a}_{kj}$$ or (slower) the smallest angle min $${{_{(1)}}\underline{1}^{T}_n}{_1}{c}^{*}$$.

If $$c^Tx^* = c^Tx$$ is repeated, perturb, which means relax, all constraints with $${b}_{i} = 0$$ by $$d \in {}^{\omega}\mathbb{R}_{>0}$$. For $${b}_{i} := ||{a}_{i}||$$, drop $$d$$ in the tableau. If another multiple vertex is encountered for once, simply add $$||{a}_{i}||$$ to every $${b}_{i}$$. Leaving it, after which the relaxation is reverted, may require to solve an LP with $$c > 0$$ and $$b = 0$$.

This task can also finally arise, if further solutions of the LPs are to be determined for at least two $$c_j = 0$$. This avoids having to compute several minimal relaxations that differ strongly but are not selected according to optimality points of view, as with the lexicographical method. Bland’s rule is not optimal, too.

The rectangle rule simply allows to compute $${c}_{j}^{*}, {a}_{ij}^{*}$$ and $${b}_{i}^{*}$$2cf. Vanderbei, Robert J.: Linear Programming; 3rd Ed.; 2008; Springer; New York, p. 63.. Despite the diameter theorem for polytopes, no pivoting rule can prevent the simplex method from being exponential as the case may be: An exponential “drift” for e. g. Klee-Minty or Jeroslow polytopes may always provide an unfavourable edge. The result follows in accordance with the state of research.$$\square$$

Remarks: TThe following fast and numerically easy algorithm could be extremely important. It was developed in 35 years and answers (e. g. with the help of Gomory cuts) Hilbert’s tenth problem positively. But simplex methods allow simple and precise calculations with rational numbers if the initial problems are appropriately formulated and smaller (for suitably approximated $$||{a}_{i}||$$).

Theorem: Intex methods solve every solvable LP with inter-/extrapolations in $$\mathcal{O}({\vartheta}^3)$$.

Proof and algorithm: Let $$z := m + n$$ and $$d \in [0, 1]$$ the density of $$A$$. First, normalise and scale $${b}^{T}y – {c}^{T}x \le 0, Ax \le b$$ as well as $${A}^{T}y \ge c$$. Let $$P_r := \{(x, y)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{z} : {b}^{T}y – {c}^{T}x \le r \in [0, \breve{r}], Ax – b \le \underline{r}_m, c – {A}^{T}y \le \underline{r}_n\}$$ have the radius $$\breve{r} := s|\min \; \{b_1, …, b_m, -c_1, …, -c_n\}|$$ and the scaling factor $$s \in [1, 2]$$. It follows $$\underline{0}_{z} \in \partial P_{\breve{r}}$$. By the strong duality theorem3loc. cit., p. 60 – 65., LP min $$\{ r \in [0, \breve{r}] : (x, y)^T \in P_r\}$$ solves LPs max $$\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\}$$ and min $$\{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\}$$.

Its solution is the geometric centre $$g$$ of the polytope $$P_0$$. For $$p_k^* := \text{min}\,\check{p}_k + \text{max}\,\check{p}_k$$ and $$k = 1, …, \grave{z}$$ approximate $$g$$ by $$p_0 := (x_0, y_0, r_0)^T$$ until $$||\Delta p||_1$$ is sufficiently small. The solution $$t^o(x^o, y^o, r^o)^T$$ of the two-dimensional LP min $$\{ r \in [0, \breve{r}] : t \in {}^{\omega}\mathbb{R}_{> 0}, t(x_0, y_0)^T \in P_r\}$$ approximates $$g$$ better and achieves $$r \le \tilde{2}\breve{r}$$. Repeat this for $$t^o(x^o, y^o)^T$$ until $$g \in P_0$$ is computed in $$\mathcal{O}({}_2\breve{r}^2dmn)$$ if it exists.

Solving all two-dimensional LPs $$\text{min}_k r_k$$ by bisection methods for $$r_k \in {}^{\omega}\mathbb{R}_{\ge 0}$$ and $$k = 1, …, z$$ in $$\mathcal{O}({\vartheta}^2)$$ each time determines $$q \in {}^{\omega}\mathbb{R}^k$$ where $$q_k := \Delta p_k \Delta r_k/r$$ and $$r := \text{min}_k \Delta r_k$$. Let simplified $$|\Delta p_1| = … = |\Delta p_{z}|$$. Here min $$r_{\grave{z}}$$ for $$p^* := p + wq$$ and $$w \in {}^{\omega}\mathbb{R}_{\ge 0}$$ may be solved, too. If $$\text{min}_k \Delta r_k r = 0$$ follows, stop computing, otherwise repeat until min $$r = 0$$ or min $$r > 0$$ is sure.$$\square$$

Remarks: Numbers of length $$\mathcal{O}({\omega})$$ can only be processed in $$\mathcal{O}(\vartheta)$$ as is generally known. Intex methods make use of the fact that the calculated geometric centre and the centre of the sphere coincide in $$P_0$$. The former is secured by rotating the coordinate axes (in pairs) like $$c$$ with the aid of a simple rotation matrix in a comparable running time. If necessary, the same small modulus relaxes constraints temporarily.

Remarks: Rounding errors provoked in $$b$$ and $$c$$ make any solution unique. Intex methods have for finite $$m$$ and $$n$$ a runtime of $$\mathcal{O}(dmn)$$. Optimising them for distributed computing in $${}^{\nu}\mathbb{R}^{\nu}$$ reduces it to $$\mathcal{O}(1)$$. Integer solutions $$v_j$$ satisfying $$\acute{v}_0 \le \lfloor v_j \rfloor – v_j$$ prove them successful for (mixed) integer problems and (non-) convex (Pareto) optimisation (according to nature4Vasuki, A: Nature-Inspired Optimization Algorithms; 1st Ed.; 2020; CRC Press; Boca Raton.).

Remarks: Simplex method and face algorithm5Pan, Ping-Qi: Linear Programming Computation; 1st Ed.; 2014; Springer; New York., p. 580 f. solve LPs faster for small $$m$$ and $$n$$. Intex methods surpass known (worst-case) LP-algorithms in $$\mathcal{O}(\omega^{37/18}\vartheta)$$. Details are only published when no misuse for non-transparent or bad decisions must be feared. Transfers to complex numbers are easy. This holds also for the following results.

Conclusion: LP max $$\{{||x – {x}^{o}||}_{1} : {c}^{T}x = {c}^{T}{x}^{o}, Ax \le b, x – {x}^{o} \in {[-1, 1]}^{n}, x \in {}^{\omega}\mathbb{R}_{\ge 0}^{n}\}$$ can determine for the first solution $${x}^{o}$$ any second one in $$\mathcal{O}({\omega}^2\vartheta)$$ where $${y}^{o}$$ may be treated analogously.$$\square$$

Conclusion: LP max $$\{\nu|\lambda_j| + ||x_j||_1: Ax_j = \lambda_j x_j, |\lambda_j| \in [0, r_j], x_j \in {[-1, 1]}^{n*}\}$$ can determine for every $$j = 1, …, n$$ the eigenvalue $$\lambda_j \in {}^{\omega}\mathbb{R}$$ and the eigenvector $$x_j \in {}^{\omega}\mathbb{R}^{n}$$ of matrix $$A \in {}^{\omega}\mathbb{R}^{n \times n}$$ in $$\mathcal{O}({\omega}^2\vartheta).\square$$

Conclusion: LPs max $$\{{x}_{j} : Ax = 0\}$$ yield all solutions to $$Ax = b$$ in $$\mathcal{O}({\omega}^2\vartheta)$$. Matrix $$A$$ is regular if and only if LP max $$\{{||x||}_{1} : Ax = 0\} = 0.\square$$

Conclusion: Let $${\alpha }_{j} := {A}_{.j}^{-1}$$ for matrix $${A}^{-1} \in {}^{\omega}\mathbb{R}^{n \times n}$$ and let $${\delta}_{ij}$$ the Kronecker delta where $$j = 1, …, n$$. A regular $$A$$ has an eigenproduct $$\ne 0$$ and allows every $${A \alpha }_{j} = {({\delta}_{1j}, …, {\delta}_{nj})}^{T}$$ to be solved in $$\mathcal{O}({\omega}^2\vartheta)$$.$$\square$$

Corollary: Intex methods and two-dimensional bisection or Newton’s methods may solve every solvable convex programme min $$\{{f}_{1}(x) : x \in {}^{\omega}\mathbb{R}^{n}, {({f}_{2}(x), …, {f}_{m}(x))}^{T} \le 0\}$$ where all $${f}_{i} \in {}^{\omega}\mathbb{R}$$ are convex functions for $$i = 1, …, m$$ in polynomial runtime, if the number of operands $${x}_{j}$$ of all $${f}_{i}$$ is $$\le {\omega}^{\nu-3}$$ and if an $$x$$ exists such that $${f}_{i}(x) < 0$$ for all $$i > 1$$6cf. Bertsekas, Dimitri P.: Nonlinear Programming; 3rd Ed.; 2016; Athena Scientific; Belmont., p. 589 ff..$$\square$$

Corollary: LP max $$\{x : y_{\acute{m}} – xy_m = b_{\acute{m}}, y_n = y_0 = 0, x \le x_0 \in {}^{\omega}\mathbb{R}\}$$ can determine for $$m = 1, …, n$$ by Horner scheme an $$x \in {}^{\omega}\mathbb{R}$$ solving $$n$$-polynomial $$b_{\acute{n}}x^{\acute{n}} + … + b_1x + b_0 = 0$$ in $$\mathcal{O}(\omega\vartheta)$$. Aborting TS can analogously solve every continuous inequation system with convex solution set.$$\square$$