### Growing sol basically done

parent 5a3680fb
 ... ... @@ -22,96 +22,96 @@ Then the maximum possible score is: which for our benchmark $15 \times 15$ field results in an upper bound of $1624$. \section{Gärtner Bound} TODO: This is wrong? If a pixel has a negative neighborhood the border leads to us underestimating. Think of $R(c_1, c_2) = -1$. The basic bound is rather optimistic. If our $R$ function assigns no two crops the value of $1$ the basic bound still assumes it will. Instead we can take the $R$ function into account when calculating a bound. Consider the following $3 \times 3$ subfield: \FloatBarrier \begin{figure}[h] \centering \begin{tikzpicture}[auto,node distance=1cm,thick,main node/.style={circle,draw,font=\sffamily\Large}] \node[main node] (1) {}; \node[main node] (2) [right of=1] {}; \node[main node] (3) [right of=2] {}; \node[main node] (4) [below of=1]{}; \node[main node] (5) [right of=4] {}; \node[main node] (6) [right of=5] {}; \node[main node] (7) [below of=4] {}; \node[main node] (8) [right of=7] {}; \node[main node] (9) [right of=8] {}; \path[every node/.style={font=\sffamily\small}] (1) edge[red] (2); \path[every node/.style={font=\sffamily\small}] (2) edge[red] (3); \path[every node/.style={font=\sffamily\small}] (4) edge[red] (5); \path[every node/.style={font=\sffamily\small}] (5) edge[red] (6); \path[every node/.style={font=\sffamily\small}] (7) edge[red] (8); \path[every node/.style={font=\sffamily\small}] (8) edge[red] (9); \path[every node/.style={font=\sffamily\small}] (1) edge[red] (4); \path[every node/.style={font=\sffamily\small}] (4) edge[red] (7); \path[every node/.style={font=\sffamily\small}] (2) edge[red] (5); \path[every node/.style={font=\sffamily\small}] (5) edge[red] (8); \path[every node/.style={font=\sffamily\small}] (3) edge[red] (6); \path[every node/.style={font=\sffamily\small}] (6) edge[red] (9); \path[every node/.style={font=\sffamily\small}] (1) edge[blue] (5); \path[every node/.style={font=\sffamily\small}] (4) edge[blue] (2); \path[every node/.style={font=\sffamily\small}] (2) edge[blue] (6); \path[every node/.style={font=\sffamily\small}] (5) edge[blue] (3); \path[every node/.style={font=\sffamily\small}] (4) edge[blue] (8); \path[every node/.style={font=\sffamily\small}] (7) edge[blue] (5); \path[every node/.style={font=\sffamily\small}] (5) edge[blue] (9); \path[every node/.style={font=\sffamily\small}] (8) edge[blue] (6); \end{tikzpicture} \end{figure} \FloatBarrier where every vertex represents a pixel and edges represent neighborhood relationships. Red edges are non-diagonal neighborhood relationships, while blue edges are diagonal relationships. Using exhaustive search we can find for each crop an optimal solution where that crop is in the center of the subfield above. Then we multiply the value of all of the red edges with $\frac{1}{6}$ and that of the blue edges with $\frac{1}{4}$. Then let $b(c)$ denote the sum of all of the now multiplied edges in the $3 \times 3$ field for crop $c$. The optimal score is then bounded by the number of pixels that are of this crop type multiplied by $b(c)$. \begin{displaymath} \sum_{c \in \Cps} X \cdot Y \cdot D(c) \cdot b(x) \end{displaymath} The multipliers are chosen to ensure that all edges/ crop relationships are counted. A red edge could be counted by up to $6$ pixels, while a blue edge could be counted by up to $4$. When they are not counted $6$ respectively $4$ times, it is because they no longer represent valid relationships within our field. This bound therefore takes into account that there might be crop types that just do not have good local neighborhood solutions. We do however count some non-existant relationships over the field border, but this may be offset by the other gains the bound brings. This bound therefore performs well for pixel farming problems that don't have solutions with high scores relative to their field size. Our first benchmark problem is bounded here by .... while our second problem is bounded by % \section{Gärtner Bound} % TODO: This is wrong? If a pixel has a negative % neighborhood the border leads to us underestimating. % Think of $R(c_1, c_2) = -1$. % The basic bound is rather optimistic. If % our $R$ function assigns no two crops % the value of $1$ the basic bound still assumes % it will. Instead we can take the $R$ function % into account when calculating a bound. Consider the % following $3 \times 3$ subfield: % \FloatBarrier % \begin{figure}[h] % \centering % \begin{tikzpicture}[auto,node distance=1cm,thick,main node/.style={circle,draw,font=\sffamily\Large}] % \node[main node] (1) {}; % \node[main node] (2) [right of=1] {}; % \node[main node] (3) [right of=2] {}; % \node[main node] (4) [below of=1]{}; % \node[main node] (5) [right of=4] {}; % \node[main node] (6) [right of=5] {}; % \node[main node] (7) [below of=4] {}; % \node[main node] (8) [right of=7] {}; % \node[main node] (9) [right of=8] {}; % \path[every node/.style={font=\sffamily\small}] (1) edge[red] (2); % \path[every node/.style={font=\sffamily\small}] (2) edge[red] (3); % \path[every node/.style={font=\sffamily\small}] (4) edge[red] (5); % \path[every node/.style={font=\sffamily\small}] (5) edge[red] (6); % \path[every node/.style={font=\sffamily\small}] (7) edge[red] (8); % \path[every node/.style={font=\sffamily\small}] (8) edge[red] (9); % \path[every node/.style={font=\sffamily\small}] (1) edge[red] (4); % \path[every node/.style={font=\sffamily\small}] (4) edge[red] (7); % \path[every node/.style={font=\sffamily\small}] (2) edge[red] (5); % \path[every node/.style={font=\sffamily\small}] (5) edge[red] (8); % \path[every node/.style={font=\sffamily\small}] (3) edge[red] (6); % \path[every node/.style={font=\sffamily\small}] (6) edge[red] (9); % \path[every node/.style={font=\sffamily\small}] (1) edge[blue] (5); % \path[every node/.style={font=\sffamily\small}] (4) edge[blue] (2); % \path[every node/.style={font=\sffamily\small}] (2) edge[blue] (6); % \path[every node/.style={font=\sffamily\small}] (5) edge[blue] (3); % \path[every node/.style={font=\sffamily\small}] (4) edge[blue] (8); % \path[every node/.style={font=\sffamily\small}] (7) edge[blue] (5); % \path[every node/.style={font=\sffamily\small}] (5) edge[blue] (9); % \path[every node/.style={font=\sffamily\small}] (8) edge[blue] (6); % \end{tikzpicture} % \end{figure} % \FloatBarrier % where every vertex represents a pixel and edges represent % neighborhood relationships. Red edges are non-diagonal neighborhood % relationships, while blue edges are diagonal relationships. % Using exhaustive search we can find for each crop % an optimal solution where that crop is in the center % of the subfield above. Then we multiply the value % of all of the red edges with $\frac{1}{6}$ and that % of the blue edges with $\frac{1}{4}$. Then let % $b(c)$ denote the sum of all of the now multiplied % edges in the $3 \times 3$ field for crop $c$. % The optimal score is then bounded by the number % of pixels that are of this crop type multiplied by % $b(c)$. % \begin{displaymath} % \sum_{c \in \Cps} X \cdot Y \cdot D(c) \cdot b(x) % \end{displaymath} % The multipliers are chosen to ensure that all % edges/ crop relationships are counted. A red edge % could be counted by up to $6$ pixels, while a blue % edge could be counted by up to $4$. When they are % not counted $6$ respectively $4$ times, it is because % they no longer represent valid relationships % within our field. % This bound therefore takes into account that there % might be crop types that just do not have good % local neighborhood solutions. We do however count % some non-existant relationships over the field % border, but this may be offset by the other gains % the bound brings. This bound therefore performs well % for pixel farming problems that don't have solutions % with high scores relative to their field size. % Our first benchmark problem is bounded here by % .... while our second problem is bounded by \section{LP bound} ... ...
 ... ... @@ -12,7 +12,7 @@ solutions may not be optimal, but in some situations may lead to good quality results quickly. For the sake of a clearer presentation we present a method that stretches the we demonstrate a method that stretches the solution by the same factor in both the $x$ and $y$ direction. Consider an algorithm that is given a small optimal solution $sol$ ... ... @@ -25,10 +25,10 @@ for cleaner formulas: sol'_{i, j} &= sol_{a, b} & & i \in \{0, \dots, f \cdot (X-1)\} & j \in \{0, \dots, f \cdot (Y-1)\} \end{align*} \begin{displaymath} a = \lfloor i / f \rfloor + ((i + \lfloor i / f \rfloor) \text{ mod } 2) f \in \{ 2i + 1 \ | \ i \in \N \} \end{displaymath} \begin{displaymath} b = \lfloor j / f \rfloor + ((j + \lfloor j / f \rfloor) \text{ mod } 2) a = \lfloor i / f \rfloor + ((i + \lfloor i / f \rfloor) \text{ mod } 2) \quad \quad b = \lfloor j / f \rfloor + ((j + \lfloor j / f \rfloor) \text{ mod } 2) \end{displaymath} which is a bit puzzling in its formal form. The idea is that we want to multiply the ... ... @@ -79,37 +79,74 @@ is: \begin{displaymath} \frac{s}{8XY - 6X - 6Y + 4} \end{displaymath} Further we can bound the score $s'$ for the grown field by: \begin{align*} s' &\geq f^2 \cdot s - 3 \cdot (f \cdot (X-1) + 1) - 3 \cdot (f \cdot (Y-1) + 1) + 4\\ &\geq f^2 \cdot s - 3 \cdot (f \cdot X - f + 1) - 3 \cdot (f \cdot Y - f + 1) + 4\\ &\geq f^2 \cdot s - 3f \cdot X + 3f - 3 - 3f \cdot Y + 3f - 3 + 4\\ &\geq f^2 \cdot s - 3f \cdot (X + Y - 2) -2\\ &\geq f^2 \cdot s - 3f \cdot (X + Y - 1) \end{align*} which means our new relative score can be bounded from below by: \begin{align*} &\frac{s'}{8 \cdot (f (X-1) + 1) \cdot (f (Y-1) + 1) - 6 \cdot (f (X-1) + 1) - 6 \cdot (f (Y-1) + 1) + 4}\\ &= \frac{s'}{8 \cdot (f X - f + 1) \cdot (f Y - f + 1) - 6 \cdot (f X - f + 1) - 6 \cdot (f Y - f + 1) + 4}\\ &= \frac{s'}{8f^2XY - 8f^2X - 8f^2Y + 2fX + 2fY - 4f + 8f^2}\\ &= \frac{s'}{8f^2XY - 8f^2X - 8f^2Y + 2f \cdot (X + Y - 2) + 8f^2}\\ &= \frac{s'}{8f^2 (XY - X - Y + 1) + 2f \cdot (X + Y - 2)}\\ &\geq \frac{f^2 \cdot s - 3f \cdot (X + Y - 1)}{8f^2 (XY - X - Y + 1) + 2f \cdot (X + Y - 2)} \end{align*} Which in turn allows us to find the following condition on $f$ that guarantees our property: \begin{align*} \frac{s}{8XY - 6X - 6Y + 4} &\leq \frac{f^2 \cdot s - 3f \cdot (X + Y - 1)}{8f^2 (XY - X - Y + 1) + 2f \cdot (X + Y - 2)}\\ \frac{s}{8XY - 6X - 6Y + 4} &\leq \frac{f \cdot s - 3 \cdot (X + Y - 1)}{8f (XY - X - Y + 1) + 2 \cdot (X + Y - 2)}\\ -2fsX -2fsY + 4fs &\leq (3 - 3X -3Y) \cdot (8XY - 6X - 6Y + 4) - 2s \cdot (X + Y - 2)\\ -2fs \cdot (X + Y - 2) &\leq - (3X + 3Y - 3) \cdot (8XY - 6X - 6Y + 4) - 2s \cdot (X + Y - 2)\\ 2fs \cdot (X + Y - 2) &\geq (3X + 3Y - 3) \cdot (8XY - 6X - 6Y + 4) + 2s \cdot (X + Y - 2)\\ f &\geq \frac{(3X + 3Y - 3) \cdot (8XY - 6X - 6Y + 4)}{2s \cdot (X + Y - 2)} + 1 \end{align*} We are now able to formulate the following theorem: \begin{theorem} For any solution to a field there exists a factor by which we can scale up the field such that the relative score doesn't worsen and we only slightly violate the distribution constraint. \end{theorem} \begin{proof} We first bound the score for the grown field $s'$ from below: \begin{align*} s' &\geq f^2 \cdot s - 2 \cdot 3 \cdot (f \cdot (X-1) + 1) - 2 \cdot 3 \cdot (f \cdot (Y-1) + 1) + 4\\ &\geq f^2 \cdot s - 6 \cdot (f \cdot X - f + 1) - 6 \cdot (f \cdot Y - f + 1) + 4\\ &\geq f^2 \cdot s - 6f \cdot X + 6f - 6 - 6f \cdot Y + 6f - 6 + 4\\ &\geq f^2 \cdot s - 6f \cdot (X + Y - 2) -8\\ &\geq f^2 \cdot s - 6f \cdot (X + Y) \end{align*} which means our new relative score can be bounded from below by: \begin{align*} &\frac{s'}{8 \cdot (f (X-1) + 1) \cdot (f (Y-1) + 1) - 6 \cdot (f (X-1) + 1) - 6 \cdot (f (Y-1) + 1) + 4}\\ &= \frac{s'}{8 \cdot (f X - f + 1) \cdot (f Y - f + 1) - 6 \cdot (f X - f + 1) - 6 \cdot (f Y - f + 1) + 4}\\ &= \frac{s'}{8f^2XY - 8f^2X - 8f^2Y + 2fX + 2fY - 4f + 8f^2}\\ &= \frac{s'}{8f^2XY - 8f^2X - 8f^2Y + 2f \cdot (X + Y - 2) + 8f^2}\\ &= \frac{s'}{8f^2 (XY - X - Y + 1) + 2f \cdot (X + Y - 2)}\\ &\geq \frac{f^2 \cdot s - 6f \cdot (X + Y)}{8f^2 (XY - X - Y + 1) + 2f \cdot (X + Y - 2)} \end{align*} Which in turn allows us to find the following condition on $f$ that guarantees our property: \begin{align*} \frac{s}{8XY - 6X - 6Y + 4} &\leq \frac{f^2 \cdot s - 6f \cdot (X + Y)}{8f^2 (XY - X - Y + 1) + 2f \cdot (X + Y - 2)}\\ \frac{s}{8XY - 6X - 6Y + 4} &\leq \frac{f \cdot s - 6 \cdot (X + Y)}{8f (XY - X - Y + 1) + 2 \cdot (X + Y - 2)}\\ - 2fsX - 2fsY + 4fs + 2s \cdot (X + Y - 2) &\leq - 6 \cdot (X + Y) \cdot (8XY - 6X - 6Y + 4)\\ f \cdot 2s \cdot (X + Y - 2) - 2s \cdot (X + Y - 2) &\geq 6 \cdot (X + Y) \cdot (8XY - 6X - 6Y + 4)\\ f &\geq \frac{6 \cdot (X + Y) \cdot (8XY - 6X - 6Y + 4)}{2s \cdot (X + Y - 2)} + 1 \end{align*} which means if $f$ is at least what we bound it by here the relative score will not worsen. Lastly we consider that we slightly violate the distribution constraint for the crops that are planted in pixels on the border. Concretely every crop planted in a corner would need to be planted: \begin{displaymath} f^2 - (f-1)^2 = 2f - 1 \end{displaymath} more times and crops planted in other places on the border would need to be planted $f$ more times. It should be stated that this would also require the grown field to be made even larger to accommodate the extra plants. It is worth noting however that the violation in the problem constraints is only linear, while in general the total number of planted pixels grows quadratically. Hence it is somewhat reasonable to ignore these violation on fields that are grown by very large factors in order to receive the guarantee of not worsening the relational score. \end{proof} \section{Creating integer solutions from fractional solutions} ... ...
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 ... ... @@ -2,11 +2,13 @@ \contentsline {proof}{{Proof}{1}{}}{7}{proof.1}% \contentsline {theorem}{{Theorem}{4.{2}}{}}{9}{theorem.4.3.2}% \contentsline {proof}{{Proof}{2}{}}{9}{proof.2}% \contentsline {theorem}{{Theorem}{7.{1}}{}}{21}{theorem.7.2.1}% \contentsline {proof}{{Proof}{3}{}}{21}{proof.3}% \contentsline {theorem}{{Theorem}{7.{2}}{}}{22}{theorem.7.2.2}% \contentsline {proof}{{Proof}{4}{}}{22}{proof.4}% \contentsline {theorem}{{Theorem}{8.{1}}{}}{29}{theorem.8.1.1}% \contentsline {proof}{{Proof}{5}{}}{29}{proof.5}% \contentsline {theorem}{{Theorem}{8.{2}}{}}{29}{theorem.8.1.2}% \contentsline {proof}{{Proof}{6}{}}{29}{proof.6}% \contentsline {theorem}{{Theorem}{7.{1}}{}}{18}{theorem.7.1.1}% \contentsline {proof}{{Proof}{3}{}}{18}{proof.3}% \contentsline {theorem}{{Theorem}{7.{2}}{}}{20}{theorem.7.2.2}% \contentsline {proof}{{Proof}{4}{}}{20}{proof.4}% \contentsline {theorem}{{Theorem}{7.{3}}{}}{21}{theorem.7.2.3}% \contentsline {proof}{{Proof}{5}{}}{21}{proof.5}% \contentsline {theorem}{{Theorem}{8.{1}}{}}{27}{theorem.8.1.1}% \contentsline {proof}{{Proof}{6}{}}{27}{proof.6}% \contentsline {theorem}{{Theorem}{8.{2}}{}}{27}{theorem.8.1.2}% \contentsline {proof}{{Proof}{7}{}}{27}{proof.7}%
 ... ... @@ -15,17 +15,16 @@ \contentsline {chapter}{\chapternumberline {5}Benchmark Problems}{11}{chapter.5}% \contentsline {chapter}{\chapternumberline {6}Bounds}{15}{chapter.6}% \contentsline {section}{\numberline {6.1}Basic Bound}{15}{section.6.1}% \contentsline {section}{\numberline {6.2}Gärtner Bound}{15}{section.6.2}% \contentsline {section}{\numberline {6.3}LP bound}{17}{section.6.3}% \contentsline {chapter}{\chapternumberline {7}Helpful Statements}{19}{chapter.7}% \contentsline {section}{\numberline {7.1}Growing solutions}{19}{section.7.1}% \contentsline {section}{\numberline {7.2}Creating integer solutions from fractional solutions}{21}{section.7.2}% \contentsline {subsection}{\numberline {7.2.1}The standard method}{21}{subsection.7.2.1}% \contentsline {subsection}{\numberline {7.2.2}The advanced method}{22}{subsection.7.2.2}% \contentsline {chapter}{\chapternumberline {8}The Linear Programming Method}{27}{chapter.8}% \contentsline {section}{\numberline {8.1}Problem Setup}{27}{section.8.1}% \contentsline {section}{\numberline {8.2}Method}{30}{section.8.2}% \contentsline {chapter}{\chapternumberline {9}Gradient Descent}{33}{chapter.9}% \contentsline {chapter}{\chapternumberline {10}Conclusion}{35}{chapter.10}% \contentsline {appendix}{\chapternumberline {A}Calculations Appendix}{37}{appendix.A}% \contentsline {chapter}{Bibliography}{39}{appendix*.4}% \contentsline {section}{\numberline {6.2}LP bound}{15}{section.6.2}% \contentsline {chapter}{\chapternumberline {7}Helpful Statements}{17}{chapter.7}% \contentsline {section}{\numberline {7.1}Growing solutions}{17}{section.7.1}% \contentsline {section}{\numberline {7.2}Creating integer solutions from fractional solutions}{19}{section.7.2}% \contentsline {subsection}{\numberline {7.2.1}The standard method}{19}{subsection.7.2.1}% \contentsline {subsection}{\numberline {7.2.2}The advanced method}{21}{subsection.7.2.2}% \contentsline {chapter}{\chapternumberline {8}The Linear Programming Method}{25}{chapter.8}% \contentsline {section}{\numberline {8.1}Problem Setup}{25}{section.8.1}% \contentsline {section}{\numberline {8.2}Method}{28}{section.8.2}% \contentsline {chapter}{\chapternumberline {9}Gradient Descent}{31}{chapter.9}% \contentsline {chapter}{\chapternumberline {10}Conclusion}{33}{chapter.10}% \contentsline {appendix}{\chapternumberline {A}Calculations Appendix}{35}{appendix.A}% \contentsline {chapter}{Bibliography}{37}{appendix*.4}%
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