Commit 3e1b1a8e authored by Michael Keller's avatar Michael Keller
Browse files

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}
......
......@@ -11,17 +11,16 @@
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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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