Commit 19e919fd authored by Michael Keller's avatar Michael Keller
Browse files

worke

parent 06f54e4b
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print("Sum: ", sum(x0))
for c in range(10):
s = 0
for i, v in enumerate(x0):
if i % 10 == c:
s += v
print("Crop ", c, ", Sum: ", s)
\ No newline at end of file
This source diff could not be displayed because it is too large. You can view the blob instead.
......@@ -155,13 +155,15 @@ def iteration(x0):
pixels = []
random.shuffle(all_pixels)
for p in all_pixels:
neighbors = False
for q in pixels:
if abs(p[0] - q[0]) <= 1 and abs(p[1] - q[1]) <= 1:
neighbors = True
break
if not neighbors:
if random.choice([True, False, False, False]):
pixels.append(p)
# neighbors = False
# for q in pixels:
# if abs(p[0] - q[0]) <= 1 and abs(p[1] - q[1]) <= 1:
# neighbors = True
# break
# if not neighbors:
# pixels.append(p)
# generate objective function
obj = []
......@@ -270,14 +272,14 @@ def main():
print("Score: ", scr)
# if score(x0) > best_x0_score:
# best_x0 = copy.deepcopy(x0)
# best_x0_score = score(x0)
# print("Write to file")
# f = open("best-score-benchmark-1.txt", "a")
# f.write("Score: " + str(best_x0_score) + "\n x0 = " + str(best_x0) + "\n\n")
# f.close()
if score(x0) > best_x0_score:
best_x0 = copy.deepcopy(x0)
best_x0_score = score(x0)
print("Write to file")
f = open("debug.txt", "a")
f.write("Score: " + str(best_x0_score) + "\n x0 = " + str(best_x0) + "\n\n")
f.close()
print("Terminated")
print("Score: ", score(x0))
......
......@@ -18,36 +18,36 @@ def maxScore(R, sol):
# larger example
# maxscore bounded by 950
x = 15
y = 15
R = lambda a, b: [[-1, 0, 0.5, -0.5, 1, 0, 0.5, 1, 0.5, 0],
[0, -1, 0, 1, 0, 0, 0.5, 0.5, 0.5, 0],
[0.5, 0, -1, 0, 0, 0, 0.5, 0.5, 0, -0.5],
[-0.5, 1, 0, -1, 0, 0, -1, 0.5, 0.5, 0.5],
[0, 0, 0, 0, -1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, -1, 1, 0.5, 0, 0],
[0.5, 0.5, 0.5, -0.5, 0, 0.5, -1, 0.5, 0.5, 0.5],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0.5, 0.5, 0, 1, 0, 0, 0.5, 0.5, -1, 0],
[0, 0, -0.5, 0.5, 0, 0, 1, 0.5, 0.5, -1]][a][b]
dist = [23, 23, 23, 23, 23, 22, 22, 22, 22, 22]
# Cotrini Data (Sum = 26)
# x = 15
# y = 15
# R = lambda a, b: [
# [-0.5, 1, 1, 1, 1, 1, -1, 0.5, 0.5, 1],
# [1, -0.5, -1, 1, 1, 0.5, -1, 1, 1, 1],
# [1, -1, -0.5, -1, 1, 0, 0.5, 0, 0, 0],
# [1, 1, -1, -0.5, 0, 0, -1, 1, -1, 0],
# [1, 1, 1, 0, -0.5, 0, 1, 1, 1, 1],
# [1, 0.5, 0, 0, 0, -0.5, 0, 0, 0, 0],
# [-1, -1, 0.5, -1, 1, 0, -0.5, 0, 0.5, 0],
# [0.5, 1, 0, 1, 1, 0, 0, -0.5, 1, 0],
# [0.5, 1, 0, -1, 1, 0, 0.5, 1, -0.5, 1],
# [1, 1, 0, 0, 1, 0, 0, 0, 1, -0.5]][a][b]
# R = lambda a, b: [[-1, 0, 0.5, -0.5, 1, 0, 0.5, 1, 0.5, 0],
# [0, -1, 0, 1, 0, 0, 0.5, 0.5, 0.5, 0],
# [0.5, 0, -1, 0, 0, 0, 0.5, 0.5, 0, -0.5],
# [-0.5, 1, 0, -1, 0, 0, -1, 0.5, 0.5, 0.5],
# [0, 0, 0, 0, -1, 0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0, -1, 1, 0.5, 0, 0],
# [0.5, 0.5, 0.5, -0.5, 0, 0.5, -1, 0.5, 0.5, 0.5],
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [0.5, 0.5, 0, 1, 0, 0, 0.5, 0.5, -1, 0],
# [0, 0, -0.5, 0.5, 0, 0, 1, 0.5, 0.5, -1]][a][b]
# dist = [23, 23, 23, 23, 23, 22, 22, 22, 22, 22]
# Cotrini Data (Sum = 26)
x = 15
y = 15
R = lambda a, b: [
[-0.5, 1, 1, 1, 1, 1, -1, 0.5, 0.5, 1],
[1, -0.5, -1, 1, 1, 0.5, -1, 1, 1, 1],
[1, -1, -0.5, -1, 1, 0, 0.5, 0, 0, 0],
[1, 1, -1, -0.5, 0, 0, -1, 1, -1, 0],
[1, 1, 1, 0, -0.5, 0, 1, 1, 1, 1],
[1, 0.5, 0, 0, 0, -0.5, 0, 0, 0, 0],
[-1, -1, 0.5, -1, 1, 0, -0.5, 0, 0.5, 0],
[0.5, 1, 0, 1, 1, 0, 0, -0.5, 1, 0],
[0.5, 1, 0, -1, 1, 0, 0.5, 1, -0.5, 1],
[1, 1, 0, 0, 1, 0, 0, 0, 1, -0.5]][a][b]
dist = [23, 23, 23, 23, 23, 22, 22, 22, 22, 22]
# auto calculate some vars
numb_crops = len(dist)
......@@ -139,15 +139,11 @@ def main():
rhs_ub = []
for c in range(numb_crops):
constrA = []
constrB = []
for i in range(numb_crops):
for j in range(i):
constrA += [1 if i == c or j == c else 0]
constrB += [-1 if i == c or j == c else 0]
lhs_ub += [constrA]
lhs_ub += [constrB]
rhs_ub += [8 * dist[c]]
rhs_ub += [-3 * dist[c]]
......
......@@ -116,7 +116,9 @@ Our first benchmark problem is bounded here by
\section{LP bound}
We can also use the solution to the
following Linear Program as a bound:
following Linear Program as a bound.
The variables are all of the possible
crop pairings:
\begin{align*}
x &= \begin{bmatrix}
x_{1, 1}\\
......@@ -125,10 +127,10 @@ following Linear Program as a bound:
x_{1, C}\\
x_{2, 2}\\
\vdots\\
x_{C-1, C}
x_{C, C}
\end{bmatrix}
&
x_{i, j} &= \text{How much of crop } c_i \text{ is planted next to crop } c_j
x_{i, j} &= \text{How much of crop } c_i \text{ is planted next to } c_j \text{ in the field}
\end{align*}
where the objective function is the sum of the relationship
function for both involved crops the variable has:
......@@ -140,7 +142,7 @@ function for both involved crops the variable has:
R(c_1, c_C) + R(c_C, c_1)\\
R(c_2, c_2) + R(c_2, c_2)\\
\vdots\\
R(c_{C-1}, c_C) + R(c_C, c_{C-1})
R(c_{C}, c_C) + R(c_C, c_{C})
\end{bmatrix}
\end{align*}
and we are trying to maximize $c^T \cdot x$ under the
......@@ -150,7 +152,18 @@ total relationships between pixels:
\sum_{x_i \in x} 2 \cdot x_i = \text{Basic Bound}
\end{align*}
as well as that we don't plant too much of a
specific crop:
specific crop. Here we simply enforce an upper bound
on the number of possible connections a crop could have:
\begin{align*}
1
\end{align*}
\ No newline at end of file
\mathcal{A}_c = \{ x_{i, j} \|\ x_{i, j} \in x \land \left( i = c \lor j = c \right) \}\\
\forall c \in \{1, \dots, C\} \quad \sum_{a \in \mathcal{A}_c} a \leq 8 \cdot X \cdot Y \cdot D(c_c)
\end{align*}
As an optimal solution for a pixel farming problem
meets these constraints, this is a valid bound. It should
be noted that this bound does not enforce any constraints
on the shape of the field, or indeed that this score
is achievable with only one field (and not multiple disjoint
fields).
The bound for the first benchmark problem is $1624$, for
the second benchmark problem $950$.
\ No newline at end of file
\chapter{Gradient Descent}
\ No newline at end of file
\chapter{Gradient Descent}
Gradient Descent is a popular method
for searching through large search spaces.
As from the helper theorems \ref{makeInt1} and
\ref{makeInt2} we can from fractional fields
build almost entirely integer fields,
we can actually use Gradient Descent
for the solution space consisting of all
fractional solutions.
......@@ -21,6 +21,7 @@ First we introduce a method that has good results
on large fields and requires no conditions of $R$.
\begin{theorem}
\label{makeInt1}
A fractional pixel farming solution can be
transformed to a fractional solution with at least
the same score where up to $4 \cdot \binom{C}{2}$
......@@ -97,9 +98,6 @@ on large fields and requires no conditions of $R$.
there are four colors among which we swap, this means
in the worst case we end up with $4 \cdot \binom{C}{2}$
pixels that are still fractional.
TODO: QED SYMBOL MISSING?
\end{proof}
......@@ -112,6 +110,7 @@ of different crops. Here the following slightly stronger
statement with a condition on $R$ might be helpful:
\begin{theorem}
\label{makeInt2}
If every crop prefers all other crops over itself,
then a fractional pixel farming solution can be
transformed to a fractional solution with at least
......@@ -130,4 +129,5 @@ statement with a condition on $R$ might be helpful:
met to ensure that swapping in either direction
even among neighboring pixels is never detrimental.
l
\end{proof}
\ No newline at end of file
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