... ... @@ -49,7 +49,8 @@ size $X \cdot Y$ and number of crops $C$. It is known that finding Hamiltonian Paths is NP complete. Thus, reducing the decision version complete \cite[Page 199, GT39]{alma990005774300205503}. Thus, reducing the decision version of hamiltonian paths (explained below) to the decision version of pixel farming demonstrates that pixel farming is NP complete. ... ... @@ -112,10 +113,13 @@ The following example illustrates this: \end{minipage} \end{figure} \FloatBarrier Now we pass the Pixel Farming problem formulated above to our algorithm that solves pixel farming problems in polynomial time. It should be noted that is problem can be built in polynomial time in relation to the field size and number of crops $C$. Now we pass the pixel farming problem formulated above to our algorithm that solves the decision version of the pixel farming problem. If we get the answer that no solution with value at least $s$ exists we return that no longest path of with value at least $s$ exists, we return that no longest path of length $\abs{V} - 1$ exists. If on the other hand a solution of at least $s$ exists, we return that a hamiltonian path must exist in $G$. ... ... @@ -124,9 +128,8 @@ The rationale for this is as follows. Let us assume a solution of value at least $s$ exists. This would imply that every pixel likes all of their neighbors. As "crop $a$ likes crop $b$" is the same function as "vertex $a$ is connected to vertex $b$" and we have a value for our arrangement of $2 \cdot (\abs{V} - 1)$, we must have found $n-1$ edges that connect our $\abs{V}$ vertices. Because we also "vertex $a$ is connected to vertex $b$", we must have found $\abs{V}-1$ edges that connect our $\abs{V}$ vertices. Because we also know that every crop can only appear once in our field and all crops have two neighbors (except the ends) we must have found a path of length $\abs{V} - 1$. ... ...
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