this is for holding javascript data
Mazdak Farrokhzad deleted file proof-b.tex
about 10 years ago
Commit id: 43f07a5e534b588ab4963c9345fc73c605d33975
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\subsection{b)}
Consider the proof for task a), but now with $I_1$ as the first interval ending after
an arbitrary starting point $P$. If $start(I_1) \geq P$, the case is equivalent with
above, and it is possible to find the optimal solution. If $start(I_1) < P$, it
is possible that there exists one or more intervals $I_n$ such that
$start(I_1) mod{2\pi} \leq end(I_n) < P$. If the algorithm adds the ray $R_n$ with an angle
in the interval $[start(I_1), end(I_n)]$, it is possible that the ray $R_1$ added at
$end(I_1)$ will not be needed, and could possibly be removed. The rest of the rays
will always be needed, since removing any of them would cause at least one circle
not to be intersected any more. This means that the result might have one more
ray than optimal, but not more.
