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Antti H S Laaksonen 2017-02-21 21:59:03 +02:00
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luku07.tex
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@ -983,7 +983,9 @@ $2^m$ distinct rows and the time complexity is
$O(n 2^{2m})$. $O(n 2^{2m})$.
As a final note, there is also a surprising direct formula As a final note, there is also a surprising direct formula
for calculating the number of tilings \cite{kas61,tem61}: for calculating the number of tilings\footnote{Surprisingly,
this formula was discovered independently
by \cite{kas61} and \cite{tem61} in 1961.}:
\[ \prod_{a=1}^{\lceil n/2 \rceil} \prod_{b=1}^{\lceil m/2 \rceil} 4 \cdot (\cos^2 \frac{\pi a}{n + 1} + \cos^2 \frac{\pi b}{m+1})\] \[ \prod_{a=1}^{\lceil n/2 \rceil} \prod_{b=1}^{\lceil m/2 \rceil} 4 \cdot (\cos^2 \frac{\pi a}{n + 1} + \cos^2 \frac{\pi b}{m+1})\]
This formula is very efficient, because it calculates This formula is very efficient, because it calculates
the number of tilings in $O(nm)$ time, the number of tilings in $O(nm)$ time,

4
luku20.tex
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@ -431,8 +431,6 @@ by using one of the following techniques:
\index{Edmonds–Karp algorithm} \index{Edmonds–Karp algorithm}
The \key{Edmonds–Karp algorithm} \cite{edm72} The \key{Edmonds–Karp algorithm} \cite{edm72}
is a variant of the
Ford–Fulkerson algorithm that
chooses each path so that the number of edges chooses each path so that the number of edges
on the path is as small as possible. on the path is as small as possible.
This can be done by using breadth-first search This can be done by using breadth-first search
@ -443,7 +441,7 @@ of the algorithm is $O(m^2 n)$.
\index{scaling algorithm} \index{scaling algorithm}
The \key{scaling algorithm} uses depth-first The \key{scaling algorithm} \cite{ahu91} uses depth-first
search to find paths where each edge weight is search to find paths where each edge weight is
at least a threshold value. at least a threshold value.
Initially, the threshold value is Initially, the threshold value is