\thm\id{itjarthm}%
The Iterated Jarn\'\i{}k's algorithm finds the MST of the input graph in time
-$\O(m\timesbeta(m,n))$, where $\beta(m,n):=\min\{ i: \log^{(i)}n < m/n \}$.
+$\O(m\timesbeta(m,n))$, where $\beta(m,n):=\min\{ i: \log^{(i)}n \le m/n \}$.
\proof
Phases are finite and in every phase at least one edge is contracted, so the outer
\n{$\log n$}{a binary logarithm of the number~$n$}
\n{$f^{(i)}$}{function~$f$ iterated $i$~times: $f^{(0)}(x):=x$, $f^{(i+1)}(x):=f(f^{(i)}(x))$}
\n{$2\tower n$}{the tower function (iterated exponential): $2\tower 0:=1$, $2\tower (n+1):=2^{2\tower n}$}
-\n{$\log^* n$}{the iterated logarithm: $\log^*n := \min\{i: \log^{(i)}n < 1\}$; the inverse of~$2\tower n$}
-\n{$\beta(m,n)$}{$\beta(m,n) := \min\{i: \log^{(i)}n < m/n \}$ \[itjarthm]}
+\n{$\log^* n$}{the iterated logarithm: $\log^*n := \min\{i: \log^{(i)}n \le 1\}$; the inverse of~$2\tower n$}
+\n{$\beta(m,n)$}{$\beta(m,n) := \min\{i: \log^{(i)}n \le m/n \}$ \[itjarthm]}
\n{$W$}{word size of the RAM \[wordsize]}
\n{$\(x)$}{number~$x\in{\bb N}$ written in binary \[bitnota]}
\n{$\(x)_b$}{$\(x)$ zero-padded to exactly $b$ bits \[bitnota]}