Updated the section on the Ackermann's function.

diff --git a/notation.tex b/notation.tex
index b63e59588a1e816c96e8a802273fea5280dd4a57..efc401473c2207a2cf5cf3ecd5462d5de641f90c 100644 (file)
--- a/notation.tex
+++ b/notation.tex
@@ -161,10 +161,10 @@ Various sources tend to differ in the exact definition of both the Ackermann's
 function and its inverse, but most of the definitions differ in factors that
 are negligible when compared with the asymptotic growth of the function.
 We will use the definition by double induction given by Tarjan \cite{tarjan:setunion},
-which are predominant in the literature on graph algorithms:
+which is predominant in the literature on graph algorithms:
 
 \defn
-The \df{Ackermann's function} $A(x,y)$ is a~function on non-negative integers defined by:
+The \df{Ackermann's function} $A(x,y)$ is a~function on non-negative integers defined as:
 $$\eqalign{
 A(0,y) &:= 2y, \cr
 A(x,0) &:= 0, \cr
@@ -172,7 +172,7 @@ A(x,1) &:= 2 \quad \hbox{for $x\ge 1$}, \cr
 A(x,y) &:= A(x-1, A(x,y-1)) \quad \hbox{for $x\ge 1$, $y\ge 2$}. \cr
 }$$
 Sometimes, a~single-parameter version of this function is also used. It is defined
-as the diagonal of the previous function, i.e., $A(n):=A(n,n)$.
+as the diagonal of the previous function, i.e., $A(x):=A(x,x)$.
 
 \example
 We can try evaluating $A(x,y)$ in some points:
@@ -186,18 +186,19 @@ A(4,3) &= A(3,A(4,2)) = A(3,4) = A(2,A(3,3)) = A(2,A(2,A(3,2))) = \cr
 }$$
 
 \defn
-The \df{Inverse Ackermann's function} $\alpha(x,y)$ is defined by:
+The \df{Inverse Ackermann's function} $\alpha(x,y)$ is defined as:
 $$
 \alpha(x,n) := \min\{ y \mid A(x,y) > \log n \}.
 $$
 Again, a~single-parameter version is occasionally considered:
 $$
-\alpha(n) = \min\{ x \mid A(x,x) > \log n \}.
+\alpha(n) := \min\{ x \mid A(x,x) > \log n \}.
 $$
 
 \example
 $\alpha(1,n) = \O(\log\log n)$, $\alpha(2,n) = \O(\log^* n)$, $\alpha(3,n)$ grows even slower.
 
-\FIXME{Relationship between single- and double-parameter versions.}
+\obs
+As the rows of the function~$A$ are increasing, we have $A(x,y) \ge A(x,x) = A(x)$ for $y\ge x$
 
 \endpart