#118 apply the master method more precisely

chapter4/sections/1/3.tex

@@ -4,4 +4,4 @@
 \]
 To show it has the same solution as the original recurrence, let's use the master method from Section 4.5.
 The above is a master recurrence with $a=8$, $b=2$, and driving function $f(n)=\Theta(n^2)$.
-Since $n^{\log_ba}=n^{\lg8}=n^3$ and $f(n)=O(n^{3-\epsilon})$ for any $\epsilon\le1$, we can apply case~1 of the master theorem to get $T(n)=\Theta(n^3)$.
+Since $n^{\log_ba}=n^{\lg8}=n^3$ and $f(n)=O(n^{3-\epsilon})$ for any $0<\epsilon\le1$, we can apply case~1 of the master theorem to get $T(n)=\Theta(n^3)$.

chapter4/sections/2/3.tex

@@ -6,8 +6,10 @@
 Let $T(n)$ be the worst-case time to multiply two $n\times n$ matrices by this algorithm.
 If we did the partitioning using index calculations, then we get the recurrence
 \[
-    T(n) = kT(n/3)+\Theta(1)
+    T(n) = kT(n/3)+\Theta(1).
 \]
-which, by the master method, has the solution $T(n)=\Theta(n^{\log_3k})$.
+To solve it, we'll use the master method.
+We have $a=k$, $b=3$, and driving function $f(n)=\Theta(1)$.
+Since $n^{\log_ba}=n^{\log_3k}$ and $f(n)=O(n^{\log_3k-\epsilon})$ for $0<\epsilon\le\log_3k$, we apply case~1 of the master theorem to get $T(n)=\Theta(n^{\log_3k})$.
 According to \refProblem{3-1}(d), $T(n)=o(n^{\lg7})$ if $\log_3k<\lg7$, which gives us $k<3^{\lg7}\approx21.85$.
 Hence, the largest integer value for $k$ is 21 and then the running time of the algorithm is $T(n)=\Theta(n^{\log_321})=O(n^{2.78})$.