#118 4.2-3

chapter4/sections/2/3.tex

@@ -1 +1,13 @@
-\workinprogress % TODO
+For convenience, we'll assume that $n$ is an exact power of 3.
+Similarly to equation (4.2), let's partition the matrices into nine $n/3\times n/3$ submatrices.
+We can then treat the input matrices as two $3\times3$ matrices that we multiply using $k$ multiplications.
+Each of those is in fact a multiplication of two $n/3\times n/3$ matrices and can be performed by running the algorithm recursively.
+
+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
+\[
+    T(n) = kT(n/3)+\Theta(1)
+\]
+which, by the master method, has the solution $T(n)=\Theta(n^{\log_3k})$.
+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 $T(n)=\Theta(n^{\log_321})=O(n^{2.78})$.