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--- courses:cs211:winter2012:journals:jeanpaul:chapter_fivesection_i [2012/03/06 16:14] – [Unrolling the Mergesort Recurrence] mugabej
+++ courses:cs211:winter2012:journals:jeanpaul:chapter_fivesection_i [2012/03/06 22:23] (current) – mugabej
@@ Line 59: / Line 59: @@
 >>>>>>>>>>>>>>>>>>>>>> = (cn log<sub>2</sub>n) - cn + cn\\
 >>>>>>>>>>>>>>>>>>>>>> = cn log<sub>2</sub>n\\
+>>>>>>>>>>>>>>> This argument establishes the bound T(n) we want, assuming it holds for smaller values m<n, and thus completes our induction proof.\\
+\\
+** An Approach Using Partial Substitution**:\\
+\\
+With this approach:
+>>>>>>>>>>>>>>> Guess the overall form of the solution without pinning down the exact values of all the constants and other parameters.\\
+>>>>>>>>>>>>>>> Suppose we believe that T(n) = O(nlogn) but we're  not sure of the constant inside the T(.) notation. So,\\
+>>>>>>>>>>>>>>> T(n)≤ kn log<sub>b</sub>nfor some constant k and base b that we solve for.\\
+>>>>>>>>>>>>>>> We try to use the induction as follows:\\
+>>>>>>>>>>>>>>> T(n)≤ 2T(n/2) + cn ≤ 2k(n/2) log<sub>b</sub>(n/2) + cn\\
+>>>>>>>>>>>>>>> Let's now choose 2 as our base since it can help us simplify our expression.So,\\
+>>>>>>>>>>>>>>> T(n) ≤ 2k(n/2)log<sub>2</sub>(n/2) + cn\\
+>>>>>>>>>>>>>>>>>>>>>>>>>> = 2k(n/2)[(log<sub>2</sub>n)-1] + cn\\
+>>>>>>>>>>>>>>>>>>>>>>>>>> = kn[(log<sub>2</sub>n)-1] + cn\\
+>>>>>>>>>>>>>>>>>>>>>>>>>> = (knlog<sub>2</sub>n) -kn + cn\\
+>>>>>>>>>>>>>>> Finally we ask ourselves if there is a choice of k that will make this expression be bounded by knlog<sub>2</sub>n\\
+>>>>>>>>>>>>>>> From the expression,it's obvious that we just need to choose a k that is at least as large as c, which gives us:\\
+>>>>>>>>>>>>>>> T(n) ≤ (knlog<sub>2</sub>n) -kn + cn ≤ knlog<sub>2</sub>n\\
+>>>>>>>>>>>>>>> Thus our proof is completed.\\
+\\
+As useful as it is, our this section was really short and straightforward. I give a 9/10.