deletions | additions       diff --git a/a) Number of blits.tex b/a) Number of blits.tex  index 92c427a..97006de 100644  --- a/a) Number of blits.tex  +++ b/a) Number of blits.tex 
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\begin{verbatim} \section{Number of blits}  T(n) = The recursive algorithm can be expressed as having the cost:  \[T(n) &=  (n = 2) ? 5 : 5 + 4 T(n / 2) 2)\]  T(2^1) For $n  = 5 2^x$ we try the algorithm for incremental sizes of $x$ to find a pattern for how the algorithm behaves.  \begin{subequation}  \begin{split}  T(2^1) &= 5\\  T(2^2) = &=  5 + 5 * 2^2 2^2\\  T(2^3) = &=  5 + 5 * 2^2 + 5 * 2^4 2^4\\  T(2^4) = &=  5 + 5 * 2^2 + 5 * 2^4 + 5 * 2^6 2^6\\  T(2^5) = &=  5 + 5 * 2^2 + 5 * 2^4 + 5 * 2^6 + 5 * 2^8 \end{split}  \end{equation}  Given that:  T(2^n) = sum(5 * 4^i, i = 0, n-1) = 5 sum(4^i, i = 0, n-1) = 5 * ((1 - 4^n)/(1 - 4)) = (5 - 5 * 4^n)/(-3) = (5 * 4^n - 5) / 3 
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T(2^(n+1)) = 5 + 4 ((5 * 4^n - 5) / 3) = 5 + (5 * 4^(n+1) - 4*5) / 3 = (5 * 4^(n+1) + 3 * 5 - 4 * 5) / 3 = (5 * 4^(n+1) - 5) / 3  T(n) = (5 * 4^log(n) - 5) / 3 = (5 * (2^2)^log(n) - 5) / 3 = (5 * n^2 - 5) / 3  T(n) = O(n^2)\end{verbatim}