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$1 - N{1-m} = m^{h+1}$  $lg_{m}(Nm - N + 1) -1 = h$  So looking at large values of N and M, this formula is basically     $lg_{m}(Nm) = h$ because $Nm$ is the dominating term. So if $m^'=m^2$, then the height of $m^' is double M$.     $lg(Nm) = h$     $lg(N(m^')^2 = h$     $2lg(N) = h$     $lg(N) = \frac{h}{2}$  \item  Using Master Thoerem, we will be able to write the recursive cost of this function as:  \[T(n) = a + (n/b) (\frac{n}{b})  + O(n^a)\] The branching factor $a$ is 1 because we are only left with one subproblem at each iteration. Also doing $\frac{y}{2}$ takes constant time because $y$ is in the order of $2^n$, so dividing by 2 is just a simple bitshift. The subproblem is $n/2$ because you only have to deal with a number with half the bitsize. THe cost of combining the subproblems back to the larger problem is the cost of squaring a number and then take mod on a number order $2^0$.