Divide and Conquer

Master Theorem

Given any recurrence of the form $T(n) = a T(\frac{n}{b}) + c n^k$ for all $n > b$, we have:

• If $a > b^k$, then $T(n) = \Theta(n^{\log_b a})$
• If $a < b^k$, then $T(n) = \Theta(n^k)$
• If $a = b^k$, then $T(n) = \Theta(n^k \log n)$

Root Finding

Given a continuous function $f$ and two points $a < b$ such that $f(a) \cdot f(b) < 0$, there exists a root of $f$ in the interval $\lbrack a, b \rbrack$ by the intermediate value theorem. Since said root may be irrational, we aim to approximate it with an arbitrary precision $\epsilon$.

• Algorithm: $Bisect(a, b, \epsilon)$
• If $b - a < \epsilon$, $a$ is a suitable approximation
• Otherwise, calculate the midpoint $m = (a + b)/2$
• If $f(m) \le 0$ then return $Bisect(m, b, \epsilon)$
• else return $Bisect(a, m, \epsilon)$
• Time: $T(n) = T(\frac{n}{2}) + O(1) = O(\log(\frac{b - a}{\epsilon}))$
• Proof:
• $P(k) =$ For any $a, b$ such that $k\epsilon \le |a - b| \le (k + 1)\epsilon$, if $f(a)f(b) \le 0$, then we find an $\epsilon$ approx to a root using $\log k$ queries to $f$.
• $P(1)$: Output $a + \epsilon$, since the whole interval is at most $epsilon$. This requires $0$ calls to $f$.
• Suppose $P(k)$ and consider an arbitrary $a$, $b$ s.t. $2k\epsilon \le |a - b| \le (2k + 1)\epsilon$.
• If $f(a + k\epsilon) = 0$, output $a + k\epsilon$.
• If $f(a)f(a + k\epsilon) < 0$, solve on the interval $\lbrack a, a + k\epsilon \rbrack$. By I.H. this takes at most $\log(k)$ queries of $f$.
• Otherwise, we have $f(b)f(a + k\epsilon) < 0$, since $f(a)f(b) < 0$ and $f(a)f(a + k\epsilon) \ge 0$. Solve the interval $\lbrack a + k\epsilon, b \rbrack$.
• In any case, we used at most $\log(k) + 1 =$\log(2k)$queries to$f$.

kth Smallest Element

• Algorithm: $f(S \in \mathbb{R}^n, k \in \mathbb{R})$
• Select an approximate median element $w$ using median of $\frac{n}{5} medians with subarrays of size$5$
• Partition each element into three sets, $S_{>}, S_{<}, S_{=}$
• If $k \le |S_{<}|$, recurse on $f(S_{<}, k)$
• Else, if $k \le |S_{<}| + |S_{=}|$, return $w$
• Else, recurse on $f(S_{>}, k - |S_{<}| - |S_{=}|)$