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# Divide and Conquer

## 基本思想

Divide-and-conquer:

Break up problem into several parts.

Solve each part recursively.

Combine solutions to sub-problems into overall solution.

Most common usage.

• Break up problem of size n into two equal parts of size ½ n.
• Solve two parts recursively.
• Combine two solutions into overall solution in linear time.

Consequence.

• Brute force: $n^2$.
• Divide-and-conquer: $n log n$.

## Mergesort

• Divide array into two halves.
• Recursively sort each half.
• Merge two halves to make sorted whole.

Challenge for the bored. In-place merge. [Kronrud, 1969]

using only a constant amount of extra storage

Def. T(n) = number of comparisons to mergesort an input of size n.

Mergesort recurrence：
$$T(n)=\left{ \begin{array}{lr} \text{0,}&\quad\text{if n = 1,}\ T(\lceil \frac{n}{2}) + T(\lfloor \frac{n}{2} \rfloor) + n, &\quad\text{otherwise.} \end{array} \right.$$
$T(n) = O(n log_2n).$

## Counting Inversions

Divide-and-conquer.

Divide: separate list into two pieces.

Conquer: recursively count inversions in each half.

Combine: count inversions where $a_i$ and $a_j$ are in different halves, and return sum of three quantities

\begin{align*} &Sort-and-Count(L)\quad{ \ &\qquad\text{if list L has one element}\ &\qquad\qquad \text{return 0 and the list L}\ &\qquad \text{Divide the list into two halves A and B}\ &\qquad (r_A, A) \leftarrow Sort-and-Count(A)\ &\qquad (r_B, B) \leftarrow Sort-and-Count(B)\ &\qquad (r_B, L) \leftarrow Merge-and-Count(A, B)\ } \end{align*}

## Closest Pair of Points

【鸽巢原理】【分治】

Closest pair: Given n points in the plane, find a pair with smallest Euclidean distance between them

Algorithm:

• Divide: draw vertical line L so that roughly ½ n points on each side.
• Conquer: find closest pair in each side recursively
• Combine: find closest pair with one point in each side.【利用鸽巢原理剪枝，减少需要判断的数量】
• Return best of 3 solutions.

Find closest pair with one point in each side, assuming that $distance < \delta = min{左右两个部分的最近点}$.

• So, only need to consider points within $\delta$ of line L

Running Time
\begin{align*} T(n) &\leq 2T(\frac{n}{2}) + O(nlogn)\ &=O(nlog^2n) \end{align*}
More optimal:

• Don't sort points in strip from scratch each time.
• Each recursive returns two lists: all points sorted by y coordinate,and all points sorted by x coordinate.
• Sort by merging two pre-sorted lists.

\begin{align*} T(n) &\leq 2T(\frac{n}{2}) + O(n)\ &=O(nlogn) \end{align*}

## Convolution and FFT

FFT: Fast way to convert between time-domain and frequency-domain.

Alternate viewpoint. Fast way to multiply and evaluate polynomials.

### coefficient representation

Polynomial.
$$A(x) = a_0 + a_1x + a_2x^2 + \dots + a_{n-1}x^{n-1}\ B(x) = b_0 + b_1x + b_2x^2 + \dots + b_{n-1}x^{n-1}$$
Add. $O(n)$ arithmetic operations.
$$A(x) + B(x) = (a_0+b_0) + (a_1 + b_1)x + \dots + (a_{n-1} + b_{n-1})x^{n-1}$$
Evaluate. $O(n)$ using Horner's method.
$$A(x) = a_0 + x(a_1 + x(a_2 + \dots + x(a_{n-2} + x(a_{n-1}))\dots))$$
Multiply (convolve): $O(n^2)$ using brute force
$$A(x) \times B(x) = \sum_{i=0}^{2n-2}c_ix^i,\qquad where, c_i = \sum_{j=0}^{i}a_jb_{i-j}$$

### Point-Value Representation

Fundamental theorem of algebra. [Gauss, PhD thesis]

A degree n polynomial with complex coefficients has exactly n complex roots.
Corollary. A degree n-1 polynomial A(x) is uniquely specified by its evaluation at n distinct values of x.

Polynomial:
$$A(x): (x_0, y_0), \dots , (x_{n-1}, y_{n-1})\ B(x): (x_0, z_0), \dots , (x_{n-1}, z_{n-1})$$
Add: $O(n)$ arithmetic operations.
$$A(x) + B(x): (x_0, y_0 + z_0), \dots , (x_{n-1}, y_{n-1} + z_{n-1})$$
Evaluate: $O(n^2)$ using Lagrange's formula.
$$A(x) = \sum_{k=0}^{n-1}y_k\frac{\prod_{j \neq k}(x-x_j)}{\prod_{j\neq k}(x_k - x_j)}$$
Multiply (convolve). $O(n)$, but need 2n-1 points
$$A(x) \times B(x): (x_0, y_0 \times z_0), \dots , (x_{n-1}, y_{n-1} \times z_{n-1})$$

### Converting Between Two Representations

Coefficient => point-value: $O(n^2)$ for matrix-vector multiply (or $O(n)$ Horner's).
Point-value => coefficient: $O(n^3)$ for Gaussian elimination.

Improve? => Divide and conquer

Divide. Break polynomial up into even and odd powers
\begin{align*} A(x) &= a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5 + a_6x^6 + a_7x^7\ A_{even}(x) &= a_0 + a_2x^2 + a_4x^4 + a_6x^6\ A_{odd}(x) &= a_1x + a_3x^3 + a_5x^5 + a_7x^7\ A(x) &= A_{even}(x^2) + x A_{odd}(x^2)\ A(-x) &= A_{even}(x^2) - x A_{odd}(x^2) \end{align*}
Intuition: Choose four complex points to be ±1, ±i.
\begin{align*} A(1) &= A_{even}(1) + A_{odd}(1)\ A(-1) &= A_{even}(1) - A_{odd}(1)\ A(i) &= A_{even}(-1) + i A_{odd}(-1)\ A(-i) &= A_{even}(-1) - i A_{odd}(-1). \end{align*}

Can evaluate polynomial of degree $\leq$ n at 4 points by evaluating two polynomials of degree $\leq \frac{2}{n}$ at 2 points.

### Fast Fourier Transform (FFT)

Roots of Unity:

• Def:  An $n_{th}$ root of unity is a complex number x such that $x_n$ = 1.
• Fact. The $n_{th}$ roots of unity are: $\omega_0, \omega_1, \dots , \omega_{n-1}$ where $\omega = e^{\frac{2\pi i}{n}}$
• Pf. $(\omega^k)^n = (e^{2\pi i k /n} ) ^n = (e^{\pi i}) ^{2k} = (-1)^{2k} = 1$
• Fact. The $\frac{n}{2}_{th}$ roots of unity are: $v^0, v^1, \dots, v^{n/2-1}$ where $v = \omega^2 = e^{4\pi i / n}$.

Goal. Evaluate a degree n-1 polynomial $A(x) = a_0 + a_1x + a_2x^2 + \dots + a_{n-1}x^{n-1}$ at its $n_{th}$ roots of unity: $\omega_0, \omega_1, \dots , \omega_{n-1}$

Divide: Break up polynomial into even and odd powers.
\begin{align*} A_{even}(x) &= a_0 + a_2x^1 + a_4x^2 + \dots + a_{n-2}x^{n/2-2}\ A_{odd}(x) &= a_1x + a_3x^1 + a_5x^2 + \dots + a_{n-1}x^{n/2-1}\ A(x) &= A_{even}(x^2) + x A_{odd}(x^2) \end{align*}
Conquer: Evaluate $A_{even}(x)$ and $A_{odd}(x)$ at the $\frac{n}{2}_{th}$ roots of unity: $v^0, v^1, \dots, v^{n/2-1}$.

Combine:

• $A(\omega^k) \qquad= A_{even}(v^k) + \omega^kA_{odd}(v^k), \qquad 0 \leq k \leq n/2$
• $A(\omega^{k+1/2n}) = A_{even}(v^k) - \omega^kA_{odd}(v^k), \qquad 0 \leq k \leq n/2$

Running Time:
\begin{align*} T(n) &= 2T(\frac{n}{2}) + \Theta(n) \ &= \Theta(nlogn) \end{align*}