Here is a visualization of such a sequence $\{x_i \bmod p\}$ with $n = 2206637$, $p = 317$, $x_0 = 2$ and $f(x) = x^2 + 1$.

From the form of the sequence you can see very clearly why the algorithm is called Pollard's $\rho$ algorithm.

Yet, there is still an open question.

How can we exploit the properties of the sequence $\{x_i \bmod p\}$ to our advantage without even knowing the number $p$ itself?

In the following image you can see a visualization of the algorithm for computing all prime numbers in the range $[1; 16]$. It can be seen, that quite often we mark numbers as composite multiple times.

The idea behind is this:

A number is prime, if none of the smaller prime numbers divides it.

The following image shows a possible interpretation of the Fenwick tree as tree.

The nodes of the tree show the ranges they cover.

Geometrically it is product of the length of the first vector by the length of the projection of the second vector onto the first one.

As you may see from the image below this projection is nothing but $|\mathbf a| \cos \theta$ where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$. Thus $\mathbf a\cdot \mathbf b = |\mathbf a| \cos \theta \cdot |\mathbf b|$.

The dot product holds some notable properties:

You can see that this set of points is exactly the set of points for which the projection onto $\mathbf a$ is the point $C \cdot \dfrac{\mathbf a}{|\mathbf a|}$ and they form a hyperplane orthogonal to $\mathbf a$.

You can see the vector $\mathbf a$ alongside with several such vectors having same dot product with it in 2D on the picture below:

In 2D these vectors will form a line, in 3D they will form a plane.

Note that this result allows us to define a line in 2D as $\mathbf r\cdot \mathbf n=C$ or $(\mathbf r - \mathbf r_0)\cdot \mathbf n=0$ where $\mathbf n$ is vector orthogonal to the line and $\mathbf r_0$ is any vector already present on the line and $C = \mathbf r_0\cdot \mathbf n$.

Assume you have three vectors $\mathbf a$, $\mathbf b$ and $\mathbf c$ in 3D space joined in a parallelepiped as in the picture below:

How would you calculate its volume?

From school we know that we should multiply the area of the base with the height, which is projection of $\mathbf a$ onto direction orthogonal to base.

That means that if we define $\mathbf b \times \mathbf c$ as the vector which is orthogonal to both $\mathbf b$ and $\mathbf c$ and which length is equal to the area of the parallelogram formed by $\mathbf b$ and $\mathbf c$ then $|\mathbf a\cdot (\mathbf b\times\mathbf c)|$ will be equal to the volume of the parallelepiped.

For integrity we will say that $\mathbf b\times \mathbf c$ will be always directed in such way that the rotation from the vector $\mathbf b$ to the vector $\mathbf c$ from the point of $\mathbf b\times \mathbf c$ is always counter-clockwise (see the picture below).

This defines the cross (or vector) product $\mathbf b\times \mathbf c$ of the vectors $\mathbf b$ and $\mathbf c$ and the triple product $\mathbf a\cdot(\mathbf b\times \mathbf c)$ of the vectors $\mathbf a$, $\mathbf b$ and $\mathbf c$.

Actually it would be a bit more convenient to consider them not as linear functions, but as points $(k;b)$ on the plane such that we will have to find the point which has the least dot product with a given point $(x;1)$, that is, for this point $kx+b$ is minimized which is the same as initial problem.

Such minimum will necessarily be on lower convex envelope of these points as can be seen below:

One has to keep points on the convex hull and normal vectors of the hull's edges.

When you have a $(x;1)$ query you'll have to find the normal vector closest to it in terms of angles between them, then the optimum linear function will correspond to one of its endpoints.

During the algorithm $D$ will be stored inside the quad-edge data structure. This structure is described in the picture:

In the algorithm we will use the following functions on edges:

Let's draw a vertical line $x = -\infty$ mentally and start moving this line to the right.

In the course of its movement, this line will meet with segments, and at each time a segment intersect with our line it intersects in exactly one point (we will assume that there are no vertical segments).

Thus, for each segment, at some point in time, its point will appear on the sweep line, then with the movement of the line, this point will move, and finally, at some point, the segment will disappear from the line.

We are interested in the **relative order of the segments** along the vertical.

Namely, we will store a list of segments crossing the sweep line at a given time, where the segments will be sorted by their $y$-coordinate on the sweep line.

This order is interesting because intersecting segments will have the same $y$-coordinate at least at one time:

We formulate key statements:

This amount is the same as the number of points such that $0 < y \leq (k - \lfloor k \rfloor) \cdot x + (b - \lfloor b \rfloor)$.

So we reduced our problem to $k'= k - \lfloor k \rfloor$, $b' = b - \lfloor b \rfloor$ and both $k'$ and $b'$ less than $1$ now.

Here is a picture, we just summed up blue points and subtracted the blue linear function from the black one to reduce problem to smaller values for $k$ and $b$:

- $k < 1$ and $b < 1$.

which returns us back to the case $k>1$.

You can see new reference point $O'$ and axes $X'$ and $Y'$ in the picture below:

As you see, in new reference system linear function will have coefficient $\tfrac 1 k$ and its zero will be in the point $\lfloor k\cdot n + b \rfloor-(k\cdot n+b)$ which makes formula above correct.

Defined this way, the distance corresponds to the so-called [Manhattan (taxicab) geometry](https://en.wikipedia.org/wiki/Taxicab_geometry), in which the points are considered intersections in a well designed city, like Manhattan, where you can only move on the streets horizontally or vertically, as shown in the image below:

This images show some of the smallest paths from one black point to the other, all of them with length $12$.

Here's an image to help visualizing the transformation:

Here is a nice visualization, which may help you understand what is going on.

One of the most common applications of Minkowski sum is computing the distance between two convex polygons (or simply checking whether they intersect).

At the first glance it may seem that finding faces of a disconnected graph is not much harder because we can run the same algorithm for each connected component. However, the components may be drawn in a nested way, forming **holes** (see the image below). In this case the inner face of some component becomes the outer face of some other components and has a complex disconnected border. Dealing with such cases is quite hard, one possible approach is to identify nested components with [point location](point-location.md) algorithms.

The following implementation returns a vector of vertices for each face, outer face goes first.

Firstly, for each query point $p\ (x_0, y_0)$ we want to find such an edge that if the point belongs to any edge, the point lies on the edge we found, otherwise this edge must intersect the line $x = x_0$ at some unique point $(x_0, y)$ where $y < y_0$ and this $y$ is maximum among all such edges.

The following image shows both cases.

We will solve this problem offline using the sweep line algorithm. Let's iterate over x-coordinates of query points and edges' endpoints in increasing order and keep a set of edges $s$. For each x-coordinate we will add some events beforehand.

For each x-coordinate we will sort the events by their types in order (_vertical_, _get_, _remove_, _add_).

The following image shows all events in sorted order for each x-coordinate.

We will keep two sets during the sweep-line process.

A set $t$ for all non-vertical edges, and one set $vert$ especially for the vertical ones.

The number of common tangents to two circles can be **0,1,2,3,4** and **infinite**.

Look at the images for different cases.

Here, we won't be considering **degenerate** cases, i.e *when the circles coincide (in this case they have infinitely many common tangents), or one circle lies inside the other (in this case they have no common tangents, or if the circles are tangent, there is one common tangent).*

Here is a graphic representation of the three cases.

Finally we should remark on processing all the additions of $1$ or $-1$ on all stripes in $[x_1, x_2]$. For each addition of $w$ on $[x_1, x_2]$ we can create events $(x_1, w),\ (x_2, -w)$

and process all these events with a sweep line.

You can see the implication graph in the following image:

It is worth paying attention to the property of the implication graph:

if there is an edge $a \Rightarrow b$, then there also is an edge $\lnot b \Rightarrow \lnot a$.

As we can check easily, neither of the four components contain a vertex $x$ and its negation $\lnot x$, therefore the example has a solution.

We will learn in the next paragraphs how to compute a valid assignment, but just for demonstration purposes the solution $a = \text{false}$, $b = \text{false}$, $c = \text{false}$ is given.

Now we construct the algorithm for finding the solution of the 2-SAT problem on the assumption that the solution exists.