On a circle, N arbitrary points are chosen on its circumference. The complete graph formed with those N points would divide the area of the circle into many pieces.
What is the maximum number of pieces of area that the circle will get divided into when the points are chosen along its circumference?
Examples:
2 points => 2 pieces
4 points => 8 pieces
Any ideas how to go about this?
This is known as Moser's circle problem.
The solution is:
i.e.
The proof is quite simple:
Consider each intersection inside the circle. It must be defined by the intersection of two lines, and each line has two points, so every intersection inside the circle defines 4 unique sets of points on the circumference. Therefore, there are at most n choose 4 inner vertices, and obviously there are n vertices on the circumference.
Now, how many edges does each vertex touch? Well, it's a complete graph, so each vertex on the outside touches n - 1 edges, and of course each vertex on the inside touches 4 edges. So the number of edges is given by (n(n - 1) + 4(n choose 4))/2 (we divide by two because otherwise each edge would be counted twice by its two vertices).
The final step is to use Euler's formula for the number of faces in a graph, i.e.: v - e + f = 1 (the Euler characteristic is 1 in our case).
Solving for f gives the formulae above :-)
Related
So this is the task I need help with:
The blueprint of a house is made on a unit square grid sheet. All rooms must be rectangular. So far, N rooms have been drawn on the blueprint. Each room is defined by the upper left and lower right corners. One field of the square grid is given by the x and y coordinates, the coordinates of the upper left field (0,0). The x-coordinates increase horizontally and the y-coordinates increase vertically. The designer wants to calculate how many new rectangular rooms can be added if the two sides of any two new rooms cannot have a common part, and all four sides are adjacent or existing, or the side of the house. The rooms planned so far are such that every free space is rectangular. Make a program that tells you what the largest possible new room area can be in your plan.
Input:
The first line of the standard input has the number of rooms in the design (1≤N≤10,000) and the coordinates of the upper left (FX, FY) and lower right (AX, AY) corners of the house (0 ≤ FX < AX ≤ 1000, 0 ≤ FY < AY ≤ 1000), separated by a space. Each of the following N lines has coordinates (FX ≤ LFXi ≤ RAXi ≤ AX, FY ≤ LFYi ≤ RAYi ≤ AY) of the upper left (LFXi, LFYi) and lower right (RAXi, RAYi) of each room, separated by a space.
Output:
The first line of the standard output should be the area of the largest new room!
Theres an example in the link and its the basic input and output that you can test the program with.
I translated this task from hungarian so the Példa=example , Bemenet=input , Kimenet=output. I would be really happy to get some help with this taks because for me it is a little bit too hard.
Basic input example
I am both new to this website and new to C. I need a program to find the average 'jumps' it takes from all points.
The idea is this: Find "jump" distance from 1 to 2, 1 to 3, 1 to 4 ... 1 to 9, or find 2 to 1, 2 to 3, 2 to 4 2 to 5 etc.
Doing them on the first row is simple, just (2-1) or (3-1) and you get the correct number. But if I want to find the distance between 1 and 4 or 1 to 8 then I have absolutely no idea.
The dimensions of the matrix should potentially be changeable. But I just want help with a 3x3 matrix.
Anyone could show me how to find it?
Jump means vertical or horizontal move from one point to another. from 1 to 2 = 1, from 1 to 9 = 4 (shortest path only)
The definition of "distance" on this kind of problems is always tricky.
Imagine that the points are marks on a field, and you can freely walk all over it. Then, you could take any path from one point to the other. The shortest route then would be a straight line; its length would be the length of the vector that joins the points, which happens to be the difference vector among two points' positions. This length can be computed with the help of Pythagora's theorem: dist = sqrt((x2-x1)^2 + (y2-y1)^2). This is known as the Euclidian distance between the points.
Now imagine that you are in a city, and each point is a building. You can't walk over a building, so the only options are to go either up/down or left/right. Then, the shortest distance is given by the sum of the components of the difference vector; which is the mathematical way of saying that "go down 2 blocks and then one block to the left" means walking 3 blocks' distance: dist = abs(x2-x1) + abs(y2-y1). This is known as the Manhattan distance between the points.
In your problem, however, it looks like the only possible move is to jump to an adjacent point, in a single step, diagonals allowed. Then the problem gets a bit trickier, because the path is very irregular. You need some Graph Theory here, very useful when modeling problems with linked elements, or "nodes". Each point would be a node, connected to their neighbors, and the problem would be to find the shortest path to another given point. If jumps had different weights (for instance, is jumping in diagonal was harder), an easy way to solve this is would be with the Dijkstra's Algorithm; more details on implementation at Wikipedia.
If the cost is always the same, then the problem is reduced to counting the number of jumps in a Breadth-First Search of the destination point from the source.
Let's define the 'jump' distance : "the number of hops required to reach from Point A [Ax,Ay] to Point B [Bx,By]."
Now there can be two ways in which the hops are allowed :
Horizontally/VerticallyIn this case, you can go up/down or left/right. As you have to travel X axis and Y axis independently, your ans is:jumpDistance = abs(Bx - Ax) + abs(By - Ay);
Horizontally/Vertically and also Diagonally
In this case, you can go up/down or left/right and diagonally as well. How it differs from Case 1 is that now you have the ability to change your X axis and Y axis together at the cost of only one jump . Your answer now is:jumpDistance = Max(abs(Bx - Ax),abs(By - Ay));
What is the definition of "jump-distance" ?
If you mean how many jumps a man needs to go from square M to N, if he can only jumps vertically and horizontally, one possibility can:
dist = abs(x2 - x1) + abs(y2 - y1);
For example jump-distance between 1 and 9 is: |3-1|+|3-1| = 4
There are two ways to calculate jump distance.
1) when only horizontal and vertical movements are allowed, in that case all you need to do is form a rectangle in between the two points and calculate the length of two adjacent side. Like if you want to move from 1 to 9 then first move from 1 to 3 and then move from 3 to 9. (Convert it to code)
2) when movements in all eight directions are allowed, things get tricky. Like if you want to move from 1 to 6 suppose. What you'll need to do is you'll have to more from 1 to 5. And then from 5 to 6. The way of doing it in code is to find the maximum in between the difference in x and y coordinates. In this example, in x coordinate, difference is 2 (3-1) and in y coordinate, difference is 1 (2-1). So the maximum of this is 2. So here's the answer. (Convert to code)
I am new to Matlab.
Lets say I have an array a = [1:1:1000]
I have to divide this into 50 parts 1-20; 21-40 .... 981-1000.
I am trying to do it this way.
E=1000X
a=[1:E]
n=50
d=E/n
b=[]
for i=0:n
b(i)=a[i:d]
end
But I am unable to get the result.
And the second part I am working on is, depending on another result, say if my answer is 3, the first split array should have a counter and that should be +1, if the answer is 45 the 3rd split array's counter should be +1 and so on and in the end I have to make a histogram of all the counters.
You can do all of this with one function: histc. In your situation:
X = (1:1:1000)';
Edges = (1:20:1000)';
Count = histc(X, Edges);
Essentially, Count contains the number of elements in X that fall into the categories defined in Edges, where Edges is a monotonically increasing vector whose elements define the boundaries of sequential categories. A more common example might be to construct X using a probability density, say, the uniform distribution, eg:
X = 1000 * rand(1000, 1);
Play around with specifications for X and Edges and you should get the idea. If you want the actual histogram plot, look into the hist function.
As for the second part of your question, I'm not really sure what you're asking.
I posted this on twitter a while ago but seeing how none of my followers appears to be a math/programming genius, I'll try my luck here as well. I got here because I found this which might contain part of my solution.
I described my problem in the following pdf document, containing a picture of what I'm trying to achieve.
To give some more details, I divided the pentagon's of a dodecahedron (12 pentagons) into triangles (5/pentagon, 60 triangles in total), then collected a set of data points relative to each of these triangles.
The idea is to generate terrain meshes for each individual triangle.
To do so, the data must be represented flat, in a 32K x 32K square (idTech4 Megatexture)
I have vaguely heard of transformation matrices, which when set up properly, could do the trick of passing all the data points trough them to have them show up in the right place.
I looked at this source code here but I don't understand how I'm supposed to get the points in and/or out of there, not to mention how to do the setup so I can present each point in turn and get the result point back.
I got as fas as identifying the point that belongs in the back right corner. All my 3D points are originally stored in latitude / longitude pairs. I retrieve the 3D vectors this way:
coord getcoord(point* p)
{
coord c;
c.x=cos(p->lat*pi/180.l) * cos(p->lon*pi/180.l);
c.y=cos(p->lat*pi/180.l) * sin(p->lon*pi/180.l);
c.z=sin(p->lat*pi/180.l);
return c;
};
My thought is that if I can find the center of my triangle, and discover how to offset my angles so the vector from the center of my sphere to the middle of the triangle moves to 90N then my points would already be in the right plane if I rotated them all along the same angles. If I then convert them all to 3d and subtracti the radius from y, they'll be at the correct y position as well.
Then all I'd need to do is the rotation, the scaling, and the moving to the final position.
There are several kinds of 'centers' for a triangle, I think the one I need is the one that is equidistant to the corners of the triangle (Circumcenter?)
But then there might be an easier approach to the whole problem so while I continue my own research, perhaps some of you can help pointing me in the right direction.
It appears as if some sample data is in order, here are a few of these triangles in obj file format:
v 0.000000 0.000000 3396.000000
v 2061.582356 0.000000 2698.646733
v 637.063983 1960.681333 2698.646733
f 1 2 3
And another:
v -938.631230 2888.810129 1518.737455
v 637.063983 1960.681333 2698.646733
v 1030.791271 3172.449325 637.064076
f 1 2 3
You will notice that each point is at a distance of 3396 from 0,0,0
I mentioned 'on the sphere' meaning that the face away from the center of the sphere is the face that needs to become the 'top' when translated into the square.
Theoretically all these triangles should in fact have identical sizes, but due to rounding errors in the math that generated them, this might not be entirely true.
If I'm not mistaken I already took measures to ensure that the first point you see here is always the one opposite the longest border, so it's the one that should go in the far left corner (testing the above 2 samples confirms this, but I'm measuring anyway just to be sure)
Both legs leading away from this point should theoretically have the same length as well, but again rounding errors might slightly offset that.
If I've done it correctly then the longer side is 1,113587 times longer than the 2 shorter sides. Assuming those are identical, then doing some goal seeking in excel, I can deduct that the final points, assuming I was just translating this triangle, should look like:
v 16384.000000 0.000000 16384.000000
v -16384.000000 0.000000 9916.165306
v 9916.165306 0.000000 -16384.000000
f 1 2 3
So I need to setup the matrix to do this transformation, preferably using the 4x4 matrix as explained below.
I would recommend using transform matrices. The 3d transform matrix is a 4x4 data structure which describes a translation and rotation (and possibly a scale). Once you have a matrix you can transform a point like so
result.x = (tmp->pt.x * m->element[0][0]) +
(tmp->pt.y * m->element[1][0]) +
(tmp->pt.z * m->element[2][0]) +
m->element[3][0];
result.y = (tmp->pt.x * m->element[0][1]) +
(tmp->pt.y * m->element[1][1]) +
(tmp->pt.z * m->element[2][1]) +
m->element[3][1];
result.z = (tmp->pt.x * m->element[0][2]) +
(tmp->pt.y * m->element[1][2]) +
(tmp->pt.z * m->element[2][2]) +
m->element[3][2];
int w = (tmp->pt.x * m->element[0][3]) + (tmp->pt.y * m->element[1][3])
+ (tmp->pt.z * m->element[2][3]) + m->element[3][3];
if (w!=0 || w!=1)
result.x/=w; result.y/=w; result.z/=w;
This will transform the 3D point pt by the matrix m. If you now a little matrix math you'll see i'm just multiplying my origin point as a vector against the matrix (and doing a little normalization if it is a skew matrix.) Matrices can be multiplied together to form complicated transformations so they are very useful.
For details on making matrices suggest reading this link.
http://en.wikipedia.org/wiki/Transformation_matrix
The problem set asks me to create two triangles, defining them using points, and then checking if they're similar.
I did first part: created a struct point and a struct triangle, as the profesor told us to. To solve the problem of checking similarity, I thought I could use the points to define vectors, and them use the law of cosines to calculate its angles, together with some if sentences to check if the triangles are similar.
Which codes could help me achieve that? I could not find anything that I'd be able to turn into a partial solution.
What you said does the trick!
For the first triangle, take some measures, like as you said: an angle (or its cosine - easy to calculate with a dot product) on any vertex and the lengths of the sides next to it.
For another triangle, use if-conditions to see if the angle (or its cosine) is the same, and if the ratios of the lengths are also the same. You'd have to do this check from all 3 vertices in this way (if at least one fits, then the triangles are similar).
A faster way would be to always start with (for instnace) the vertex with the smallest angle, then you'd need to only compare once.
Now go code it! :-)
You are given coordinates of all three points of each triangle. Let us consider two triangles T1 A(a1,a2) B(b1,b2) C(c1, c2), T2 P(p1,p2) Q(q1,q2) R(r1,r2).
a = length of opposite side of vertex A
b = length of opposite side of vertex B
c = length of opposite side of vertex C
similarly p,q,r of triangle T2
So, for the two triangles to be similar, it has to follow the following conditions
1. AB = PQ; BC = QR; CA = RP
(We don't need their directions, So I am considering only magnitudes)
2. angle (A) = angle(B) i.e angle(BAC) = angle(QPR);
angle(B) = angle(Q) i.e angle(CBA) = angle (RQP) and
angle(C) = angle(R).
Now, you got to use coordinate geometry/ spherical geometry here.
COS (A) = ( b^2 + c^2 - a^2 )/2bc
COS (B) = ( c^2 + a^2 - b^2 )/2ac
COS (C) = (a^2 + b^2 - c^2)/2ab
Note:: As cosine is periodic with 2*pi, please make sure that you have exact angle. So, why don't you think of using inverse cosine functions where you get principle angles.(I am not sure of them, as how they work. please do check)
(Similarly for P,Q,R of triangle T2).
Actually there is another rule by which its easy to do.
law: a/sin(A) = b/sin(B) = c/sin(C).
I think you have to go through Spherical Geometry
I hope this helps you to do the program.
How to do the program:
Actually, its fine if you want to use structures. Create a structure with fields of 3 sides and 3 angles. Thus you need to take two variables under structure type and compare those quantities mentioned above.
If they satisfy, they are similar triangles.
I hope this helps you.