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Project Euler – Problem 264

July 28, 2017 Leave a comment

Today I managed to solve problem 264 from Project Euler. This is my highest rating problem until now: 85%. You can click the link for the full text of the problem. The main idea is to find all triangles ABC with vertices having integer coordinates such that

  • the circumcenter O of each of the triangles is the origin
  • the orthocenter H (the intersection of the heights) is the point of coordinates (0,5)
  • the perimeter is lower than a certain bound

I will not give detailed advice or codes. You can already find a program online for this problem (I won’t tell you where) and it can serve to verify the final code, before going for the final result. Anyway, following the hints below may help you get to a solution.

The initial idea has to do with a geometric relation linking the points A, B, C, O and H. Anyone who did some problems with vectors and triangles should have come across the needed relation at some time. If not, just search for important lines in triangles, especially the line passing through O and H (and another important point).

Once you find this vectorial relation, it is possible to translate it in coordinates. The fact that points A, B, C are on a circle centered in O shows that their coordinates satisfy an equation of the form x^2+y^2=n, where n is a positive integer, not necessarily a square… It is possible to enumerate all solutions to the following equation for fixed n, simply by looping over x and y. This helps you find all lattice points on the circle of radius \sqrt{n}.

Once these lattice points are found one needs to check the orthocenter condition. The relations are pretty simple and in the end we have two conditions to check for the sum of the x and y coordinates. The testing procedure is a triple loop. We initially have a list of points on a circle, from the previous step. We loop over them such that we dont count triangles twice: i from 1 to m, j from i+1 to m, k from j+1 to m, etc. Once a suitable solution is found, we compute the perimeter using the classical distance formula between two points given in coordinates. Once the perimeter is computed we add it to the total.

Since the triple loop has cubic complexity, one could turn it in a double loop. Loop over pairs and construct the third point using the orthocenter condition. Then just check if the point is also on the circle. I didn’t manage to make this double loop without overcounting things, so I use it as a test: use double loops to check every family of points on a given circle. If you find something then use a triple loop to count it properly. It turns out that cases where the triple loop is needed are quite rare.

So now you have the ingredients to check if on a circle of given radius there are triangles with the desired properties. Now we just iterate over the square of the radius. The problem is to find the proper upper bound for this radius in order to get all the triangles with perimeter below the bound. It turns out that a simple observation can get you close to a near optimal bound. Since in the end the radii get really large and the size of the triangles gets really large, the segment OH becomes small, being of fixed length 5. When OH is very small, the triangle is almost equilateral. Just use the upper bound for the radius for an equilateral triangle of perimeter equal to the upper bound of 100000 given in the problem.

Using these ideas you can build a bruteforce algorithm. Plotting the values of the radii which give valid triangles will help you find that you only need to loop over a small part of the radii values. Factoring these values will help you reduce even more the search space. I managed to  solve the problem in about 5 hours in Pari GP. This means things could be improved. However, having an algorithm which can give the result in “reasonable” time is fine by me.

I hope this will help you get towards the result.

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Project Euler 607

June 11, 2017 2 comments

If you like solving Project Euler problems you should try Problem number 607. It’s not very hard, as it can be reduced to a small optimization problem. The idea is to find a path which minimizes time, knowing that certain regions correspond to different speeds. A precise statement of the result can be found on the official page. Here’s an image of the path which realizes the shortest time:

euler607

 

Project Euler tips

March 28, 2017 Leave a comment

A few years back I started working on Project Euler problems mainly because it was fun from a mathematical point of view, but also to improve my programming skills. After solving about 120 problems I seem to have hit a wall, because the numbers involved in some of the problems were just too big for my simple brute-force algorithms.

Recently, I decided to try and see if I can do some more of these problems. I cannot say that I’ve acquired some new techniques between 2012-2016 concerning the mathematics involved in these problems. My research topics are usually quite different and my day to day programming routines are more about constructing new stuff which works fast enough than optimizing actual code. Nevertheless, I have some experience coding in Matlab, and I realized that nested loops are to be avoided. Vectorizing the code can speed up things 100 fold.

So the point of Project Euler tasks is making things go well for large numbers. Normally all problems are tested and should run within a minute on a regular machine. This brings us to choosing the right algorithms, the right simplifications and finding the complexity of the algorithms involved.

Read more…

Linear programming #1 – Project Euler 185

January 28, 2017 2 comments

I was recently faced with a very nice challenge from Project Euler – Problem 185. The idea is to find a number with a fixed number of digits by making guesses. Each time you are told how many digits you got right. By got right, I mean guessing the right digit on the right position. If you find a digit which is in the number, but on a different position, you don’t know…

There is a test case with 5 digits:

90342 ;2 correct
70794 ;0 correct
39458 ;2 correct
34109 ;1 correct
51545 ;2 correct
12531 ;1 correct

with unique solution 39542 and the difficult case with 16 digits

5616185650518293 ;2 correct
3847439647293047 ;1 correct
5855462940810587 ;3 correct
9742855507068353 ;3 correct
4296849643607543 ;3 correct
3174248439465858 ;1 correct
4513559094146117 ;2 correct
7890971548908067 ;3 correct
8157356344118483 ;1 correct
2615250744386899 ;2 correct
8690095851526254 ;3 correct
6375711915077050 ;1 correct
6913859173121360 ;1 correct
6442889055042768 ;2 correct
2321386104303845 ;0 correct
2326509471271448 ;2 correct
5251583379644322 ;2 correct
1748270476758276 ;3 correct
4895722652190306 ;1 correct
3041631117224635 ;3 correct
1841236454324589 ;3 correct
2659862637316867 ;2 correct

While the small case could be tackled using a pure brute force approach (only 10^5) cases to check, the second case becomes intractable. Looping through all 10^{16} cases would take ages. One could try genetic algorithms or other algorithms based on randomness. Recursive approaches are also possible.

There is however a way to express this problem which makes it solvable right away using a linear programming solver: that is an algorithm which finds valid solutions given some constraints. How do we find these constraints? There are essentially two types of constraints in our case:

  • each of the positions of the result contains one digit from 0 to 9
  • in each of the guesses given above we have a fixed number of correct digits

Read more…

About Problem 587 – Project Euler

January 27, 2017 Leave a comment

I started looking again at some problems on the Project Euler site. It’s not often that I manage to solve a recent problem. Below are a few hints about the math needed to provide an answer to problem 587.

I’ll not repeat the statement of the problem here, so please read it by following the above link. The idea is to find the ratio of the areas of some triangular shapes made of two segments and one circle arc. The breakdown of the problem is the following:

  • Note that the size of the square does not matter, since we need the result as a ratio, so we can consider that the square has side equal to 1. First note that the L-shape in the corner of the triangle has area equal to 1/8-\pi/32. We’ll need this in the end, to compute the ratio.
  • Secondly, the shape whose area we need to compute can be decomposed into a right-angled triangle and a complement of a part of the circle. In order to have precise information about these shapes we compute the intersection of the line y=x/n and the circle (x-0.5)^2+(y-0.5)^2=0.25. It is possible to find the value explicitly.
  • Once we have the coordinates of the point of intersection we can compute the area of the triangle. The area of the rounded part which remains is the difference between a rectangle and the circle portion PQON. In order to do find this difference it suffices to find the angle PSQ.
  • Once the mathematical part is done, the programming job is not that tough, since the numbers involved are not that high. I managed to write a Matlab solution in no time.fig587

Identifying edges and boundary points – 2D Mesh – Matlab

April 21, 2015 1 comment

A triangulation algorithm often gives as output a list of points, and a list of triangle. Each triangle is represented by the indexes of the points which form it. Sometimes we need extra information on the structure of the triangulation, like the list of edges, or the list of boundary points. I will present below two fast algorithms for doing this.

Finding the list of edges is not hard. The idea is to go through each triangle, and extract all edges. The algorithm proposed below creates the adjacency matrix in the following way: for each triangle {[i,j,k]} we set the elements {a_{ij},a_{jk},a_{ik}} (and their symmetric correspondents) to be equal to {1}.

In order to find the points on the boundary (in two dimensions), it is enough to look for the edges which are sides to only one triangle. We can do this using the adjacency matrix. Note that if {A} is the adjacency matrix, then {A^2=(b_{ik})} stores the number of paths of length {2} (two sides) between two points of the triangulation. Note that any edge which is not on the boundary will contain the starting and ending point of two paths of length {2}. If {[i,j,k]} is a triangle such that points {i,j} are on the boundary, then {b_{i,j}=1} (there is one path of length {2} going through {i,k,j}. We also have {a_{i,j} = 1}. Conversely, if {a_{i,j} = 1} and {b_{i,j} = 1} then {i,j} are connected, and there is a unique path of length {2} going from {i} to {j}. Thus, {i,j} is an edge on the boundary. Therefore, we just need to identify the indexes {i} such that there exists {j} with {a_{i,j} b_{i,j} = 1}.

Below are two short Matlab codes doing these two algorithms. I guess they are close to being optimal, since only sparse and vectorized operations are used.


%p is the list of points
%T is the list of triangles, ap is the number of points
%this computes the adjacency matrix
A = min(sparse(T(:,1),T(:,2),1,ap,ap)+sparse(T(:,2),T(:,3),1,ap,ap)+sparse(T(:,3),T(:,1),1,ap,ap),1);
A = min(A+A',1);
% this finds the boundary points, whose indexes are stored in Ibord
B = A^2.*A==1;
Ibord = find(sum(B,2)>0);
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