I have the following question on a practice exam:
I need to use MATLAB to solve it. The problem is, I have not seen a problem like this before and I'm struggling to get started.
I have my 1x1 grid, split into 10x10. I know I can calculate the whole bottom row besides the corners using 1/10 * x*2. I also know I can calculate the entire right row using (1/10)(1+t)^2. However, I cannot figure out how to get enough points to be able to fill in the values for the entire grid. I know it must have something to do with the partial derivatives given in the problem, but I'm not quite sure where they come into play (especially the u_x equation). Can someone help me get a start here?
I don't need the whole solution. Once I have enough points I can easily write a matlab program to solve the rest. Really, I think I just need the x=0 axis solved, then I just fill in the middle of the grid.
I have calculated the bottom row, minus the two corners, to be 0.001, 0.004, 0.009, 0.016, 0.025, 0.036, 0.049, 0.064, 0.081. And similarly, the entire right row is trival to calculate using the given boundry condition. I just can't piece together where to go from there.
Edit: the third boundry condition equation was mistyped. it should read:
u_x(0,t) = 1/5t, NOT u(0,t) = 1/5t
First realise that the equation you have to solve is the linear wave equation, and the numerical scheme you are given can be rewritten as
( u^(n+1)_m - 2u^n_m + u^(n-1)_m )/k^2 = ( u^n_(m-1) - 2u^n_m + u^n_(m+1) )/h^2
where k is the time step and h is the delta x in space.
The reformulated numerical scheme makes clear that the left- and right-hand sides are the second order centred finite difference approximations of u_tt and u_xx respectively.
To solve the problem numerically, however, you need to use the form given to you because it is the explicit update formula that you need to implement numerically: it gives you the solution at time n+1 as a function of the previous two times n and n-1. You need to start from the initial condition and march the solution in time.
Observe that the solution is assigned on the boundaries of the domain (x=0 and x=1), so the values of the discretized solution u^(n)_0 and u^(n)_10 are known for any n (t=n*k). At the nth time step your unknown is the vector [u^(n+1)_1, u^(n+1)_2, ..., u^(n+1)_9].
Observe also that to use the update formula to find the solution at the n+1 step, requires the knowledge of the solution at two previous steps. So, how do you start from n=0 if you need information from two previous times? This is where the initial conditions come into play.
You have the solution at n=0 (t=0), but you also have u_t at t=0. These two pieces of information combined can give you both u^0 and u^1 and get you started.
I would use the following start-up scheme:
u^0_m = u(h*m,0) // initial condition on u
(u^2_m - u^0_m)/(2k) = u_t(h*m,0) // initial condition on u_t
that combined with the numerical scheme used with n=1 gives you everything you need to define a linear system for both u^1_m and u^2_m for m=1,...,9.
To summarize:
--use the start-up scheme to find solution at n=1 and n=2 simultaneously.
--from there on march in time using the numerical scheme you are given.
If you are completely lost check out things like: finite difference schemes, finite difference schemes for advection equations, finite difference schemes for hyperbolic equations, time marching.
EDITING:
For the boundary condition on u_x you typically use the ghost cell method:
Introduce a ghost cell at m=-1, i.e. a fictitious (or auxiliary) grid point that is used to deal with boundary condition, but that is not part of the solution.
The first node m=0 is back into your unknown vector, i.e. you are now working with [u_0 u_1 ... u_9].
Use the left side boundary condition to close the system.
Specifically, by writing down the centered approx of the boundary condition
u^n_(1) - u^n_(-1) = 2*h*u_x(0,k*n)
The above equation allows you to express the solution on the ghost node in terms on the solution on an internal, real node. Therefore you can apply the time-marching numerical scheme (the one you are given) to the m=0 node. (The numerical scheme applied to m=0 would contain contributions from the m=-1 ghost node, but now you have that expressed in terms of the m=1 node.)
Related
I'm coming from biology's field and thus I have some difficulties in understanding (intuitively?) some of the ideas of that paper. I really tried my best to decipher it step by step by using a lot of google and youtube, but now I feel, it's the time to refer to the professionals in that field.
Before filling out the whole universe with (unordered) questions, let me put the whole thing down and try to introduce you to the subject while at the same time explain to you what I got so far from my research on that.
Microarrays
For those that do not have any idea of what this is, you can imagine, that it is literally an array (matrix) where each cell of it contains a probe for a specific gene. Making the long story short, by the end of the microarray experiment, you have a matrix (in computational terms) with each column representing a sample, each line a different gene while the contents of the matrix represent the expression values of the genes for each sample.
Pathways
In biology pathway / gene-set they call a set of genes that interact with each other forming a small network responsible for a specific function.These pathways are not isolated but they talk/interact with each other too. What that paper does on the first hand, is to expand the initial pathway (let us call it target pathway), by including some other genes from other pathways that might interact with that.
Procedure
1.
Let's assume now that we have a matrix G x S. Where G for genes and S for Samples. We construct a gene co-expression network (G x G) using as weights the Pearson's correlation coefficients between genes' pairs (a). This could also be represented as an undirected weighted graph. .
2.
For each gene (row OR column) we calculate the weighted degree (d) which is nothing more than the sum of all correlation coefficients of that gene.
3.
From the two previous matrices, they construct the transition matrix producing the probabilities (P) to transit from one gene to another by using the
formula
Q1. Why do they call this transition probability? Is there any intuitive way to see this as a probability in the biological context?
4.
Since we have the whole transition matrix, we can define a subnetwork of the initial one, that we want to expand it and it consisted out of let's say 15 genes. In that step, they used formula number 3 (on the paper) which transforms the values of the initial transition matrix as it says. They set the probability of 1 on the nodes that are part of the selected subnetwork because they define them as absorbing states.
Q2. In that same formula (3), I cannot understand what the second condition does. When should the probability be 0? Intuitively, in my opinion, all nodes that didn't exist in subnetwork, should have the P_ij value as a probability.
5.
After that, the newly constructed transition matrix is showed at formula (4) in the paper and I managed to understand it using this excellent article.
6.
Here is where everything is getting more blur for me and where I need the most of the help. What I imagine at that step, is that the algorithm starts randomly from one node and keep walking around the network. In order to construct a relevance function (What that exactly means?), they firstly calculate a probability called joint probability of visiting one node/edge E(i,j) and noted as :
From the other hand they seem to calculate another probability called probability of a walk of length L starting in x and denoted as :
7.
In the next step, they divide the previously calculated probabilities and calculate the number of times a random walk starts in x using the transition from i to j that I don't really understand what this means.
After that step, I lost their reasoning at all :-P.
I'm not expecting an expert to come open my mind and give me understand that procedure. What I'm expecting is some guidelines, hints, ideas, useful resources or more intuitive approaches to understanding the whole procedure. Then when I fully understand it I will try to implement it on R or python.
So any idea / critics is welcome.
Thanks.
I am attempting to create a puzzle to let players piece together an n×n grid using n connected n-minos (definition: connected piece of n 1×1 blocks, e.g. each of the Tetris pieces is a 4-mino). However, generating a way to cut the grid first proves to be a challenge despite seemingly easy enough for a human.
example board
For human, generating such a solution is a relative easy task by recursively following the following logic/pseudo-code:
:start_of_recursion:
Start with a random "least connected" piece (end, corner, edge pieces that has the fewest member connecting to it) to be the starting mino block
:start_of_recursion:
Make a "grow" in a random available direction from a random piece in the current mino
If "grow" results in a "separated" remaining board(, if the separated region isn't a multiple of n), try some other location and direction
if all location and direction has been attempted, revert to previous board configuration (shouldn't really occur?)
If size-n has been reached, exit recursion
:end_of_recursion:
if board has been filled, exit recursion
:end_of_recursion:
Performing this routine seem to generate an O(n^2) method of solution generation, however the condition checks prove to be really expensive for computers. In order to determine whether the board to be connected, a human simply checks for any "gap" inside the remaining region, and is processed in almost O(1) fashion for a simple non-overlapping graph, whereas my code implementation need to "spread" from a point on the graph into its neighboring territories and check after the spreading is complete to check whether if any points lies outside of reach (O(n) at best). Since this check is to be performed every time in the innermost iteration, it degenerates the complexity into an O(n^(3+)) problem and becomes really inefficient.
Is there a method to check for "gap" in a manner similar to that of human cognition? Or can the problem be fundamentally thought of and simplified into a problem easier for computer to solve?
Your problem sounds like a variant of bin packing problem. I would approach this by constraint satisfaction method. Below I'll use Minizinc pseudo-code.
A board consists of cells, each cell could be colored into one color from several. We can represent it as follows:
int: rows;
int: cols;
int: colors_num;
array [1...colors_num] of int: colors;
array [1..rows,1..cols] of var colors: board;
Next, we add constraints. For example, if a cell has color A then atleast 1 adjacent cell must have the same color A:
constraint forall (c in colors) (
if board[i, j] == c then
at_least (1, [board[i-1, j], board[i+1, j], board[i, j-1], board[i,j+1]], c)
else
true
endif
You can describe all allowed/prohibited shapes as constraints or use some other smart approach introducing possible cuts.
Constraint satisfaction should be much more efficient than your recursive approach. However, it's not very scalable - if you'll try to generate a game for a gigantic board (hundreds or thousands of cells/colors), it will take quite some time and memory to generate minos.
I need to find the value of a variable s such that the taylor expansion of an expression involving s:
Has a minimum (preferably zero, but due to binary minimum is sufficient) in as many coefficients other than 0th order as possible (preferably more than that one minimum coefficient, but 2nd and 3rd have priority).
reports the best n values of s that fulfill the condition within the region (ie show me the 3 best values of s and what the coefficients look like for each).
I have no idea how to even get the output of a Series[] command into any other mathematica command without receiving an error, much less how to actually solve the problem. The equation I am working with is too complex to post here (multi-regional but continuous polynomial expression that can be expanded). Does anyone know what commands to use for this?
The first thing you should realize is that the output of Series is not a sum but a a SeriesData object. To convert it into a sum you have to wrap it in Normal[Series[...]]. Since the question doesn't provide details, I can't say more.
TrialPay posted a programming question about a snake cube puzzle on their blog.
Recently, one of our engineers introduced us to the snake cube. A snake cube is a puzzle composed of a chain of cubelets, connected by an elastic band running through each cubelet. Each cubelet can rotate 360° about the elastic band, allowing various structures to be built depending upon the way in which the chain is initially constructed, with the ultimate goal of arranging the cubes in such a way to create a cube.
Example:
This particular arrangement contains 17 groups of cubelets, composed of 8 groups of two cubelets and 9 groups of three cubelets. This arrangement can be expressed in a variety of ways, but for the purposes of this exercise, let '0' denote pieces whose rotation does not change the orientation of the puzzle, or may be considered a "straight" piece, while '1' will denote pieces whose rotation changes the puzzle configuration, or "bend" the snake. Using that schema, the snake puzzle above could be described as 001110110111010111101010100.
Challenge:
Your challenge is to write a program, in any language of your choosing, that takes the cube dimensions (X, Y, Z) and a binary string as input, and outputs '1' (without quotes) if it is possible to solve the puzzle, i.e. construct a proper XYZ cube given the cubelet orientation, and '0' if the current arrangement cannot be solved.
I posted a semi-detailed explanation of the solution, but how do I determine if the program solves the problem? I thought about getting more test cases, but I ran into some problems:
The snake cube example from TrialPay's blog has the same combination as the picture on Wikipedia's Snake Cube page and www.mathematische-basteleien.de.
It's very tedious to manually convert an image into a string.
I tried to make a program that would churn out a lot of combinations:
#We should start at the binary representation of 16777216 (00100...), because
#lesser numbers have more than 2 consecutive 0s (000111...)
i = 16777216
solved = []
while i <= 2**27:
s = str(bin(i))[2:]
#Add 0s
if len(s) < 27:
s = '0'*(27-len(s)) + s
#Check if there are more than 2 consecutive 0s
print s
if s.find("000") != -1:
if snake_cube_solution(3, 3, 3, s) == 1:
solved.append(s)
i += 1
But it just takes forever to finish executing. Is there a better way to verify the program?
Thanks in advance!
TL;DR: This isn't a programming problem, but a mathematical one. You may be better served at math.stackexchange.com.
Since the cube size and snake length are passed as input, the space of inputs a checker program would need to verify is essentially infinite. Even though checking the solutions's answer for a single input is reasonable, brute forcing this check across the entire input space is clearly not.
If your solution fails on certain cases, your checker program can help you find these. However it can't establish your program's correctness: if your solution is actually correct the checker will simply run forever and leave you wondering.
Unfortunately (or not, depending on your tastes), what you are looking for is not a program but a mathematical proof.
(Proving) Algorithm correctness is itself an entire field of study, and you can spend a long time in it. That said, proof by induction is often applicable (especially for recursive algorithms.)
Other times, navigating between state configurations can be restated as optimizing a utility function. Proving things about the space being optimized (such as it has only one extrema) can then translate to a proof of program correctness.
Your state configurations in this second approach could be snake orientations, or they might be some deeper structure. For example, the general strategy underneath solving a Rubik's cube
isn't usually stated on literal cube states, but on expressions of a group of relevant symmetries. This is what I personally expect your solution will eventually play out as.
EDIT: Years later, I feel I should point out that for a given, fixed cube size and snake length, of course the search space is actually finite. You can write a program to brute-force check all combinations. If you were clever, you could even argue that the times to check a set of cases can be treated as a set of independent random variables. From this you could build a reasonable progress bar to estimate how (very) long your wait would be.
I think your assertion that there can not be three consecutive 0's is false. Consider this arrangement:
000
100
101
100
100
101
100
100
100
One of the problems I'm having with this puzzle is the notation. A 1 indicates that the cubelet can change the puzzle's orientation, but about which axis? In my example above, assume that the Y axis is vertical and the X axis is horizontal. A 1 on the left indicates the ability to rotate about the cubelet's Y axis, and a 1 on the right indicates the ability to rotate about the cubelet's X axis.
I think it's possible to construct an arrangement similar to that above, but with three 000 groups. But I don't have the notation for it. Clearly, the example above could be modified so that the first three lines are:
001
000
101
With the first segment's 1 indicating rotation about the Y axis.
I wrote a Java application for the same problem not long ago.
I used the backtracking algorithm for this.
You just have to do an recursive search through the whole cube checking what directions are possible. If you have found one, you can stop and print the solution (I chose to print out all solutions).
For the 3x3x3 cubes my program solved them in under a second, for the bigger ones it takes about five seconds up to 15 minutes.
I'm sorry I couldn't find any code right now.
I need to generate n random points in general position in the plane, i.e. no three points can lie on a same line. Points should have coordinates that are integers and lie inside a fixed square m x m. What would be the best algorithm to solve such a problem?
Update: square is aligned with the axes.
Since they're integers within a square, treat them as points in a bitmap. When you add a point after the first, use Bresenham's algorithm to paint all pixels on each of the lines going through the new point and one of the old ones. When you need to add a new point, get a random location and check if it's clear; otherwise, try again. Since each pair of pixels gives a new line, and thus excludes up to m-2 other pixels, as the number of points grows you will have several random choices rejected before you find a good one. The advantage of the approach I'm suggesting is that you only pay the cost of going through all lines when you have a good choice, while rejecting a bad one is a very quick test.
(if you want to use a different definition of line, just replace Bresenham's with the appropriate algorithm)
Can't see any way around checking each point as you add it, either by (a) running through all of the possible lines it could be on, or (b) eliminating conflicting points as you go along to reduce the possible locations for the next point. Of the two, (b) seems like it could give you better performance.
Similar to #LaC's answer. If memory is not a problem, you could do it like this:
Add all points on the plane to a list (L).
Shuffle the list.
For each point (P) in the list,
For each point (Q) previously picked,
Remove every point from L which are linear to P-Q.
Add P to the picked list.
You could continue the outer loop until you have enough points, or run out of them.
This might just work (though might be a little constrained on being random). Find the largest circle you can draw within the square (this seems very doable). Pick any n points on the circle, no three will ever be collinear :-).
This should be an easy enough task in code. Say the circle is centered at origin (so something of the form x^2 + y^2 = r^2). Assuming r is fixed and x randomly generated, you can solve to find y coordinates. This gives you two points on the circle for every x which are diametrically opposite. Hope this helps.
Edit: Oh, integer points, just noticed that. Thats a pity. I'm going to keep this solution up though - since I like the idea
Both #LaC's and #MizardX's solution are very interesting, but you can combine them to get even better solution.
The problem with #LaC's solution is that you get random choices rejected. The more points you have already generated the harder it gets to generate new ones. If there is only one available position left you have slight chance of randomly choosing it (1/(n*m)).
In the #MizardX's solution you never get rejected choices, however if you directly implement the "Remove every point from L which are linear to P-Q." step you'll get worse complexity (O(n^5)).
Instead it would be better to use a bitmap to find which points from L are to be removed. The bitmap would contain a value indicating whether a point is free to use and what is its location on the L list or a value indicating that this point is already crossed out. This way you get worst-case complexity of O(n^4) which is probably optimal.
EDIT:
I've just found that question: Generate Non-Degenerate Point Set in 2D - C++
It's very similar to this one. It would be good to use solution from this answer Generate Non-Degenerate Point Set in 2D - C++. Modifying it a bit to use radix or bucket sort and adding all the n^2 possible points to the P set initially and shufflying it, one can also get worst-case complexity of O(n^4) with a much simpler code. Moreover, if space is a problem and #LaC's solution is not feasible due to space requirements, then this algorithm will just fit in without modifications and offer a decent complexity.
Here is a paper that can maybe solve your problem:
"POINT-SETS IN GENERAL POSITION WITH MANY
SIMILAR COPIES OF A PATTERN"
by BERNARDO M. ABREGO AND SILVIA FERNANDEZ-MERCHANT
um, you don't specify which plane.. but just generate 3 random numbers and assign to x,y, and z
if 'the plane' is arbitrary, then set z=o every time or something...
do a check on x and y to see if they are in your m boundary,
compare the third x,y pair to see if it is on the same line as the first two... if it is, then regenerate the random values.