Hauptinhalt
Results for
Twitch built an entire business around letting you watch over someone's shoulder while they play video games. I feel like we should be able to make at least a few videos where we get to watch over someone's shoulder while they solve Cody problems. I would pay good money for a front-row seat to watch some of my favorite solvers at work. Like, I want to know, did Alfonso Nieto-Castonon just sit down and bang out some of those answers, or did he have to think about it for a while? What was he thinking about while he solved it? What resources was he drawing on? There's nothing like watching a master craftsman at work.
I can imagine a whole category of Cody videos called "How I Solved It". I tried making one of these myself a while back, but as far as I could tell, nobody else made one.
Here's the direct link to the video: https://www.youtube.com/watch?v=hoSmO1XklAQ
I hereby challenge you to make a "How I Solved It" video and post it here. If you make one, I'll make another one.
The Ans Hack is a dubious way to shave a few points off your solution score. Instead of a standard answer like this
function y = times_two(x)
y = 2*x;
end
you would do this
function ans = times_two(x)
2*x;
end
The ans variable is automatically created when there is no left-hand side to an evaluated expression. But it makes for an ugly function. I don't think anyone actually defends it as a good practice. The question I would ask is: is it so offensive that it should be specifically disallowed by the rules? Or is it just one of many little hacks that you see in Cody, inelegant but tolerable in the context of the surrounding game?
Incidentally, I wrote about the Ans Hack long ago on the Community Blog. Dealing with user-unfriendly code is also one of the reasons we created the Head-to-Head voting feature. Some techniques are good for your score, and some are good for your code readability. You get to decide with you care about.
The study of the dynamics of the discrete Klein - Gordon equation (DKG) with friction is given by the equation :
In the above equation, W describes the potential function:
to which every coupled unit 
adheres. In Eq. (1), the variable $
$ is the unknown displacement of the oscillator occupying the n-th position of the lattice, and 
is the discretization parameter. We denote by h the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient 
, while
is the coefficient of the nonlinear cubic term.
adheres. In Eq. (1), the variable $$ is the unknown displacement of the oscillator occupying the n-th position of the lattice, and is the discretization parameter. We denote by h the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient , whileis the coefficient of the nonlinear cubic term.For the DKG chain (1), we will consider the problem of initial-boundary values, with initial conditions
and Dirichlet boundary conditions at the boundary points 
and 
, that is,
and , that is,
Therefore, when necessary, we will use the short notation 
for the one-dimensional discrete Laplacian
for the one-dimensional discrete Laplacian
Now we want to investigate numerically the dynamics of the system (1)-(2)-(3). Our first aim is to conduct a numerical study of the property of Dynamic Stability of the system, which directly depends on the existence and linear stability of the branches of equilibrium points.
For the discussion of numerical results, it is also important to emphasize the role of the parameter 
. By changing the time variable 
, we rewrite Eq. (1) in the form
. By changing the time variable , we rewrite Eq. (1) in the form
. We consider spatially extended initial conditions of the form:
where 
is the distance of the grid and 
is the amplitude of the initial condition We also assume zero initial velocity:
the following graphs for 
and 
and % Parameters
L = 200; % Length of the system
K = 99; % Number of spatial points
j = 2; % Mode number
omega_d = 1; % Characteristic frequency
beta = 1; % Nonlinearity parameter
delta = 0.05; % Damping coefficient
% Spatial grid
h = L / (K + 1);
n = linspace(-L/2, L/2, K+2); % Spatial points
N = length(n);
omegaDScaled = h * omega_d;
deltaScaled = h * delta;
% Time parameters
dt = 1; % Time step
tmax = 3000; % Maximum time
tspan = 0:dt:tmax; % Time vector
% Values of amplitude 'a' to iterate over
a_values = [2, 1.95, 1.9, 1.85, 1.82]; % Modify this array as needed
% Differential equation solver function
function dYdt = odefun(~, Y, N, h, omegaDScaled, deltaScaled, beta)
U = Y(1:N);
Udot = Y(N+1:end);
Uddot = zeros(size(U));
% Laplacian (discrete second derivative)
for k = 2:N-1
Uddot(k) = (U(k+1) - 2 * U(k) + U(k-1)) ;
end
% System of equations
dUdt = Udot;
dUdotdt = Uddot - deltaScaled * Udot + omegaDScaled^2 * (U - beta * U.^3);
% Pack derivatives
dYdt = [dUdt; dUdotdt];
end
% Create a figure for subplots
figure;
% Initial plot
a_init = 2; % Example initial amplitude for the initial condition plot
U0_init = a_init * sin((j * pi * h * n) / L); % Initial displacement
U0_init(1) = 0; % Boundary condition at n = 0
U0_init(end) = 0; % Boundary condition at n = K+1
subplot(3, 2, 1);
plot(n, U0_init, 'r.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title('$t=0$', 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-3 3]);
grid on;
% Loop through each value of 'a' and generate the plot
for i = 1:length(a_values)
a = a_values(i);
% Initial conditions
U0 = a * sin((j * pi * h * n) / L); % Initial displacement
U0(1) = 0; % Boundary condition at n = 0
U0(end) = 0; % Boundary condition at n = K+1
Udot0 = zeros(size(U0)); % Initial velocity
% Pack initial conditions
Y0 = [U0, Udot0];
% Solve ODE
opts = odeset('RelTol', 1e-5, 'AbsTol', 1e-6);
[t, Y] = ode45(@(t, Y) odefun(t, Y, N, h, omegaDScaled, deltaScaled, beta), tspan, Y0, opts);
% Extract solutions
U = Y(:, 1:N);
Udot = Y(:, N+1:end);
% Plot final displacement profile
subplot(3, 2, i+1);
plot(n, U(end,:), 'b.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title(['$t=3000$, $a=', num2str(a), '$'], 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-2 2]);
grid on;
end
% Adjust layout
set(gcf, 'Position', [100, 100, 1200, 900]); % Adjust figure size as needed
Dynamics for the initial condition , 
, for 
, for different amplitude values. By reducing the amplitude values, we observe the convergence to equilibrium points of different branches from 
and the appearance of values 
for which the solution converges to a non-linear equilibrium point 
Parameters: 
, for , for different amplitude values. By reducing the amplitude values, we observe the convergence to equilibrium points of different branches from and the appearance of values for which the solution converges to a non-linear equilibrium point Parameters: 
Detection of a stability threshold 
: For 
, the initial condition , 
, converges to a non-linear equilibrium point
.
: For , the initial condition , , converges to a non-linear equilibrium point.Characteristics for 
, with corresponding norm 
where the dynamics appear in the first image of the third row, we observe convergence to a non-linear equilibrium point of branch 
This has the same norm and the same energy as the previous case but the final state has a completely different profile. This result suggests secondary bifurcations have occurred in branch 
, with corresponding norm where the dynamics appear in the first image of the third row, we observe convergence to a non-linear equilibrium point of branch This has the same norm and the same energy as the previous case but the final state has a completely different profile. This result suggests secondary bifurcations have occurred in branch By further reducing the amplitude, distinct values of 
are discerned: 1.9, 1.85, 1.81 for which the initial condition 
with norms 
respectively, converges to a non-linear equilibrium point of branch 
This equilibrium point has norm 
and energy 
. The behavior of this equilibrium is illustrated in the third row and in the first image of the third row of Figure 1, and also in the first image of the third row of Figure 2. For all the values between the aforementioned a, the initial condition 
converges to geometrically different non-linear states of branch 
as shown in the second image of the first row and the first image of the second row of Figure 2, for amplitudes 
and 
respectively.
are discerned: 1.9, 1.85, 1.81 for which the initial condition with norms respectively, converges to a non-linear equilibrium point of branch This equilibrium point has norm and energy . The behavior of this equilibrium is illustrated in the third row and in the first image of the third row of Figure 1, and also in the first image of the third row of Figure 2. For all the values between the aforementioned a, the initial condition converges to geometrically different non-linear states of branch as shown in the second image of the first row and the first image of the second row of Figure 2, for amplitudes and respectively. Refference:
There are a host of problems on Cody that require manipulation of the digits of a number. Examples include summing the digits of a number, separating the number into its powers, and adding very large numbers together.
If you haven't come across this trick yet, you might want to write it down (or save it electronically):
digits = num2str(4207) - '0'
That code results in the following:
digits =
4 2 0 7
Now, summing the digits of the number is easy:
sum(digits)
ans =
13
Hello and a warm welcome to everyone! We're excited to have you in the Cody Discussion Channel. To ensure the best possible experience for everyone, it's important to understand the types of content that are most suitable for this channel.
Content that belongs in the Cody Discussion Channel:
- Tips & tricks: Discuss strategies for solving Cody problems that you've found effective.
- Ideas or suggestions for improvement: Have thoughts on how to make Cody better? We'd love to hear them.
- Issues: Encountering difficulties or bugs with Cody? Let us know so we can address them.
- Requests for guidance: Stuck on a Cody problem? Ask for advice or hints, but make sure to show your efforts in attempting to solve the problem first.
- General discussions: Anything else related to Cody that doesn't fit into the above categories.
Content that does not belong in the Cody Discussion Channel:
- Comments on specific Cody problems: Examples include unclear problem descriptions or incorrect testing suites.
- Comments on specific Cody solutions: For example, you find a solution creative or helpful.
Please direct such comments to the Comments section on the problem or solution page itself.
We hope the Cody discussion channel becomes a vibrant space for sharing expertise, learning new skills, and connecting with others.
Spring is here in Natick and the tulips are blooming! While tulips appear only briefly here in Massachusetts, they provide a lot of bright and diverse colors and shapes. To celebrate this cheerful flower, here's some code to create your own tulip!
Check out this episode about PIVLab: https://www.buzzsprout.com/2107763/15106425
Join the conversation with William Thielicke, the developer of PIVlab, as he shares insights into the world of particle image velocimetery (PIV) and its applications. Discover how PIV accurately measures fluid velocities, non invasively revolutionising research across the industries. Delve into the development journey of PI lab, including collaborations, key features and future advancements for aerodynamic studies, explore the advanced hardware setups camera technologies, and educational prospects offered by PIVlab, for enhanced fluid velocity measurements. If you are interested in the hardware he speaks of check out the company: Optolution.
One of the starter prompts is about rolling two six-sided dice and plot the results. As a hobby, I create my own board games. I was able to use the dice rolling prompt to show how a simple roll and move game would work. That was a great surprise!
Let's talk about probability theory in Matlab.
Conditions of the problem - how many more letters do I need to write to the sales department to get an answer?
To get closer to the problem, I need to buy a license under a contract. Maybe sometimes there are responsible employees sitting here who will give me an answer.
Thank you
In the MATLAB description of the algorithm for Lyapunov exponents, I believe there is ambiguity and misuse.
The lambda(i) in the reference literature signifies the Lyapunov exponent of the entire phase space data after expanding by i time steps, but in the calculation formula provided in the MATLAB help documentation, Y_(i+K) represents the data point at the i-th point in the reconstructed data Y after K steps, and this calculation formula also does not match the calculation code given by MATLAB. I believe there should be some misguidance and misunderstanding here.
According to the symbol regulations in the algorithm description and the MATLAB code, I think the correct formula might be y(i) = 1/dt * 1/N * sum_j( log( ||Y_(j+i) - Y_(j*+i)|| ) )
Drumlin Farm has welcomed MATLAMB, named in honor of MathWorks, among ten adorable new lambs this season!
Are you local to Boston?
Shape the Future of MATLAB: Join MathWorks' UX Night In-Person!
When: June 25th, 6 to 8 PM
Where: MathWorks Campus in Natick, MA
🌟 Calling All MATLAB Users! Here's your unique chance to influence the next wave of innovations in MATLAB and engineering software. MathWorks invites you to participate in our special after-hours usability studies. Dive deep into the latest MATLAB features, share your valuable feedback, and help us refine our solutions to better meet your needs.
🚀 This Opportunity Is Not to Be Missed:
- Exclusive Hands-On Experience: Be among the first to explore new MATLAB features and capabilities.
- Voice Your Expertise: Share your insights and suggestions directly with MathWorks developers.
- Learn, Discover, and Grow: Expand your MATLAB knowledge and skills through firsthand experience with unreleased features.
- Network Over Dinner: Enjoy a complimentary dinner with fellow MATLAB enthusiasts and the MathWorks team. It's a perfect opportunity to connect, share experiences, and network after work.
- Earn Rewards: Participants will not only contribute to the advancement of MATLAB but will also be compensated for their time. Plus, enjoy special MathWorks swag as a token of our appreciation!
👉 Reserve Your Spot Now: Space is limited for these after-hours sessions. If you're passionate about MATLAB and eager to contribute to its development, we'd love to hear from you.
I found this plot of words said by different characters on the US version of The Office sitcom. There's a sparkline for each character from pilot to finale episode.
A high school student called for help with this physics problem:
- Car A moves with constant velocity v.
- Car B starts to move when Car A passes through the point P.
- Car B undergoes...
- uniform acc. motion from P to Q.
- uniform velocity motion from Q to R.
- uniform acc. motion from R to S.
- Car A and B pass through the point R simultaneously.
- Car A and B arrive at the point S simultaneously.
Q1. When car A passes the point Q, which is moving faster?
Q2. Solve the time duration for car B to move from P to Q using L and v.
Q3. Magnitude of acc. of car B from P to Q, and from R to S: which is bigger?
Well, it can be solved with a series of tedious equations. But... how about this?
Code below:
%% get images and prepare stuffs
figure(WindowStyle="docked"),
ax1 = subplot(2,1,1);
hold on, box on
ax1.XTick = [];
ax1.YTick = [];
A = plot(0, 1, 'ro', MarkerSize=10, MarkerFaceColor='r');
B = plot(0, 0, 'bo', MarkerSize=10, MarkerFaceColor='b');
[carA, ~, alphaA] = imread('https://cdn.pixabay.com/photo/2013/07/12/11/58/car-145008_960_720.png');
[carB, ~, alphaB] = imread('https://cdn.pixabay.com/photo/2014/04/03/10/54/car-311712_960_720.png');
carA = imrotate(imresize(carA, 0.1), -90);
carB = imrotate(imresize(carB, 0.1), 180);
alphaA = imrotate(imresize(alphaA, 0.1), -90);
alphaB = imrotate(imresize(alphaB, 0.1), 180);
carA = imagesc(carA, AlphaData=alphaA, XData=[-0.1, 0.1], YData=[0.9, 1.1]);
carB = imagesc(carB, AlphaData=alphaB, XData=[-0.1, 0.1], YData=[-0.1, 0.1]);
txtA = text(0, 0.85, 'A', FontSize=12);
txtB = text(0, 0.17, 'B', FontSize=12);
yline(1, 'r--')
yline(0, 'b--')
xline(1, 'k--')
xline(2, 'k--')
text(1, -0.2, 'Q', FontSize=20, HorizontalAlignment='center')
text(2, -0.2, 'R', FontSize=20, HorizontalAlignment='center')
% legend('A', 'B') % this make the animation slow. why?
xlim([0, 3])
ylim([-.3, 1.3])
%% axes2: plots velocity graph
ax2 = subplot(2,1,2);
box on, hold on
xlabel('t'), ylabel('v')
vA = plot(0, 1, 'r.-');
vB = plot(0, 0, 'b.-');
xline(1, 'k--')
xline(2, 'k--')
xlim([0, 3])
ylim([-.3, 1.8])
p1 = patch([0, 0, 0, 0], [0, 1, 1, 0], [248, 209, 188]/255, ...
EdgeColor = 'none', ...
FaceAlpha = 0.3);
%% solution
v = 1; % car A moves with constant speed.
L = 1; % distances of P-Q, Q-R, R-S
% acc. of car B for three intervals
a(1) = 9*v^2/8/L;
a(2) = 0;
a(3) = -1;
t_BatQ = sqrt(2*L/a(1)); % time when car B arrives at Q
v_B2 = a(1) * t_BatQ; % speed of car B between Q-R
%% patches for velocity graph
p2 = patch([t_BatQ, t_BatQ, t_BatQ, t_BatQ], [1, 1, v_B2, v_B2], ...
[248, 209, 188]/255, ...
EdgeColor = 'none', ...
FaceAlpha = 0.3);
p3 = patch([2, 2, 2, 2], [1, v_B2, v_B2, 1], [194, 234, 179]/255, ...
EdgeColor = 'none', ...
FaceAlpha = 0.3);
%% animation
tt = linspace(0, 3, 2000);
for t = tt
A.XData = v * t;
vA.XData = [vA.XData, t];
vA.YData = [vA.YData, 1];
if t < t_BatQ
B.XData = 1/2 * a(1) * t^2;
vB.XData = [vB.XData, t];
vB.YData = [vB.YData, a(1) * t];
p1.XData = [0, t, t, 0];
p1.YData = [0, vB.YData(end), 1, 1];
elseif t >= t_BatQ && t < 2
B.XData = L + (t - t_BatQ) * v_B2;
vB.XData = [vB.XData, t];
vB.YData = [vB.YData, v_B2];
p2.XData = [t_BatQ, t, t, t_BatQ];
p2.YData = [1, 1, vB.YData(end), vB.YData(end)];
else
B.XData = 2*L + v_B2 * (t - 2) + 1/2 * a(3) * (t-2)^2;
vB.XData = [vB.XData, t];
vB.YData = [vB.YData, v_B2 + a(3) * (t - 2)];
p3.XData = [2, t, t, 2];
p3.YData = [1, 1, vB.YData(end), v_B2];
end
txtA.Position(1) = A.XData(end);
txtB.Position(1) = B.XData(end);
carA.XData = A.XData(end) + [-.1, .1];
carB.XData = B.XData(end) + [-.1, .1];
drawnow
end
is there any sites available online free ai course learning except: coursera.org
Are you a Simulink user eager to learn how to create apps with App Designer? Or an App Designer enthusiast looking to dive into Simulink?
Don't miss today's article on the Graphics and App Building Blog by @Robert Philbrick! Discover how to build Simulink Apps with App Designer, streamlining control of your simulations!
Northern lights captured from this weekend at MathWorks campus ✨
Did you get a chance to see lights and take some photos?
Hi to all.
I'm trying to learn a bit about trading with cryptovalues. At the moment I'm using Freqtrade (in dry-run mode of course) for automatic trading. The tool is written in python and it allows to create custom strategies in python classes and then run them.
I've written some strategy just to learn how to do, but now I'd like to create some interesting algorithm. I've a matlab license, and I'd like to know what are suggested tollboxes for following work:
- Create a criptocurrency strategy algorythm (for buying and selling some crypto like BTC, ETH etc).
- Backtesting the strategy with historical data (I've a bunch of json files with different timeframes, downloaded with freqtrade from binance).
- Optimize the strategy given some parameters (they can be numeric, like ROI, some kind of enumeration, like "selltype" and so on).
- Convert the strategy algorithm in python, so I can use it with Freqtrade without worrying of manually copying formulas and parameters that's error prone.
- I'd like to write both classic algorithm and some deep neural one, that try to find best strategy with little neural network (they should run on my pc with 32gb of ram and a 3080RTX if it can be gpu accelerated).
What do you suggest?
Dear MATLAB contest enthusiasts,
I believe many of you have been captivated by the innovative entries from Zhaoxu Liu / slanderer, in the 2023 MATLAB Flipbook Mini Hack contest.
Ever wondered about the person behind these creative entries? What drives a MATLAB user to such levels of skill? And what inspired his participation in the contest? We were just as curious as you are!
We were delighted to catch up with him and learn more about his use of MATLAB. The interview has recently been published in MathWorks Blogs. For an in-depth look into his insights and experiences, be sure to read our latest blog post: Community Q&A – Zhaoxu Liu.
But the conversation doesn't end here! Who would you like to see featured in our next interview? Drop their name in the comments section below and let us know who we should reach out to next!
The study of the dynamics of the discrete Klein - Gordon equation (DKG) with friction is given by the equation :
above equation, W describes the potential function :
The objective of this simulation is to model the dynamics of a segment of DNA under thermal fluctuations with fixed boundaries using a modified discrete Klein-Gordon equation. The model incorporates elasticity, nonlinearity, and damping to provide insights into the mechanical behavior of DNA under various conditions.
% Parameters
numBases = 200; % Number of base pairs, representing a segment of DNA
kappa = 0.1; % Elasticity constant
omegaD = 0.2; % Frequency term
beta = 0.05; % Nonlinearity coefficient
delta = 0.01; % Damping coefficient
- Position: Random initial perturbations between 0.01 and 0.02 to simulate the thermal fluctuations at the start.
- Velocity: All bases start from rest, assuming no initial movement except for the thermal perturbations.
% Random initial perturbations to simulate thermal fluctuations
initialPositions = 0.01 + (0.02-0.01).*rand(numBases,1);
initialVelocities = zeros(numBases,1); % Assuming initial rest state
The simulation uses fixed ends to model the DNA segment being anchored at both ends, which is typical in experimental setups for studying DNA mechanics. The equations of motion for each base are derived from a modified discrete Klein-Gordon equation with the inclusion of damping:
% Define the differential equations
dt = 0.05; % Time step
tmax = 50; % Maximum time
tspan = 0:dt:tmax; % Time vector
x = zeros(numBases, length(tspan)); % Displacement matrix
x(:,1) = initialPositions; % Initial positions
% Velocity-Verlet algorithm for numerical integration
for i = 2:length(tspan)
% Compute acceleration for internal bases
acceleration = zeros(numBases,1);
for n = 2:numBases-1
acceleration(n) = kappa * (x(n+1, i-1) - 2 * x(n, i-1) + x(n-1, i-1)) ...
- delta * initialVelocities(n) - omegaD^2 * (x(n, i-1) - beta * x(n, i-1)^3);
end
% positions for internal bases
x(2:numBases-1, i) = x(2:numBases-1, i-1) + dt * initialVelocities(2:numBases-1) ...
+ 0.5 * dt^2 * acceleration(2:numBases-1);
% velocities using new accelerations
newAcceleration = zeros(numBases,1);
for n = 2:numBases-1
newAcceleration(n) = kappa * (x(n+1, i) - 2 * x(n, i) + x(n-1, i)) ...
- delta * initialVelocities(n) - omegaD^2 * (x(n, i) - beta * x(n, i)^3);
end
initialVelocities(2:numBases-1) = initialVelocities(2:numBases-1) + 0.5 * dt * (acceleration(2:numBases-1) + newAcceleration(2:numBases-1));
end
% Visualization of displacement over time for each base pair
figure;
hold on;
for n = 2:numBases-1
plot(tspan, x(n, :));
end
xlabel('Time');
ylabel('Displacement');
legend(arrayfun(@(n) ['Base ' num2str(n)], 2:numBases-1, 'UniformOutput', false));
title('Displacement of DNA Bases Over Time');
hold off;
The results are visualized using a plot that shows the displacements of each base over time . Key observations from the simulation include :
- Wave Propagation: The initial perturbations lead to wave-like dynamics along the segment, with visible propagation and reflection at the boundaries.
- Damping Effects: The inclusion of damping leads to a gradual reduction in the amplitude of the oscillations, indicating energy dissipation over time.
- Nonlinear Behavior: The nonlinear term influences the response, potentially stabilizing the system against large displacements or leading to complex dynamic patterns.
% 3D plot for displacement
figure;
[X, T] = meshgrid(1:numBases, tspan);
surf(X', T', x);
xlabel('Base Pair');
ylabel('Time');
zlabel('Displacement');
title('3D View of DNA Base Displacements');
colormap('jet');
shading interp;
colorbar; % Adds a color bar to indicate displacement magnitude
% Snapshot visualization at a specific time
snapshotTime = 40; % Desired time for the snapshot
[~, snapshotIndex] = min(abs(tspan - snapshotTime)); % Find closest index
snapshotSolution = x(:, snapshotIndex); % Extract displacement at the snapshot time
% Plotting the snapshot
figure;
stem(1:numBases, snapshotSolution, 'filled'); % Discrete plot using stem
title(sprintf('DNA Model Displacement at t = %d seconds', snapshotTime));
xlabel('Base Pair Index');
ylabel('Displacement');
% Time vector for detailed sampling
tDetailed = 0:0.5:50; % Detailed time steps
% Initialize an empty array to hold the data
data = [];
% Generate the data for 3D plotting
for i = 1:numBases
% Interpolate to get detailed solution data for each base pair
detailedSolution = interp1(tspan, x(i, :), tDetailed);
% Concatenate the current base pair's data to the main data array
data = [data; repmat(i, length(tDetailed), 1), tDetailed', detailedSolution'];
end
% 3D Plot
figure;
scatter3(data(:,1), data(:,2), data(:,3), 10, data(:,3), 'filled');
xlabel('Base Pair');
ylabel('Time');
zlabel('Displacement');
title('3D Plot of DNA Base Pair Displacements Over Time');
colorbar; % Adds a color bar to indicate displacement magnitude
Sie können auch eine Website aus der folgenden Liste auswählen:
Amerika
- América Latina (Español)
- Canada (English)
- United States (English)
Europa
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom(English)
Asien-Pazifik
- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)