ans = 
Results for
Hi Creative Coders!
I've been working my way through the problem set (and enjoying all the references), but the final puzzle has me stumped. I've managed to get 16/20 of the test cases to the right answer, and the rest remain very unsolvable for my current algorithm. I know there's some kind of leap of logic I'm missing, but can't figure out quite what it is. Can any of you help?
What I've Done So Far
My current algorithm looks a bit like this. The code is doing its best to embody spaghetti at the moment, so I'll refrain from posting the whole thing to spare you all from trying to follow my thought processes.
Step 1: Go through all the turns and fill out tables of 'definitely', 'maybe', and 'clue' based on the information provided in a single run through the turns. This means that the case mentioned in the problem where information from future turns affecting previous turns does not matter yet. 'Definitely' information is for when I know a card must belong to a certain player (or to no-one). 'Maybe' starts off with all players in all cells, and when a player is found to not be able to have a card, their number is removed from the cell. Think of Sudoku notes where someone has helpfully gone through the grid and put every single possible number in each cell. 'Clue' contains information about which cards players were hinted about.
Example from test case 1:
definitelyTable =
6×3 table
G1 G2 G3
____________ ____________ ____________
{[ 0]} {0×0 double} {0×0 double}
{0×0 double} {[ -1]} {[ 1]}
{0×0 double} {[ 6]} {[ 0]}
{[ 3]} {[ 4]} {0×0 double}
{0×0 double} {[ 0]} {0×0 double}
{[ 5]} {[ 3]} {0×0 double}
maybeTable =
6×3 table
G1 G2 G3
_________ _______ _______
{[ 0]} {[2 5]} {[1 2]}
{[ 4]} {[ 0]} {[ 0]}
{[2 4 6]} {[ 0]} {[ 0]}
{[ 0]} {[ 0]} {[1 4]}
{[ 1 4]} {[ 0]} {[ 1]}
{[ 0]} {[ 0]} {[2 4]}
clueTable =
6×3 table
G1 G2 G3
____________ ____________ ____________
{0×0 double} {[ 5 6]} {[ 2 4]}
{[ 4 6]} {[ 4 6]} {0×0 double}
{[ 2 6]} {[ 5 6]} {0×0 double}
{0×0 double} {[ 4]} {[ 4 5 6]}
{[ 4]} {0×0 double} {[ 1 4 6]}
{[ 2 5]} {0×0 double} {[ 2 4 5 6]}
(-1 indicates the card is in the envelope. 0 indicates the card is commonly known.)
Step 2: While a solution has not yet been found, loop through all the turns again. This is the part where future turn info can now be fed back into previous turns, and where my sticky test cases loop forever. I've coded up each of the implementation tips from the problem statement for this stage.
Where It All Comes Undone
The problem is, for certain test cases (e.g., case 5), I reach a point where going through all turns yields no new information. I either end up with an either-or scenario, where the potential culprit card is one of two choices, or with so little information it doesn't look like there is anywhere left to turn.
I solved some of the either-or cases by adding a snippet that assumes one of the values and then tries to solve the problem based on that new information. If it can't solve it, then it tries the other option and goes round again. Unfortunately, however, this results in an infinite flip-flop for some cases as neither guess resolves the puzzle.
Essentially guessing the solution and following through to a logical inconsistency for however many combinations of players and cards sounds a) inefficient and b) not the way this was intended to be solved. Does anyone have any hints that might get me on track to solve this mystery?
% Recreation of Saturn photo
figure('Color', 'k', 'Position', [100, 100, 800, 800]);
ax = axes('Color', 'k', 'XColor', 'none', 'YColor', 'none', 'ZColor', 'none');
hold on;
% Create the planet sphere
[x, y, z] = sphere(150);
% Saturn colors - pale yellow/cream gradient
saturn_radius = 1;
% Create color data based on latitude for gradient effect
lat = asin(z);
color_data = rescale(lat, 0.3, 0.9);
% Plot Saturn with smooth shading
planet = surf(x*saturn_radius, y*saturn_radius, z*saturn_radius, ...
color_data, ...
'EdgeColor', 'none', ...
'FaceColor', 'interp', ...
'FaceLighting', 'gouraud', ...
'AmbientStrength', 0.3, ...
'DiffuseStrength', 0.6, ...
'SpecularStrength', 0.1);
% Use a cream/pale yellow colormap for Saturn
cream_map = [linspace(0.4, 0.95, 256)', ...
linspace(0.35, 0.9, 256)', ...
linspace(0.2, 0.7, 256)'];
colormap(cream_map);
% Create the ring system
n_points = 300;
theta = linspace(0, 2*pi, n_points);
% Define ring structure (inner radius, outer radius, brightness)
rings = [
1.2, 1.4, 0.7; % Inner ring
1.45, 1.65, 0.8; % A ring
1.7, 1.85, 0.5; % Cassini division (darker)
1.9, 2.3, 0.9; % B ring (brightest)
2.35, 2.5, 0.6; % C ring
2.55, 2.8, 0.4; % Outer rings (fainter)
];
% Create rings as patches
for i = 1:size(rings, 1)
r_inner = rings(i, 1);
r_outer = rings(i, 2);
brightness = rings(i, 3);
% Create ring coordinates
x_inner = r_inner * cos(theta);
y_inner = r_inner * sin(theta);
x_outer = r_outer * cos(theta);
y_outer = r_outer * sin(theta);
% Front side of rings
ring_x = [x_inner, fliplr(x_outer)];
ring_y = [y_inner, fliplr(y_outer)];
ring_z = zeros(size(ring_x));
% Color based on brightness
ring_color = brightness * [0.9, 0.85, 0.7];
fill3(ring_x, ring_y, ring_z, ring_color, ...
'EdgeColor', 'none', ...
'FaceAlpha', 0.7, ...
'FaceLighting', 'gouraud', ...
'AmbientStrength', 0.5);
end
% Add some texture/gaps in the rings using scatter
n_particles = 3000;
r_particles = 1.2 + rand(1, n_particles) * 1.6;
theta_particles = rand(1, n_particles) * 2 * pi;
x_particles = r_particles .* cos(theta_particles);
y_particles = r_particles .* sin(theta_particles);
z_particles = (rand(1, n_particles) - 0.5) * 0.02;
% Vary particle brightness
particle_colors = repmat([0.8, 0.75, 0.6], n_particles, 1) .* ...
(0.5 + 0.5*rand(n_particles, 1));
scatter3(x_particles, y_particles, z_particles, 1, particle_colors, ...
'filled', 'MarkerFaceAlpha', 0.3);
% Add dramatic outer halo effect - multiple layers extending far out
n_glow = 20;
for i = 1:n_glow
glow_radius = 1 + i*0.35; % Extend much farther
alpha_val = 0.08 / sqrt(i); % More visible, slower falloff
% Color gradient from cream to blue/purple at outer edges
if i <= 8
glow_color = [0.9, 0.85, 0.7]; % Warm cream/yellow
else
% Gradually shift to cooler colors
mix = (i - 8) / (n_glow - 8);
glow_color = (1-mix)*[0.9, 0.85, 0.7] + mix*[0.6, 0.65, 0.85];
end
surf(x*glow_radius, y*glow_radius, z*glow_radius, ...
ones(size(x)), ...
'EdgeColor', 'none', ...
'FaceColor', glow_color, ...
'FaceAlpha', alpha_val, ...
'FaceLighting', 'none');
end
% Add extensive glow to rings - make it much more dramatic
n_ring_glow = 12;
for i = 1:n_ring_glow
glow_scale = 1 + i*0.15; % Extend farther
alpha_ring = 0.12 / sqrt(i); % More visible
for j = 1:size(rings, 1)
r_inner = rings(j, 1) * glow_scale;
r_outer = rings(j, 2) * glow_scale;
brightness = rings(j, 3) * 0.5 / sqrt(i);
x_inner = r_inner * cos(theta);
y_inner = r_inner * sin(theta);
x_outer = r_outer * cos(theta);
y_outer = r_outer * sin(theta);
ring_x = [x_inner, fliplr(x_outer)];
ring_y = [y_inner, fliplr(y_outer)];
ring_z = zeros(size(ring_x));
% Color gradient for ring glow
if i <= 6
ring_color = brightness * [0.9, 0.85, 0.7];
else
mix = (i - 6) / (n_ring_glow - 6);
ring_color = brightness * ((1-mix)*[0.9, 0.85, 0.7] + mix*[0.65, 0.7, 0.9]);
end
fill3(ring_x, ring_y, ring_z, ring_color, ...
'EdgeColor', 'none', ...
'FaceAlpha', alpha_ring, ...
'FaceLighting', 'none');
end
end
% Add diffuse glow particles for atmospheric effect
n_glow_particles = 8000;
glow_radius_particles = 1.5 + rand(1, n_glow_particles) * 5;
theta_glow = rand(1, n_glow_particles) * 2 * pi;
phi_glow = acos(2*rand(1, n_glow_particles) - 1);
x_glow = glow_radius_particles .* sin(phi_glow) .* cos(theta_glow);
y_glow = glow_radius_particles .* sin(phi_glow) .* sin(theta_glow);
z_glow = glow_radius_particles .* cos(phi_glow);
% Color particles based on distance - cooler colors farther out
particle_glow_colors = zeros(n_glow_particles, 3);
for i = 1:n_glow_particles
dist = glow_radius_particles(i);
if dist < 3
particle_glow_colors(i,:) = [0.9, 0.85, 0.7];
else
mix = (dist - 3) / 4;
particle_glow_colors(i,:) = (1-mix)*[0.9, 0.85, 0.7] + mix*[0.5, 0.6, 0.9];
end
end
scatter3(x_glow, y_glow, z_glow, rand(1, n_glow_particles)*2+0.5, ...
particle_glow_colors, 'filled', 'MarkerFaceAlpha', 0.05);
% Lighting setup
light('Position', [-3, -2, 4], 'Style', 'infinite', ...
'Color', [1, 1, 0.95]);
light('Position', [2, 3, 2], 'Style', 'infinite', ...
'Color', [0.3, 0.3, 0.4]);
% Camera and view settings
axis equal off;
view([-35, 25]); % Angle to match saturn_photo.jpg - more dramatic tilt
camva(10); % Field of view - slightly wider to show full halo
xlim([-8, 8]); % Expanded to show outer halo
ylim([-8, 8]);
zlim([-8, 8]);
% Material properties
material dull;
title('Saturn - Left click: Rotate | Right click: Pan | Scroll: Zoom', 'Color', 'w', 'FontSize', 12);
% Enable interactive camera controls
cameratoolbar('Show');
cameratoolbar('SetMode', 'orbit'); % Start in rotation mode
% Custom mouse controls
set(gcf, 'WindowButtonDownFcn', @mouseDown);
function mouseDown(src, ~)
selType = get(src, 'SelectionType');
switch selType
case 'normal' % Left click - rotate
cameratoolbar('SetMode', 'orbit');
rotate3d on;
case 'alt' % Right click - pan
cameratoolbar('SetMode', 'pan');
pan on;
end
end
Fittingly for a Creative Coder, @Vasilis Bellos clearly enjoyed the silliness I put into the problems. If you've solved the whole problem set, don't forget to help out your teammates with suggestions, tips, tricks, etc. But also, just for fun, I'm curious to see which of my many in-jokes and nerdy references you noticed. Many of the problems were inspired by things in the real world, then ported over into the chaotic fantasy world of Nedland.
I guess I'll start with the obvious real-world reference: @Ned Gulley (I make no comment about his role as insane despot in any universe, real or otherwise.)
Experimenting with Agentic AI
44%
I am an AI skeptic
0%
AI is banned at work
11%
I am happy with Conversational AI
44%
9 votes
The Cody Contest 2025 is underway, and it includes a super creative problem group which many of us have found fascinating. The central theme of the problems, expertly curated by @Matt Tearle, humorously revolves around the whims of the capricious dictator Lord Ned, as he goes out of his way to complicate the lives of his subjects and visitors alike. We cannot judge whether or not there's any truth to the rumors behind all the inside jokes, but it's obvious that the team had a lot of fun creating these; and we had even more fun solving them.
Today I want to showcase a way of graphically solving and visualizing one of those problems which I found very elegant, The Bridges of Nedsburg.
To briefly reiterate the problem, the number of islands and the arrangement of bridges of the city of Nedsburg are constantly changing. Lord Ned has decided to take advantage of this by charging visitors with an increasingly expensive n-bridge pass which allows them to cross up to n bridges in one journey. Given the Connectivity Matrix C, we are tasked with calculating the minimum n needed so that there is a path from every island to every other island in n steps or fewer.
Matt kindly provided us with some useful bit of math in the description detailing how to calculate the way to get from one island to another in an number of m steps. However, he has also hidden an alternative path to the solution in plain sight, in one of the graphs he provided. This involves the extremely useful and versatile class digraph, representing directed graphs, which have directional edges connecting the nodes. Here's some further great documentation and other cool resources on the topic for those who are interested in learning more about it:
Let's start using this class to explore a graphical solution to Lord Ned's conundrum. We will use the unit tests included in the problem to visualize the solution. We can retrieve the connectivity matrix for each case using the following function:
function C = getConnectivityMatrix(unit_test)
% Number of islands and bridge arrangement
switch unit_test
case 1
m = 3; idx = [3;4;8];
case 2
m = 3; idx = [3;4;7;8];
case 3
m = 4; idx = [2;7;8;10;13];
case 4
m = 4; idx = [4;5;7;8;9;14];
case 5
m = 5; idx = [5;8;11;12;14;18;22;23];
case 6
m = 5; idx = [2;5;8;14;20;21;24];
case 7
m = 6; idx = [3;4;7;11;18;23;24;26;30;32];
case 8
m = 6; idx = [3;11;12;13;18;19;28;32];
case 9
m = 7; idx = [3;4;6;8;13;14;20;21;23;31;36;47];
case 10
m = 7; idx = [4;11;13;14;19;22;23;26;28;30;34;35;37;38;45];
case 11
m = 8; idx = [2;4;5;6;8;12;13;17;27;39;44;48;54;58;60;62];
case 12
m = 8; idx = [3;9;12;20;24;29;30;31;33;44;48;50;53;54;58];
case 13
m = 9; idx = [8;9;10;14;15;22;25;26;29;33;36;42;44;47;48;50;53;54;55;67;80];
case 14
m = 9; idx = [8;10;22;32;37;40;43;45;47;53;56;57;62;64;69;70;73;77;79];
case 15
m = 10; idx = [2;5;8;13;16;20;24;27;28;36;43;49;53;62;71;75;77;83;86;87;95];
case 16
m = 10; idx = [4;9;14;21;22;35;37;38;44;47;50;51;53;55;59;61;63;66;69;76;77;84;85;86;90;97];
end
C = zeros(m);
C(idx) = 1;
end
The case in the example refers to unit test case 2.
unit_test = 2;
C = getConnectivityMatrix(unit_test);
disp(C)
D = digraph(C);
figure
p = plot(D,'LineWidth',1.5,'ArrowSize',10);
This is the same as the graph provided in the example. Another very useful method of digraph is shortestpath. This allows us to calculate the path and distance from one single node to another. For example:
% Path and distance from node 1 to node 2
[path12,dist12] = shortestpath(D,1,2);
fprintf('The shortest path from island %d to island %d is: %s. The minimum number of steps is: n = %d\n', 1, 2, join(string(path12), ' -> '),dist12)
% Path and distance from node 2 to node 1
[path21,dist21] = shortestpath(D,2,1);
fprintf('The shortest path from island %d to island %d is: %s. The minimum number of steps is: n = %d\n', 2, 1, join(string(path21), ' -> '),dist21)
figure
p = plot(D,'LineWidth',1.5,'ArrowSize',10);
highlight(p,path12,'EdgeColor','r','NodeColor','r','LineWidth',2)
highlight(p,path21,'EdgeColor',[0 0.8 0],'LineWidth',2)
But that's not all! digraph can also provide us with a matrix of the distances d, i.e. the steps needed to travel from island i to island j, where i and j are the rows and columns of d respectively. This is accomplished by using its distances method. The distance matrix can be visualized as:
d = distances(D);
figure
% Using pcolor w/ appending matrix workaround for convenience
pcolor([d,d(:,end);d(end,:),d(end,end)])
% Alternatively you can use imagesc(d), but you'll have to recreate the grid manually
axis square
set(gca,'YDir','reverse','XTick',[],'YTick',[])
[X,Y] = meshgrid(1:height(d));
text(X(:)+0.5,Y(:)+0.5,string(d(:)),'FontSize',11)
colormap(interp1(linspace(0,1,4), [1 1 1; 0.7 0.9 1; 0.6 0.7 1; 1 0.3 0.3], linspace(0,1,8)))
clim([-0.5 7+0.5])
This confirms what we saw before, i.e. you need 1 step to go from island 1 to island 2, but 2 steps for vice versa. It also confirms that the minimum number of steps n that you need to buy the pass for is 2 (which also occurs for traveling from island 3 to island 2). As it's not the point of the post to give the full solution to the problem but rather present the graphical way of visualizing it I will not include the code of how to calculate this, but I'm sure that by now it's reduced to a trivial problem which you have already figured out how to solve.
That being said, now that we have the distance matrix, let's continue with the visualizations. First, let's plot the corresponding paths for each of these combinations:
figure
tiledlayout(size(C,1),size(C,2),'TileSpacing','tight','Padding','tight');
for i = 1:size(C,1)
for j = 1:size(C,2)
nexttile
p = plot(D,'ArrowSize',10);
highlight(p,shortestpath(D,i,j),'EdgeColor','r','NodeColor','r','LineWidth',2)
lims = axis;
text(lims(1)+diff(lims(1:2))*0.05,lims(3)+diff(lims(3:4))*0.9,sprintf('n = %d',d(i,j)))
end
end
This allows us to go from the distance matrix to visualizing the paths and number of steps for each corresponding case. Things are rather simple for this 3-island example case, but evil Lord Ned is just getting started. Let's now try to solve the problem for all provided unit test cases:
% Cell array of connectivity matrices
C = arrayfun(@getConnectivityMatrix,1:16,'UniformOutput',false);
% Cell array of corresponding digraph objects
D = cellfun(@digraph,C,'UniformOutput',false);
% Cell array of corresponding distance matrices
d = cellfun(@distances,D,'UniformOutput',false);
% id of solutions: Provided as is to avoid handing out the code to the full solution
id = [2, 2, 9, 3, 4, 6, 16, 4, 44, 43, 33, 34, 7, 18, 39, 2];
First, let's plot the distance matrix for each case:
figure
tiledlayout('flow','TileSpacing','compact','Padding','compact');
% Vary this to plot different combinations of cases
plot_cases = 1:numel(C);
for i = plot_cases
nexttile
pcolor([d{i},d{i}(:,end);d{i}(end,:),d{i}(end,end)])
axis square
set(gca,'YDir','reverse','XTick',[],'YTick',[])
title(sprintf('Case %d',i),'FontWeight','normal','FontSize',8)
end
c = colorbar('Ticks',0:7,'TickLength',0,'Limits',[-0.5 7+0.5],'FontSize',8);
c.Layout.Tile = 'East';
c.Label.String = 'Number of Steps';
c.Label.FontSize = 8;
colormap(interp1(linspace(0,1,4), [1 1 1; 0.7 0.9 1; 0.6 0.7 1; 1 0.3 0.3], linspace(0,1,8)))
clim(findobj(gcf,'type','axes'),[-0.5 7+0.5])
We immediately notice some inconsistencies, perhaps to be expected of the eccentric and cunning dictator. Things are pretty simple for the configurations with a small number of islands, but the minimum number of steps n can increase sharply and disproportionally to the additional number of islands. Cases 8 and 9 specifically have a particularly large n (relative to their grid dimensions), and case 14 has the largest n, almost double that of case 16 despite the fact that the latter has one extra island.
To visualize how this is possible, let's plot the path corresponding to the largest n for each case (though note that there might be multiple possible paths for each case):
figure
tiledlayout('flow','TileSpacing','tight','Padding','tight');
for i = plot_cases
nexttile
% Changing the layout to circular so we can better visualize the paths
p = plot(D{i},'ArrowSize',10,'Layout','Circle');
% Alternatively we could use the XData and YData properties if the positions of the islands were provided
axis([-1.5 1.5 -1.5 1.75])
[row,col] = ind2sub(size(d{i}),id(i));
highlight(p,shortestpath(D{i},row,col),'EdgeColor','r','NodeColor','r','LineWidth',2)
lims = axis;
text(lims(1)+diff(lims(1:2))*0.05,lims(3)+diff(lims(3:4))*0.9,sprintf('n = %d',d{i}(row,col)))
end
And busted! Unraveled! Exposed! Lord Ned has clearly been taking advantages of the tectonic forces by instructing his corrupt civil engineer lackeys to design the bridges to purposely force the visitors to go around in circles in order to drain them of their precious savings. In particular, for cases 8 and 9, he would have them go through every single island just to get from one island to another, whereas for case 14 they would have to visit 8 of the 9 islands just to get to their destination. If that's not diabolical then I don't know what is!
Ned jokes aside, I hope you enjoyed this contest just as much as I did, and that you found this article useful. I look forward to seeing more creative problems and solutions in the future.
It’s exciting to dive into a new dataset full of unfamiliar variables but it can also be overwhelming if you’re not sure where to start. Recently, I discovered some new interactive features in MATLAB live scripts that make it much easier to get an overview of your data. With just a few clicks, you can display sparklines and summary statistics using table variables, sort and filter variables, and even have MATLAB generate the corresponding code for reproducibility.
The Graphics and App Building blog published an article that walks through these features showing how to explore, clean, and analyze data—all without writing any code.
If you’re interested in streamlining your exploratory data analysis or want to see what’s new in live scripts, you might find it helpful:
If you’ve tried these features or have your own tips for quick data exploration in MATLAB, I’d love to hear your thoughts!
I am Prof Ansar Interested in coding challenge taker inmatlab
From my experience, MATLAB's Deep Learning Toolbox is quite user-friendly, but it still falls short of libraries like PyTorch in many respects. Most users tend to choose PyTorch because of its flexibility, efficiency, and rich support for many mathematical operators. In recent years, the number of dlarray-compatible mathematical functions added to the toolbox has been very limited, which makes it difficult to experiment with many custom networks. For example, svd is currently not supported for dlarray inputs.
This link (List of Functions with dlarray Support - MATLAB & Simulink) lists all functions that support dlarray as of R2026a — only around 200 functions (including toolbox-specific ones). I would like to see support for many more fundamental mathematical functions so that users have greater freedom when building and researching custom models. For context, the core MATLAB mathematics module contains roughly 600 functions, and many application domains build on that foundation.
I hope MathWorks will prioritize and accelerate expanding dlarray support for basic math functions. Doing so would significantly increase the Deep Learning Toolbox's utility and appeal for researchers and practitioners.
Thank you.
Run MATLAB using AI applications by leveraging MCP. This MCP server for MATLAB supports a wide range of coding agents like Claude Code and Visual Studio Code.
Check it out and share your experiences below. Have fun!
GitHub repo: https://github.com/matlab/matlab-mcp-core-server
Yann Debray's blog post: https://blogs.mathworks.com/deep-learning/2025/11/03/releasing-the-matlab-mcp-core-server-on-github/
Hey Creative Coders! 😎
Let’s get to know each other. Drop a quick intro below and meet your teammates! This is your chance to meet teammates, find coding buddies, and build connections that make the contest more fun and rewarding!
You can share:
- Your name or nickname
- Where you’re from
- Your favorite coding topic or language
- What you’re most excited about in the contest
Let’s make Team Creative Coders an awesome community—jump in and say hi! 🚀
Welcome to the Cody Contest 2025 and the Creative Coders team channel! 🎉
You think outside the box. Where others see limitations, you see opportunities for innovation. This is your space to connect with like-minded coders, share insights, and help your team win. To make sure everyone has a great experience, please keep these tips in mind:
- Follow the Community Guidelines: Take a moment to review our community standards. Posts that don’t follow these guidelines may be flagged by moderators or community members.
- Ask Questions About Cody Problems: When asking for help, show your work! Include your code, error messages, and any details needed to reproduce your results. This helps others provide useful, targeted answers.
- Share Tips & Tricks: Knowledge sharing is key to success. When posting tips or solutions, explain how and why your approach works so others can learn your problem-solving methods.
- Provide Feedback: We value your feedback! Use this channel to report issues or share creative ideas to make the contest even better.
Have fun and enjoy the challenge! We hope you’ll learn new MATLAB skills, make great connections, and win amazing prizes! 🚀
I just learned you can access MATLAB Online from the following shortcut in your web browser: https://matlab.new
Thanks @Yann Debray
From his recent blog post: pip & uv in MATLAB Online » Artificial Intelligence - MATLAB & Simulink
I'm working on training neural networks without backpropagation / automatic differentiation, using locally derived analytic forms of update rules. Given that this allows a direct formula to be derived for the update rule, it removes alot of the overhead that is otherwise required from automatic differentiation.
However, matlab's functionalities for neural networks are currently solely based around backpropagation and automatic differentiation, such as the dlgradient function and requiring everything to be dlarrays during training.
I have two main requests, specifically for functions that perform a single operation within a single layer of a neural network, such as "dlconv", "fullyconnect", "maxpool", "avgpool", "relu", etc:
- these functions should also allow normal gpuArray data instead of requiring everything to be dlarrays.
- these functions are currently designed to only perform the forward pass. I request that these also be designed to perform the backward pass if user requests. There can be another input user flag that can be "forward" (default) or "backward", and then the function should have all the necessary inputs to perform that operation (e.g. for "avgpool" forward pass it only needs the avgpool input data and the avgpool parameters, but for the "avgpool" backward pass it needs the deriviative w.r.t. the avgpool output data, the avgpool parameters, and the original data dimensions). I know that there is a maxunpool function that achieves this for maxpool, but it has significant issues when trying to use it this way instead of by backpropagation in a dlgradient type layer, see (https://www.mathworks.com/matlabcentral/answers/2179587-making-a-custom-way-to-train-cnns-and-i-am-noticing-that-avgpool-is-significantly-faster-than-maxpo?s_tid=srchtitle).
I don't know how many people would benefit from this feature, and someone could always spend their time creating these functionalities themselves by matlab scripts, cuDNN mex, etc., but regardless it would be nice for matlab to have this allowable for more customizable neural net training.
Edit 15 Oct 2025: Removed incorrect code. Replaced symmatrix2sym and symfunmatrix2symfun with sym and symfun respectively (latter supported as of 2024b).
The Symbolic Math Toolbox does not have its own dot and and cross functions. That's o.k. (maybe) for garden variety vectors of sym objects where those operations get shipped off to the base Matlab functions
x = sym('x',[3,1]); y = sym('y',[3,1]);
which dot(x,y)
dot(x,y)
which cross(x,y)
cross(x,y)
clearvars
x = symmatrix('x',[3,1]); y = symmatrix('y',[3,1]);
z = symmatrix('z',[1,1]);
sympref('AbbreviateOutput',false);
dot() expands the result, which isn't really desirable for exposition.
eqn = z == dot(x,y)
Also, dot() returns the the result in terms of the conjugate of x, which can't be simplifed away at the symmatrix level
assumeAlso(sym(x),'real')
class(eqn)
try
eqn = z == simplify(dot(x,y))
catch ME
ME.message
end
To get rid of the conjugate, we have to resort to sym
eqn = simplify(sym(eqn))
but again we are in expanded form, which defeats the purpose of symmatrix (et. al.)
But at least we can do this to get a nice equation
eqn = z == x.'*y
dot errors with symfunmatrix inputs
clearvars
syms t real
x = symfunmatrix('x(t)',t,[3,1]); y = symfunmatrix('y(t)',t,[3,1]);
try
dot(x,y)
catch ME
ME.message
end
Cross works (accidentally IMO) with symmatrix, but expands the result, which isn't really desirable for exposition
clearvars
x = symmatrix('x',[3,1]); y = symmatrix('y',[3,1]);
z = symmatrix('z',[3,1]);
eqn = z == cross(x,y)
And it doesn't work at all if an input is a symfunmatrix
syms t
w = symfunmatrix('w(t)',t,[3,1]);
try
eqn = z == cross(x,w);
catch ME
ME.message
end
In the latter case we can expand with
eqn = z == cross(sym(x),symfun(w)) % x has to be converted
But we can't do the same with dot (as shown above, dot doesn't like symfun inputs)
try
eqn = z == dot(sym(x),symfun(w))
catch ME
ME.message
end
Looks like the only choice for dot with symfunmatrix is to write it by hand at the matrix level
x.'*w
or at the sym/symfun level
sym(x).'*symfun(w) % assuming x is real
Ideally, I'd like to see dot and cross implemented for symmatrix and symfunmatrix types where neither function would evaluate, i.e., expand, until both arguments are subs-ed with sym or symfun objects of appropriate dimension.
Also, it would be nice if symmatrix could be assumed to be real. Is there a reason why being able to do so wouldn't make sense?
try
assume(x,'real')
catch ME
ME.message
end
Automating Parameter Identifiability Analysis in SimBiology
Is it possible to develop a MATLAB Live Script that automates a series of SimBiology model fits to obtain likelihood profiles? The goal is to fit a kinetic model to experimental data while systematically fixing the value of one kinetic constant (e.g., k1) and leaving the others unrestricted.
The script would perform the following:
Use a pre-configured SimBiology project where the best fit to the experimental data has already been established (including dependent/independent variables, covariates, the error model, and optimization settings).
Iterate over a defined sequence of fixed values for a chosen parameter.
For each fixed value, run the estimation to optimize the remaining parameters.
Record the resulting Sum of Squared Errors (SSE) for each run.
The final output would be a likelihood profile—a plot of SSE versus the fixed parameter value (e.g., k1)—to assess the practical identifiability of each model parameter.
What if you had no isprime utility to rely on in MATLAB? How would you identify a number as prime? An easy answer might be something tricky, like that in simpleIsPrime0.
simpleIsPrime0 = @(N) ismember(N,primes(N));
But I’ll also disallow the use of primes here, as it does not really test to see if a number is prime. As well, it would seem horribly inefficient, generating a possibly huge list of primes, merely to learn something about the last member of the list.
Looking for a more serious test for primality, I’ve already shown how to lighten the load by a bit using roughness, to sometimes identify numbers as composite and therefore not prime.
https://www.mathworks.com/matlabcentral/discussions/tips/879745-primes-and-rough-numbers-basic-ideas
But to actually learn if some number is prime, we must do a little more. Yes, this is a common homework problem assigned to students, something we have seen many times on Answers. It can be approached in many ways too, so it is worth looking at the problem in some depth.
The definition of a prime number is a natural number greater than 1, which has only two factors, thus 1 and itself. That makes a simple test for primality of the number N easy. We just try dividing the number by every integer greater than 1, and not exceeding N-1. If any of those trial divides leaves a zero remainder, then N cannot be prime. And of course we can use mod or rem instead of an explicit divide, so we need not worry about floating point trash, as long as the numbers being tested are not too large.
simpleIsPrime1 = @(N) all(mod(N,2:N-1) ~= 0);
Of course, simpleIsPrime1 is not a good code, in the sense that it fails to check if N is an integer, or if N is less than or equal to 1. It is not vectorized, and it has no documentation at all. But it does the job well enough for one simple line of code. There is some virtue in simplicity after all, and it is certainly easy to read. But sometimes, I wish a function handle could include some help comments too! A feature request might be in the offing.
simpleIsPrime1(9931)
simpleIsPrime1(9932)
simpleIsPrime1 works quite nicely, and seems pretty fast. What could be wrong? At some point, the student is given a more difficult problem, to identify if a significantly larger integer is prime. simpleIsPrime1 will then cause a computer to grind to a distressing halt if given a sufficiently large number to test. Or it might even error out, when too large a vector of numbers was generated to test against. For example, I don't think you want to test a number of the order of 2^64 using simpleIsPrime1, as performing on the order of 2^64 divides will be highly time consuming.
uint64(2)^63-25
Is it prime? I’ve not tested it to learn if it is, and simpleIsPrime1 is not the tool to perform that test anyway.
A student might realize the largest possible integer factors of some number N are the numbers N/2 and N itself. But, if N/2 is a factor, then so is 2, and some thought would suggest it is sufficient to test only for factors that do not exceed sqrt(N). This is because if a is a divisor of N, then so is b=N/a. If one of them is larger than sqrt(N), then the other must be smaller. That could lead us to an improved scheme in simpleIsPrime2.
simpleIsPrime2 = @(N) all(mod(N,2:sqrt(N)));
For an integer of the size 2^64, now you only need to perform roughly 2^32 trial divides. Maybe we might consider the subtle improvement found in simpleIsPrime3, which avoids trial divides by the even integers greater than 2.
simpleIsPrime3 = @(N) (N == 2) || (mod(N,2) && all(mod(N,3:2:sqrt(N))));
simpleIsPrime3 needs only an approximate maximum of 2^31 trial divides even for numbers as large as uint64 can represent. While that is large, it is still generally doable on the computers we have today, even if it might be slow.
Sadly, my goals are higher than even the rather lofty limit given by UINT64 numbers. The problem of course is that a trial divide scheme, despite being 100% accurate in its assessment of primality, is a time hog. Even an O(sqrt(N)) scheme is far too slow for numbers with thousands or millions of digits. And even for a number as “small” as 1e100, a direct set of trial divides by all primes less than sqrt(1e100) would still be practically impossible, as there are roughly n/log(n) primes that do not exceed n. For an integer on the order of 1e50,
1e50/log(1e50)
It is practically impossible to perform that many divides on any computer we can make today. Can we do better? Is there some more efficient test for primality? For example, we could write a simple sieve of Eratosthenes to check each prime found not exceeding sqrt(N).
function [TF,SmallPrime] = simpleIsPrime4(N)
% simpleIsPrime3 - Sieve of Eratosthenes to identify if N is prime
% [TF,SmallPrime] = simpleIsPrime3(N)
%
% Returns true if N is prime, as well as the smallest prime factor
% of N when N is composite. If N is prime, then SmallPrime will be N.
Nroot = ceil(sqrt(N)); % ceil caters for floating point issues with the sqrt
TF = true;
SieveList = true(1,Nroot+1); SieveList(1) = false;
SmallPrime = 2;
while TF
% Find the "next" true element in SieveList
while (SmallPrime <= Nroot+1) && ~SieveList(SmallPrime)
SmallPrime = SmallPrime + 1;
end
% When we drop out of this loop, we have found the next
% small prime to check to see if it divides N, OR, we
% have gone past sqrt(N)
if SmallPrime > Nroot
% this is the case where we have now looked at all
% primes not exceeding sqrt(N), and have found none
% that divide N. This is where we will drop out to
% identify N as prime. TF is already true, so we need
% not set TF.
SmallPrime = N;
return
else
if mod(N,SmallPrime) == 0
% smallPrime does divide N, so we are done
TF = false;
return
end
% update SieveList
SieveList(SmallPrime:SmallPrime:Nroot) = false;
end
end
end
simpleIsPrime4 does indeed work reasonably well, though it is sometimes a little slower than is simpleIsPrime3, and everything is hugely faster than simpleIsPrime1.
timeit(@() simpleIsPrime1(111111111))
timeit(@() simpleIsPrime2(111111111))
timeit(@() simpleIsPrime3(111111111))
timeit(@() simpleIsPrime4(111111111))
All of those times will slow to a crawl for much larger numbers of course. And while I might find a way to subtly improve upon these codes, any improvement will be marginal in the end if I try to use any such direct approach to primality. We must look in a different direction completely to find serious gains.
At this point, I want to distinguish between two distinct classes of tests for primality of some large number. One class of test is what I might call an absolute or infallible test, one that is perfectly reliable. These are tests where if X is identified as prime/composite then we can trust the result absolutely. The tests I showed in the form of simpleIsPrime1, simpleIsPrime2, simpleIsPrime3 and aimpleIsprime4, were all 100% accurate, thus they fall into the class of infallible tests.
The second general class of test for primality is what I will call an evidentiary test. Such a test provides evidence, possibly quite strong evidence, that the given number is prime, but in some cases, it might be mistaken. I've already offered a basic example of a weak evidentiary test for primality in the form of roughness. All primes are maximally rough. And therefore, if you can identify X as being rough to some extent, this provides evidence that X is also prime, and the depth of the roughness test influences the strength of the evidence for primality. While this is generally a fairly weak test, it is a test nevertheless, and a good exclusionary test, a good way to avoid more sophisticated but time consuming tests.
These evidentiary tests all have the property that if they do identify X as being composite, then they are always correct. In the context of roughness, if X is not sufficiently rough, then X is also not prime. On the other side of the coin, if you can show X is at least (sqrt(X)+1)-rough, then it is positively prime. (I say this to suggest that some evidentiary tests for primality can be turned into truth telling tests, but that may take more effort than you can afford.) The problem is of course that is literally impossible to verify that degree of roughness for numbers with many thousands of digits.
In my next post, I'll look at the Fermat test for primality, based on Fermat's little theorem.
Something that I periodically wonder about is whether an integration with the Rubi integration rules package would improve symbolic integration in Matlab's Symbolic Toolbox. The project is open-source with an MIT-licensed, has a Mathematica implementation, and supposedly SymPy is working on an implementation. Much of my intrigue comes from this 2022 report that compared the previous version of Rubi (4.16.1) against various CAS systems, including Matlab 2021a (Mupad):
While not really an official metric for Rubi, this does "feel" similar to my experience computing symbolic integrals in Matlab Symbolic Toolbox vs Maple/Mathematica. What do y'all think?
I saw an interesting problem on a reddit math forum today. The question was to find a number (x) as close as possible to r=3.6, but the requirement is that both x and 1/x be representable in a finite number of decimal places.
The problem of course is that 3.6 = 18/5. And the problem with 18/5 has an inverse 5/18, which will not have a finite representation in decimal form.
In order for a number and its inverse to both be representable in a finite number of decimal places (using base 10) we must have it be of the form 2^p*5^q, where p and q are integer, but may be either positive or negative. If that is not clear to you intuitively, suppose we have a form
2^p*5^-q
where p and q are both positive. All you need do is multiply that number by 10^q. All this does is shift the decimal point since you are just myltiplying by powers of 10. But now the result is
2^(p+q)
and that is clearly an integer, so the original number could be represented using a finite number of digits as a decimal. The same general idea would apply if p was negative, or if both of them were negative exponents.
Now, to return to the problem at hand... We can obviously adjust the number r to be 20/5 = 4, or 16/5 = 3.2. In both cases, since the fraction is now of the desired form, we are happy. But neither of them is really close to 3.6. My goal will be to find a better approximation, but hopefully, I can avoid a horrendous amount of trial and error. It would seem the trick might be to take logs, to get us closer to a solution. That is, suppose I take logs, to the base 2?
log2(3.6)
I used log2 here because that makes the problem a little simpler, since log2(2^p)=p. Therefore we want to find a pair of integers (p,q) such that
log2(3.6) + delta = p + log2(5)*q
where delta is as close to zero as possible. Thus delta is the error in our approximation to 3.6. And since we are working in logs, delta can be viewed as a proportional error term. Again, p and q may be any integers, either positive or negative. The two cases we have seen already have (p,q) = (2,0), and (4,-1).
Do you see the general idea? The line we have is of the form
log2(3.6) = p + log2(5)*q
it represents a line in the (p,q) plane, and we want to find a point on the integer lattice (p,q) where the line passes as closely as possible.
[Xl,Yl] = meshgrid([-10:10]);
plot(Xl,Yl,'k.')
hold on
fimplicit(@(p,q) -log2(3.6) + p + log2(5)*q,[-10,10,-10,10],'g-')
plot([2 4],[0,-1],'ro')
hold off
Now, some might think in terms of orthogonal distance to the line, but really, we want the vertical distance to be minimized. Again, minimize abs(delta) in the equation:
log2(3.6) + delta = p + log2(5)*q
where p and q are integer.
Can we do that using MATLAB? The skill about about mathematics often lies in formulating a word problem, and then turning the word problem into a problem of mathematics that we know how to solve. We are almost there now. I next want to formulate this into a problem that intlinprog can solve. The problem at first is intlinprog cannot handle absolute value constraints. And the trick there is to employ slack variables, a terribly useful tool to emply on this class of problem.
Rewrite delta as:
delta = Dpos - Dneg
where Dpos and Dneg are real variables, but both are constrained to be positive.
prob = optimproblem;
p = optimvar('p',lower = -50,upper = 50,type = 'integer');
q = optimvar('q',lower = -50,upper = 50,type = 'integer');
Dpos = optimvar('Dpos',lower = 0);
Dneg = optimvar('Dneg',lower = 0);
Our goal for the ILP solver will be to minimize Dpos + Dneg now. But since they must both be positive, it solves the min absolute value objective. One of them will always be zero.
r = 3.6;
prob.Constraints = log2(r) + Dpos - Dneg == p + log2(5)*q;
prob.Objective = Dpos + Dneg;
The solve is now a simple one. I'll tell it to use intlinprog, even though it would probably figure that out by itself. (Note: if I do not tell solve which solver to use, it does use intlinprog. But it also finds the correct solution when I told it to use GA offline.)
solve(prob,solver = 'intlinprog')
The solution it finds within the bounds of +/- 50 for both p and q seems pretty good. Note that Dpos and Dneg are pretty close to zero.
2^39*5^-16
and while 3.6028979... seems like nothing special, in fact, it is of the form we want.
R = sym(2)^39*sym(5)^-16
vpa(R,100)
vpa(1/R,100)
both of those numbers are exact. If I wanted to find a better approximation to 3.6, all I need do is extend the bounds on p and q. And we can use the same solution approch for any floating point number.
Have you ever been enrolled in a course that uses an LMS and there is an assignment that invovles posting a question to, or answering a question in, a discussion group? This discussion group is meant to simulate that experience.
The functionality would allow report generation straight from live scripts that could be shared without exposing the code. This could be useful for cases where the recipient of the report only cares about the results and not the code details, or when the methodology is part of a company know how, e.g. Engineering services companies.
In order for it to be practical for use it would also require that variable values could be inserted into the text blocks, e.g. #var_name# would insert the value of the variable "var_name" and possibly selecting which code blocks to be hidden.
